case study questions class 10 maths chapter 5

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CBSE Class 10 Maths Case Study Questions for Maths Chapter 5 - Arithmetic Progression (Published by CBSE)

Case study questions on cbse class 10 maths chapter 5 - arithmetic progression are provided here. these questions are published by cbse to help students prepare for their maths exam..

CBSE Class 10 Case Study Questions for Maths Chapter 5 - Arithmetic Progression are available here with answers. All the questions have been published by the CBSE board. Students must practice all these questions to prepare themselves for attempting the case study based questions with absolute correctness and obtain a high score in their Maths Exam 2021-22.

Case Study Questions for Class 10 Maths Chapter 5 - Arithmetic Progression

CASE STUDY 1:

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

Based on the above information, answer the following questions:

1. Find the production during first year.

2. Find the production during 8th year.

3. Find the production during first 3 years.

4. In which year, the production is Rs 29,200.

5. Find the difference of the production during 7th year and 4th year.

2. Production during 8th year is (a+7d) = 5000 + 2(2200) = 20400

3. Production during first 3 year = 5000 + 7200 + 9400 = 21600

4. N = 12 5.

Difference = 18200 - 11600 = 6600

CASE STUDY 2:

Your friend Veer wants to participate in a 200m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.

1. Which of the following terms are in AP for the given situation

a) 51,53,55….

b) 51, 49, 47….

c) -51, -53, -55….

d) 51, 55, 59…

Answer: b) 51, 49, 47….

2. What is the minimum number of days he needs to practice till his goal is achieved

Answer: c) 11

3. Which of the following term is not in the AP of the above given situation

Answer: b) 30

4. If nth term of an AP is given by an = 2n + 3 then common difference of an AP is

Answer: a) 2

5. The value of x, for which 2x, x+ 10, 3x + 2 are three consecutive terms of an AP

Answer: a) 6

CASE STUDY 3:

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month , answer the following:

1. The amount paid by him in 30th installment is

Answer: a) 3900

2. The amount paid by him in the 30 installments is

Answer: b) 73500

3. What amount does he still have to pay offer 30th installment?

Answer: c) 44500

4. If total installments are 40 then amount paid in the last installment?

Answer: a) 4900

5. The ratio of the 1st installment to the last installment is

Answer: b) 10:49

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

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Case Study Questions Class 10 Maths Arithmetic Progressions

Case study questions class 10 maths chapter 5 arithmetic progressions.

CBSE Class 10 Case Study Questions Maths Arithmetic Progressions. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Arithmetic Progressions.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 10 Maths Arithmetic Progressions

Case Study – 1

Q.1) Aditya is celebrating his birthday. He invited his friends. He bought a packet of toffees/candies which contains 120 candies. He arranges the candies such that in the first row there are 3 candies, in second there are 5 candies, in third there are 7 candies and so on.

[ KVS Raipur 2021 – 22 ]

On the basis of the above information, answer any four of the following questions:

(i) Find the total number of rows of candies.

(ii) Find the difference in number of candies placed in 7th and 3rd rows.

(i) There is an AP: 3,5,7,

a = 3 & d = 2 so apply let there are n rows sum of n terms = n/2 {2 + ( − 1)}

120 = n/2 {2 × 3 + (n − 1) 2}

we get n 2 + 2n − 120 = 0 then n = 10 & − 12 So there are 10 rows

(ii) 7th row=3+(7-1)X2 = 3 + 12 = 15 and 3rd row = 3+(3-1) X 2 = 7 their diff.= 8

CASE STUDY 2:

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, answer the following:

[ CBSE Academic Question Bank ]

(1) The amount paid by him in 30th installment is

Answer: (a) 3900

(2) The amount paid by him in the 30 installments is

Answer: (b) 73500

(3) What amount does he still have to pay offer 30th installment?

Answer: (c) 44500

(4) If total installments are 40 then amount paid in the last installment?

Answer: (a) 4900

(5) The ratio of the 1st installment to the last installment is

Answer: (b) 10:49

Case Study – 3

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

1.) In which year, the production is Rs 29,200.

2.) Find the production during first year.

Ans: Rs – 5000

3.) Find the difference of the production during 7th year and 4th year.

Ans: Difference= 18200-11600=6600

4.) Find the production during first 3 years.

Ans: Production during first 3 year= 5000 + 7200 + 9400 = 21600

5.) Find the production during 8th year.

Ans: Production during 8th year is (a+7d) = 5000 + 2 (2200) = 20400

We hope that above case study questions will help you for your upcoming exams. To see more click below – 

• CBSE Class 10 Maths (standard)
• CBSE Class 10 Maths (Basic)

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Chapter 5 Class 10 Arithmetic Progressions

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Updated for new NCERT Book - 2023-24 Edition

Get solutions of all NCERT Questions with examples of Chapter 5 Class 10 Arithmetic Progressions (AP). Video of all questions are also available. 

In this chapter, we will learn

• What is an AP - and what is First term (a) and Common Difference (d) of an Arithmetic Progression
• Finding n th term of an AP (a n )
• Finding n using a n formula
• Finding AP when some terms are given
• Finding n th term from the last term
• Finding Sum of n terms of an AP ( S n ) - both formulas
• Finding number of terms when Sum is given
• If n th term is given, finding S
• Some statement questions using a n and S n formulas, including optional exercise

Check out the answers to the exercises (including examples and optional) by clicking a link below, or learn from the concepts.

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
• Assertion and Reason Questions of Class 10th Science
• Assertion and Reason Questions of Class 10th Social Science

Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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Class 10 Maths Chapter 5 Case Based Questions - Arithmetic Progressions

Case study - 1.

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, answer the following:

Case Study - 2

Case Study - 3

Based on the above information, answer the following questions:

Q1: Find the production during first year. Ans: We are given that the production of TV sets in the factory is increasing uniformly. This means that we can apply the formula of an arithmetic progression, where the nth term (a_n) is given by a + (n-1)d. Here, 'a' is the first term, 'n' is the term number, and 'd' is the common difference. We know that the 6th year production (a_6) is 16000 and the 9th year production (a_9) is 22600. We can subtract the 6th year production from the 9th year to find the total increase over 3 years, which is 6600. This means that the common difference ('d') is 6600/3 = 2200 sets per year. Substituting the values into the nth term formula: 16000 = a + (6-1)*2200 16000 = a + 11000 => a = 16000 - 11000 => a = 5000 So, the production during the first year was 5000 sets.   Q2: Find the production during 8th year. Ans: We can use the nth term formula again to find the production in the 8th year. a_8 = a + (8-1)*d = 5000 + 7*2200 = 5000 + 15400 = 20400 So, the production during the 8th year was 20400 sets.   Q3: Find the production during first 3 years. Ans:  To find the total production over the first 3 years, we sum the production over each year. In an arithmetic progression, the sum of the first n terms (S_n) is given by n/2 * (2a + (n-1)d). S_3 = 3/2 * (2*5000 + (3-1)*2200) = 1.5 * (10000 + 4400) = 1.5 * 14400 = 21600 So, the total production during the first 3 years was 21600 sets.   Q4: In which year, the production is Rs 29,200. Ans: To find the year when the production was 29200, we can use the nth term formula and solve for n. 29200 = 5000 + (n-1)2200 => 24200 = (n-1)2200 => n-1 = 24200/2200 => n-1 = 11 => n = 12 So, the production was 29200 sets in the 12th year.   Q5: Find the difference of the production during 7th year and 4th year. Ans: We can find the production in the 7th and 4th years using the nth term formula, and then subtract the two. a_7 = 5000 + 6*2200 = 18200 a_4 = 5000 + 3*2200 = 11600 Difference = a_7 - a_4 = 18200 - 11600 = 6600 So, the difference in production between the 7th and 4th years was 6600 sets.  

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Case Study Class 10 Maths

If you are looking for the CBSE Case Study class 10 Maths in PDF, then you are in the right place. CBSE 10th Class Case Study for the Maths Subject is available here on this website. These Case studies can help the students to solve the different types of questions that are based on the case study or passage.

CBSE Board will be asking case study questions based on Maths subjects in the upcoming board exams. Thus, it becomes an essential resource to study. 

The Case Study Class 10 Maths Questions cover a wide range of chapters from the subject. Students willing to score good marks in their board exams can use it to practice questions during the exam preparation. The questions are highly interactive and it allows students to use their thoughts and skills to solve the given Case study questions.

Download Class 10 Maths Case Study Questions and Answers PDF (Passage Based)

Download links of class 10 Maths Case Study questions and answers pdf is given on this website. Students can download them for free of cost because it is going to help them to practice a variety of questions from the exam perspective.

Case Study questions class 10 Maths include all chapters wise questions. A few passages are given in the case study PDF of Maths. Students can download them to read and solve the relevant questions that are given in the passage.

Students are advised to access Case Study questions class 10 Maths CBSE chapter wise PDF and learn how to easily solve questions. For gaining the basic knowledge students can refer to the NCERT Class 10th Textbooks. After gaining the basic information students can easily solve the Case Study class 10 Maths questions.

Case Study Questions Class 10 Maths Chapter 1 Real Numbers

Case Study Questions Class 10 Maths Chapter 2 Polynomials

Case Study Questions Class 10 Maths Chapter 3 Pair of Equations in Two Variables

Case Study Questions Class 10 Maths Chapter 4 Quadratic Equations

Case Study Questions Class 10 Maths Chapter 5 Arithmetic Progressions

Case Study Questions Class 10 Maths Chapter 6 Triangles

Case Study Questions Class 10 Maths Chapter 7 Coordinate Geometry

Case Study Questions Class 10 Maths Chapter 8. Introduction to Trigonometry

Case Study Questions Class 10 Maths Chapter 9 Some Applications of Trigonometry

Case Study Questions Class 10 Maths Chapter 10 Circles

Case Study Questions Class 10 Maths Chapter 12 Areas Related to Circles

Case Study Questions Class 10 Maths Chapter 13 Surface Areas & Volumes

Case Study Questions Class 10 Maths Chapter 14 Statistics

Case Study Questions Class 10 Maths Chapter 15 Probability

How to Solve Case Study Based Questions Class 10 Maths?

In order to solve the Case Study Based Questions Class 10 Maths students are needed to observe or analyse the given information or data. Students willing to solve Case Study Based Questions are required to read the passage carefully and then solve them. 

While solving the class 10 Maths Case Study questions, the ideal way is to highlight the key information or given data. Because, later it will ease them to write the final answers. 

Case Study class 10 Maths consists of 4 to 5 questions that should be answered in MCQ manner. While answering the MCQs of Case Study, students are required to read the paragraph as they can get some clue in between related to the topics discussed.

Also, before solving the Case study type questions it is ideal to use the CBSE Syllabus to brush up the previous learnings.

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Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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• CBSE Reduced Syllabus Class 10 (2020-21)
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• NCERT Solutions
• NCERT Solutions for Class 10
• NCERT Solutions for Class 10 Maths
• Chapter 5 Arithmetic Progressions

NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions

Ncert solutions for class 10 maths chapter 5 – cbse free pdf download.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression is presented here for the benefit of the students preparing for the CBSE Class 10 Maths examination. It is very important for the students to get well versed with these solutions of NCERT to get a good score in the Class 10 examination. These NCERT problems are solved by the experts at BYJU’S. These solutions will help you understand and master different types of questions on arithmetic progressions. NCERT Solutions help you to attain perfection in solving different kinds of questions.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 5 Arithmetic Progressions

Download most important questions for class 10 maths chapter – 5 arithmetic progressions.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability
• Exercise 5.1
• Exercise 5.2
• Exercise 5.3
• Exercise 5.4

NCERT solutions for Class 10 Maths chapter 5 Arithmetic Progression

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Access Answers of Maths NCERT solutions for Class 10 Chapter 5 – Arithmetic Progressions

Exercise 5.1 page: 99.

1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

We can write the given condition as;

Taxi fare for 1 km = 15

Taxi fare for first 2 kms = 15+8 = 23

Taxi fare for first 3 kms = 23+8 = 31

Taxi fare for first 4 kms = 31+8 = 39

And so on……

Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Let the volume of air in a cylinder, initially, be  V  litres.

In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time. Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.

Therefore, volumes will be  V , 3 V /4 , (3 V /4) 2  , (3 V /4) 3 …and so on

Clearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A.P.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

Cost of digging a well for first metre = Rs.150

Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200

Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250

Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300

And so on..

Clearly, 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

We know that if Rs. P is deposited at  r % compound interest per annum for n years, the amount of money will be:

P(1+r/100) n

Therefore, after each year, the amount of money will be;

10000(1+8/100), 10000(1+8/100) 2 , 10000(1+8/100) 3 ……

Clearly, the terms of this series do not have the common difference between them. Therefore, this is not an A.P.

2. Write first four terms of the A.P. when the first term a and the common difference are given as follows :

(i)  a  = 10,  d  = 10 (ii)  a  = -2,  d  = 0 (iii)  a  = 4,  d  = – 3 (iv)  a  = -1  d  = 1/2 (v)  a  = – 1.25,  d  = – 0.25

(i)  a  = 10,  d  = 10

Let us consider, the Arithmetic Progression series be  a 1 , a 2 , a 3 , a 4 , a 5  …

a 1  =  a  = 10

a 2  =  a 1 + d  = 10+10 = 20

a 3  =  a 2 + d  = 20+10 = 30

a 4  =  a 3 + d  = 30+10 = 40

a 5  =  a 4 + d  = 40+10 = 50

Therefore, the A.P. series will be 10, 20, 30, 40, 50 …

And First four terms of this A.P. will be 10, 20, 30, and 40.

(ii)  a  = – 2,  d  = 0

a 1  =  a  = -2

a 2  =  a 1 + d  = – 2+0 = – 2

a 3  =  a 2 +d = – 2+0 = – 2

a 4  =  a 3 + d  = – 2+0 = – 2

Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …

And, First four terms of this A.P. will be – 2, – 2, – 2 and – 2.

(iii)  a  = 4,  d  = – 3

a 1  =  a  = 4

a 2  =  a 1 + d  = 4-3 = 1

a 3  =  a 2 + d  = 1-3 = – 2

a 4  =  a 3 + d  = -2-3 = – 5

Therefore, the A.P. series will be 4, 1, – 2 – 5 …

And, first four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv)  a  = – 1,  d  = 1/2

a 2  =  a 1 + d  = -1+1/2 = -1/2

a 3  =  a 2 + d  = -1/2+1/2 = 0

a 4  =  a 3 + d  = 0+1/2 = 1/2

Thus, the A.P. series will be-1, -1/2, 0, 1/2

And First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v)  a  = – 1.25,  d  = – 0.25

a 1  =  a  = – 1.25

a 2  =  a 1  +  d  = – 1.25-0.25 = – 1.50

a 3  =  a 2  +  d  = – 1.50-0.25 = – 1.75

a 4  =  a 3  +  d  = – 1.75-0.25 = – 2.00

Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..

And first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

3. For the following A.P.s, write the first term and the common difference. (i) 3, 1, – 1, – 3 … (ii) -5, – 1, 3, 7 … (iii) 1/3, 5/3, 9/3, 13/3 …. (iv) 0.6, 1.7, 2.8, 3.9 …

(i) Given series,

3, 1, – 1, – 3 …

First term,  a  = 3

Common difference,  d  = Second term – First term

⇒  1 – 3 = -2

(ii) Given series, – 5, – 1, 3, 7 …

First term,  a  = -5

⇒ ( – 1)-( – 5) = – 1+5 = 4

(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….

First term,  a  = 1/3

⇒ 5/3 – 1/3 = 4/3

(iv) Given series, 0.6, 1.7, 2.8, 3.9 …

First term,  a  = 0.6

⇒ 1.7 – 0.6

4. Which of the following are APs? If they form an A.P. find the common difference  d  and write three more terms.

(i) 2, 4, 8, 16 … (ii) 2, 5/2, 3, 7/2 …. (iii) -1.2, -3.2, -5.2, -7.2 … (iv) -10, – 6, – 2, 2 … (v) 3, 3 + √2, 3 + 2√2, 3 + 3√2 (vi) 0.2, 0.22, 0.222, 0.2222 …. (vii) 0, – 4, – 8, – 12 … (viii) -1/2, -1/2, -1/2, -1/2 …. (ix) 1, 3, 9, 27 … (x)  a , 2 a , 3 a , 4 a  … (xi)  a ,  a 2 ,  a 3 ,  a 4  … (xii) √2, √8, √18, √32 … (xiii) √3, √6, √9, √12 … (xiv) 1 2 , 3 2 , 5 2 , 7 2  … (xv) 1 2 , 5 2 , 7 2 , 7 3  …

(i) Given to us,

2, 4, 8, 16 …

Here, the common difference is;

a 2  –  a 1  = 4 – 2 = 2

a 3  –  a 2  = 8 – 4 = 4

a 4  –  a 3  = 16 – 8 = 8

Since,  a n +1  –  a n or the common difference is not the same every time.

Therefore, the given series are not forming an A.P.

(ii) Given, 2, 5/2, 3, 7/2 ….

a 2  –  a 1  = 5/2-2 = 1/2

a 3  –  a 2  = 3-5/2 = 1/2

a 4  –  a 3  = 7/2-3 = 1/2

Since, a n +1  –  a n  or the common difference is same every time.

Therefore,  d  = 1/2 and the given series are in A.P.

The next three terms are;

a 5  = 7/2+1/2 = 4

a 6  = 4 +1/2 = 9/2

a 7  = 9/2 +1/2 = 5

(iii) Given, -1.2, – 3.2, -5.2, -7.2 …

a 2  –  a 1  = (-3.2)-(-1.2) = -2

a 3  –  a 2  = (-5.2)-(-3.2) = -2

a 4  –  a 3  = (-7.2)-(-5.2) = -2

Since, a n +1  –  a n  or common difference is same every time.

Therefore,  d  = -2 and the given series are in A.P.

Hence, next three terms are;

a 5  = – 7.2-2 = -9.2

a 6  = – 9.2-2 = – 11.2

a 7  = – 11.2-2 = – 13.2

(iv) Given, -10, – 6, – 2, 2 … Here, the terms and their difference are;

a 2  –  a 1  = (-6)-(-10) = 4

a 3  –  a 2  = (-2)-(-6) = 4

a 4  –  a 3  = (2 -(-2) = 4

Since,  a n +1  –  a n  or the common difference is same every time.

Therefore,  d  = 4 and the given numbers are in A.P.

a 5  = 2+4 = 6

a 6  = 6+4 = 10

a 7  = 10+4 = 14

(v) Given, 3, 3+√2, 3+2√2, 3+3√2 Here,

a 2  –  a 1  = 3+√2-3 = √2

a 3  –  a 2  = (3+2√2)-(3+√2) = √2

a 4  –  a 3  = (3+3√2) – (3+2√2) = √2

Therefore,  d  = √2 and the given series forms a A.P.

a 5  = (3+√2) +√2 = 3+4√2

a 6  = (3+4√2)+√2 = 3+5√2

a 7  = (3+5√2)+√2 = 3+6√2

(vi) 0.2, 0.22, 0.222, 0.2222 …. Here,

a 2  –  a 1  = 0.22-0.2 = 0.02

a 3  –  a 2  = 0.222-0.22 = 0.002

a 4  –  a 3  = 0.2222-0.222 = 0.0002

Since,  a n +1  –  a n  or the common difference is not same every time.

Therefore, and the given series doesn’t forms a A.P.

(vii) 0, -4, -8, -12 … Here,

a 2  –  a 1  = (-4)-0 = -4

a 3  –  a 2  = (-8)-(-4) = -4

a 4  –  a 3  = (-12)-(-8) = -4

Therefore,  d  = -4 and the given series forms a A.P.

a 5  = -12-4 = -16

a 6  = -16-4 = -20

a 7  = -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 …. Here,

a 2  –  a 1  = (-1/2) – (-1/2) = 0

a 3  –  a 2  = (-1/2) – (-1/2) = 0

a 4  –  a 3  = (-1/2) – (-1/2) = 0

Therefore,  d  = 0 and the given series forms a A.P.

a 5  = (-1/2)-0 = -1/2

a 6  = (-1/2)-0 = -1/2

a 7  = (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 … Here,

a 2  –  a 1  = 3-1 = 2

a 3  –  a 2  = 9-3 = 6

a 4  –  a 3  = 27-9 = 18

Therefore, and the given series doesn’t form a A.P.

(x)  a , 2 a , 3 a , 4 a  …

a 2  –  a 1  = 2 a – a  =  a

a 3  –  a 2  = 3 a -2 a  =  a

a 4  –  a 3  = 4 a -3 a  =  a

Therefore,  d  =  a  and the given series forms a A.P.

a 5  = 4 a + a  = 5 a

a 6  = 5 a + a  = 6 a

a 7  = 6 a + a  = 7 a

(xi)  a ,  a 2 ,  a 3 ,  a 4  …

a 2  –  a 1  =  a 2 – a  = a( a -1)

a 3  –  a 2  =  a 3  –   a 2  =  a 2 ( a -1)

a 4  –  a 3  =  a 4  –  a 3  =  a 3 ( a -1)

Therefore, the given series doesn’t forms a A.P.

(xii) √2, √8, √18, √32 … Here,

a 2  –  a 1  = √8-√2  = 2√2-√2 = √2

a 3  –  a 2  = √18-√8 = 3√2-2√2 = √2

a 4  –  a 3  = 4√2-3√2 = √2

a 5  = √32+√2 = 4√2+√2 = 5√2 = √50

a 6  = 5√2+√2 = 6√2 = √72

a 7  = 6√2+√2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

a 2  –  a 1  = √6-√3 = √3×√2-√3 = √3(√2-1)

a 3  –  a 2  = √9-√6 = 3-√6 = √3(√3-√2)

a 4  –  a 3  = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)

Therefore, the given series doesn’t form a A.P.

(xiv) 1 2 , 3 2 , 5 2 , 7 2  …

Or, 1, 9, 25, 49 …..

a 2  −  a 1  = 9−1 = 8

a 3  −  a 2  = 25−9 = 16

a 4  −  a 3  = 49−25 = 24

(xv) 1 2 , 5 2 , 7 2 , 73 …

Or 1, 25, 49, 73 …

a 2  −  a 1  = 25−1 = 24

a 3  −  a 2  = 49−25 = 24

a 4  −  a 3  = 73−49 = 24

Therefore,  d  = 24 and the given series forms a A.P.

a 5  = 73+24 = 97

a 6  = 97+24 = 121

a 7  = 121+24 = 145

Exercise 5.2 Page: 105

1. Fill in the blanks in the following table, given that  a  is the first term,  d  the common difference and  a n  the  n th  term of the A.P.

(i)  Given,

First term, a  = 7

Common difference,  d  = 3

Number of terms,  n  = 8,

We have to find the nth term, a n  = ?

As we know, for an A.P.,

a n  =  a +( n −1) d

Putting the values,

=> 7+(8 −1) 3

=> 7+(7) 3

=> 7+21 = 28

Hence,  a n  = 28

(ii) Given,

First term, a  = -18

Common difference,  d  = ?

Number of terms,  n  = 10

Nth term, a n  = 0

0 = − 18 +(10−1) d

d  = 18/9 = 2

Hence, common difference,  d  = 2

(iii) Given,

First term, a  = ?

Common difference,  d  = -3

Number of terms,  n  = 18

Nth term, a n  = -5

−5 =  a +(18−1) (−3)

−5 =  a +(17) (−3)

−5 =  a −51

a  = 51−5 = 46

Hence,  a  = 46

(iv)  Given,

First term, a  = -18.9

Common difference,  d  = 2.5

Number of terms,  n  = ?

Nth term, a n  = 3.6

a n  =  a  +( n  −1) d

3.6 = − 18.9+( n  −1)2.5

3.6 + 18.9 = ( n −1)2.5

22.5 = ( n −1)2.5

( n  – 1) = 22.5/2.5

Hence,  n  = 10

(v)  Given,

First term, a  = 3.5

Common difference,  d  = 0

Number of terms,  n  = 105

Nth term, a n  = ?

a n  =  a +( n  −1) d

a n  = 3.5+(105−1) 0

a n  = 3.5+104×0

Hence,  a n  = 3.5

2. Choose the correct choice in the following and justify: (i) 30 th  term of the A.P: 10,7, 4, …, is (A) 97 (B) 77 (C) −77 (D) −87

(i) Given here,

A.P. = 10, 7, 4, …

Therefore, we can find,

First term,  a  = 10

Common difference,  d  =  a 2  −  a 1  = 7−10 = −3

a n  =  a  +( n −1) d

Putting the values;

a 30  = 10+(30−1)(−3)

a 30  = 10+(29)(−3)

a 30  = 10−87 = −77

Hence, the correct answer is option C.

(ii) Given here,

A.P. = -3, -1/2, ,2 …

First term  a  = – 3

Common difference,  d  =  a 2  −  a 1  = (-1/2) -(-3)

⇒(-1/2) + 3 = 5/2

a 11  = -3+(11-1)(5/2)

a 11  = -3+(10)(5/2)

a 11  = -3+25

Hence, the answer is option B.

3. In the following APs find the missing term in the boxes.

(i) For the given A.P., 2,2 , 26

The first and third term are;

Therefore, putting the values here,

a 3  = 2+(3-1) d

a 2  = 2+(2-1)12

Therefore, 14 is the missing term.

(ii) For the given A.P., , 13, ,3

a 2  = 13 and

a n  =  a +( n −1)  d

a 2  =  a  +(2-1) d

13 =  a + d  ……………….  (i)

a 4  =  a +(4-1) d

3 =  a +3 d  …………..  (ii)

On subtracting equation  (i)  from  (ii) , we get,

From equation  (i) , putting the value of d,we get

13 =  a +(-5)

a 3  = 18+(3-1)(-5)

= 18+2(-5) = 18-10 = 8

Therefore, the missing terms are 18 and 8 respectively.

(iii) For the given A.P.,

a 4  = 19/2

19/2 =  5 +3d

(19/2) – 5 = 3d

a 2  =  a +(2-1) d

a 2  =  5 +3/2

a 2  = 13/2

a 3  =  a +(3-1) d

a 3  =  5 +2×3/2

Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For the given A.P.,

a  = −4 and

a n  =  a  +( n −1)  d

a 6  = a+(6−1)d

6 = − 4+5 d

a 2  =  a + d  = − 4+2 = −2

a 3  =  a +2 d  = − 4+2(2) = 0

a 4  =  a +3 d  = − 4+ 3(2) = 2

a 5  =  a +4 d  = − 4+4(2) = 4

Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v) For the given A.P.,

a 2  =  a +(2−1) d

38 =  a + d  …………………….  (i)

a 6  =  a +(6−1) d

−22 =  a +5 d  ………………….  (ii)

On subtracting equation  (i)  from  (ii) , we get

− 22 − 38 = 4 d

a  =  a 2  −  d  = 38 − (−15) = 53

a 3  =  a  + 2 d  = 53 + 2 (−15) = 23

a 4  =  a  + 3 d  = 53 + 3 (−15) = 8

a 5  =  a  + 4 d  = 53 + 4 (−15) = −7

Therefore, the missing terms are 53, 23, 8, and −7 respectively.

4. Which term of the A.P. 3, 8, 13, 18, … is 78?

Given the A.P. series as3, 8, 13, 18, …

First term, a = 3

Common difference, d = a 2  − a 1  = 8 − 3 = 5

Let the n th  term of given A.P. be 78. Now as we know,

78 = 3+( n  −1)5

75 = ( n −1)5

( n −1) = 15

Hence, 16 th  term of this A.P. is 78.

5. Find the number of terms in each of the following A.P.

(i) 7, 13, 19, …, 205

(i) Given, 7, 13, 19, …, 205 is the A.P

Common difference, d  =  a 2  −  a 1  = 13 − 7 = 6

Let there are  n  terms in this A.P.

a n  =  a  + ( n  − 1)  d

Therefore, 205 = 7 + ( n  − 1) 6

198 = ( n  − 1) 6

33 = ( n  − 1)

Therefore, this given series has 34 terms in it.

First term, a = 18

d = (31-36)/2 = -5/2

Let there are n terms in this A.P.

a n  = a+(n−1)d

-47 = 18+(n-1)(-5/2)

-47-18 = (n-1)(-5/2)

-65 = (n-1)(-5/2)

(n-1) = -130/-5

Therefore, this given A.P. has 27 terms in it.

6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …

For the given series, A.P. 11, 8, 5, 2..

First term, a  = 11

Common difference, d  =  a 2 − a 1  = 8−11 = −3

Let −150 be the  n th  term of this A.P.

-150 = 11+( n  -1)(-3)

-150 = 11-3 n  +3

-164 = -3 n

Clearly,  n  is not an integer but a fraction.

Therefore, – 150 is not a term of this A.P.

7. Find the 31 st  term of an A.P. whose 11 th  term is 38 and the 16 th  term is 73.

Given that,

11 th term, a 11  = 38

and 16 th term, a 16  = 73

We know that,

a 11  = a+(11−1)d

38 = a+10d ……………………………….  (i)

In the same way,

a 16  =  a  +(16−1) d

73 =  a +15 d  …………………………………………  (ii)

On subtracting equation  (i)  from  (ii) , we get

From equation  (i) , we can write,

38 =  a +10×(7)

38 − 70 = a

a 31  = a +(31−1) d

= − 32 + 30 (7)

= − 32 + 210

Hence, 31 st  term is 178.

8. An A.P. consists of 50 terms of which 3 rd  term is 12 and the last term is 106. Find the 29 th  term.

Solution: Given that,

3 rd term, a 3  = 12

50 th term, a 50  = 106

a 3  =  a +(3−1) d

12 =  a +2 d  …………………………….  (i)

a 50  =  a +(50−1) d

106 =  a +49 d  ………………………….  (ii)

d  = 2 = common difference

From equation  (i) , we can write now,

12 =  a +2(2)

a  = 12−4 = 8

a 29  =  a +(29−1)  d

a 29  = 8+(28)2

a 29  = 8+56 = 64

Therefore, 29 th  term is 64.

9. If the 3 rd  and the 9 th  terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero. Solution:

3 rd term, a 3  = 4

and 9 th term, a 9  = −8

4 =  a +2 d  ………………………………………  (i)

a 9  =  a +(9−1) d

−8 =  a +8 d  …………………………………………………  (ii)

On subtracting equation  (i)  from  (ii) , we will get here,

4 =  a +2(−2)

Let  n th  term of this A.P. be zero.

a n  =  a +( n −1) d

0 = 8+( n −1)(−2)

0 = 8−2 n +2

Hence, 5 th  term of this A.P. is 0.

10. If 17 th  term of an A.P. exceeds its 10 th  term by 7. Find the common difference.

We know that, for an A.P series;

a 17  =  a +(17−1) d

a 17  =  a  +16 d

a 10  =  a +9 d

As it is given in the question,

a 17  −  a 10  = 7

( a  +16 d )−( a +9 d ) = 7

Therefore, the common difference is 1.

11. Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54 th  term?

Given A.P. is 3, 15, 27, 39, …

first term, a  = 3

common difference, d  =  a 2  −  a 1  = 15 − 3 = 12

a 54  =  a +(54−1) d

⇒3+(53)(12)

⇒3+636 = 639

a 54 = 639+132=771

We have to find the term of this A.P. which is 132 more than a 54, i.e.771.

Let  n th  term be 771.

771 = 3+( n  −1)12

768 = ( n −1)12

( n  −1) = 64

Therefore, 65 th  term was 132 more than 54 th  term.

Or another method is;

Let  n th  term be 132 more than 54 th  term.

n  = 54 + 132/2

= 54 + 11 = 65 th  term

12. Two APs have the same common difference. The difference between their 100 th  term is 100, what is the difference between their 1000 th  terms?

Let, the first term of two APs be  a 1  and  a 2  respectively

And the common difference of these APs be  d .

For the first A.P.,we know,

a 100  =  a 1 +(100−1) d

=  a 1  + 99d

a 1000  =  a 1 +(1000−1) d

a 1000  =  a 1 +999 d

For second A.P., we know,

a 100  =  a 2 +(100−1) d

=  a 2 +99 d

a 1000  =  a 2 +(1000−1) d

=  a 2 +999 d

Given that, difference between 100 th  term of the two APs = 100

Therefore, ( a 1 +99 d ) − ( a 2 +99 d ) = 100

a 1 − a 2  = 100………………………………………………………………..  (i)

Difference between 1000 th  terms of the two APs

( a 1 +999 d ) − ( a 2 +999 d ) =  a 1 − a 2

From equation  (i) ,

This difference,  a 1 − a 2  = 100

Hence, the difference between 1000 th  terms of the two A.P. will be 100.

13. How many three digit numbers are divisible by 7?

First three-digit number that is divisible by 7 are;

First number = 105

Second number = 105+7 = 112

Third number = 112+7 =119

Therefore, 105, 112, 119, …

All are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

As we know, the largest possible three-digit number is 999.

When we divide 999 by 7, the remainder will be 5.

Therefore, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.

Now the series is as follows.

105, 112, 119, …, 994

Let 994 be the nth term of this A.P.

first term, a = 105

common difference, d = 7

As we know,

994 = 105+(n−1)7

889 = (n−1)7

(n−1) = 127

Therefore, 128 three-digit numbers are divisible by 7.

14. How many multiples of 4 lie between 10 and 250?

The first multiple of 4 that is greater than 10 is 12.

Next multiple will be 16.

Therefore, the series formed as;

12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.

The series is as follows, now;

12, 16, 20, 24, …, 248

Let 248 be the  n th  term of this A.P.

first term, a  = 12

common difference, d  = 4

248 = 12+( n -1)×4

236/4 = n-1

Therefore, there are 60 multiples of 4 between 10 and 250.

15. For what value of  n , are the  n th  terms of two APs 63, 65, 67, and 3, 10, 17, … equal?

Given two APs as; 63, 65, 67,… and 3, 10, 17,….

Taking first AP,

63, 65, 67, …

First term, a = 63

Common difference, d = a 2 −a 1  = 65−63 = 2

We know, n th  term of this A.P. = a n  = a+(n−1)d

a n = 63+( n −1)2 = 63+2 n −2

a n  = 61+2 n  ……………………………………….  (i)

Taking second AP,

3, 10, 17, …

Common difference, d = a 2  − a 1  = 10 − 3 = 7

n th  term of this A.P. = 3+( n −1)7

a n  = 3+7 n −7

a n  = 7 n −4 ………………………………………………………..  (ii)

Given,  n th  term of these A.P.s are equal to each other.

Equating both these equations, we get,

61+2 n  = 7 n −4

Therefore, 13 th  terms of both these A.P.s are equal to each other.

16. Determine the A.P. whose third term is 16 and the 7 th  term exceeds the 5 th  term by 12.

Third term, a 3  = 16

a  +(3−1) d  = 16

a +2 d  = 16 ………………………………………. (i)

It is given that, 7 th  term exceeds the 5 th  term by 12.

a 7  −  a 5  = 12

( a +6 d )−( a +4 d ) = 12

From equation (i), we get,

a +2(6) = 16

Therefore, A.P. will be4, 10, 16, 22, …

17. Find the 20 th  term from the last term of the A.P. 3, 8, 13, …, 253.

Given A.P. is3, 8, 13, …, 253

Common difference, d= 5.

Therefore, we can write the given AP in reverse order as;

253, 248, 243, …, 13, 8, 5

Now for the new AP,

first term, a = 253

and common difference, d = 248 − 253 = −5

Therefore, using nth term formula, we get,

a 20  =  a +(20−1) d

a 20  = 253+(19)(−5)

a 20  = 253−95

Therefore, 20 th  term from the last term of the AP 3, 8, 13, …, 253 . is 158.

18. The sum of 4 th  and 8 th  terms of an A.P. is 24 and the sum of the 6 th  and 10 th  terms is 44. Find the first three terms of the A.P .

We know that, the nth term of the AP is;

a 4  =  a +(4−1) d

a 4  =  a +3 d

In the same way, we can write,

a 8  =  a +7 d

a 6  =  a +5 d

a 4 +a 8  = 24

a+3d+a+7d = 24

2a+10d = 24

a+5d = 12 ……………………………………………………  (i)

a 6 +a 10  = 44

a +5d+a+9d = 44

2a+14d = 44

a+7d = 22 ……………………………………..  (ii)

2d = 22 − 12

From equation  (i) , we get,

a +5 d  = 12

a +5(5) = 12

a 2  =  a + d  = − 13+5 = −8

a 3  =  a 2 + d  = − 8+5 = −3

Therefore, the first three terms of this A.P. are −13, −8, and −3.

19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

It can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.

Therefore, after 1995, the salaries of each year are;

5000, 5200, 5400, …

Here, first term, a  = 5000

and common difference, d  = 200

Let after  n th  year, his salary be Rs 7000.

Therefore, by the n th term formula of AP,

7000 = 5000+( n −1)200

200( n −1)= 2000

( n −1) = 10

Therefore, in 11th year, his salary will be Rs 7000.

20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the  n th  week, her weekly savings become Rs 20.75, find  n.

Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.

First term, a = 5

and common difference, d = 1.75

Also given,

a n  = 20.75

Find, n = ?

As we know, by the n th term formula,

20.75 = 5+( n  -1)×1.75

15.75 = ( n  -1)×1.75

( n  -1) = 15.75/1.75 = 1575/175

Hence,  n  is 10.

Exercise 5.3 Page: 112

1. Find the sum of the following APs.

(i) 2, 7, 12 ,…., to 10 terms. (ii) − 37, − 33, − 29 ,…, to 12 terms (iii) 0.6, 1.7, 2.8 ,…….., to 100 terms (iv) 1/15, 1/12, 1/10, …… , to 11 terms

(i) Given, 2, 7, 12 ,…, to 10 terms

For this A.P.,

first term, a = 2

And common difference, d = a 2  − a 1  = 7−2 = 5

We know that, the formula for sum of nth term in AP series is,

S n  = n/2 [2a +(n-1)d]

S 10  = 10/2 [2(2)+(10 -1)×5]

= 5[4+(9)×(5)]

= 5 × 49 = 245

(ii) Given, −37, −33, −29 ,…, to 12 terms

first term, a = −37

And common difference, d = a 2 − a 1

d= (−33)−(−37)

= − 33 + 37 = 4

S n  = n/2 [2a+(n-1)d]

S 12  = 12/2 [2(-37)+(12-1)×4]

= 6[-74+11×4]

= 6[-74+44]

= 6(-30) = -180

(iii) Given, 0.6, 1.7, 2.8 ,…, to 100 terms

first term, a = 0.6

Common difference, d = a 2  − a 1  = 1.7 − 0.6 = 1.1

S n  = n/2[2a +(n-1)d]

S 12  = 50/2 [1.2+(99)×1.1]

= 50[1.2+108.9]

= 50[110.1]

(iv) Given, 1/15, 1/12, 1/10, …… , to 11 terms

First term, a = 1/5

Common difference, d = a 2 –a 1 = (1/12)-(1/5) = 1/60

And number of terms n = 11

S n  =  n /2 [2a + ( n  – 1)  d ]

= 11/2(2/15 + 10/60)

= 11/2 (9/30)

2. Find the sums given below :

First term, a = 7

n th term, a n = 84

Let 84 be the  n th  term of this A.P., then as per the n th term formula,

a n = a(n-1)d

84 = 7+(n – 1)×7/2

77 = (n-1)×7/2

We know that, sum of n term is;

S n  = n/2 (a + l) , l = 84 S n  = 23/2 (7+84)

S n   = (23×91/2) = 2093/2

(ii) Given, 34 + 32 + 30 + ……….. + 10

first term, a = 34

common difference, d = a 2 −a 1  = 32−34 = −2

n th term, a n = 10

Let 10 be the  n th  term of this A.P., therefore,

a n = a +(n−1)d

10 = 34+(n−1)(−2)

−24 = (n −1)(−2)

We know that, sum of n terms is;

S n  = n/2 (a +l) , l = 10

= 13/2 (34 + 10)

= (13×44/2) = 13 × 22

(iii)  Given, (−5) + (−8) + (−11) + ………… + (−230)

First term, a = −5

nth term, a n = −230

Common difference, d = a 2 −a 1  = (−8)−(−5)

⇒d = − 8+5 = −3

Let −230 be the  n th  term of this A.P., and by the n th term formula we know,

a n =  a +( n −1) d

−230 = − 5+( n −1)(−3)

−225 = ( n −1)(−3)

( n −1) = 75

And, Sum of n term,

S n  =  n /2 ( a  +  l )

= 76/2 [(-5) + (-230)]

3. In an AP (i) Given  a  = 5,  d  = 3,  a n  = 50, find  n  and  S n . (ii) Given  a  = 7,  a 13  = 35, find  d  and  S 13 . (iii) Given  a 12  = 37,  d  = 3, find  a  and  S 12 . (iv) Given  a 3  = 15,  S 10  = 125, find  d  and  a 10 . (v) Given  d  = 5,  S 9  = 75, find  a  and  a 9 . (vi) Given  a  = 2,  d  = 8,  S n  = 90, find  n  and  a n . (vii) Given  a  = 8,  a n  = 62,  S n  = 210, find  n  and  d . (viii) Given  a n  = 4,  d  = 2,  S n  = − 14, find  n  and  a . (ix) Given  a  = 3,  n  = 8,  S  = 192, find  d . (x) Given  l  = 28,  S  = 144 and there are total 9 terms. Find  a .

(i) Given that,  a  = 5,  d  = 3,  a n  = 50

As we know, from the formula of the nth term in an AP,

a n  =  a  +( n  −1) d ,

Therefore, putting the given values, we get,

⇒ 50 = 5+( n  -1)×3

⇒ 3( n  -1) = 45

⇒  n  -1 = 15

Now, sum of n terms,

S n  =  n /2 ( a  + a n )

S n  = 16/2 (5 + 50) = 440

(ii)  Given that,  a  = 7,  a 13  = 35

a n  =  a +( n −1) d ,

⇒ 35 = 7+(13-1) d

⇒ 12 d  = 28

⇒  d  = 28/12 = 2.33

Now,  S n  =  n /2 ( a + a n )

S 13  = 13/2 (7+35) = 273

(iii) Given that,  a 12  = 37,  d  = 3

As we know, from the formula of the n th term in an AP,

a n  =  a +( n  −1) d ,

⇒  a 12  =  a +(12−1)3

⇒ 37 =  a +33

Now, sum of nth term,

S n  =  n /2 ( a + a n )

S n  =  12 /2 (4+37)

(iv) Given that,  a 3  = 15,  S 10  = 125

a n  =  a  +( n −1) d ,

15 =  a +2 d  …………………………..  (i)

Sum of the nth term,

S n  =  n /2 [2 a +( n -1) d ]

S 10  = 10/2 [2 a +(10-1) d ]

125 = 5(2 a +9 d )

25 = 2 a +9 d  ………………………..  (ii)

On multiplying equation  (i)  by  (ii) , we will get;

30 = 2 a +4 d  ……………………………….  (iii)

By subtracting equation  (iii)  from  (ii) , we get,

15 =  a +2(−1)

a  = 17 = First term

a 10  =  a +(10−1) d

a 10  = 17+(9)(−1)

a 10  = 17−9 = 8

(v) Given that,  d  = 5,  S 9  = 75

As, sum of n terms in AP is,

S n  =  n /2 [2 a  +( n  -1) d ]

Therefore, the sum of first nine terms are;

S 9  = 9/2 [2 a  +(9-1) 5 ]

25 = 3( a +20)

25 = 3 a +60

3 a  = 25−60

As we know, the n th term can be written as;

a 9  =  a +(9−1)(5)

= -35/3+8(5)

= (35+120/3) = 85/3

(vi) Given that,  a  = 2,  d  = 8,  S n  = 90

As, sum of n terms in an AP is,

90 =  n /2 [2 a  +( n  -1) d ]

⇒ 180 =  n (4+8 n  -8) =  n (8 n -4) = 8 n 2 -4 n

⇒ 8 n 2 -4 n – 180 = 0

⇒ 2 n 2 – n -45 = 0

⇒ 2 n 2 -10 n +9 n -45 = 0

⇒ 2 n ( n  -5)+9( n  -5) = 0

⇒ ( n -5)(2 n +9) = 0

So,  n  = 5 (as n only be a positive integer)

∴  a 5   = 8+5×4 = 34

(vii) Given that,  a  = 8,  a n  = 62,  S n  = 210

S n  =  n /2 ( a  +  a n )

210 =  n /2 (8 +62)

⇒ 35 n  = 210

⇒  n  = 210/35 = 6

Now, 62 = 8+5 d

⇒ 5 d  = 62-8 = 54

⇒  d  = 54/5 = 10.8

(viii) Given that, n th term, a n  = 4, common difference, d  = 2, sum of n terms, S n  = −14.

4 =  a +( n  −1)2

4 =  a +2 n −2

a +2 n  = 6

a  = 6 − 2 n  ………………………………………….  (i)

As we know, the sum of n terms is;

-14 =  n /2 ( a + 4 )

−28 =  n  ( a +4)

−28 =  n  (6 −2 n  +4) {From equation  (i) }

−28 =  n  (− 2 n  +10)

−28 = − 2 n 2 +10 n

2 n 2  −10 n  − 28 = 0

n 2  −5 n  −14 = 0

n 2  −7 n+ 2 n  −14 = 0

n  ( n −7)+2( n  −7) = 0

( n  −7)( n  +2) = 0

Either  n  − 7 = 0 or  n  + 2 = 0

n  = 7 or  n  = −2

However,  n  can neither be negative nor fractional.

Therefore,  n  = 7

From equation  (i) , we get

a  = 6−2(7)

(ix) Given that, first term, a  = 3,

Number of terms,  n  = 8

And sum of n terms, S  = 192

S n  =  n /2 [2 a +( n  -1) d ]

192 = 8/2 [2×3+(8 -1) d ]

192 = 4[6 +7 d ]

(x) Given that,  l  = 28, S  = 144 and there are total of 9 terms.

Sum of n terms formula,

144 = 9/2( a +28)

(16)×(2) =  a +28

32 =  a +28

4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Let there be  n  terms of the AP. 9, 17, 25 …

First term, a  = 9

Common difference, d  =  a 2 − a 1  = 17−9 = 8

As, the sum of n terms, is;

636 =  n /2 [2× a +(8-1)×8]

636 =  n /2 [18+( n -1)×8]

636 =  n  [9 +4 n  −4]

636 =  n  (4 n  +5)

4 n 2  +5 n  −636 = 0

4 n 2  +53 n  −48 n  −636 = 0

n  (4 n  + 53)−12 (4 n  + 53) = 0

(4 n  +53)( n  −12) = 0

Either 4 n +53 = 0 or  n −12 = 0

n  = (-53/4) or  n  = 12

n  cannot be negative or fraction, therefore,  n  = 12 only.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

first term, a  = 5

last term, l  = 45

Sum of the AP, S n  = 400

As we know, the sum of AP formula is;

S n  =  n /2 ( a + l )

400 =  n /2(5+45)

400 =  n /2(50)

Number of terms, n  =16

As we know, the last term of AP series can be written as;

l = a+ ( n  −1) d

45 = 5 +(16 −1) d

Common difference, d  = 40/15 = 8/3

6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

First term, a  = 17

Last term, l  = 350

Common difference, d  = 9

Let there be  n  terms in the A.P., thus the formula for last term can be written as;

350 = 17+( n  −1)9

333 = ( n −1)9

( n −1) = 37

S 38  = 38/2 (17+350)

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

7. Find the sum of first 22 terms of an AP in which  d  = 7 and 22 nd  term is 149. Solution:

Given, Common difference, d = 7

22 nd term, a 22  = 149

Sum of first 22 term, S 22  = ?

By the formula of nth term,

a 22  =  a +(22−1) d

149 =  a +21×7

149 =  a +147

a  = 2 = First term

Sum of n terms,

S n  =  n /2( a + a n )

S 22 = 22/2 (2+149)

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. Solution:

Second term, a 2  = 14

Third term, a 3  = 18

Common difference, d  =  a 3 − a 2  = 18−14 = 4

a 2  =  a + d

a  = 10 = First term

Sum of n terms;

S n  =  n /2 [2 a  + ( n  – 1) d ]

S 51  = 51/2 [2×10 (51-1) 4]

= 51/2 [20+(50)×4]

= 51 × 220/2

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first  n  terms.

S 17  = 289

We know, Sum of n terms;

S 7 =  7 /2 [2 a  +( n  -1) d ]

S 7  = 7/2 [2 a  + (7 -1) d ]

49 = 7/2 [2 a  +6 d ]

7 = ( a +3 d )

a  + 3 d  = 7 ………………………………….  (i)

S 17  = 17/2 [2 a +(17-1) d ]

289 = 17/2 (2 a  +16 d )

17 = ( a +8 d )

a  +8 d  = 17 ……………………………….  (ii)

Subtracting equation  (i)  from equation  (ii) ,

From equation  (i) , we can write it as;

a +3(2) = 7

S n  =  n /2[2 a +( n -1) d ]

=  n /2[2(1)+( n  – 1)×2]

=  n /2(2+2 n -2)

=  n /2(2 n )

10. Show that  a 1 ,  a 2  … ,  a n  , … form an AP where  a n  is defined as below

(i)  a n  = 3+4 n (ii)  a n  = 9−5 n Also find the sum of the first 15 terms in each case.

(i)  a n  = 3+4 n

a 1  = 3+4(1) = 7

a 2  = 3+4(2) = 3+8 = 11

a 3  = 3+4(3) = 3+12 = 15

a 4  = 3+4(4) = 3+16 = 19

We can see here, the common difference between the terms are;

a 2  −  a 1  = 11−7 = 4

a 3  −  a 2  = 15−11 = 4

a 4  −  a 3  = 19−15 = 4

Hence,  a k  + 1 − a k  is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

Now, we know, the sum of nth term is;

S n  =  n /2[2 a +( n  -1) d ]

S 15  = 15/2[2(7)+(15-1)×4]

= 15/2[(14)+56]

(ii)  a n  = 9−5 n

a 1  = 9−5×1 = 9−5 = 4

a 2  = 9−5×2 = 9−10 = −1

a 3  = 9−5×3 = 9−15 = −6

a 4  = 9−5×4 = 9−20 = −11

a 2  −  a 1  = −1−4 = −5

a 3  −  a 2  = −6−(−1) = −5

a 4  −  a 3  = −11−(−6) = −5

Hence, a k  + 1  −  a k  is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

S n  =  n /2 [2 a  +( n -1) d ]

S 15  = 15/2[2(4) +(15 -1)(-5)]

= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

11. If the sum of the first  n  terms of an AP is 4 n  −  n 2 , what is the first term (that is  S 1 )? What is the sum of first two terms? What is the second term? Similarly find the 3 rd , the10 th  and the  n th  terms.

S n  = 4 n − n 2

First term,  a  =  S 1  = 4(1) − (1) 2  = 4−1 = 3

Sum of first two terms =  S 2 = 4(2)−(2) 2  = 8−4 = 4

Second term,  a 2  =  S 2  −  S 1  = 4−3 = 1

Common difference, d  =  a 2 − a  = 1−3 = −2

N th term, a n  =  a +( n −1) d 

= 3+( n  −1)(−2)

= 3−2 n  +2

Therefore,  a 3  = 5−2(3) = 5-6 = −1

a 10  = 5−2(10) = 5−20 = −15

Hence, the sum of first two terms is 4. The second term is 1.

The 3 rd , the 10 th , and the n th  terms are −1, −15, and 5 − 2 n  respectively.

12. Find the sum of first 40 positive integers divisible by 6.

The positive integers that are divisible by 6 are 6, 12, 18, 24 ….

We can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.

By the formula of sum of n terms, we know,

S n  =  n /2 [2 a  +( n  – 1) d ]

Therefore, putting n = 40, we get,

S 40  = 40/2 [2(6)+(40-1)6]

= 20[12+(39)(6)]

= 20(12+234)

13. Find the sum of first 15 multiples of 8.

The multiples of 8 are 8, 16, 24, 32…

The series is in the form of AP, having first term as 8 and common difference as 8.

Therefore,  a  = 8

By the formula of sum of nth term, we know,

S 15   = 15/2 [2(8) + (15-1)8]

= 15/2[16 +(14)(8)]

= 15/2[16 +112]

= 15(128)/2

14. Find the sum of the odd numbers between 0 and 50.

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.

Therefore, we can see that these odd numbers are in the form of A.P.

First term, a = 1

Common difference, d = 2

Last term, l = 49

By the formula of last term, we know,

l  =  a +( n −1)  d

49 = 1+( n −1)2

48 = 2( n  − 1)

n  − 1 = 24

n  = 25 = Number of terms

S n  =  n /2( a  + l )

S 25  = 25/2 (1+49)

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

We can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.

Therefore, a  = 200 and d  = 50

Penalty that has to be paid if contractor has delayed the work by 30 days =  S 30

S 30 = 30/2[2(200)+(30 – 1)50]

= 15[400+1450]

Therefore, the contractor has to pay Rs 27750 as penalty.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Let the cost of 1 st  prize be Rs. P .

Cost of 2 nd  prize = Rs. P  − 20

And cost of 3 rd  prize = Rs. P  − 40

We can see that the cost of these prizes are in the form of A.P., having common difference as −20 and first term as  P .

Thus, a  =  P and d  = −20

Given that,  S 7  = 700

7/2 [2 a  + (7 – 1) d ] = 700

a  + 3(−20) = 100

a  −60 = 100

Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

It can be observed that the number of trees planted by the students is in an AP.

1, 2, 3, 4, 5………………..12

First term,  a  = 1

Common difference,  d  = 2−1 = 1

S 12   = 12/2 [2(1)+(12-1)(1)]

Therefore, number of trees planted by 1 section of the classes = 78

Number of trees planted by 3 sections of the classes = 3×78 = 234

Therefore, 234 trees will be planted by the students.

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

Perimeter of a semi-circle = π r

P 1  = π(0.5) = π/2 cm

P 2  = π(1) = π cm

P 3  = π(1.5) = 3π/2 cm

Where, P 1,   P 2 ,  P 3  are the lengths of the semi-circles.

Hence we got a series here, as,

π/2, π, 3π/2, 2π, ….

P 1   = π/2 cm

P 2  = π cm

Common difference, d  =  P 2 – P 1 = π – π/2 = π/2

First term = P 1 =  a  = π/2 cm

By the sum of n term formula, we know,

Therefor, Sum of the length of 13 consecutive circles is;

S 13  =  13 /2 [2(π/2) + (13 – 1)π/2]

=   13 /2 [π + 6π]

=13 /2 (7π)

=  13 /2 × 7 ×  22 /7

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

We can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18…

For the given A.P.,

First term, a  = 20 and common difference, d  =  a 2 − a 1  = 19−20 = −1

Let a total of 200 logs be placed in  n  rows.

Thus, S n  = 200

By the sum of nth term formula,

S 12   = 12/2 [2(20)+( n  -1)(-1)]

400 =  n  (40− n +1)

400 =  n  (41- n )

400 = 41 n − n 2

n 2 −41 n  + 400 = 0

n 2 −16 n −25 n +400 = 0

n ( n  −16)−25( n  −16) = 0

( n  −16)( n  −25) = 0

Either ( n  −16) = 0 or  n −25 = 0

n  = 16 or  n  = 25

By the nth term formula,

a 16  = 20+(16−1)(−1)

a 16  = 20−15

Similarly, the 25 th term could be written as;

a 25  = 20+(25−1)(−1)

a 25  = 20−24

It can be seen, the number of logs in 16 th  row is 5 as the numbers cannot be negative.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16 th  row is 5.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? [Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2×5+2×(5+3)]

The distances of potatoes from the bucket are 5, 8, 11, 14…, which is in the form of AP.

Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.

Therefore, distances to be run w.r.t distances of potatoes, could be written as;

10, 16, 22, 28, 34,……….

Hence, the first term, a  = 10 and d  = 16−10 = 6

S 10   = 10/2 [2(10)+(10 -1)(6)]

Therefore, the competitor will run a total distance of 370 m.

Exercise 5.4 Page: 115

1. Which term of the AP: 121, 117, 113, . . ., is its first negative term? [Hint: Find n for a n < 0]

Given the AP series is 121, 117, 113, . . .,

Thus, first term, a = 121

Common difference, d = 117-121= -4

a n  = 121+(n−1)(-4)

To find the first negative term of the series, a n < 0

125-4n < 0

125 < 4n

Therefore, the first negative term of the series is 32 nd term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

From the given statements, we can write,

a 3 + a 7 = 6 …………………………….(i)

a 3 ×a 7 = 8 ……………………………..(ii)

Third term, a 3 = a+(3 -1)d

a 3 = a + 2d………………………………(iii)

And Seventh term, a7= a+(7-1)d

a 7 = a + 6d ………………………………..(iv)

From equation (iii) and (iv), putting in equation(i), we get,

a+2d +a+6d = 6

a = 3–4d …………………………………(v)

Again putting the eq.(iii) and (iv), in eq. (ii), we get,

(a+2d)×(a+6d) = 8

Putting the value of a from equation (v), we get,

(3–4d +2d)×(3–4d+6d) = 8

(3 –2d)×(3+2d) = 8

3 2 – 2d 2 = 8

9 – 4d 2 = 8

d = 1/2 or -1/2

Now, by putting both the values of d, we get,

a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1, when d = 1/2

a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5, when d = -1/2

We know, the sum of nth term of AP is;

So, when a = 1 and d=1/2

Then, the sum of first 16 terms are;

S 16  =  16 /2 [2 +(16-1)1/2] = 8(2+15/2) = 76

And when a = 5 and d= -1/2

S 16  =  16 /2 [2(5)+(16-1)(-1/2)] = 8(5/2)=20

Distance between the rungs of the ladder is 25cm.

Therefore, total number of rungs = 250/25 + 1 = 11

As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude now, that the rungs are decreasing in an order of AP.

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.

First term, a = 45

Last term, l = 25

Number of terms, n = 11

Now, as we know, sum of nth terms is equal to,

S n = n/2(a+ l )

S n = 11/2(45+25) = 11/2(70) = 385 cm

Hence, the length of the wood required for the rungs is 385cm.

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint :Sx – 1 = S49 – Sx ]

Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

First term, a = 1

Common difference, d=1

Let us say the number of x th houses can be represented as;

Sum of nth term of AP = n/2[2a+(n-1)d]

Sum of number of houses beyond x house = S x-1

= (x-1)/2[2(1)+(x-1-1)1]

= (x-1)/2 [2+x-2]

= x(x-1)/2 ………………………………………(i)

By the given condition, we can write,

S 49 – S x = {49/2[2(1)+(49-1)1]}–{x/2[2(1)+(x-1)1]}

= 25(49) – x(x + 1)/2 ………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

x(x-1)/2 = 25(49) – x(x+1)/2

As we know, the number of houses cannot be a negative number. Hence, the value of x is 35.

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1 4 m and a tread of 1 2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = ¼ ×1/2 ×50 m 3 .]

As we can see from the given figure, the first step is ½ m wide, 2 nd step is 1m wide and 3 rd step is 3/2m wide. Thus we can understand that the width of step by ½ m each time when height is ¼ m. And also, given length of the steps is 50m all the time. So, the width of steps forms a series AP in such a way that;

½ , 1, 3/2, 2, ……..

Volume of steps = Volume of Cuboid

= Length × Breadth Height

Volume of concrete required to build the first step = ¼ ×1/2 ×50 = 25/4

Volume of concrete required to build the second step =¼ ×1×50 = 25/2

Volume of concrete required to build the second step = ¼ ×3/2 ×50 = 75/4

Now, we can see the volumes of concrete required to build the steps, are in AP series;

25/4 , 25/2 , 75/4 …..

Thus, applying the AP series concept,

First term, a = 25/4

Common difference, d = 25/2 – 25/4 = 25/4

S n = n/2[2a+(n-1)d] = 15/2(2×(25/4 )+(15/2 -1)25/4)

Upon solving, we get,

S n = 15/2 (100)

Hence, the total volume of concrete required to build the terrace is 750 m 3 .

This chapter comes under Unit 3 algebra, and this unit has 20 marks allotted in the examination. Students can expect an average of 3 questions from arithmetic progressions. Along with Class 10 examinations, this topic is very important from the point of competitive exams.

Sub-topics of Class 10 Chapter 5 Arithmetic Progression

5.1 Introduction

In this chapter, we shall discuss patterns which we come across in our day-to-day life, in which succeeding terms are obtained by adding a fixed number to the preceding terms. We shall also see how to find their nth terms and the sum of n consecutive terms, and use this knowledge to solve some day-to-day problems.

5.2 Arithmetic Progressions

The topic describes Arithmetic Progressions, its definition and relatable terms, along with fine examples. You will also learn about Finite Arithmetic Progressions and Infinite Arithmetic Progressions. The general form of AP is a, a + d, a + 2d, a + 3d,…

5.3 nth Term of an AP This topic discusses various methods to determine the n th term of an AP. The concepts are explained with different types of problems solving techniques and finding the n th term of an AP. The examples mentioned in the chapter will help you while solving the exercise problems.

5.4 Sum of First n Terms of an AP The topics discuss different techniques to find the sum of the first n terms of an AP. It also provides suitable examples, which show different techniques to find the sum of the first n terms of AP.

5.5 Summary It gives an overview of the entire chapter and the important topics explained in the chapter. By going through the summary part, you can cover the entire chapter in a few points, which helps in memorising the essential concepts.

List of Exercises from Class 10 Maths Chapter 5 Arithmetic progression

Exercise 5.1 – 4 questions 1 MCQ and 3 descriptive type questions Exercise 5.2 – 20 questions, 1 fill in the blanks, 2 MCQs, 7 Short answer questions and 10 Long answer questions Exercise 5.3 – 20 Questions 3 fill in the blanks, 4 daily life examples, and 13 descriptive-type questions Exercise 5.4 5 Questions- 5 Long answer questions

This NCERT Solutions for Class 10 Maths is a perfect study material that will help you solve different kinds of problems. Solving these NCERT Solutions will help you understand the topic completely and help you lay a greater foundation for future studies.

In this chapter, students will discuss patterns in succeeding terms obtained by adding a fixed number to the preceding terms. They also see how to find nth terms and the sum of n consecutive terms. Students will learn arithmetic progression effectively when they solve daily life problems.

This chapter has Arithmetic Progression Derivation of the n th term and sum of the first n terms of an A.P. and their application in solving daily life problems. This is one of the important chapters from the point of the Class 10 examination. An arithmetic progression is a very basic and important topic to study, as almost all the competitive exams will ask questions on arithmetic progression.

Key Features of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

• Has answers to different types of questions such as MCQs and long answer questions.
• Solving this NCERT Solutions will make you well versed with important formulas.
• Acts as a basis to solve arithmetic progression problems asked in competitive examination.
• Has answers to all the exercise questions provided in the NCERT textbook
• Provides you with the necessary practice of solving questions
• You can solve different types of questions with varying difficulty levels
• Different examples taken from day to day life will help you understand the topic thoroughly.

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• RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions

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NCERT Solutions for Class 10 Maths Chapter 5 AP Arithmetic Progression

NCERT solutions for class 10 Maths chapter 5 all exercises of AP Arithmetic Progression in Hindi and English Medium updated for academic session 2024-25. Class 10 Maths solution is updated according to new textbooks published for CBSE board 2024 exams.

10th Maths Chapter 5 Exercise 5.1 Solutions

• Class 10th Maths Exercise 5.1 in English
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10th Maths Chapter 5 Exercise 5.2 Solutions

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Class 10 Maths Chapter 5 solutions are useful for not only for CBSE Board but UP Board, MP Board, Gujrat Board, etc. also who are using NCERT Textbooks as a course books. Uttar Pradesh Madhyamik Shiksha parishad, Prayagraj has implemented NCERT Books for Class 10 students. So, UP Board High school students can download UP Board Solutions for Class 10 Maths Chapter 5 in Hindi and English Medium here.

All the Online and Offline Apps , NCERT solutions are based on latest NCERT Books. These apps are applicable for UP Board (Higher Secondary) with CBSE Board and other boards who are using NCERT Books 2024-25 in Hindi and English medium. Download Class 10 Maths Apps based on updated NCERT Solutions for new academic session 2024-25.

NCERT Solutions for class 10 Maths chapter 5 all exercises are given to free use. NCERT Solutions 2024-25 and NCERT books are in PDF format to study online/offline by downloading in your device. Go through this page completely to know more about arithmetic series.

Extra Questions on 10th Maths Chapter 5?

What is an arithmetic progression (ap).

An arithmetic progression (AP) is a list (or pattern or series) of numbers in which each next term is obtained by adding or subtracting a fixed number to the preceding term except the first term. This fixed number is called the common difference of the AP, it may be positive, negative or zero.

What are the objective of studying Arithmetic Progression?

Objective of studying Arithmetic Progression – AP To identify arithmetic progression from a given list of numbers, to determine the general term of an arithmetic progression and to find the sum of first n terms of an arithmetic progression.

What is general form of Arithmetic Progression?

a, a + d, a + 2d, a + 3d, . . . is the general form of Arithmetic Progression where a is first term and d is common difference.

Be connected with us to get the latest update regarding to CBSE solutions, Sample Papers, board questions and all other study material related to class 10 Maths solutions of AP.

TWO MARKS QUESTIONS Find how many integers between 200 and 500 are divisible by 8. [CBSE 2017] THREE MARKS QUESTIONS 1. Find the sum of n terms of the series (4 – 1/n) + (4 – 2/n) + (4 – 3/n) + …. [CBSE 2017] 2. If the mth term of an AP is 1/n and nth term is 1/m then show that its (mn)th term is 1. [CBSE 2017] FOUR MARKS QUESTIONS 1. If the sum of first m terms of an A. P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero. [CBSE 2017] 2. The ratio of the sums of first m and first n terms of an AP is m²:n². Show that the ratio of its mth and nth terms is (2m – 1):(2n – 1). [CBSE 2017]

About Arithmetic Progression – AP

a, a + d, a + 2d, a + 3d, . . . is called the general form of an AP, where a is the first term and d the common difference. If there are a finite number of terms in the AP, then it is called a finite AP. The general formula for finding nth term is given by a + (n-1)d and the sum of n terms is given by n/2[2a + (n-1)d]. The last term is denoted by l and given by a + (n-1)d.

Historical Facts!

1. Leonardo Pisano Bigollo also known as Leonardo of Pisano, Leonardo Bonacci, Leonardo Fibonacci was an Italian mathematician. He gave Fibonacci series on the basis of, how fast rabbits could breed in ideal circumstances. Fibonacci Series: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 … Here, every next term is sum of previous two terms. For the solutions of other maths chapters of class 10, click here. 2. Once, when famous mathematician Carl Friedrich Gauss (1777 – 1855) misbehaved in primary school, his teacher I.G. Buttner gave him a task to add a list of integers from 1 to 100. Gauss’s method was to realise that pairwise addition of terms from opposite ends of the list yielded identical intermediate sum: 1 + 100 = 2 + 99 = 3 + 98 = … = 50 + 51 = 101. So, here 1 + 2+ 3 + 4 + …. + 100 = the sum of 50 sums each equal to 101. Therefore, the sum is 5050. Finally he gave the answer in seconds.

How to Practice Arithmetic Progression in Class 10 Maths?

We all know the current state of education infrastructure of the country after the COVID-19 impact. We say education is evolving from Didactic methodology to concept-based education. However, it is still very expensive for most of the students to gain access. The disproportionate effects of the COVID-19 pandemic affected brilliant and average minds also. Undoubtedly, students and teachers are sailing the same boat to make the preparations effective. Some of the students are still finding ways to score 100% in the Maths board exam for 10th grade. We selected some of the suggestions that can help you to score well in Chapter 4 AP of class 10th.

Step 1: Learn class 10 Maths chpater 5 with basics and Examples before starting.

Step 2: to solve sums of 10th maths chpater 5 take help from online learning platform., step 3: class 10 maths chapter 5 needs regular homework and revision session., step 4: once go through the chapter 5 ncert 10th maths before it starts in the class., step 5: be confident with the help of regular practice in ap..

Which exercise of 10th Maths chapter 5 need less efforts?

Class 10th Maths exercise 5.2 and 5.3 are easier to solve and the scoring also. The questions given in these two exercise are based on just to formulae – nth term and sum to n terms. Most of the questions get solved just by putting the values in proper formula. So, a student can score more easily in these two exercises.

How many exercises are there in 10th Maths Chapter 5 AP?

There are 4 exercises in chapter 4 of class 10 Maths subject. Out of these exercise 5.4 is an optional exercise, which is not important for examination point of view. It is given just to practice and enhance the knowledge in arithmetic progression.

Is chapter 5 Arithmetic Progression of 10th Maths interesting?

Yes, chapter 5 of 10th Maths is very interesting chapter because you can observe Arithmetic Progressions around you at various points. Examples of AP around you: 1) The heights (in cm) of some students of a school standing in a queue in the morning assembly are 147, 148, 149, . . ., 157. 2) The cash prizes (in Rs) given by a school to the toppers of Classes I to XII are, respectively, 200, 250, 300, 350, . . ., 750.

Is there any optional exercise in chapter 5 Class 10th Math?

Yes, there is optional exercise in chapter 5 of 10th Maths. Fourth exercise (exercise 5.4) is the optional exercise. Exercise 5.4 is not an easy exercise because this exercise contains very tricky questions.

Why chapter 5 of 10th Math important?

Chapter 5 (Arithmetic Progression) of 10th Math is important because students have to study Arithmetic Progression in class 11 maths also. Arithmetic Progression(chapter 5) of 10th maths is base of Arithmetic Progression of class 11th maths.

« Chapter 4: Quadratic Equations

Chapter 6: triangles ».

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• Important Questions for CBSE Class 10 Maths Chapter 5 - Arithmetic Progressions

CBSE Class 10 Maths Important Questions Chapter 5 - Arithmetic Progressions - Free PDF Download

Class 10th plays an essential role in the career of everyone. The class has a board examination, and its mark sheet is also referred to in the future life for various purposes. Math is a crucial subject and is also a comparatively tougher one. The students of Class 10th are required to prepare all the topics well and frequently practice the concepts for a better understanding. Students are advised to practice the problems regularly for all the chapters, and especially for Chapter 5, they must refer to the set of Arithmetic Progression important questions.

Class 10 Maths Chapter 5 Important questions carry all the frequently asked questions and other related ones for Arithmetic Progression. Vedantu experts develop these Arithmetic Progression Class 10 important questions to help them get a better resource for the examination period revision and know the type of questions asked in the examinations.

Vedantu is a platform that provides free CBSE Solutions and other study materials for students. You can also Download NCERT Solutions Class 10 Maths and NCERT Solution Class 10 Science to help you to revise complete Syllabus and score more marks in your examinations.

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Class 10 Maths Chapter 5 Extra Questions – PDF Download

1. The fourth term of an AP is zero. Prove that its \[{{25}^{th}}\] term is triple its \[{{11}^{th}}\] term.

Ans: Given that \[{{a}_{4}}=0~\]

\[\Rightarrow a+3d=0\]

\[\Rightarrow a=-3d\]                                                  …(1)

We have to prove that \[{{a}_{25}}=3{{a}_{11}}\]

\[\Rightarrow a+24d=3\left( a+10d \right)\]

\[\Rightarrow a+24d=3a+30d\]

From equation (1),

\[\Rightarrow -3d+24d=3\left( -3d \right)+30d\]

\[\Rightarrow 21d=21d\]

\[\therefore LHS=RHS\]

Hence proved. 

2.Find the \[{{20}^{th}}\] term from the end of the AP \[3,8,13,...,253\]. 

Ans: Given AP \[3,8,13,...,253\]

Last term is \[253\]

\[{{a}_{20}}\] from end is \[I-(n-1)d\]

Putting the values we get,

\[\Rightarrow 253-(20-1)5\]

\[\Rightarrow 158\] 

3.If the \[{{p}^{th}}\] , \[{{q}^{th}}\] and \[{{r}^{th}}\] term of an AP is \[x,y\] and \[z\] respectively, show that \[x\left( q-r \right)+y\left( r-p \right)+z\left( p-q \right)=0\] .

Ans: \[{{\text{p}}^{th}}\text{term}\Rightarrow x=A+\left( p-1 \right)D\]

\[{{q}^{th}}\text{term}\Rightarrow y=A+\left( q-1 \right)D\]

\[{{r}^{th}}\text{term}\Rightarrow z=A+\left( r-1 \right)D\]

We have to prove:

\[x\left( q-r \right)+y\left( r-p \right)+z\left( p-q \right)=0\] 

\[\Rightarrow \left\{ A+\left( p-1 \right)D \right\}\left( q-r \right)+\left\{ A+\left( q-1 \right)D \right\}\left( r-p \right)~+\left\{ A+\left( r-1 \right)D \right\}\left( p-q \right)~=0\]

\[\Rightarrow A\left\{ \left( q-r \right)+\left( r-p \right)+\left( p-q \right) \right\}+D\{\left( p-1 \right)\left( q-r \right)~+\left( r-1 \right)\left( r-p \right)+\left( r-1 \right)\left( p-q \right)\}=0\]

\[\Rightarrow A\cdot 0+D\{p\left( q-r \right)+q\left( r-p \right)+r\left( p-q \right)-\left( q-r \right)\left( r-p \right)-\left( p-q \right)\}=0\]

\[\Rightarrow A\cdot 0+D\cdot 0=0\]

4. Find the sum of first \[40\] positive integers divisible by \[6\] also find the sum of first \[20\] positive integers divisible by \[5\] or \[6\] . 

Ans: Numbers which are divisible by \[6\] are \[6,12,18,...,240\]

Using the formula,

\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]

\[\Rightarrow {{S}_{40}}=\dfrac{40}{2}\left[ 12+39\left( 6 \right) \right]\]

\[\therefore {{S}_{40}}=4920\]

Numbers div by \[5\] or \[6\]  are 

\[30,60,...600\]

\[\Rightarrow {{S}_{20}}=\dfrac{20}{2}\left[ 60+19\left( 30 \right) \right]\]

\[\therefore {{S}_{20}}=6300\]

5. A man arranges to pay a debt of Rs. \[3600\] in \[40\] monthly installments which are in an AP. When \[30\] installments are paid he dies leaving one-third of the debt unpaid. Find the value of the first installment. 

Ans: Let the value of instalment be \[{{S}_{40}}=3600\] 

\[\Rightarrow \dfrac{40}{2}[2a+39d]=3600\]

\[\Rightarrow [2a+39d]=180\]                                 …(1)

If thirtyinstallments are paid then,

\[\Rightarrow {{\text{S}}_{30}}=\dfrac{30}{2}[2a+29d]\]

\[\Rightarrow \dfrac{30}{2}[2a+29d]=\dfrac{2}{3}\times 3600\]

\[\Rightarrow 2a+29d=160\]                                    …(2)

Solving equations (1) and (2) we get,

Therefore, the first instalment is \[Rs.51\] . 

6. Find the sum of all three-digit numbers which leave remainder \[3\] when divided by \[5\].

    Ans: Three-digit numbers which leave remainder \[3\] when divided by \[5\] are \[103,108,...,998\]

    To find \[n\] ,

\[{{a}_{n}}=a+\left( n-1 \right)d\]

\[\Rightarrow 998=103+5\left( n-1 \right)\]

\[\therefore n=180\]

To find sum use the formula,

\[\Rightarrow {{S}_{180}}=\dfrac{180}{2}\left[ 2\left( 103 \right)+5\left( 180-1 \right) \right]\]

\[\therefore {{S}_{180}}=99090\]  

7.Find the value of \[x\] if \[2x+1,{{x}^{2}}+x+1,3{{x}^{2}}-3x+3\] are consecutive terms of an AP.

Ans: We know that consecutive terms of an AP have a same common difference

\[{{a}_{2}}\text{ }{{a}_{1}}=\text{ }{{a}_{3}}\text{ }{{a}_{2}}~\]

\[\Rightarrow {{x}^{2}}\text{+}x+1-2x-1=3{{x}^{2}}-3x+3-{{x}^{2}}-x-1\]

\[\Rightarrow {{x}^{2}}-3x+2=0~\]

\[\Rightarrow \left( x-1 \right)\left( x2 \right)=0\]

\[\therefore x=1,2\] 

8. Raghav buys a shop for Rs. \[1,20,000\] . He pays half the balance of the amount in cash and agrees to pay the balance in \[12\] annual installments of Rs. \[5000\] each. If the rate of interest is \[12%\] and he pays with the installment the interest is due for the unpaid amount. Find the total cost of the shop. 

Ans: Given that balance is Rs. \[60,000\] in \[12\]installments of Rs. \[5000\] each. 

Amount of first instalment is \[5000+\left( \dfrac{12}{100}\times 60,000 \right)=12200\]

Amount of second instalment is \[5000+\left( \dfrac{12}{100}\times 55,000 \right)=11600\] 

Amount of third instalment is \[5000+\left( \dfrac{12}{100}\times 50,000 \right)=11000\] 

∴ AP is \[12200,11600,11000,...\]

Here, \[d=600\]

Cost of shop is = \[60,000\] + [sum of \[12\] instalment] 

\[\Rightarrow 60,000+\dfrac{12}{2}[24,400-6600]=1,66,800\]

The total cost of the shop is Rs. \[1,66,800\] .

9. Prove that  $ {{a}_{m+n}}+{{a}_{m-n}}=2{{a}_{m}} $   .

Ans: We know that,

\[{{a}_{m+n}}={{a}_{1}}+\left( m+n-1 \right)d\]                              …(1)

\[{{a}_{m-n}}={{a}_{1}}+\left( m-n-1 \right)d\]                              …(2)

\[{{a}_{m}}={{a}_{1}}+\left( m-1 \right)d\]                                       …(3)

Adding equations (1) and (2),

\[{{a}_{m+n}}+{{a}_{m-n}}={{a}_{1}}+\left( m+n-1 \right)d+{{a}_{1}}+\left( m-n-1 \right)d\]

\[{{a}_{m+n}}+{{a}_{m-n}}=2[{{a}_{1}}+\left( m-1 \right)d]~\]

From equation (3) we can conclude that,

\[{{a}_{m+n}}+{{a}_{m-n}}=2{{a}_{m}}\]

Hence proved.

10. If the roots of the equation  $ (b-c){{x}^{2}}+(c-a)x+(a-b)=0 $   are equal show that \[a,b,c\] are in A.P

Ans: If the roots of the equation \[a{{x}^{2}}+bx+c=0\] are equal then,

\[{{b}^{2}}-4ac=0\]

Therefore, for equation  $ (b-c){{x}^{2}}+(c-a)x+(a-b)=0 $ 

\[\Rightarrow {{\left( c-a \right)}^{2}}-4\left( b-c \right)\left( a-b \right)=0\]

\[\Rightarrow {{c}^{2}}+{{a}^{2}}-2ca-4\left( ab-{{b}^{2}}-ca+cb \right)=0\]

\[\Rightarrow {{c}^{2}}+{{a}^{2}}+4{{b}^{2}}+2ca-4ba-4bc=0\]

\[\Rightarrow {{\left( a-2b+c \right)}^{2}}=0\]

\[\Rightarrow \left( a-2b+c \right)=0\]

\[\therefore \left( a+c \right)=2b\]

Hence proved that \[a,b,c\] are in AP.

11.  Balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second two balls, and so on. If \[669\] more balls are added, then all the balls can be arranged in the shape of a square and each of its sides then contains \[8\] balls less than each side of the triangle. Find the initial number of balls. 

        Ans: Let there be \[n\] balls in each side of the triangle 

∴ No. of ball (in ∆),

\[1+2+3+...=\dfrac{n\left( n+1 \right)}{2}\] 

No. of balls in each side square is \[n-8\]

No. of balls in square is \[{{\left( n-8 \right)}^{2}}\]

From the given data,

\[\dfrac{n\left( n+1 \right)}{2}+669={{\left( n-8 \right)}^{2}}\]

\[\Rightarrow {{n}^{2}}+n+1338=2({{n}^{2}}-16n+64)\]

\[\Rightarrow {{n}^{2}}-33n-1210=0\]

\[\Rightarrow \left( n-55 \right)\left( n+22 \right)=0\]

\[\Rightarrow n=55,-22\]

Negative number is not possible,

Number of balls is \[55\] .

12.  Find the sum of  $ \left( 1-\dfrac{1}{n} \right)+\left( 1-\dfrac{2}{n} \right)+\left( 1-\dfrac{3}{n} \right)+...nterms $  .

       Ans: We have to find sum of  $ \left( 1-\dfrac{1}{n} \right)+\left( 1-\dfrac{2}{n} \right)+\left( 1-\dfrac{3}{n} \right)+...nterms $ 

 $ \Rightarrow \left[ 1+1+...nterms \right]-\left[ \left( \dfrac{1}{n} \right)+\left( \dfrac{2}{n} \right)+\left( \dfrac{3}{n} \right)+...nterms \right] $ 

\[\Rightarrow {{S}_{n}}=n-\dfrac{n}{2}\left[ 2\left( \dfrac{1}{n} \right)+\left( n-1 \right)\left( \dfrac{1}{n} \right) \right]\]

\[\Rightarrow {{S}_{n}}=n-1-\dfrac{n}{2}+\dfrac{1}{2}\]

\[\therefore {{S}_{n}}=\dfrac{n-1}{2}\]

13.  If the following terms form a AP. Find the common difference & write the next \[3\] terms \[3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...\] 

Ans: Common difference is \[d=3+\sqrt{2}-3\]

  \[\therefore d=\sqrt{2}\]

  Next three terms are  $ 3+4\sqrt{2},3+5\sqrt{2},3+6\sqrt{2}.... $  

 14.  Find the sum of \[a+b,a-b,a-3b,...22terms\] .

Ans: Using the formula,

\[\Rightarrow {{S}_{22}}=\dfrac{22}{2}\left[ 2\left( a+b \right)+\left( 22-1 \right)\left( -2b \right) \right]\]

\[\Rightarrow {{S}_{22}}=11\left[ 2a-40b \right]\]

\[\therefore {{S}_{22}}=22a-440b\] 

15. Write the next two terms \[\sqrt{12,}\sqrt{27},\sqrt{48},\sqrt{75},...\]

Ans: Here, \[a=\sqrt{12}\] and 

\[d=\sqrt{27}-\sqrt{12}\]

\[\Rightarrow d=3\sqrt{3}-2\sqrt{3}\]

\[\therefore d=\sqrt{3}\]

Next two terms are  $ \sqrt{108},\sqrt{147} $

16. If the \[{{p}^{th}}\] term of an AP is \[q\] and the \[{{q}^{th}}\] term is \[p\] . P.T its \[{{n}^{th}}\] term is \[p+q-n\] . 

Ans: Given that 

\[{{a}_{p}}=q\]

\[\Rightarrow a+\left( p-1 \right)d=q\]                                  …(1)

\[{{a}_{q}}=p\]

\[\Rightarrow a+\left( q-1 \right)d=p\]                                  …(2)

From equations (1) and (2),

\[\Rightarrow d\left[ pq \right]=qp\text{ }\]

\[\therefore d=-1\]

Putting it in equation (1) we get,

\[a=q+p-1~\]

We know that,

\[\Rightarrow {{a}_{n}}=\left( q+p1 \right)-1\left( n1 \right)\]

\[\therefore {{a}_{n}}=\left( q+p-n \right)\]

17.  If  $ \dfrac{1}{x+2},\dfrac{1}{x+3},\dfrac{1}{x+5} $    are in AP find \[x\] . 

Ans:  We have to find \[x\] if  $ \dfrac{1}{x+2},\dfrac{1}{x+3},\dfrac{1}{x+5} $  are in AP. 

\[\Rightarrow \dfrac{1}{x+2}+\dfrac{1}{x+5}=\dfrac{2}{\left( x+3 \right)}\]

\[\Rightarrow \dfrac{2x+7}{{{x}^{2}}+7x+10}=\dfrac{2}{x+3}\]

\[\Rightarrow \left( 2x+7 \right)\left( x+3 \right)=2\left( {{x}^{2}}+7x+10 \right)\]

\[\Rightarrow \left( 2{{x}^{2}}+13x+21 \right)=\left( 2{{x}^{2}}+14x+20 \right)\]

\[\therefore x=1\]

18.  Find the middle term of the AP \[1,8,15,...,505\] . 

Ans: Total number of terms,

\[\Rightarrow 505=1+7\left( n-1 \right)\]

\[\therefore n=73\]

\[\therefore n=36,37\] are middle terms

\[{{a}_{36}}=1+7\left( 36-1 \right)\]

\[\therefore {{a}_{36}}=246\] and

 \[{{a}_{37}}=253\]

19. Find the common difference of an AP whose first term is \[100\] and the sum of whose first \[6\] terms is \[5\] times the sum of next \[6\] terms. 

Ans: Here \[a=100\] and 

\[{{a}_{1}}+{{a}_{2}}+...+{{a}_{6}}=5({{a}_{7}}+\ldots +{{a}_{12}})~\]

\[\Rightarrow 6\left( \dfrac{{{a}_{1}}+{{a}_{6}}}{2} \right)=5\times 6\left( \dfrac{{{a}_{7}}+{{a}_{12}}}{2} \right)\]

\[\Rightarrow a+a+5d=5\left[ a+6d+a+11d \right]\]

\[\Rightarrow 800+80d=0\]

\[\therefore d=-10\] 

20. Find the sum of all natural no. between \[101\] and \[304\] which are divisible by \[3\] or \[5\] . Find their sum.

Ans: The AP formed by numbers that are divisible by \[3\] or \[5\] is

\[105,120,135,....300\]

The sum of this AP is:

\[\Rightarrow {{S}_{14}}=\dfrac{14}{2}\left[ 300+13\left( 15 \right) \right]\]

\[\Rightarrow {{S}_{14}}=19035\]

21. The ratio of the sum of first \[n\] terms of two AP’s is \[7n+1:4n+27\] . Find the ratio of their \[{{11}^{th}}\] terms. 

Ans: Using the formula \[{{a}_{n}}=a+\left( n-1 \right)d\] ,  

\[\dfrac{\dfrac{m}{2}\left[ 2{{a}_{1}}+(n-1){{d}_{1}} \right]}{\dfrac{m}{2}\left[ 2{{a}_{2}}+(n-1){{d}_{2}} \right]}=\dfrac{7n+1}{4n+27}\]

\[\Rightarrow \dfrac{2{{a}_{1}}+(n-1){{d}_{1}}}{2{{a}_{2}}+(n-1){{d}_{2}}}=\dfrac{7n+1}{4n+27}\]

Here \[n=21\] ,

\[\Rightarrow \dfrac{2{{a}_{1}}+20{{d}_{1}}}{2{{a}_{2}}+20{{d}_{2}}}=\dfrac{7\times 21+1}{4(21)+27}\]

\[\Rightarrow \dfrac{{{a}_{1}}+10{{d}_{1}}}{{{a}_{2}}+10{{d}_{2}}}=\dfrac{148}{111}\]

\[\therefore \dfrac{{{a}_{1}}+10{{d}_{1}}}{{{a}_{2}}+10{{d}_{2}}}=\dfrac{4}{3}\]

Ratio of their \[{{11}^{th}}\] terms is \[4:3\] 

22. If there are \[2n+1\] terms in an AP, prove that the ratio of the sum of odd terms and the sum of even terms is \[n+1:n\] .

Ans: Use the formula \[{{a}_{n}}=a+\left( n-1 \right)d\] ,  

Sum of odd terms: 

\[{{s}_{1}}=\text{ }{{a}_{1}}+\text{ }{{a}_{3}}+\text{ }\ldots \ldots \ldots \text{ }{{a}_{2n+1}}\] 

\[{{s}_{1}}=\dfrac{n+1}{2}[2{{a}_{1}}+2nd]\] 

\[\therefore {{s}_{1}}=\left( n+1 \right)\left( a+nd \right)~\]

Sum of even terms:

\[{{s}_{2}}=\text{ }{{a}_{2}}+\text{ }{{a}_{4}}+\text{ }\ldots ..\text{ }{{a}_{2n}}~\] 

 $ {{s}_{2}}=\dfrac{n}{2}[a+d+a+(2n-1)d] $ 

\[\therefore {{s}_{2}}=n[a+nd]\]

\[\therefore \dfrac{{{s}_{1}}}{{{s}_{2}}}=\dfrac{(n+1)(a+nd)}{n(a+nd)}\]

\[\therefore \dfrac{{{s}_{1}}}{{{s}_{2}}}=\dfrac{(n+1)}{n}\]

23. Find the sum of all natural numbers amongst the first one thousand numbers which are neither divisible \[2\] nor by \[5\] .

Ans: Sum of all natural numbers in first thousand integers which are not divisible by two i.e. sum of odd integers. 

1 + 3 + 5 + ………. + 999 

n = 500 

 $ {{S}_{500}}=\dfrac{500}{2}[1+999] $  

 $ \therefore {{S}_{500}}=250000 $ 

Odd No’s which are divisible by five 

5 + 15 + 25 …….. + 995 

n = 100 

 $ {{S}_{100}}=\dfrac{100}{2}[5+995] $  

 $ \therefore {{S}_{100}}=50000 $ 

Required sum is  $ 250000-50000=20,000 $ 

Very Short Answer Questions (1 Mark)

1. The next term of the AP in  $ {{1}^{2}},{{5}^{2}},{{7}^{2}},73,.... $  is 

  $ 97 $ 

  $ 92 $ 

  $ 99 $ 

  $ 95 $ 

Ans: (a)  $ 97 $ 

2. The \[{{10}^{th}}\] term of the AP in \[2,7,12,...\] is 

 \[45\]

 \[47\] 

 \[48\]

 \[50\]

Ans: (b) \[47\]

3. If the sum of the circumferences of two circles with radii \[{{R}_{1}}\] and \[{{R}_{2}}\] is equal to the circumference of a circle of Radius, then \[R\]  

 \[{{R}_{1}}+{{R}_{2}}=R\] 

 \[{{R}_{1}}+{{R}_{2}}>R\] 

 \[{{R}_{1}}+{{R}_{2}}<R\]  

 None of these 

Ans: (a) \[{{R}_{1}}+{{R}_{2}}=R\] 

4. If the perimeter of a circle is equal to that of a square, then the ratio of their area is 

 \[22:7\]

 \[7:22\]

 \[11:14\]

Ans: (b) \[14:11\]

5. Area of a sector of angle \[{{p}^{\circ }}\] of a circle with radius \[R\] is 

  \[\dfrac{P}{100}\times 2\pi R\] 

  \[\dfrac{P}{180}\times \pi {{R}^{2}}\] 

  \[\dfrac{P}{360}\times 2\pi R\] 

  \[\dfrac{P}{720}\times 2\pi {{R}^{2}}\] 

Ans: (d) \[\dfrac{P}{720}\times 2\pi {{R}^{2}}\]

6. Area of the sector of angle \[{{60}^{\circ }}\] of a circle with radius \[10\] cm is 

  \[52\dfrac{5}{21}c{{m}^{2}}\] 

  \[52\dfrac{8}{21}c{{m}^{2}}\] 

  \[52\dfrac{4}{21}c{{m}^{2}}\] 

 none of there 

Ans: (b)   \[52\dfrac{8}{21}c{{m}^{2}}\] 

7. \[{{11}^{th}}\] term of the AP \[-3,-\dfrac{1}{2},2,....\]  is 

 \[28\]

 \[22\]

 \[-38\]

  \[-48\dfrac{1}{2}\] 

Ans: (b) \[22\]

8. If \[{{17}^{th}}\] term of an AP exceeds its \[{{10}^{th}}\] term by \[7\] . The common difference is

 \[2\]

 \[-1\]

 \[3\]

 \[1\]

Ans: (d) \[1\]

9. Which of the following list of no. form an AP? 

 \[2,4,8,16,...\]

 \[2,\dfrac{5}{2},3,\dfrac{7}{2},...\]

 \[0.2,0.22,0.222,...\]

 \[1,3,9,27,...\]

Ans: (b) \[2,\dfrac{5}{2},3,\dfrac{7}{2},...\]

10.  The \[{{n}^{th}}\] term of the AP in \[2,5,8,...\] is 

 \[3n-1\]

 \[2n-1\]

 \[3n-2\]

 \[2n-3\]

Ans: (a) \[3n-1\]

11.  If  \[a,(a-2),3a\] are in AP, then value of \[a\] is 

 \[-3\]

 \[-2\]

Ans: (b) \[-2\]

12.  The sum of first \[n\] positive integers is given by 

 \[\dfrac{n(n-1)}{2}\] 

 \[\dfrac{n(2n+1)}{2}\] 

 \[\dfrac{n(n+1)}{2}\] 

none of these 

Ans: (c) \[\dfrac{n(n+1)}{2}\]

Short Answer Questions (2 Marks)

1. Find the missing variable from \[a,d,n\] and \[{{a}_{n}}\] , where \[a\] is the first term, \[d\] is the common difference and \[{{a}_{n}}\] is the \[{{n}^{th}}\] term of AP. 

(i).  \[a=7,d=3,n=8\]

Ans: (i)  \[a=7,d=3,n=8\]

We need to find \[{{a}_{n}}\] here. 

Using formula \[{{a}_{n}}=a+\left( n-1 \right)d\] ,

Putting the given values, 

\[{{a}_{8}}=7+3\left( 8-1 \right)\]

\[\therefore {{a}_{8}}=28\]

(ii). \[a=-18,n=10,{{a}_{n}}=0\]

Ans: (ii)  \[a=-18,n=10,{{a}_{n}}=0\]

\[0=-18+d\left( 10-1 \right)\]

\[\therefore d=2\]

(iii). \[d=-3,n=18,{{a}_{n}}=-5\]

Ans: (iii)  \[d=-3,n=18,{{a}_{n}}=-5\]

\[-5=a-3\left( 18-1 \right)\]

\[\therefore a=46\]

(iv). \[a=-18.9,d=2.5,{{a}_{n}}=3.6\]

Ans: (iv)  \[a=-18.9,d=2.5,{{a}_{n}}=3.6\]

\[3.6=-18.9+2.5\left( n-1 \right)\]

\[\therefore n=10\]

(v). \[a=3.5,d=0,n=105\]

Ans: (v)  \[a=3.5,d=0,n=105\]

\[{{a}_{n}}=3.5+0\left( 105-1 \right)\]

\[\therefore {{a}_{n}}=3.5\]

2. Choose the correct choice in the following and justify: 

(i). \[{{30}^{th}}\] term of the AP: \[10,7,4,...\] is 

 \[97\]

 \[77\]

\[-77\] 

Ans: (i) Given AP \[10,7,4,...\]

\[{{a}_{30}}=10-3\left( 29 \right)\]

\[\therefore {{a}_{30}}=-77\]

Therefore, the answer is (C). 

(ii). \[{{11}^{th}}\] term of the AP: \[-3,-\dfrac{1}{2},2,...\] is 

 \[-48\dfrac{1}{2}\] 

Ans: (ii) Given AP \[-3,-\dfrac{1}{2},2,...\]

\[{{a}_{11}}=-3-2.5\left( 11-1 \right)\]

\[\therefore {{a}_{11}}=22\]

Therefore, the answer is (B). 

3. Which term of the AP: \[3,8,13,18,...\] is \[78\] ? 

Ans: Here, \[a=3,d=5,{{a}_{n}}=78\]

\[78=3+5\left( n-1 \right)\]

\[\therefore n=16\]

It means \[{{16}^{th}}\] term of the given AP is equal to \[78\] . 

4. Find the number of terms in each of the following APs: 

(i). \[7,13,19,...,205\]

Ans: (i)  Here, \[a=7,d=6,{{a}_{n}}=205\]

\[205=7+6\left( n-1 \right)\]

\[\therefore n=34\]

Therefore, there are \[34\] terms in the given arithmetic progression. 

(ii). \[18,15\dfrac{1}{2},13,...,-47\]

 Ans: (ii) Here, \[a=18,d=-2.5,{{a}_{n}}=-47\]

\[-47=18-2.5\left( n-1 \right)\]

\[\therefore n=26\]

Therefore, there are \[26\] terms in the given arithmetic progression.

5. Check whether \[-150\] is a term of the AP: \[11,,8,5,2,...\]

Ans: Here, \[a=11,d=-3,{{a}_{n}}=-150\]

\[-150=11-3\left( n-1 \right)\]

\[\therefore n=\dfrac{164}{3}\] 

But, \[n\] cannot be in fraction. 

Therefore, our supposition is wrong. \[-150\] cannot be a term in AP. 

6. An AP consists of \[50\] terms of which \[{{3}^{rd}}\] term is \[12\] and the last term is \[106\] . Find the \[{{29}^{th}}\] term. 

\[{{a}_{3}}=12\] and \[{{a}_{50}}=106\]

\[\Rightarrow 12=a+2d\]                                      …(1)

\[\Rightarrow 106=a+49d\]                                …(2) 

\[\Rightarrow a=106-49d\]                                …(3)

These are equations consisting of two variables. 

Putting value of \[a\] in the equation (1), 

\[12=106-49d+2d\]

\[\Rightarrow d=2\] 

Putting it in equation (3),

\[\Rightarrow a=10649\left( 2 \right)\]

\[\therefore a=8\] 

Therefore, First term is \[a=8\] and Common difference is \[d=2\] 

To find \[{{29}^{th}}\] term use the formula \[{{a}_{n}}=a+\left( n-1 \right)d\] ,

\[\Rightarrow {{a}_{29}}=8+2\left( 29-1 \right)\]

\[\therefore {{a}_{29}}=64\]

7. How many multiples of \[4\] lie between \[10\] and \[250\]? 

Ans: Given AP \[12,16,20,...,248\]

Here, \[a=12,d=4,{{a}_{n}}=248\]

\[248=12+4\left( n-1 \right)\]

\[\therefore n=60\]

Therefore, sixty multiples of \[4\] lie between \[10\] and \[250\] .

8. Which term of the AP: \[121,117,113,...\] is its first negative term? 

Ans: Given AP: \[121,117,113,...\]

Here, \[a=121,d=117-121=-4\]

\[\Rightarrow {{a}_{n}}=121-4\left( n-1 \right)\]

\[\Rightarrow {{a}_{n}}=125-4n\]  

For the first negative term, \[{{a}_{n}}<0\]

\[\Rightarrow 125-4n<0\]

\[\Rightarrow n>\dfrac{125}{4}\]

\[\Rightarrow n>31\dfrac{1}{4}\]  

\[n\] is an integer and \[n>31\dfrac{1}{4}\] 

Hence, the first negative term is \[{{32}^{nd}}\] term 

9. The sum of the third and the seventh terms of an AP is \[6\] and their product is \[8\]. Find the sum of sixteen terms of the AP. 

Ans: Let the AP be \[a-4d,a-3d,a-2d,a-d,a,a+d,a+2d,a+3d,...\] 

Then, 

\[{{a}_{3}}=a-2d,{{a}_{7}}=a+2d\]

\[\Rightarrow {{a}_{3}}+{{a}_{7}}=a-2d+a+2d=6\]

\[\Rightarrow a=3\]                                               …(1)

\[(a-2d)(a+2d)=8\]

\[\Rightarrow {{a}^{2}}-4{{d}^{2}}=8\]

\[\Rightarrow {{d}^{2}}=\dfrac{1}{4}\]

\[\therefore d=\pm \dfrac{1}{2}\]  

Taking \[d=\dfrac{1}{2}\] ,

\[{{S}_{16}}=\dfrac{16}{2}[2\times (a-4d)+(16-1)d]\]

\[\Rightarrow {{S}_{16}}=\dfrac{16}{2}\left[ 2\times \left( 3-4\left( \dfrac{1}{2} \right) \right)+(16-1)\left( \dfrac{1}{2} \right) \right]\]

\[\therefore {{S}_{16}}=76\] 

Taking \[d=-\dfrac{1}{2}\]  , 

\[\Rightarrow {{S}_{16}}=\dfrac{16}{2}\left[ 2\times \left( 3-4\left( -\dfrac{1}{2} \right) \right)+(16-1)\left( -\dfrac{1}{2} \right) \right]\]

\[\therefore {{S}_{16}}=20\]

The sum of sixteen terms of the AP is \[{{S}_{16}}=20\]or \[{{S}_{16}}=76\] 

10. A ladder has rungs \[25\] cm apart (see figure). The rungs decrease uniformly in length from \[45\] cm, at the bottom to \[25\] cm at the top. If the top and the bottom rungs are \[2\dfrac{1}{2}\] m apart, what is the length of the wood required for the rungs? 

Ans: Number of rungs, \[(n)=\dfrac{2\dfrac{1}{2}\times 100}{25}\]

The length of the wood required for rungs = sum of ten rungs 

\[\Rightarrow \dfrac{10}{2}\left[ 25+45 \right]=350cm\]

11. The houses of a row are numbered consecutively from \[1\] to \[49\]. Show that there is a value of \[x\] such that the sum of the numbers of the houses preceding the house numbered \[x\] is equal to the sum of the numbers of the houses following it. Find this value of \[x\] . 

Ans: Here \[a=1,d=1\] 

\[{{S}_{x-1}}=\dfrac{x-1}{2}[2\times 1+(x-1-1)\times 1]\]

\[\Rightarrow {{S}_{x-1}}=\dfrac{{{x}^{2}}-x}{2}\]

\[{{S}_{x}}=\dfrac{x}{2}\left[ 2\times 1+(x-1)\times 1 \right]\]

\[\Rightarrow {{S}_{x}}=\dfrac{{{x}^{2}}+x}{2}\]

Now, \[{{S}_{49}}=\dfrac{49}{2}[2\times 1+(49-1)\times 1]\]

\[\Rightarrow {{S}_{49}}=49\times 25\] 

According to question, 

\[{{S}_{x-1}}={{S}_{49}}-{{S}_{x}}\]

\[\Rightarrow \dfrac{{{x}^{2}}-x}{2}=49\times 25-\dfrac{{{x}^{2}}+x}{2}\]

\[\Rightarrow \dfrac{{{x}^{2}}-x}{2}+\dfrac{{{x}^{2}}+x}{2}=49\times 25\]

\[\Rightarrow {{x}^{2}}=49\times 25\]

\[\Rightarrow x=\pm 35\] 

Since, \[x\] is a counting number, so negative value will be neglected. 

\[\therefore x=35\]

12. Find the first term and the common difference \[\dfrac{1}{3},\dfrac{5}{3},\dfrac{9}{3}\] .

Ans: Given AP: \[\dfrac{1}{3},\dfrac{5}{3},\dfrac{9}{3}\]

First term: \[a=\dfrac{1}{3}\]   

Common difference: \[d=\dfrac{5}{3}-\dfrac{1}{3}=\dfrac{4}{3}\] 

13.  Does \[\sqrt{3},\sqrt{6},\sqrt{9},....\]  form an AP? 

Ans: Here, \[{{a}_{1}}=\sqrt{3},{{a}_{2}}=\sqrt{6},{{a}_{3}}=\sqrt{9}\]

\[{{d}_{1}}=\sqrt{6}-\sqrt{3}\]

\[\Rightarrow {{d}_{1}}=\sqrt{3}(\sqrt{2}-1)\]

\[{{d}_{2}}=\sqrt{9}-\sqrt{6}\]

\[\Rightarrow {{d}_{2}}=3-\sqrt{6}\] 

Since \[{{d}_{1}}\ne {{d}_{2}}\] , 

Hence, it is not an AP. 

14. Which is the next term of the AP \[\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},....\] 

Ans:  Given AP is \[\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},....\]

Here, \[d=\sqrt{8}-\sqrt{2}\]

\[\Rightarrow d=\sqrt{2}\]

\[{{a}_{5}}=a+(5-1)d\]

\[\therefore {{a}_{5}}=5\sqrt{2}\] 

Next term is \[5\sqrt{2}\] or \[\sqrt{50}\] 

15. Find the \[{{11}^{th}}\] term from the last term of the AP. \[10,7,4,...,-62\]. 

Ans: Here, we consider the AP in reverse

\[\therefore a=-62,d=-(7-10)=3\]

\[{{a}_{11}}=a+10d\]

\[\therefore {{a}_{11}}=-32\]

Therefore, \[{{11}^{th}}\] term from the last is \[-32\] .

16.  If \[x+1,3x\] and \[4x+2\] are in A.P, find the value of \[x\] . 

Ans: Since \[x+1,3x\] and \[4x+2\] are in AP 

\[2(3x)=x+1+4x+2\]

\[\therefore x=3\]

17.  Find the sum of first \[n\] odd natural numbers. 

Ans: AP of odd numbers are \[1,3,5,7,.....\]

\[a=1,d=3-1=2\]

\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2\times 1+(n-1)2 \right]\]

\[\therefore {{S}_{n}}={{n}^{2}}\]

18.  Find the \[{{12}^{th}}\] term of the AP \[\sqrt{2},3\sqrt{2},5\sqrt{2}....\] 

Ans: Here, \[a=\sqrt{2},d=3\sqrt{2}-\sqrt{2}=2\sqrt{2}\]

\[{{a}_{12}}=a+11d\]

\[\Rightarrow {{a}_{12}}=\sqrt{2}+11\left( 2\sqrt{2} \right)\]

\[\therefore {{a}_{12}}=23\sqrt{2}\]

19.  Find the sum of first \[11\] terms of AP \[2,6,10,...\]

Ans: Here, \[a=2,d=6-2=4\]

\[\Rightarrow {{S}_{11}}=\dfrac{11}{2}\left[ 2\times 2+(11-1)\times 4 \right]\]

\[\therefore {{S}_{11}}=242\]

20.Find the sum of the first hundred even natural numbers divisible by \[5\] .

Ans: Even natural no. divisible by \[5\] are \[10,20,30,...\]

Here, \[a=10,d=10,n=100\]

\[{{S}_{100}}=\dfrac{100}{2}\left[ 2\left( 10 \right)+10\left( 100-1 \right) \right]\]

\[\therefore {{S}_{100}}=50500\]

21.  Find \[{{a}_{30}}-{{a}_{20}}\] for the A.P \[-9,-14,-19,-24,...\]

Ans:  Here, \[a=-9\]

\[d=-14+9\]

\[\Rightarrow d=-5\]

\[\Rightarrow {{a}_{30}}-{{a}_{20}}=a+29d-a-19d\]

\[\Rightarrow {{a}_{30}}-{{a}_{20}}=10d\]

\[\therefore {{a}_{30}}-{{a}_{20}}=-50\]

22. Find the common difference and write the next two terms of the AP \[{{1}^{2}},{{5}^{2}},{{7}^{2}},73,...\]

Ans: Given AP \[{{1}^{2}},{{5}^{2}},{{7}^{2}},73,...\]

\[\Rightarrow 1,25,49,73,...\]

\[d={{a}_{2}}-{{a}_{1}}\]

\[\Rightarrow {{a}_{2}}-{{a}_{1}}=25-1\]

\[\Rightarrow {{a}_{2}}-{{a}_{1}}=24\]

Next two terms are,

\[{{a}_{5}}=97\]

\[{{a}_{6}}=121\]

23.  Show that sequence defined by \[{{a}_{n}}=3+2n\] is an AP. 

Ans: Here, \[{{a}_{1}}=5,{{a}_{2}}=7,{{a}_{3}}=9,{{a}_{4}}=11\]

Common difference:

\[\therefore d=2\] 

Hence, it is AP. 

24. The first term of an AP is \[-7\] and common difference \[5\] . Find its general term.

 Ans: Here, \[a=-7,d=5\]

\[{{a}_{n}}=-7+5\left( n-1 \right)\]

\[{{a}_{n}}=5n-12\]

25.  How many terms are there in A.P? \[18,15\dfrac{1}{2},13,....,-47\] 

Ans: Here, \[a=18,d=\dfrac{31}{2}-\dfrac{18}{1}=\dfrac{-5}{2}\]

\[{{a}_{n}}=-47\]

\[\Rightarrow -47=18+\left( n-1 \right)\left( \dfrac{-5}{2} \right)\]

\[\therefore n=27\]

26. In an AP, the sum of first \[n\] terms is \[\dfrac{3{{n}^{2}}}{2}+\dfrac{13}{2}n\] find its \[{{2}^{nd}}\] term. 

Ans:  Given \[{{S}_{n}}=\dfrac{3{{n}^{2}}}{2}+\dfrac{13}{2}n\]  

Put \[n=1,2,3,...\]

\[{{S}_{1}}=\dfrac{16}{2}=8\]

\[{{S}_{2}}=19\]

\[{{a}_{1}}={{S}_{1}}=8\]

\[{{a}_{2}}={{S}_{2}}-{{S}_{1}}\]

\[\therefore {{a}_{2}}=11\]

27. Show that the progression \[4,7\dfrac{1}{2},10\dfrac{1}{2},13\dfrac{3}{4},17,..\] is an AP. 

Ans: Common difference: \[{{d}_{1}}=\dfrac{29}{4}-\dfrac{4}{1}\]

\[\Rightarrow {{d}_{1}}=\dfrac{13}{4}\]

\[{{d}_{2}}=\dfrac{21}{2}-\dfrac{29}{4}\]

\[\Rightarrow {{d}_{2}}=\dfrac{13}{4}\] 

\[\because {{d}_{1}}={{d}_{2}}\]

Hence, it is an AP. 

Short Answer Questions (3 Marks)

1. Write first four terms of the AP, when the first term \[a\] and common difference \[d\] are given as follows: 

(i) \[a=10,d=10\]

Ans: (i) Given: \[a=10,d=10\]

Using these values we get,

First term: \[a=10\]

Second term: \[a+d=20\]

Third term: \[a+2d=30\]

Fourth term: \[a+3d=40\]

Therefore, first four terms are: \[10,20,30,40\]

(ii) \[a=-2,d=0\]

Ans: (ii) Given: \[a=-2,d=0\]

First term: \[a=-2\]

Second term: \[a+d=-2\]

Third term: \[a+2d=-2\]

Fourth term: \[a+3d=-2\]

Therefore, first four terms are: \[-2,-2,-2,-2\]

(iii) \[a=4,d=-3\]

Ans: (iii) Given: \[a=4,d=-3\]

First term: \[a=4\]

Second term: \[a+d=1\]

Fourth term: \[a+3d=-5\]

Therefore, first four terms are: \[4,1,-2,-5\]

(iv) \[a=-1,d=\dfrac{1}{2}\]

Ans: (iv) Given: \[a=-1,d=\dfrac{1}{2}\]

First term: \[a=-1\]

Second term: \[a+d=\dfrac{-1}{2}\]

Third term: \[a+2d=0\]

Fourth term: \[a+3d=\dfrac{1}{2}\]

Therefore, first four terms are: \[-1,\dfrac{-1}{2},0,\dfrac{1}{2}\]

(v) \[a=-1.25,d=-0.25\]

Ans: (v) Given: \[a=-1.25,d=-0.25\]

First term: \[a=-1.25\]

Second term: \[a+d=-1.50\]

Third term: \[a+2d=-1.75\]

Fourth term: \[a+3d=-2.00\]

Therefore, first four terms are: \[-1.25,-1.50,-1.75,-2.00\]

2. Find the \[{{31}^{st}}\] term of an AP whose \[{{11}^{th}}\] term is \[38\] and \[{{16}^{th}}\] term is \[73\] .

Ans: Given that: \[{{a}_{11}}=38\] and \[{{a}_{16}}=73\]

Putting the given values,

\[\Rightarrow 38=a+\left( 11-1 \right)d\] 

\[\therefore ~38=a+10d\]                                      …(1)

\[\Rightarrow ~73=a+\left( 16-1 \right)d\]

\[\therefore ~73=a+15d\]                                  …(2)

Solving equation (1) and (2) we get,

Common difference: \[d=7\] 

And first term: \[a=-32\] 

Again by using formula \[{{a}_{n}}=a+\left( n-1 \right)d\] 

\[{{a}_{31}}=-32+7\left( 31-1 \right)\]

\[\therefore {{a}_{31}}=178\]

Therefore, \[{{31}^{st}}\] term of AP is \[178\] . 

3. If the third and the ninth terms of an AP are \[4\] and \[-8\] respectively, which term of this AP is zero? 

Ans: Given that: \[{{a}_{3}}=4\] and \[{{a}_{9}}=-8\]

\[\Rightarrow 4=a+\left( 3-1 \right)d\] 

\[\therefore ~4=a+2d\]                                      …(1)

\[\Rightarrow ~-8=a+\left( 9-1 \right)d\]

\[\therefore ~-8=a+8d\]                                  …(2)

Common difference: \[d=-2\] 

And first term: \[a=8\] 

\[0=8-2\left( n-1 \right)\]

\[\therefore n=5\]

Therefore, \[{{5}^{th}}\] term of AP is \[0\] . 

4. Two AP's have the same common difference. The difference between their \[{{100}^{th}}\] terms is \[100\] , what is the difference between their \[{{1000}^{th}}\] terms. 

Ans: Let first term of \[{{1}^{st}}\] AP be \[a\]

Let first term of \[{{2}^{nd}}\] AP be \[a'\]

It is given that their common difference is same. 

Let their common difference be \[d\]  

It is given that difference between their terms \[{{100}^{th}}\] is \[100\] . 

\[a+\left( 100-1 \right)d\text{ }\left[ a\prime +\left( 100-1 \right)d \right]\]

\[\Rightarrow a+99d-a\prime -99d=100\]

\[\therefore a-a\prime =100\]                                   …(1)

We have to find difference between their \[{{1000}^{th}}\] terms,

\[\therefore a+\left( 1000-1 \right)d\text{ }\left[ a\prime +\left( 1000-1 \right)d \right]\]

\[\Rightarrow a+999d-a\prime -999d\]

\[\Rightarrow aa\prime ~\] 

Putting equation (1) in the above equation we get, 

\[aa\prime ~=100\] 

Therefore, difference between their \[{{1000}^{th}}\] terms would be equal to \[100\] . 

5. How many three-digit numbers are divisible by \[7\] ? 

Ans: From the given data we have the AP:

\[105,112,119,...,994\]

Let \[994\] be the \[{{n}^{th}}\] term of the AP. 

We need to find \[n\] here. 

Here, First term: \[a=105\], Common difference: \[d=7\] 

\[\Rightarrow 994=105+\left( n-1 \right)\left( 7 \right)\]

\[\therefore n=128\] 

Therefore, there are \[128\] terms in the given AP. 

6. A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. \[200\] for the first day, Rs \[250\] for the second day, Rs \[300\] for the third day, etc., the penalty for each succeeding day being Rs \[50\] more than for the preceding day. How much money the contractor has to pay a penalty if he has delayed the work by \[30\] days? 

Ans: Given that:  

Penalty for first day is Rs \[200\] , 

Penalty for second day is Rs \[250\] ,

Penalty for third day is Rs \[300\] .

We want to know how much money the contractor has to pay as penalty, if he has delayed the work by \[30\] days. 

So, we have an AP of the form \[200,250,300,350,....30terms\]

First term: \[a=200\] , Common difference: \[d=50\] , number of terms: \[n=30\] 

\[{{S}_{30}}=\dfrac{30}{2}\left[ 400+\left( 30-1 \right)50 \right]\]

\[\therefore {{S}_{30}}=27750\] 

Therefore, penalty for \[30\] days is Rs. \[27750\] . 

7. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g, a section of Class I will plant one tree, a section of class II will plant two trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students? 

Ans: Given that:

The number of trees planted by class I = number of sections × 1 = 3 × 1 = 3 

The number of trees planted by class II = number of sections × 2 = 3 × 2 = 6 

The number of trees planted by class III = number of sections × 3 = 3 × 3 = 9 

Therefore, we have sequence of the form \[3,6,9,...,12terms\]

To find total number of trees planted by all the students, we need to find sum of this sequence.

Here, First term: \[a=3\] , 

Common difference: \[d=3\] and

Number of terms: \[n=12\]

\[{{S}_{12}}=\dfrac{12}{2}\left[ 6+3\left( 12-1 \right) \right]\]

\[\therefore {{S}_{12}}=234\]

Total \[234\] trees will be planted by the students. 

8. A small terrace at a football ground comprises of \[15\] steps each of which is \[50\] m long and built of solid concrete. Each step has a rise of \[\dfrac{1}{4}\] m and a tread of  \[\dfrac{1}{2}\] m (see figure). Calculate the total volume of concrete required to build the terrace. 

Ans: Volume of concrete required to build the first step, second step, third step,… are:

\[\dfrac{1}{4}\times \dfrac{1}{2}\times 50,\left( 2\times \dfrac{1}{4} \right)\times \dfrac{1}{2}\times 50,\left( 3\times \dfrac{1}{4} \right)\times \dfrac{1}{2}\times 50,...\]

\[\Rightarrow \dfrac{50}{8},2\times \dfrac{50}{8},3\times \dfrac{50}{8},...\] 

To find the total volume of concrete required, 

Use the formula,

\[{{S}_{15}}=\dfrac{50}{8}\left( \dfrac{15}{2}\left[ 2+\left( 15-1 \right) \right] \right)\]

\[\therefore {{S}_{15}}=750\]  

The total volume of concrete required to build the terrace is \[750{{m}^{3}}\]. 

9. For what value of n are the \[{{n}^{th}}\] term of the following two AP’s are same \[13,19,25,...\] and \[69,68,67,...\] .

Ans: Given that: \[{{n}^{th}}\] term of \[13,19,25,...\] is equal to \[{{n}^{th}}\] term of \[69,68,67,...\]

\[\Rightarrow 7\left( n-1 \right)=56\]

\[\therefore n=9\]

10. Check whether \[301\] is a term of the list of numbers \[5,11,17,32,...\] ? 

Ans: Given AP is \[5,11,17,32,...\]

Here, \[a=5,d=6\]

\[\Rightarrow 301=5+6\left( n-1 \right)\]

\[\therefore n=\dfrac{302}{6}\]

Since \[n\] cannot be a fraction.  

Therefore, \[301\] is not a term of the given AP. 

11. Determine the AP whose third term is \[16\] and the \[{{7}^{th}}\] term exceeds the \[{{5}^{th}}\] term by \[12\] .

Ans: Given \[{{a}_{3}}=16\] and \[{{a}_{7}}={{a}_{5}}+12\]

\[\Rightarrow a+2d=16\]                                                            …(1)

And \[\Rightarrow a+6d=a+4d+12\]

\[\therefore d=6\] 

Put the value of \[d\] in equation (1), 

\[a+2\times 6=16\]

\[\therefore a=4\]

Therefore, the AP is \[4,10,16,...\] 

12. Find the sum of AP in \[-5+(-8)+(-11)+....+(-230)\] .  

Ans:  Here, \[a=-5,d=-3,{{a}_{n}}=-230\]

\[\Rightarrow -230=-5-3\left( n-1 \right)\]

\[\therefore n=76\]

\[{{S}_{76}}=\dfrac{76}{2}\left[ -10-3\left( 76-1 \right) \right]\]

\[\therefore {{S}_{76}}=-8930\]

13. In an AP, \[{{a}_{n}}=4,d=2,{{S}_{n}}=-14\] find \[n\] and \[a\] . 

Ans: Given that \[{{a}_{n}}=4,d=2,{{S}_{n}}=-14\]

\[\Rightarrow 4=a+2\left( n-1 \right)\]                                    …(1)

\[\Rightarrow -14=\dfrac{n}{2}\left[ a+4 \right]\]                                …(2)

From equation (1) and (2) we get,

\[\Rightarrow -28=n\left[ 6-2n+4 \right]\]

\[\Rightarrow {{n}^{2}}-5n-14=0\]

\[\Rightarrow n=7,-2\]

\[\therefore a=-8\]

14. Find \[{{a}_{30}}-{{a}_{20}}\] for the AP in \[-9,-14,-19,-24,...\]

Ans:  Here, \[a=-9\] and \[d=-5\]

\[{{a}_{30}}-{{a}_{20}}=(a+29d)-(a+19d)\]

\[\therefore {{a}_{30}}-{{a}_{20}}=-50\] 

15. Find the sum to \[n\] terms of the AP in \[5,2,-1,-4,-7,...\]

Ans:  Here, \[a=5\] and \[d=-3\]

\[{{S}_{n}}=\dfrac{n}{2}\left[ 2\times 5+(n-1)(-3) \right]\]

\[{{S}_{n}}=\dfrac{n}{2}\left[ 13-3n \right]\]

16. Find the sum of first \[24\] terms of the list of no. whose \[{{n}^{th}}\] term is given by \[{{a}_{n}}=3+2n\] . 

Ans:  Given that: \[{{a}_{n}}=3+2n\]

\[{{a}_{1}}=5,{{a}_{2}}=7,{{a}_{3}}=9...\]

Here, \[a=5,d=2\]

\[\Rightarrow {{S}_{24}}=\dfrac{24}{2}\left[ 2\times 5+(24-1)\times 2 \right]\]

\[\therefore {{S}_{24}}=672\] 

Long Answer Questions (4 Marks)

1. For the following APs, write the first term and the common difference. 

(i). \[3,1,-1,-3,...\]

Ans: (i) Given AP is \[3,1,-1,-3,...\]

First term: \[a=3\]

Common difference: 

\[\therefore d=-2\]

(ii). \[-5,-1,3,7,...\]

Ans: (ii) Given AP is \[-5,-1,3,7,...\]

First term: \[a=-5\]

\[d=-1-\left( -5 \right)\]

\[\therefore d=4\]

(iii). \[\dfrac{1}{3},\dfrac{5}{3},\dfrac{9}{3},\dfrac{13}{3}...\] 

Ans: (iii) Given AP is \[\dfrac{1}{3},\dfrac{5}{3},\dfrac{9}{3},\dfrac{13}{3}...\]

First term: \[a=\dfrac{1}{3}\]

\[d=\dfrac{5}{3}-\dfrac{1}{3}\]

\[\therefore d=\dfrac{4}{3}\]

(iv). \[0.6,1.7,2.8,3.9,...\]

Ans: (iv) Given AP is \[0.6,1.7,2.8,3.9,...\]

First term: \[a=0.6\]

\[d=1.7-0.6\]

\[\therefore d=1.1\]

2. The \[{{17}^{th}}\] term of an AP exceeds its \[{{10}^{th}}\] term by \[7\] . Find the common difference.

 Ans: Given that: \[{{a}_{17}}={{a}_{10}}+7\]

 Using formula \[{{a}_{n}}=a+\left( n-1 \right)d\] ,

\[a+16d=a+9d+7\]

\[\Rightarrow 7d=7\]

\[\therefore d=7\]

Common difference of the given AP is \[7\] .

3. A sum of Rs \[700\] is to be used to give seven cash prizes to students of a school for their overall academic performance. If, each prize is Rs \[20\] less than its preceding term, find the value of each of the prizes. 

Ans: It is given that sum of seven cash prizes is equal to Rs \[700\] . 

And, each prize is Rs \[20\] less than its preceding term. 

Let the value of first prize be Rs \[a\] 

Let value of second prize be Rs \[\left( a-20 \right)\]

Let value of third prize be Rs \[\left( a-40 \right)\]

So, we have sequence of the form: 

\[a,a-20,a-40,a-60,...\]

It is an arithmetic progression because the difference between consecutive terms is constant.

Here, first term is \[a\] , 

Common difference: \[d=-20\]

\[n=7\] (Because there are total of seven prizes) 

\[{{S}_{7}}=Rs700\] (given) 

 \[\Rightarrow {{S}_{7}}=\dfrac{7}{2}\left[ 2\times a+(7-1)\times -20 \right]\]

\[\Rightarrow 200=2a-120\]

\[\therefore a=160\] 

Therefore, the value of prizes is \[160,140,120,100,80,60,40\] respectively.

4. A spiral is made up of successive semicircles, with centers alternatively at \[A\] and \[B\], starting with center at \[A\], of radii \[0.5\] cm, \[1.0\] cm, \[1.5\] cm, \[2.0\] cm, ... What is the total length of such a spiral made up of thirteen consecutive semicircles. 

Ans: The Length of semi–circle is   $ \dfrac{\text{Circumferenceofcircle}}{2}=\dfrac{2\pi r}{2}=\pi r $  

The Length of semicircle of radii \[0.5\] cm is \[0.5\pi \] cm 

The Length of semicircle of radii \[1.0\] cm is \[\pi \] cm 

The Length of semicircle of radii \[1.5\] cm is \[1.5\pi \] cm 

Therefore, we have sequence of the form: 

\[0.5\pi ,1.0\pi ,1.5\pi ,....13terms\]

To find total length of the spiral, we need to find sum of the sequence.

Here, \[a=0.5,d=0.5,n=13\]

\[\Rightarrow {{S}_{13}}=\dfrac{13}{2}\left[ 1+0.5\left( 13-1 \right) \right]\]

\[\therefore {{S}_{13}}=45.5\]

Total length of spiral is \[45.5\pi =143cm\] .

5. \[200\] logs are stacked in the following manner: \[20\] logs in the bottom row, \[19\] in the next row, \[18\] in the row next to it, and so on. In how many rows are the \[200\] logs placed and how many logs are in the top row? 

The number of logs in the bottom row is \[20\]

The number of logs in the next row is \[19\]

The number of logs in the next to next row is \[18\]

Therefore, we have a sequence of the form \[20,19,18,...\] 

Here, \[a=20,d=-1\]

We need to find that how many rows make total of 200 logs. 

Putting the given values we get,

\[\Rightarrow 200=\dfrac{n}{2}\left( 40-n+1 \right)\]

\[\Rightarrow {{n}^{2}}-41n+400=0\]

\[\therefore n=25,16\]

We can discard \[n=25\] because we cannot have more than \[20\] rows in the sequence. \[\therefore n=16\] which means \[16\] rows make total number of logs equal to \[200\] .

We also need to find number of logs in the top row. 

\[{{a}_{16}}=20-15\]

\[\therefore {{a}_{16}}=5\] 

Therefore, there are \[5\] logs in the topmost row and there is a total of \[16\] rows. 

6. In a potato race, a bucket is placed at the starting point, which is \[5\] meters from the first potato, and the other potatoes are placed \[3\] meters apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? 

The distance of the first potato from the starting point is \[5\] meters 

Therefore, the distance covered by the competitor to pick up the first potato and put it in the bucket is \[5\times 2=10m\]

The distance of the Second potato from the starting point is \[5+3=8\] meters 

Therefore, the distance covered by the competitor to pick up the second potato and put it in the bucket is \[8\times 2=16\] meters 

The distance of the third potato from the starting point is \[8+3=11\] meters 

Therefore, the distance covered by the competitor to pick up third potato and put it in the bucket is \[11\times 2=22\] meters 

Therefore, we have a sequence of the form \[10,16,22,...10terms\]

(There are ten terms because there are ten potatoes) 

The total distance covered by the competitor is: 

Here, \[a=10,d=6,n=10\]

\[{{S}_{10}}=\dfrac{10}{2}\left[ 20+6\left( 10-1 \right) \right]\]

\[\therefore {{S}_{10}}=370\]

7. Which term of the sequence \[20,19\dfrac{1}{4},18\dfrac{1}{2},17\dfrac{3}{4},...\] is the first negative term?

Ans: For first negative term, \[{{a}_{n}}<0\]

Here, \[a=20,d=\dfrac{-3}{4}\]

\[\Rightarrow \dfrac{20}{1}-\dfrac{3}{4}n+\dfrac{3}{4}<0\]

\[\Rightarrow \dfrac{80-3n+3}{4}<0\]

\[\Rightarrow n>\dfrac{83}{3}\] 

Therefore, \[{{28}^{th}}\] term is the first negative term.

8. The \[{{p}^{th}}\] term of an AP is \[q\] and \[{{q}^{th}}\] term is \[p\]. Find its \[{{(p+q)}^{th}}\] term. 

Ans:  Given that

\[\Rightarrow {{a}_{p+q}}=\left( q+p1 \right)-1\left( p+q1 \right)\]

\[\therefore {{a}_{p+q}}=0\]

9. If \[m\] times the \[{{m}^{th}}\] term of an A.P is equal to \[n\] times its \[{{n}^{th}}\] term, show that the \[{{(m+n)}^{th}}\] term of the AP is zero.

Ans:  Given that: \[{{(m+n)}^{th}}\]

\[\therefore m[a+(m-1)d]=n[a+(n-1)d]\]

\[\Rightarrow ma+{{m}^{2}}d-md=na+{{n}^{2}}d-nd\]

\[\Rightarrow a(m-n)+({{m}^{2}}-{{n}^{2}})d-md+nd=0\]

\[\Rightarrow a(m-n)+(m-n)(m+n)d-(m-n)d=0\]

\[\Rightarrow (m-n)[a+(m+n-1)d]=0\]

\[\Rightarrow a+(m+n-1)d=0\]

\[\therefore {{a}_{m+n}}=0\]

Hence Proved.  

10. The sum of the \[{{4}^{th}}\] and \[{{8}^{th}}\] terms of an AP is \[24\] and the sum of the \[{{6}^{th}}\] and \[{{10}^{th}}\] terms is \[44\] . Find the first three terms of the AP. 

Ans:  We know that,

\[{{a}_{4}}+{{a}_{8}}=24\] (Given) 

\[\therefore a+3d+a+7d=24\]

\[\Rightarrow a+5d=12\]                                  …(1)

Also, \[{{a}_{5}}+{{a}_{10}}=44\]     (Given)

\[\Rightarrow a+5d+a+9d=44\]

\[\Rightarrow a+7d=22\]                                 …(2)  

On solving equations (1) and (2), 

\[d=5,a=-13\] 

Therefore, First three terms are \[-13,-8,-3\]

11. If the sum of \[n\] terms of an AP is \[3{{n}^{2}}+5n\] and its \[{{m}^{th}}\] term is \[164\], find the value of \[m\] . 

Ans: Given that: \[{{S}_{n}}=3{{n}^{2}}+5n\]  

Put \[n=1,2,3,...\] ,

\[\therefore {{S}_{1}}=8\] and

\[{{S}_{2}}=22\]

\[\therefore {{a}_{1}}={{S}_{1}}=8\] and

\[\therefore {{a}_{2}}=14\]

\[{{a}_{m}}=164\] (Given)

\[\therefore 8+(m-1)(6)=164\]

\[\therefore m=27\]

12. If the sum of three numbers in AP, be \[24\] and their product is \[440\] , find the numbers. 

Ans: Let numbers be \[a-d,a,a+d\]

\[(a-d)+a+(a+d)=24\]  (Given) 

\[\therefore a=8\]

\[(a-d)(a)(a+d)=440\]

\[\Rightarrow (8-d).8.(8-d)=440\]

\[\Rightarrow {{8}^{2}}-{{d}^{2}}=55\]

\[\therefore d=\pm 3\]          

If \[d=3\] 

Then AP is \[5,8,11,...\]

If \[d=-3\] 

Then AP is \[11,8,5,...\]

13. If \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\] are in AP, then prove that \[\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}\] are in AP. 

Ans: Given \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\] are in AP 

Then \[2{{b}^{2}}={{a}^{2}}+{{c}^{2}}\]                                  …(1)

If \[\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}\] are in AP then 

\[\dfrac{1}{c+a}-\dfrac{1}{b+c}=\dfrac{1}{a+b}-\dfrac{1}{c+a}\]

\[\Rightarrow \dfrac{(b+c)-(c+a)}{(c+a)(b+c)}=\dfrac{(c+a)-(a+b)}{(a+b)(c+a)}\]

\[\Rightarrow \dfrac{b-a}{b+c}=\dfrac{c-b}{a+b}\]

\[\therefore 2{{b}^{2}}={{c}^{2}}+{{a}^{2}}\]                                        …(2) 

From (1) and (2), 

14. If \[{{S}_{1}},{{S}_{2}},{{S}_{3}}\] be the sum of \[n,2n,3n\] terms respectively of an AP, prove that \[{{S}_{3}}=3({{S}_{2}}-{{S}_{1}})\] . 

\[\therefore {{S}_{1}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]

\[{{S}_{2}}=\dfrac{2n}{2}\left[ 2a+\left( 2n-1 \right)d \right]\]

\[{{S}_{3}}=\dfrac{3n}{2}\left[ 2a+\left( 3n-1 \right)d \right]\] 

We have to prove that \[{{S}_{3}}=3({{S}_{2}}-{{S}_{1}})\]

RHS: \[\Rightarrow 3\left( \dfrac{2n}{2}\left[ 2a+\left( 2n-1 \right)d \right]-\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right)\]

\[\Rightarrow 3\left( \dfrac{n}{2}\left[ 2a+3nd-d \right] \right)\]

\[\Rightarrow {{S}_{3}}\]

Here, \[LHS=RHS\]

15. The ratio of the sums of \[m\] and \[n\] terms of an AP is \[{{m}^{2}}:{{n}^{2}}\] , show that the ratio of the \[{{m}^{th}}\] and \[{{n}^{th}}\] term is \[(2m-1):(2n-1)\] 

\[\dfrac{{{S}_{m}}}{{{S}_{n}}}=\dfrac{\dfrac{m}{2}[2a+(m-a)d]}{\dfrac{n}{2}[2a+(n-1)d]}\]

\[\Rightarrow \dfrac{{{m}^{2}}}{{{n}^{2}}}=\dfrac{m}{n}\left[ \dfrac{2a+(m-1)d}{2a+(n-1)d} \right]\]  (Given)

\[\Rightarrow \dfrac{m}{n}=\left[ \dfrac{2a+(m-1)d}{2a+(n-1)d} \right]\]

\[\Rightarrow 2am+mnd-md=2an+mnd-nd\]

\[\Rightarrow 2am-2an-md+nd=0\]

\[\Rightarrow (m-n)(2a-d)=0\]

\[\Rightarrow 2a=d\]                                 …(1)

According to question,

\[\Rightarrow \dfrac{{{a}_{m}}}{{{a}_{n}}}=\dfrac{a+(m-1)d}{a+(n-1)d}\]

\[\Rightarrow \dfrac{{{a}_{m}}}{{{a}_{n}}}=\dfrac{a+(m-1)2a}{a+(n+1)2a}\]

\[\Rightarrow \dfrac{{{a}_{m}}}{{{a}_{n}}}=\dfrac{2m-1}{2n-1}\]

16. If the sum of first \[p\] terms of an AP is the same as the sum of its first \[q\] terms, show that the sum of the first \[(p+q)\] terms is zero. 

Ans:  Given that: \[{{S}_{P}}={{S}_{q}}\]  

\[\therefore \dfrac{p}{2}[2a+(p-1)d]=\dfrac{q}{2}[2a+(q-1)d]\]

\[\Rightarrow 2ap+{{p}^{2}}d-pd=2aq+{{q}^{2}}d-qd\]

\[\Rightarrow 2a(p-q)+(p-q)(p-q)d-(p-q)d=0\]

\[\Rightarrow (p-q)[2a+(p+q-1)d]=0\]

\[\therefore {{S}_{p+q}}=0\]

Hence Proved. 

17. For the A.P \[{{a}_{1}},{{a}_{2}},{{a}_{3}},...\] If \[\dfrac{{{a}_{4}}}{{{a}_{7}}}=\dfrac{2}{3}\] find \[\dfrac{{{a}_{6}}}{{{a}_{8}}}\] . 

Ans: Given that \[\dfrac{{{a}_{4}}}{{{a}_{7}}}=\dfrac{2}{3}\]

We know that, 

\[\therefore \dfrac{a+3d}{a+6d}=\dfrac{2}{3}\]

\[\Rightarrow 3a+9d=2a+12d\]

\[\Rightarrow a=3d\]

\[\therefore \dfrac{{{a}_{6}}}{{{a}_{7}}}=\dfrac{a+5d}{a+6d}\]

\[\Rightarrow \dfrac{{{a}_{6}}}{{{a}_{7}}}=\dfrac{3d+5d}{3d+6d}\]

\[\Rightarrow \dfrac{{{a}_{6}}}{{{a}_{7}}}=\dfrac{8}{9}\]

18. In an AP  \[{{p}^{th}}\],  \[{{q}^{th}}\]  and  \[{{r}^{th}}\]  terms  are respectively  \[a,b,c\] . Prove that \[p(b-c)+q(c-a)+r(a-b)=0\] . 

Ans:  We know that, 

\[A+(p-1)D=a\]                              …(1)

\[A+(q-1)D=b\]                              …(2)

\[A+(r-1)D=c\]                               …(3)

Equation (2) – (3),

\[\Rightarrow b-c=(q-1)D-(r-1)D\]

\[\Rightarrow b-c=D(q-r)\]

\[\Rightarrow p(b-c)=pD(q-r)\]                 …(4)  

Similarly, 

\[q(c-a)=qD(r-p)\]                        …(5)

\[r(a-b)=rD(p-q)\]                        …(6) 

Adding (4), (5) and (6) we get,

\[\Rightarrow p(b-c)+q(c-a)+r(a-b)=0\] 

19. If \[{{(p+1)}^{th}}\]  term of an A.P is twice the \[{{(q+1)}^{th}}\] term, show that \[{{(3p+1)}^{th}}\] term is twice the \[{{(p+q+1)}^{th}}\] term. 

Ans: Given that:  $ {{a}_{p+1}}=2{{a}_{q+1}} $ 

\[\therefore a+(p+1-1)d=2[a+(q+1-1)d]\]

\[\Rightarrow a+pd=2a+2qd\]

\[\Rightarrow (p-2q)d=a\]

Now, \[\dfrac{{{a}_{3p+1}}}{{{a}_{p+q+1}}}=\dfrac{a+(3p+1-1)d}{a+(p+q+1-1)d}\]

\[\Rightarrow \dfrac{(p-2q)d+3pd}{p-2q+(p+q)d}=2\]

Hence proved that \[{{(3p+1)}^{th}}\] term is twice the \[{{(p+q+1)}^{th}}\] term.  

20. The sum of four numbers in AP is \[50\] and the greatest number four times the least. Find the numbers. 

Ans: Let the numbers be \[(a-3d),(a-d),(a+d),(a+3d)\]

Given that:

\[(a-3d)+(a-d)+(a+d)+(a+3d)=50\]

\[\Rightarrow 4a=50\]

\[\therefore a=12.5\] 

\[(a+3d)=4\times (a-3d)\]

\[\Rightarrow a=5d\]

\[\therefore d=2.5\] 

Numbers are \[5,10,15,20\]

21. Find the sum of all integers between \[84\] and \[719\] which are multiples of \[5\] .

Ans: The numbers are \[85,90,95,....,715\]

Here, \[a=85,d=5,{{a}_{n}}=715\]

\[\Rightarrow 85+5(n-1)=715\]

\[\Rightarrow n=127\]

\[\Rightarrow {{S}_{127}}=\dfrac{127}{2}(85+715)\]

\[\Rightarrow {{S}_{127}}=50800\]

22. If \[{{m}^{th}}\] term of an A.P is \[\dfrac{1}{n}\] and the \[{{n}^{th}}\] term is \[\dfrac{1}{m}\] show that the sum of terms is \[\dfrac{1}{2}(mn+1)\] .

\[\Rightarrow \dfrac{1}{n}=a+(m-1)d\]

\[\Rightarrow \dfrac{1}{m}=a+(n-1)d\] 

On solving the above equations we get, 

\[a=\dfrac{1}{mn},d=\dfrac{1}{mn}\]

\[\Rightarrow {{S}_{mn}}=\dfrac{mn}{2}[2a+(mn-1)d]\]

\[\Rightarrow {{S}_{mn}}=\dfrac{mn}{2}\left[ 2\cdot \dfrac{1}{mn}+(mn-1)\cdot \dfrac{1}{mn} \right]\]

\[\Rightarrow {{S}_{mn}}=\dfrac{1}{2}(mn+1)\]

23. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? 

(i). The taxi fare after each km when the fare is Rs \[15\] for the first km and Rs \[8\] for each additional km. 

Ans: (i) Given that Taxi fare for first km is Rs \[15\], 

Taxi fare after two km is Rs \[23\] ,

Taxi fare after three km is Rs \[31\]

Taxi fare after four km is Rs \[39\] 

Therefore, the sequence is \[15,23,31,39,...\]

It is an arithmetic progression because the difference between any two consecutive terms is equal which is \[8\].

(ii). The amount of air present in a cylinder when a vacuum pump removes \[{{\dfrac{1}{4}}^{th}}\] of the air remaining in the cylinder at a time. 

Ans: (ii) Let amount of air initially present in a cylinder be \[V\]

Amount of air left after pumping out air by vacuum pump is \[V-\dfrac{V}{4}=\dfrac{4V-V}{4}=\dfrac{3V}{4}\] 

Amount of air left when vacuum pump again pumps out air is \[\dfrac{3}{4}V-\left( \dfrac{1}{4}\times \dfrac{3}{4}V \right)=\dfrac{9}{16}V\]  

So, the sequence we get is \[V,\dfrac{3}{4}V,\dfrac{9}{16}V...\] 

Checking for difference between consecutive terms, 

\[\dfrac{3}{4}V-V=-\dfrac{V}{4}\]

\[\dfrac{9}{16}V-\dfrac{3}{4}V=\dfrac{-3V}{16}\] 

The difference between consecutive terms is not equal. 

Therefore, it is not an arithmetic progression. 

(iii). The cost of digging a well after every meter of digging, when it costs Rs \[150\] for the first meter and rises by Rs \[50\] for each subsequent meter. 

Ans. (iii) Cost of digging one meter of well is Rs \[150\]

The cost of digging two meters of well is Rs \[200\]

The cost of digging three meters of well is Rs \[250\]

Therefore, we get a sequence of the form \[150,200,250,...\]

Here, the difference between any two consecutive terms which is also called a common difference is constant and equal to \[50\]. 

Therefore, it is an AP.

(iv). The amount of money in the account every year, when Rs \[10,000\] is deposited at compound Interest at \[8%\] per annum. 

Ans. (iv) Amount in bank after first year is \[10000\left( 1+\dfrac{8}{100} \right)\]  

Amount in bank after two years is \[10000{{\left( 1+\dfrac{8}{100} \right)}^{2}}\]

Amount in bank after three years is \[10000{{\left( 1+\dfrac{8}{100} \right)}^{3}}\] 

Amount in bank after four years is \[10000{{\left( 1+\dfrac{8}{100} \right)}^{4}}\]

It is not an arithmetic progression because the difference between consecutive terms is not constant.

Therefore, it is not an Arithmetic Progression. 

24. In the following AP's find the missing terms: 

(i). \[2,\_,26\]

Ans: (i) Given AP: \[2,\_,26\]

We know that difference between consecutive terms is equal in any A.P. 

Let the missing term be \[x\] ,

\[\therefore x-2=26-x\]

\[\Rightarrow x=14\] 

Therefore, missing term is \[14\] . 

(ii). \[\_,13,\_,3\]

Ans: (ii) Given AP: \[\_,13,\_,3\]

Let the missing terms be \[x\] and \[y\] ,

\[\therefore x-13=3-x\]

\[\Rightarrow x=8\] 

And \[13-x=y-13\]

But \[\because x=8\]

\[\therefore y=18\]

Therefore, missing terms are \[8,18\] . 

(iii). \[5,\_,\_,9\dfrac{1}{2}\]

Ans: (iii) Given AP: \[5,\_,\_,9\dfrac{1}{2}\]

Here, \[a=5,{{a}_{4}}=9\dfrac{1}{2}\]

Using formula \[{{a}_{n}}=a+\left( n-1 \right)d\] , 

\[\Rightarrow \dfrac{19}{2}=5+3d\]

\[\therefore d=\dfrac{3}{2}\] Second term: \[a+d=5+\dfrac{3}{2}=\dfrac{13}{2}\] 

Third term: \[a+2d=\dfrac{13}{2}+\dfrac{3}{2}=8\] 

Therefore, missing terms are \[\dfrac{13}{2}\] and \[8\] .

(iv). \[-4,\_,\_,\_,\_,6\]

Ans: (iv) Given AP: \[-4,\_,\_,\_,\_,6\]

Here, \[a=-4,{{a}_{6}}=6\]

\[\Rightarrow 6=-4+5d\]

\[\therefore d=2\] Second term: \[a+d=-4+2=-2\] 

Third term: \[a+2d=-4+4=0\] 

Fourth term: \[a+3d=-4+6=2\] 

Fifth term: \[a+4d=-4+8=4\] 

Therefore, missing terms are \[-2,0,2,4\] .

(v). \[\_,38,\_,\_,\_,-22\]

Ans: (v) Given AP: \[\_,38,\_,\_,\_,-22\]

Here, \[{{a}_{2}}=38,{{a}_{6}}=-22\]

\[\Rightarrow 38=a+d\] and

\[\Rightarrow -22=a+5d\]

Solving the above equations we get,

\[d=-15,a=53\]

First term: \[a=53\] 

Third term: \[a+2d=53-30=23\] 

Fourth term: \[a+3d=53-45=8\] 

Fifth term: \[a+4d=53-60=-7\] 

Therefore, missing terms are \[53,23,8,-7\] . 

25. The given figure depicts a racing track whose left and right ends are semi-circular. The difference between the two inner parallel line segments is \[60\] m and they are each \[106\] m long. If the track is \[10\] m wide, find: 

        (i). The distance around the track along its inner edge.

Ans: (i) The distance around the track along the inner edge:

\[\Rightarrow 106+106+(\pi \times 30+\pi \times 30)\]

\[\Rightarrow 212+\dfrac{22}{7}\times 60\]

\[\Rightarrow \dfrac{2804}{7}m\]

        (ii). The area of the track.

Ans. (ii) The area of the track:

\[\Rightarrow 106\times 80-106\times 60+2\dfrac{1}{2}\pi [{{40}^{2}}-{{30}^{2}}]\]

\[\Rightarrow 106\times 20+\pi (70)(10)\]

\[\Rightarrow 12+700\times \dfrac{22}{7}\]

\[\Rightarrow 4320{{m}^{2}}\]

26. Which of the following are APs? If they form an AP, find the common difference \[d\] and write three more terms. 

(i)  \[2,4,8,16,...\]

Ans: (i) \[2,4,8,16,...\]

It is not an AP because difference between consecutive terms is not equal. 

As \[4-2\ne 8-4\]

(ii) \[2,\dfrac{5}{2},3,\dfrac{7}{2},...\]

Ans. (ii) \[2,\dfrac{5}{2},3,\dfrac{7}{2},...\]

It is an AP because difference between consecutive terms is equal. 

\[\Rightarrow \dfrac{5}{2}-2=\dfrac{1}{2}\] 

Common difference: \[d=\dfrac{1}{2}\] 

Fifth term: \[\dfrac{7}{2}+\dfrac{1}{2}=4\]    

Sixth term: \[4+\dfrac{1}{2}=\dfrac{9}{2}\] 

Seventh term: \[\dfrac{9}{2}+\dfrac{1}{2}=5\] 

Therefore, next three terms are \[4,\dfrac{9}{2},5\] .

(iii) \[-1.2,-3.2,-5.2,-7.2,...\]

Ans. (iii) \[-1.2,-3.2,-5.2,-7.2,...\]

\[\Rightarrow -3.2+1.2=-2\] 

Fifth term: \[-7.2-2=-9.2\]    

Sixth term: \[-9.2-2=-11.2\] 

Seventh term: \[-11.2-2=-13.2\] 

Therefore, next three terms are \[9.2,-11.2,-13.2\] .

(iv) \[-10,-6,-2,2,...\]

Ans. (iv) \[-10,-6,-2,2,...\]

\[\Rightarrow -6+10=4\] 

Common difference: \[d=4\] 

Fifth term: \[2+4=6\]    

Sixth term: \[6+4=10\] 

Seventh term: \[10+4=14\] 

Therefore, next three terms are \[6,10,14\] .

(v) \[3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2}...\]

Ans. (v) \[3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2}...\]

\[\Rightarrow 3+\sqrt{2}-3=\sqrt{2}\] 

Common difference: \[d=\sqrt{2}\] 

Fifth term: \[3+3\sqrt{2}+\sqrt{2}=3+4\sqrt{2}\]    

Sixth term: \[3+4\sqrt{2}+\sqrt{2}=3+5\sqrt{2}\] 

Seventh term: \[3+5\sqrt{2}+\sqrt{2}=3+6\sqrt{2}\] 

Therefore, next three terms are \[3+4\sqrt{2},3+5\sqrt{2},3+6\sqrt{2}\] .

(vi) \[0.2,0.22,0.222,0.2222,...\]

Ans: (vi) \[0.2,0.22,0.222,0.2222,...\]

As \[0.222-0.22\ne 0.22-0.2\]

(vii) \[0,-4,-8,-12,...\]

Ans. (vii) \[0,-4,-8,-12,...\]

\[\Rightarrow -4-0=-4\] 

Common difference: \[d=-4\] 

Fifth term: \[-12-4=-16\]    

Sixth term: \[-16-4=-20\] 

Seventh term: \[-20-4=-24\] 

Therefore, next three terms are \[-16,-20,-24\] .

(viii) \[-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},...\]

Ans. (viii) \[-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},...\]

\[\Rightarrow -\dfrac{1}{2}+\dfrac{1}{2}=0\] 

Common difference: \[d=0\] 

Fifth term: \[-\dfrac{1}{2}+0=-\dfrac{1}{2}\]    

Sixth term: \[-\dfrac{1}{2}+0=-\dfrac{1}{2}\] 

Seventh term: \[-\dfrac{1}{2}+0=-\dfrac{1}{2}\] 

Therefore, next three terms are \[-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2}\] .

(ix) \[1,3,9,27,...\]

Ans: (ix) \[1,3,9,27,...\]

As \[3-1\ne 9-3\]

(x) \[a,2a,3a,4a,...\]

Ans. (x) \[a,2a,3a,4a,...\]

\[\Rightarrow 2a-a=a\] 

Common difference: \[d=a\] 

Fifth term: \[4a+a=5a\]    

Sixth term: \[5a+a=6a\] 

Seventh term: \[6a+a=7a\] 

Therefore, next three terms are \[5a,6a,7a\] .

(xi) \[a,{{a}^{\mathbf{2}}},{{a}^{\mathbf{3}}},{{a}^{\mathbf{4}}}...~\]

Ans: (xi) \[a,{{a}^{\mathbf{2}}},{{a}^{\mathbf{3}}},{{a}^{\mathbf{4}}}...~\]

As \[{{a}^{2}}-a\ne {{a}^{3}}-{{a}^{2}}\]

(xii) \[\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...\] 

Ans. (xii) \[\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...\]

\[\Rightarrow 2\sqrt{2}-\sqrt{2}=\sqrt{2}\] 

Fifth term: \[4\sqrt{2}+\sqrt{2}=\sqrt{50}\]    

Sixth term: \[5\sqrt{2}+\sqrt{2}=\sqrt{72}\] 

Seventh term: \[6\sqrt{2}+\sqrt{2}=\sqrt{98}\] 

Therefore, next three terms are \[\sqrt{50},\sqrt{72},\sqrt{98}\] .

(xiii) \[\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...\]

Ans: (xiii) \[\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...\]

As \[\sqrt{6}-\sqrt{3}\ne \sqrt{9}-\sqrt{6}\]

(xiv) \[{{1}^{2}},{{3}^{2}},{{5}^{2}},{{7}^{2}}...~\]

Ans: (xiv) \[{{1}^{2}},{{3}^{2}},{{5}^{2}},{{7}^{2}}...~\]

As \[9-1\ne 25-9\]

(xv) \[{{1}^{2}},{{5}^{2}},{{7}^{2}},73...~\]

Ans. (xv) \[{{1}^{2}},{{5}^{2}},{{7}^{2}},73...~\]

\[\Rightarrow 25-1=24\] 

Common difference: \[d=24\] 

Fifth term: \[73+24=97\]    

Sixth term: \[97+24=121\] 

Seventh term: \[121+24=145\] 

Therefore, next three terms are \[24,121,145\] .

Vedantu provides a free downloadable PDF version of Chapter 5 Maths Class 10 important questions for the students. It is easily accessible by all the students on any devices, and is also easily downloadable for future offline reference on any device, without any limitations. Along with the important questions, the PDF also carries the solutions for each problem, and some alternative solutions to help the students.

Chapter 5 of Maths Class 10 is about the Arithmetic progression that is a vast chapter and has various applications and extended studies in the higher classes. Therefore, the students need to practice all the questions available in the PDF of important questions of Chapter 5 Maths Class 10 to briefly discuss the problems that are important for all the examinations and the ones that require more practice. This is also helpful as the students can access them anytime and anywhere.

Class 10 Maths Ch 5 Important questions help the students get well-versed with all the concepts and be confident about solving various types of problems related to a given concept. This is a significant reason why it is always advised that the students practice more questions from different sets of Class 10, Chapter 5 important questions.

Important Questions for Class 10 Maths Arithmetic Progression – Related Essential Concepts

Arithmetic Progression has many topics from which the questions are asked in the examinations. Here are the major topics that cover some questions in several examinations and whose problems are also included in the set of important questions for Class 10 Maths Chapter 5.

Series, Sequence, and Progressions

A series refers to the Sum of elements in a given sequence, as the Sum of n natural numbers is the series of n natural numbers. On the other hand, a sequence refers to a list of finite or infinite numbers which follows a specific pattern, like the numbers 1, 2, 3, 4, 5,… is an endless sequence of natural numbers. Each number coming in a series or a sequence is known as a term. Furthermore, a progression refers to a sequence where the general term is expressible using a mathematical formula.

Arithmetic Progression

Arithmetic Progression refers to a progression where the difference between any two consecutive numbers of the concerned sequence is always constant. Any AP has a general form as a, a+d, a+2d, a+3d, a+4d… where a is the first term of the given sequence and d is a common difference. 

Arithmetic Progression can be finite as well as infinite. Finding the last term for a finite AP is feasible, whereas it is not for an infinite AP. 

Common Difference

For a given sequence in an AP, the difference between any two consecutive terms is always constant. This difference refers to the "common difference" (d) of the concerned AP. According to the given scenario, the common difference (d) for a given AP can be:

Positive, for an increasing AP.

Zero, for a constant AP.

Negative, for a decreasing AP.

Term and Sum of n Terms of an AP

For a given AP, the nth term is given as T n = a+ (n−1)d. Here, a refers to the AP's first term, d is a common difference, and n is the number of terms of it.

On the other hand, the Sum of n terms for a given AP is given as S n =n/2(2a+(n−1)d). Here, a refers to the first term of the concerned AP, d is the common difference and n is the number of terms for which the Sum is calculated. Sum of the given AP can also be computed using S n =n/2(a+l). Here, a is the first term of the given AP, l is its last term, and n is the total number of terms considered.

Arithmetic Mean (AM)

Arithmetic Mean of the given numbers is simply the average of them. Or a given set of numbers, 

Arithmetic Mean = Sum of terms/Number of terms

For defining the arithmetic mean of the given numbers, they are not required to be in an AP.

Sum of First n Natural Numbers

For a given range of up to n, the Sum of n natural numbers is given by:

S n =n(n+1)/2

To derive this formula of Sum of n natural numbers, one must consider the given natural numbers sequence as an AP, with the first term as one and common difference as 1.

Class 10 Maths Chapter 5 Important Questions

Based on the surveys done on previous year Boards examination papers and competitive level examinations, Vedantu experts designed Arithmetic Progression Class 10 important questions for the students. The top ten questions from the entire set of Arithmetic progression important questions are as follows:

For a given AP: 21, 18, 15, ....., which term is -81? Also, tell whether any of the terms is 0 or not? Justify the answers.

For a given AP: 11, 8, 5, 2, check whether 150 is the term associated with it or not.

For a given AP, the third term is four and the 9th term is -8. Find out whether any of its terms is zero. Also, find the term which is 0.

For a given AP: 3, 15, 27, 39, ..., write the difference between the 54th term of the AP and the number 132. Is 132 greater than the 54th term of the AP?

Write the number of multiples of 4 lying between 10 and 250.

For a given AP, the Sum of 4th and 8th term is 24, and that of 6th and 10th term is 44. Calculate the first five terms of the AP.

Shiva saved 5 INR in the first week of the year and increased her weekly savings by 1.75 INR every week. If his saving is 20.75 INR in the nth week, find n.

For a given AP: 24, 21, 18,..., how many terms are essential for making the Sum as 78?

For an AP, the first term is 5, and the last is 45. The Sum of all the terms is 400. Find the common difference and the number of terms.

For a given AP, the Sum of 3rd and 7th term is 6, and the product is 8. Find the Sum of the first sixteen terms of the given AP.

Practice Questions for Class 10 Maths Chapter 5: Arithmetic Progression

The following are some of the questions that can be taken up by students to assist them in the board preparations.

Question 1. A total of Rs 700 would be used to provide seven cash prizes to pupils for their overall academic performance at a school. Determine the value of each reward if it is Rs 20 less than the prize before it.

Answer. 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60 and Rs 40 are the correct answers.

Question 2. Find the 9th term from the end of the A.P. 5,9,13,..., 185 towards the first term.

Answer. 153 (Using the formula a n = a + (n-1)d).

Question 3. The angles of a triangle are measured in A.P., with the smallest angle being half of the largest. Determine the angles.

Answer. 40, 60, and 80 are the correct answers.

Question 4. In the series of AP 4, 9, 14, 19,..., which term is 109?

Answer. 22nd term. 

Question 5. Find the AP, if the sum of the first seven terms of an AP is 182. Given that the 4th and 17th terms are in the ratio 1: 5.

Answer. AP would be  2, 10, 18, 26, …. .

Benefits of Important Questions for Class 10 Maths Chapter 5:

Crucial in student preparation, Class 10 Maths Chapter 5 important questions serve as a pivotal study and revision resource. These essential questions enhance understanding and contribute significantly to exam success in Arithmetic Progression. The key benefits of engaging with these important questions include:

Comprehensive Preparation:

Important questions for Class 10 Maths Chapter 5 in Arithmetic Progression compile frequently asked and diverse question types.

Offer students a consolidated study material for better preparation.

Insights and Approaches:

Provide an in-depth understanding of question types and various approaches to solve them.

Equip students with knowledge on handling different types of problems effectively.

Confident Exam Appearance:

Enable students to confidently appear in exams with the desired level of practice.

Contribute to successful exam outcomes and performance with flying colors.

Versatile Practice:

Beneficial for both school and competitive level examinations.

Include alternative approaches for problem-solving in different scenarios.

Extra Load of Practice:

Serve as an additional resource for extra practice in mathematics.

Acknowledge the importance of consistent practice for successful outcomes.

Important Related Links for CBSE Class 10 Maths

Vedantu's mission is to support students nationwide in their exam preparations. All our study resources, including " Important Questions for CBSE Class 10 Maths Chapter 5 - Arithmetic Progressions," are freely available for download in PDF format. Our experts provide sample responses for each question in CBSE Class 10 test papers, enabling students to practice independently. Utilizing the Vedantu Website and App, students can access essential questions, revision notes , NCERT chapter-wise solutions, and other necessary materials related to Chapter 5. This inclusive approach ensures that valuable resources are accessible to all, fostering effective and holistic learning experiences.

FAQs on Important Questions for CBSE Class 10 Maths Chapter 5 - Arithmetic Progressions

Q1. Is Chapter 5 Arithmetic Progression of Class 10 Maths easy?

Ans: Chapter 5 Arithmetic Progression of Class 10 Maths is easy provided that you follow the right strategy and study plan to simplify your preparation for this chapter. Before solving the exercise questions, first, read the theory and obtain the basic conceptual understanding of the arithmetic progression and the related terms. Then, go through the solved examples to understand the stepwise method of answering each type of question. Next, solve all the NCERT questions and clear all doubts as soon as possible. Revise the chapter multiple times to make this chapter easy for you.

Q2. How can I prepare Chapter 5 Arithmetic Progression of Class 10 Maths ?

Ans: The first step to prepare Class 10 Maths, Chapter 5- Arithmetic Progression is to solve every question and example of this chapter from the NCERT standard Maths textbook. Clear all doubts as soon as possible to ensure a clear and smooth Maths exam preparation journey. Lastly, practice questions from the past year question papers from this chapter and give timed mock tests to build confidence and manage time well in the exam. If you follow this strategy diligently, you will master this chapter and perform well in the Maths exam.

Q3. What is Arithmetic Progression according to  Chapter 5 Arithmetic Progression of Class 10 Maths?

Ans: As discussed in the chapter, Arithmetic Progression, in simple words, refers to a mathematical sequence in a particular order in which the difference between two consecutive terms is a fixed number throughout the sequence. For instance, the series of natural numbers are in Arithmetic Progression because the difference between two consecutive terms is the same throughout the sequence. 

These solutions are available on Vedantu's official website( vedantu ) and mobile app free of cost.

Q4. What are the most important topics in Chapter 5 Arithmetic Progression of Class 10 Maths?

Ans: Class 10 Maths, Chapter 5- Arithmetic Progression, as the name suggests, deals with the study of arithmetic progression and related concepts. The most important topics in this chapter are- the difference between a sequence, a series and an arithmetic progression, the terms in an AP, the common difference between the terms in an AP, the nth term in an AP, the general form and the sum of n terms in an AP.

Q5. Where can I find the most important questions for Chapter 5 Arithmetic Progression of Class 10 Maths?

Ans: Refer to Vedantu's Important Questions for Class 10 Maths, Chapter 5- Arithmetic Progression by clicking here. It covers all the most important questions of Arithmetic Progression from an examination point of view. These questions have been prepared by the best Maths teachers in India after deeply analyzing the previous year Board exam questions, latest Maths syllabus and exam pattern. If you thoroughly prepare these questions, you will be an expert in this chapter and will surely fetch high marks in the Maths exam.

CBSE Class 10 Maths Important Questions

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Case Study Questions for Class 10 Maths

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This article covers case study questions for Class 10 Maths. Students are suggested to go through all the questions to score better in the exams.

Why Students Fear Case Study Questions for Class 10 Maths?

Students may fear Case Study questions for Class 10 Maths for the following reasons:

• Application of Concepts: Case Study questions require the application of mathematical concepts to real-world situations. Students may find it difficult to apply the concepts they have learned to solve the given scenario.
• Limited Practice: Unlike other types of questions, students may have limited practice with Case Study questions. This lack of familiarity may cause anxiety and fear among students.
• Lengthy and Complex Scenarios: Case Study questions often present lengthy and complex scenarios, which can be overwhelming for students. They may struggle to identify the relevant information and apply the appropriate mathematical concepts.
• Time Pressure: Case Study questions may require more time to solve compared to other types of questions. This can add to the stress and anxiety of students, especially during an exam.
• Fear of Making Mistakes: Since Case Study questions involve applying concepts to real-world situations, students may fear making mistakes or getting the wrong answer. This fear may cause them to avoid attempting the question altogether.

In conclusion, students may fear Case Study questions for Class 10 Maths due to the application of concepts, limited practice, lengthy and complex scenarios, time pressure, and fear of making mistakes. Teachers can help alleviate these fears by providing ample practice opportunities, breaking down the scenarios into smaller parts, and encouraging students to attempt the questions.

Best Way to Approach Case Study Questions for Class 10 Maths

Here are some tips on how to approach Case Study questions for Class 10 Maths:

• Read the scenario carefully: The first step is to read the scenario carefully and identify the key information. Pay attention to the given values, units, and any other important details.
• Identify the mathematical concepts involved: Once you have read the scenario, identify the mathematical concepts that are involved. This will help you determine which formulas or equations to apply.
• Break down the scenario into smaller parts: Some Case Study questions may have lengthy and complex scenarios. To make it easier, try to break down the scenario into smaller parts and identify the specific information that is needed to solve each part.
• Solve the problem step by step: Once you have identified the key information and the mathematical concepts involved, start solving the problem step by step. Show all the calculations and equations used.
• Check your answers: After you have solved the problem, check your answers to ensure that they are accurate and relevant to the scenario given. If possible, try to cross-check your answer using a different approach or formula.
• Practice, Practice, Practice: The more you practice Case Study questions, the more familiar you will become with the format and the types of scenarios presented. This will help you develop confidence and improve your performance.

In conclusion, approaching Case Study questions for Class 10 Maths involves careful reading of the scenario, identifying the mathematical concepts involved, breaking down the problem into smaller parts, solving the problem step by step, checking the answers, and practicing regularly.

Topics Covered in CBSE Class 10 Maths

Here are the topics covered in CBSE Class 10 Maths:

• Real Numbers: Euclid’s division lemma, HCF and LCM, irrational numbers, decimal representation of rational numbers, and the relationship between roots and coefficients of a quadratic equation.
• Polynomials: Zeros of a polynomial, relationship between zeros and coefficients of a polynomial, division algorithm for polynomials, and factorization of polynomials.
• Pair of Linear Equations in Two Variables: Graphical method of solution, algebraic methods of solution, and word problems based on linear equations.
• Quadratic Equations: Standard form of a quadratic equation, solutions of a quadratic equation by factorization and by using the quadratic formula, relationship between roots and coefficients, and nature of roots.
• Arithmetic Progressions: nth term of an AP, sum of first n terms of an AP, and word problems based on arithmetic progressions.
• Triangles: Properties of triangles, congruence of triangles, criteria for similarity of triangles, and Pythagoras theorem.
• Coordinate Geometry: Distance formula, section formula, area of a triangle, and equation of a line in different forms.
• Introduction to Trigonometry: Trigonometric ratios, trigonometric ratios of complementary angles, and word problems based on trigonometry.
• Some Applications of Trigonometry: Heights and distances.
• Circles: Tangent to a circle, number of tangents from a point on a circle, and chord properties.
• Constructions: Construction of bisectors of line segments and angles, construction of a triangle similar to a given triangle, and construction of a triangle of given perimeter and base angles.
• Areas Related to Circles: Areas of sectors and segments of a circle.
• Surface Areas and Volumes: Surface areas and volumes of spheres, cones, cylinders, and cuboids.

In conclusion, CBSE Class 10 Maths covers a wide range of topics including real numbers, polynomials, linear equations, quadratic equations, arithmetic progressions, triangles, coordinate geometry, trigonometry, circles, constructions, areas related to circles, and surface areas and volumes.

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