problem solving about geometric sequence

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Geometric Sequences Problems with Solutions

Geometric sequences are used in several branches of applied mathematics to engineering, sciences, computer sciences, biology, finance... Problems and exercises involving geometric sequences, along with answers are presented.

Review OF Geometric Sequences

The sequence shown below

Problems with Solutions

Problem 1 Find the terms a 2 , a 3 , a 4 and a 5 of a geometric sequence if a 1 = 10 and the common ratio r = - 1. Solution to Problem 1: Use the definition of a geometric sequence \( a_2 = a_1 \times r = 10 (-1) = - 10 \\ a_3 = a_2 \times r = - 10 (-1) = 10 \\ a_4 = a_3 \times r = 10 (-1) = - 10 \\ a_5 = a_4 \times r = - 10 (-1) = 10 \)

Find the 10 th term of a geometric sequence if a 1 = 45 and the common ration r = 0.2. Solution to Problem 2: Use the formula \[ a_n = a_1 \times r^{n-1} \] that gives the n th term to find a 10 as follows \( a_{10} = 45 \times 0.2^{10-1} = 2.304 \times 10^{-5} \)

Find a 20 of a geometric sequence if the first few terms of the sequence are given by

Given the terms a 10 = 3 / 512 and a 15 = 3 / 16384 of a geometric sequence, find the exact value of the term a 30 of the sequence. Solution to Problem 4: We first use the formula for the n th term to write a 10 and a 15 as follows \( a_{10} = a_1 \times r^{10-1} = a_1 r^9 = 3 / 512 \\ \\ a_{15} = a_1 \times r^{15-1} = a_1 r^{14} = 3 / 16384 \) We now divide the terms a 10 and a 15 to write \( a_{15} / a_{10} = a_1 \times r^{14} / (a_1 \times r^9) = (3 / 16384) / (3 / 512) \) Simplify expressions in the above equation to obtain. r 5 = 1 / 32 which gives r = 1/2 We now use a 10 to find a 1 as follows. \( a_{10} = 3 / 512 = a_1 (1/2)^9 \) Solve for a 1 to obtain. \( a_1 = 3 \) We now use the formula for the n th term to find a 30 as follows. \( a_{30} = 3(1/2)^{29} = 3 / 536870912 \)

Find the sum \[ S = \sum_{k=1}^{6} 3^{k - 1} \] Solution to Problem 5: We first rewrite the sum S as follows S = 1 + 3 + 9 + 27 + 81 + 243 = 364 Another method is to first note that the terms making the sum are those of a geometric sequence with a 1 = 1 and r = 3 using the formula s n = a 1 (1 - r n ) / (1 - r) with n = 6. s 6 = 1 (1 - 3 6 ) / (1 - 3) = 364

Find the sum \[ S = \sum_{i=1}^{10} 8 \times (1/4)^{i - 1} \] Solution to Problem 6: An examination of the terms included in the sum are 8 , 8× ((1/4) 1 , 8×((1/4) 2 , ... , 8×((1/4) 9 These are the terms of a geometric sequence with a 1 = 8 and r = 1/4 and therefore we can use the formula for the sum of the terms of a geometric sequence s 10 = a 1 (1 - r n ) / (1 - r) = 8 × (1 - (1/4) 10 ) / (1 - 1/4) = 10.67 (rounded to 2 decimal places)

Write the rational number 5.31313131... as the ratio of two integers. Solution to Problem 7: We first write the given rational number as an infinite sum as follows 5.313131... = 5 + 0.31 + 0.0031 + 0.000031 + .... The terms making 0.31 + 0.0031 + 0.000031 ... are those of a geometric sequence with a 1 = 0.31 and r = 0.01. Hence the use of the formula for an infinite sum of a geometric sequence S = a 1 / (1 - r) = 0.31 / (1 - 0.01) = 0.31 / 0.99 = 31 / 99 We now write 5.313131... as follows 5.313131... = 5 + 31/99 = 526 / 99

Exercises with Answers

Answer the following questions related to geometric sequences: a) Find a 20 given that a 3 = 1/2 and a 5 = 8 b) Find a 30 given that the first few terms of a geometric sequence are given by -2 , 1 , -1/2 , 1/4 ... c) Find r given that a 1 = 10 and a 20 = 10 -18 d) write the rational number 0.9717171... as a ratio of two positive integers.

a) a 20 = 2 18 b) a 30 = 1 / 2 28 c) r = 0.1 d) 0.9717171... = 481/495

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How to Solve Geometric Sequences? (+FREE Worksheet!)

Learn how to solve Geometric Sequence problems using the following step-by-step guide with detailed solutions.

Related Topics

• How to Solve Finite Geometric Series
• How to Solve Infinite Geometric Series
• How to Solve Arithmetic Sequences

Step by step guide to solve Geometric Sequence Problems

• It is a sequence of numbers where each term after the first is found by multiplying the previous item by the common ratio, a fixed, non-zero number. For example, the sequence \(2, 4, 8, 16, 32\), … is a geometric sequence with a common ratio of \(2\).
• To find any term in a geometric sequence use this formula: \(\color{blue}{x_{n}=ar^{(n – 1)}}\)
• \(a =\) the first term , \(r =\) the common ratio , \(n =\) number of items

Geometric Sequences – Example 1:

Given the first term and the common ratio of a geometric sequence find the first five terms of the sequence. \(a_1=3,r=-2\)

Use geometric sequence formula: \(\color{blue}{x_{n}=ar^{(n – 1)}}\) \(→x_{n}=0.8 .(-5)^{n-1}\) If \(n=1\) then: \(x_{1}=3 .(-2)^{1-1}=3 (1)=3\), First Five Terms: \(3,-6,12,-24,48\)

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Geometric sequences – example 2:.

Given two terms in a geometric sequence find the 8th term. \(a_{3}=10\) and \(a_{5}=40\)

Use geometric sequence formula: \(\color{blue}{x_{n}=ar^{(n – 1)}}\) \(→a_{3}=ar^{(3 – 1)}=ar^2=10\) \(x_{n}=ar^{(n – 1)}→a_5=ar^{(5 – 1)}=ar^4=40\) Now divide \(a_{5}\) by \(a_{3}\). Then: \(\frac{a_{5}}{a_{3}} =\frac{ar^4}{ar^2 }=\frac{40}{10}\), Now simplify: \(\frac{ar^4}{ar^2 }=\frac{40}{10}→r^2=4→r=2\) We can find a now: \(ar^2=12→a(2^2 )=10→a=2.5\) Use the formula to find the 8th term: \(x_{n}=ar^{(n – 1)}→a_8=(2.5) (2)^8=2.5(256)=640\)

Geometric Sequences – Example 3:

Given the first term and the common ratio of a geometric sequence find the first five terms of the sequence. \(a_{1}=0.8,r=-5\)

Use geometric sequence formula: \(\color{blue}{x_{n}=ar^{(n – 1)}}\) \(→x_{n}=0.8 .(-5)^{n-1}\) If \(n=1\) then: \(x_{1}=0.8 .(-5)^{1-1}=0.8 (1)=0.8\), First Five Terms: \(0.8,-4,20,-100,500\)

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Geometric sequences – example 4:.

Given two terms in a geometric sequence find the 8th term. \(a_3=12\) and \(a_5=48\)

Use geometric sequence formula: \(\color{blue}{x_{n}=ar^{(n – 1)}}\) \(→a_3=ar^{(3 – 1)}=ar^2=12\) \(\color{blue}{x_{n}=ar^{(n – 1)}}\) \(→a_5=ar^{(5 – 1)}=ar^4=48\) Now divide \(a_{5}\) by \(a_{3}\). Then: \(\frac{a_{5}}{a_{3} }=\frac{ar^4}{ar^2}=\frac{48}{12}\), Now simplify: \(\frac{ar^4}{ar^2}=\frac{48}{12}→r^2=4→r=2\) We can find a now: \(ar^2=12→a(2^2 )=12→a=3\) Use the formula to find the \(8^{th}\) term: \(\color{blue}{x_{n}=ar^{(n – 1)}}\) \(→a_{8}=(3) (2)^8=3(256)=768\)

Exercises for Solving Geometric Sequences

Determine if the sequence is geometric. if it is, find the common ratio..

• \(\color{blue}{1, – 5, 25, – 125, …}\)
• \(\color{blue}{– 2, – 4, – 8, – 16, …}\)
• \(\color{blue}{4, 16, 36, 64, …}\)
• \(\color{blue}{– 3, – 15, – 75, – 375, …}\)

Download Geometric Sequences Worksheet

• \(\color{blue}{r=-5}\)
• \(\color{blue}{r=2}\)
• \(\color{blue}{not \ geometric}\)
• \(\color{blue}{r=5}\)

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by: Effortless Math Team about 4 years ago (category: Articles , Free Math Worksheets )

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Geometric Sequences

Here we will learn what a geometric sequence is, how to continue a geometric sequence, how to find missing terms in a geometric sequence, and how to generate a geometric sequence.

At the end, you’ll find geometric sequence worksheets based on Edexcel, AQA, and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is a geometric sequence?

A geometric sequence (geometric progression) is an ordered set of numbers that progresses by multiplying or dividing each term by a common ratio.

If we multiply or divide by the same number each time to make the sequence, it is a geometric sequence .

The common ratio is the same for any two consecutive terms in the same sequence.

Here are a few examples,

What are geometric sequences?

Geometric sequence formula

The geometric sequence formula is,

Where, 

\pmb{ a_{n} } is the n^{th} term (general term),

\pmb{ a_{1} } is the first term,

\pmb{ n } is the term position,

and \pmb{ r } is the common ratio.

We get the geometric sequence formula by looking at the following example,

We can see the common ratio (r) is 2 , so r = 2 .

a_{1} is the first term which is 5 ,

a_{2} is the second term which is 10 ,

and a_{3} is the third term which is 20 etc.

However we can write this using the common difference of 2 ,

Related lessons on sequences

Geometric sequences  is part of our series of lessons to support revision on  sequences . You may find it helpful to start with the main sequences lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Quadratic sequences
• Arithmetic sequence
• Nth term of a sequence
• Recurrence relation
• Quadratic nth term

How to continue a geometric sequence

To continue a geometric sequence, you need to calculate the common ratio. This is the factor that is used to multiply one term to get the next term. To calculate the common ratio and continue a geometric sequence you need to:

Take two consecutive terms from the sequence.

• Divide the second term by the first term to find the common ratio r .

Multiply the last term in the sequence by the common ratio to find the next term. Repeat for each new term.

Explain how to continue a geometric sequence

Geometric sequences worksheet

Get your free geometric sequences worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Continuing a geometric sequence examples

Example 1: continuing a geometric sequence.

Calculate the next three terms for the geometric progression 1, 2, 4, 8, 16, …

Here we will take the numbers 4 and 8 .

2 Divide the second term by the first term to find the value of the common ratio, r .

3 Multiply the last term in the sequence by the common ratio to find the next term. Repeat for each new term.

The next three terms in the sequence are 32, 64, and 128 .

Example 2: continuing a geometric sequence with negative numbers

Calculate the next three terms for the sequence -2, -10, -50, -250, -1250, …

Here we will take the numbers -10 and -50 .

Divide the second term by the first term to find the common ratio, r.

The next three terms are -6250, -31250, and -156250.

Example 3: continuing a geometric sequence with decimals

Calculate the next three terms for the sequence 100, 10, 1, 0.1, 0.01, …

Here we will take the numbers 0.1 and 0.01 .

Divide the second term by the first term to find the common ratio, r .

The next three terms in the sequence are 0.001, 0.0001, and 0.00001.

Example 4: continuing a geometric sequence involving fractions

Calculate the next three terms for the sequence

Here we will take the numbers 5 and 2\frac{1}{2} .

The next three terms are

\frac{5}{16}, \frac{5}{32}, and \frac{5}{64} .

 Geometric sequence practice questions – continue the sequence

1. Write the next three terms of the sequence 0.5, 5, 50, 500, …

Choose two consecutive terms. For example, 5 and 50 .

Common ratio,

2. Write the next three terms of the sequence 0.04, 0.2, 1, 5, 25, …

Choose two consecutive terms. For example, 5 and 25 .

3. Calculate the next 3 terms of the sequence -1, -3, -9, -27, -81, …

Choose two consecutive terms. For example, -27 and -9 .

4. By finding the common ratio, state the next 3 terms of the sequence 640, 160, 40, 10, 2.5 .

Choose two consecutive terms. For example, 40 and 10 .

5. Work out the common ratio and therefore the next three terms in the sequence 36, 12, 4, \frac{4}{3}, \frac{4}{9}, …

Choose two consecutive terms. For example, 12 and 4 .

6. Find the common ratio and hence calculate the next three terms of the sequence 1, -1, 1, -1, 1, …

Choose two consecutive terms. For example, -1 and 1 .

How to find missing numbers in a geometric sequence

The common ratio can be used to find missing numbers in a geometric sequence. To find missing numbers in a geometric sequence you need to:

Calculate the common ratio between two consecutive terms.

• Multiply the term before any missing value by the common ratio.

Divide the term after any missing value by the common ratio.

Repeat Steps 2 and 3 until all missing values are calculated. You may only need to use Step 2 or 3 depending on what terms you have been given.

Explain how to find missing numbers in a geometric sequence

Finding missing numbers in a geometric sequence examples

Example 5: find the missing numbers in the geometric sequence.

Fill in the missing terms in the sequence 7, 14, …, …,112 .

  Multiply the term before any missing value by the common ratio.

The missing terms are 28 and 56 .

Note: Here, you could repeat Step 2 by using 28 \times 2 = 56.

Example 6: find the missing numbers in a geometric sequence including decimals

Find the missing values in the sequence 0.4, …, ..., 137.2, 960.4.

  Divide the term after any missing value by the common ratio.

The missing terms are 2.8 and 19.6 .

Example 7: find the missing numbers in a geometric sequence when there are multiple consecutive terms missing

Find the missing values in the sequence, -4, ..., …, -108,...

First, we need to find the factor between the two terms, -108 \div -4 = 27 .

To get from -108 to -4 , we jump 3 terms.

This means that -4 has been multiplied by the common ratio three times or -4 \times r \times r \times r = -4r^3 .

\begin{aligned} r^{3}&=27\\\\ r&=3 \end{aligned}

Note: Term -108 is already given.

The missing terms are -12, -36, and -324.

We don’t need to complete this step.

Example 8: find the missing numbers in a geometric sequence including mixed numbers

Find the missing values in the sequence

  Calculate the common ratio between two consecutive terms.

Repeat this step to find the next term.

40 \frac{1}{2} \times 3=121 \frac{1}{2}

The missing terms in the sequence are

1 \frac{1}{2}, 40 \frac{1}{2}, and 121 \frac{1}{2} .

 Geometric sequence practice questions – find missing numbers

1. Find the missing numbers in the geometric sequence 4, 2, …, 0.5, …

Choose two consecutive terms. For example, 4 and 2 .

2. Find the missing numbers in the sequence -7, -35, …, …, -4375

Choose two consecutive terms. For example, -7 and -35 .

3. Find the missing terms in the sequence 0.6, …, …, 0.075, 0.0375

Choose two consecutive terms. For example, 0.075 and 0.0375 .

4. Calculate the missing terms in the arithmetic sequence 2 \frac{1}{5}, \frac{11}{20}, \frac{11}{80}, \ldots, \ldots

Choose two consecutive terms. For example, \frac{11}{20} and \frac{11}{80} .

5. Work out the missing terms in the sequence 3, …, …, 24 .

3 has been multiplied by the common ratio, r, three times to get 24.

3 \times r \times r \times r=24 \text{ or } 3r^{3}=24 .

Solving the equation,

6. Work out the missing terms in the sequence 90, …, …, \frac{10}{3} .

90 has been multiplied by the common ratio, r, three times to get \frac{10}{3}.

90 \times r \times r \times r=\frac{10}{3} \text{ or } 90r^{3}=\frac{10}{3} .

How to generate a geometric sequence

In order to generate a geometric sequence, we need to know the n^{th} term. Using a as the first term of the sequence, r as the common ratio and n to represent the position of the term, the n^{th} term of a geometric sequence is written as ar^{n-1}.

Once we know the first term and the common ratio, we can work out any number of terms in the sequence.

The first term is found when n=1 , the second term when n=2 , the third term when n=3 and so on.

To generate a geometric sequence you need to:

• Substitute n=1 into the n^{th} term to calculate the first term.
• Substitute n=2 into the n^{th} term to calculate the second term.

Continue to substitute values for n until all the required terms of the sequence are calculated.

Explain how to generate a geometric sequence

Generating a geometric sequence examples

Example 9: generate a geometric sequence using the n th term.

Generate the first 5 terms of the sequence 4^{n-1} .

Substitute n = 1 into the n^{th} term to calculate the first term.

When n = 1,\quad 4^{1-1} = 4^{0} = 1 .

  Substitute n = 2 into the n^{th} term to calculate the second term.

When n = 2,\quad 4^{2-1 }= 4^{1} = 4 .

When n=3, \quad 4^{3-1}=4^{2}=16 .

When n=4, \quad 4^{4-1}=4^{3}=64 .

When n=5, \quad 4^{5-1}=4^{4}=256 .

The first 5 terms of the sequence are 1, 4, 16, 64, 256.

Example 10: generate a geometric sequence using a table

Complete the table for the first 5 terms of the arithmetic sequence 6 \times 2^{n-1}.

Example 11: generate larger terms in a geometric sequence

A geometric sequence has the n^{th} term \left(\frac{1}{2}\right)^{n} .

Calculate the 1^{st}, 2^{nd}, 10^{th} and 12^{th} terms in the sequence. Express your answers as fractions. 

When n=1,\quad \left(\frac{1}{2}\right)^{1}=\frac{1}{2} .

Substitute n = 2 into the n^{th} term to calculate the second term.

When n=2, \quad \left(\frac{1}{2}\right)^{2}=\frac{1}{4} .

When n=10,\quad \left(\frac{1}{2}\right)^{10}=\frac{1}{1024} .

When n=12,\quad \left(\frac{1}{2}\right)^{12}=\frac{1}{4096} .

The unknown terms are

1, \frac{1}{4}, \frac{1}{1024}, and \frac{1}{4096} .

Example 12: generate a geometric sequence with a negative common ratio

Generate the first 5 terms of the geometric sequence 2(- 3)^{n-1} .

When n=1, \quad 2(-3)^{n-1}=2(-3)^{1-1}=2(-3)^{0}=2 \times 1=2 .

When n=2,\quad 2(-3)^{n-1}=2(-3)^{2-1}=2(-3)^{1}=2 \times-3=-6 .

When n=3, \quad 2(-3)^{n-1}=2(-3)^{3-1}=2(-3)^{2}=2 \times 9=18 .

When n=4, \quad 2(-3)^{n-1}=2(-3)^{4-1}=2(-3)^{3}=2 \times-27=-54 .

When n=5, \quad 2(-3)^{n-1}=2(-3)^{5-1}=2(-3)^{4}=2 \times 81=162 .

The first 5 terms of the sequence are 2, -6, 18, -54, and 162 .

 Geometric sequence practice questions – generate a sequence

1. Generate the first 5 terms of the sequence 10^{n} .

When  n=1, 10^{1}=10 .

When n=2, 10^{2}=100 .

When n=3, 10^{3}=1000 .

When n=4, 10^{4}=10000 .

When n=5, 10^{5}=100000 .

2. Generate the first 5 terms of the sequence 5^{n-1} .

When n=1, 5^{1-1}=5^{0}=1 .

When n=2, 5^{2-1}=5^{1}=5 .

When n=3, 5^{3-1}=5^{2}=25 .

When n=4, 5^{4-1}=5^{3}=125 .

When n=5, 5^{5-1}=5^{4}=625  .

3. Generate the first 5 terms of the sequence 4 \times 3^{n-1} .

When n=1, 4 \times 3^{1-1}=4 \times 3^{0}=4 .

When n=2, 4 \times 3^{2-1}=4 \times 3^{1}=12 .

When n=3, 4 \times 3^{3-1}=4 \times 3^{2}=36 .

When n=4, 4 \times 3^{4-1}=4 \times 3^{3}=108 .

When n=5, 4 \times 3^{5-1}=4 \times 3^{4}=324  .

4. Generate the first 5 terms of the sequence \frac{3^{n}}{6} .

When n=1, \frac{3^1}{6}= \frac{1}{2} .

When n=2, \frac{3^2}{6}= \frac{9}{6} = 1 \frac{1}{2} .

When n=3, \frac{3^3}{6}= \frac{27}{6} = 4 \frac{1}{2} .

When n=4, \frac{3^4}{6}= \frac{81}{6} = 13 \frac{1}{2} .

When n=5, \frac{3^5}{6}= \frac{243}{6} = 40 \frac{1}{2}  .

5. Calculate the 1st, 3rd, 10th and 15th term of the sequence 2^{n} .

When n=1, 2^{1}= 2 .

When n=3, 2^{3}= 8 .

When n=10, 2^{10}= 1024 .

When n=15, 2^{15}= 32768 .

6. Calculate the first 5 terms of the sequence 3 \times (-5)^{n-1} .

When n=1, 3 \times (-5)^{1-1} = 3 \times (-5)^{0} = 3 .

When n=2, 3 \times (-5)^{2-1} = 3 \times (-5)^{1} = -15 .

When n=3, 3 \times (-5)^{3-1} = 3 \times (-5)^{2} = 75 .

When n=4, 3 \times (-5)^{4-1} = 3 \times (-5)^{3} = -375 .

When n=5, 3 \times (-5)^{5-1} = 3 \times (-5)^{4} = 1875 .

Geometric sequences GCSE exam questions

1. Which sequence is a geometric progression?

1, 3, 5, 7, 9,…. \quad \quad \quad 1, 3, 9, 27, 81, …..

1, 3, 6, 10, 15, …. \quad \quad 1, 0.6, 0.2, -0.2, -0.6,….

1, 3, 9, 27, 81, …..

2.  Here is a geometric progression,

1, -5, 25, …., 625, …

(a) Find the common ratio.

(b) Work out the fourth term of the sequence.

25 \div -5 = – 5

Common ratio = -5

25 \times -5

3.  A scientist is studying a type of bacteria. The number of bacteria over the first four days is shown below.

How many bacteria will there be on day 7?

180 \div 60 = 3

1620 \times 3 \times 3 \times 3

Common misconceptions

• Mixing up the common ratio with the common difference for arithmetic sequences

Although these two phrases are similar, each successive term in a geometric sequence of numbers is calculated by multiplying the previous term by a common ratio and not by adding a common difference.

• A negative value for r means that all terms in the sequence are negative

This is not always the case as when r is raised to an even power, the solution is always positive.

• The first term in a geometric sequence

The first term is a . With ar^{n-1} , the first term would occur when n = 1 and so the power of r would be equal to 0 . Anything to the power of 0 is equal to 1 , leaving a as the first term in the sequence. This is usually mistaken when a = 1 as it is not clearly noted in the question for example, 2^{n-1} is the same as 1 \times 2^{n-1} .

• Incorrect simplifying of the n th term

For example, 6 \times 3^{n-1} is incorrectly simplified to 18^{n-1} as 6 \times 3 = 18 .

• The difference between an arithmetic and a geometric sequence

Arithmetic sequences are formed by adding or subtracting the same number. Geometric sequences are formed by multiplying or dividing the same number.

Learning checklist

You have now learned how to:

• Recognise geometric sequences

The next lessons are

• Inequalities
• Functions in algebra
• Laws of indices

Still stuck?

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Geometric Sequence

Geometric sequence formula.

A geometric sequence (also known as geometric progression) is a type of sequence wherein every term except the first term is generated by multiplying the previous term by a fixed nonzero number called common ratio , r.

More so, if we take any term in the geometric sequence except the first term and divide it by the previous term, the quotient is always the same. This constant or fixed quotient is called the common ratio and is usually represented by the letter r.

How to “Derive” the Geometric Sequence Formula

To generate a geometric sequence, we start by writing the first term. Then we multiply the first term by a fixed nonzero number to get the second term of the geometric sequence.

To obtain the third sequence, we take the second term and multiply it by the common ratio. Maybe you are seeing the pattern now. To get to the next term of the sequence, you multiply the preceding term by the constant nonzero number that we used as the common multiplier.

To make a sense of this all, let’s have a concrete example. Suppose we have a geometric sequence wherein

the first term , [latex]\large{a_1}[/latex], is 3

the fixed constant or common ratio , [latex]\large{r}[/latex], is 2 .

So if the first term is [latex]3[/latex] then we have

[latex]\large{{a_1} = 3}[/latex]

The second term is generated by multiplying the first term by the fixed constant [latex]2[/latex].

[latex]\large{{a_2} = 3\left( 2 \right)^1}[/latex]

The third term is generated by multiplying the second term by the fixed constant [latex]2[/latex].

[latex]\large{{a_3} = 3\left( 2 \right)\left( 2 \right) = 3{\left( 2 \right)^2}}[/latex]

The fourth term is obtained by multiplying the third term by the fixed constant [latex]2[/latex].

[latex]\large{{a_4} = 3\left( 2 \right)\left( 2 \right)\left( 2 \right) = 3{\left( 2 \right)^3}}[/latex]

The fifth term is obtained by multiplying the fourth term by the fixed constant [latex]2[/latex].

[latex]\large{{a_5} = 3\left( 2 \right)\left( 2 \right)\left( 2 \right)\left( 2 \right) = 3{\left( 2 \right)^4}}[/latex]

So now how can we interpret and use the example above to convert it into a formula? Observe that the first term [latex]\color{red}\large{{a_1}}[/latex] is always present in every term of the sequence. In the same manner, the fixed constant [latex]\color{red}\large{r}[/latex] is also attached in every term to some power. Notice that

• if [latex]\large{n}[/latex] is [latex]1[/latex] the power of [latex]\large{r}[/latex] is [latex]0[/latex]
• if [latex]\large{n}[/latex] is [latex]2[/latex] the power of [latex]\large{r}[/latex] is [latex]1[/latex]
• if [latex]\large{n}[/latex] is [latex]3[/latex] the power of [latex]\large{r}[/latex] is [latex]2[/latex]
• if [latex]\large{n}[/latex] is [latex]4[/latex] the power of [latex]\large{r}[/latex] is [latex]3[/latex]
• if [latex]\large{n}[/latex] is [latex]5[/latex] the power of [latex]\large{r}[/latex] is [latex]4[/latex]

Note: [latex]\large{n}[/latex] is the subscript of [latex]\large{a}[/latex] as in [latex]\large{a_n}[/latex]

It means that the power of the fixed constant (also known as the common ratio) [latex]\color{red}\large{r}[/latex] is [latex]1[/latex] less than whatever [latex]\large{n}[/latex] is.

Therefore, we can now deduce that the nth term [latex]\color{red}\large{{a_n}}[/latex] of a geometrc sequence is equal to the first term [latex]\color{red}\large{{a_1}}[/latex] multiplied to the fixed constant (also known as the common ratio) [latex]\color{red}\large{{r}}[/latex] raised to 1 less than [latex]\large{{n}}[/latex].

[latex]\LARGE{{a_n} = {a_1}{\left( r \right)^{n – 1}}}[/latex]

Below is a quick illustration on how we derive the geometric sequence formula.

Breakdown of the Geometric Sequence Formula

Notes about the geometric sequence formula:

• the common ratio r cannot be zero
• n is the position of the term in the sequence. For example, the third term is [latex]n=3[/latex], the fourth term is [latex]n=4[/latex], the fifth term is [latex]n=5[/latex], and so on.

Examples of Using the Geometric Sequence Formula

To learn and get familiar with the formula quickly, we will start with easy or foundational problems then gradually progress to more challenging ones. Feel free to skip the problems that you already know and jump to the ones that you want to go over.

Example 1: Tell whether each sequence is geometric or not. Explain.

a) Sequence A: [latex]3,12,48,192,…[/latex]

b) Sequence B: [latex] – 1,2, – 4,8,…[/latex]

c) Sequence C: [latex]4,8,12,16,…[/latex]

d) Sequence D: [latex]\Large{{1 \over 3},{1 \over 2},{3 \over 4},{9 \over 8},…}[/latex]

a) Yes. Sequence A is a geometric sequence because there is a common ratio between consecutive terms. The common ratio is [latex]4[/latex].

b) Yes. Sequence B is also a geometric sequence since the adjacent terms have a common ratio which is [latex]-2[/latex].

Notice that when a geometric sequence has a negative common ratio, the sequence will have alternating signs. That means the signs of the terms are switching back and forth between positive and negative.

c) No. Sequence C is not a geometric sequence. The consecutive terms don’t have a common ratio.

I hope you recognize that this is another kind of sequence. Notice, there is a common difference between consecutive terms which is [latex]4[/latex]. That is,

[latex]8-4=4[/latex]

[latex]12-8=4[/latex]

[latex]16-12=4[/latex]

Since the common difference is [latex]4[/latex], this is in fact an arithmetic sequence .

d) Yes. Sequence D is a geometric sequence because it has a common ratio of [latex]\Large{{3 \over 2}}[/latex].

Remember that when we divide fractions, we convert the problem from division to multiplication. Take the dividend (fraction being divided) and multiply it to the reciprocal of the divisor. Then, we simplify as needed.

[latex]\Large{{1 \over 2} \div {1 \over 3} = {1 \over 2} \times {3 \over 1} = {3 \over 2}}[/latex]

[latex]\Large{{3 \over 4} \div {1 \over 2} = {3 \over 4} \times {2 \over 1} = {3 \over 2}}[/latex]

[latex]\Large{{9 \over 8} \div {3 \over 4} = {9 \over 8} \times {4 \over 3} = {3 \over 2}}[/latex]

Example 2: Write a geometric sequence with five (5) terms wherein the first term is [latex]0.5[/latex] and the common ratio is [latex]6[/latex].

The first term is given to us which is [latex]\large{{a_1} = 0.5}[/latex]. Thus, we will have to find the other four terms. We can use the common ratio to produce the next four terms. The common ratio which in this case is [latex]6[/latex] will serve as the fixed multiplier to calculate the rest of the terms in the sequence.

The first term is [latex]{0.5}[/latex]. The second term is the first term multiplied by the common ratio [latex]6[/latex] which equals [latex]3[/latex]. The third term is the second term times [latex]6[/latex], and so on.

[latex]\large{{a_1} = 0.5}[/latex]

[latex]\large{{a_2} = 0.5\left( 6 \right) = 3}[/latex]

[latex]\large{{a_3} = 3\left( 6 \right) = 18}[/latex]

[latex]\large{{a_4} = 18\left( 6 \right) = 108}[/latex]

[latex]\large{{a_5} = 108\left( 6 \right) = 648}[/latex]

Therefore, the geometric sequence that satisfies the given conditions is

[latex]{0.5\,\,,\,\,3\,\,,\,\,18\,\,,\,\,\,108\,\,,\,\,648\,\,,\,\,…}[/latex]

Example 3: Generate a geometric sequence with five (5) terms such that each term is half the previous term. Answers may vary.

There is an infinite number of geometric sequences that can satisfy this condition because we are not restricted to a particular first term. We can use any numbers as our first term. That’s why the answers to this problem will vary as we are free to choose our starting term.

The problem can be expressed algebraically as

It only means that to go from the previous term to the following term, we multiply the previous term by [latex]\Large{{1 \over 2}}[/latex] to get to the next.

If we pick [latex]48[/latex] as our first term, we multiply it by [latex]\Large{{1 \over 2}}[/latex] to produce the second term. To generate the third term, multiply the second term by [latex]\Large{{1 \over 2}}[/latex], and so on.

[latex]\large{{a_1} = 48}[/latex]

[latex]\large{{a_2} = 48\left( {{1 \over 2}} \right) = 24}[/latex]

[latex]\large{{a_3} = 24\left( {{1 \over 2}} \right) = 12}[/latex]

[latex]\large{{a_4} = 12\left( {{1 \over 2}} \right) = 6}[/latex]

[latex]\large{{a_5} = 6\left( {{1 \over 2}} \right) = 3}[/latex]

So here’s our geometric sequence with five (5) terms such that each term (except the first term) is one-half of the previous term.

[latex]\large{48\,,\,\,24\,\,,\,\,12\,\,,\,\,6\,\,,\,\,3\,\,,\,\,…}[/latex]

Example 4: The first term of the geometric sequence is [latex]7[/latex] while its common ratio is [latex]-2[/latex]. Write the nth term formula of the sequence in the standard form.

The standard formula of the geometric sequence is

This is an easy problem because the values of the first term and the common ratio are given to us. We simply substitute them into the formula and we are done.

Since [latex]{a_1} = 7[/latex] and [latex]r = – 2[/latex], we have

[latex]\large{{a_n} = 7{\left( { – 2} \right)^{n – 1}}}[/latex]

Example 5: Determine the nth term formula of the geometric sequence below.

[latex]\large{16\,,\,\,12\,\,,\,\,9\,\,,\,\,…}[/latex]

To write the nth term formula, we will need the values of the first term and the common ratio. Since we are given the geometric sequence itself, the first term [latex]\large{{a_1}}[/latex] can easily be found.

The first term of the geometric sequence is obviously [latex]16[/latex].

Divide each term by the previous term. Since the quotients are the same, then it becomes our common ratio. In this case, we have [latex]\Large{r = {3 \over 4}}[/latex].

Substituting the values of the first term and the common ratio into the formula, we get

[latex]\Large{{a_n} = {a_1}{\left( r \right)^{n – 1}}}[/latex]

[latex]\Large{{a_n} = 16{\left( {{3 \over 4}} \right)^{n – 1}}}[/latex]

Example 6: Determine the indicated terms below using the geometric sequence formula.

a) The first term is [latex]3[/latex]. Find the sixth term if the common ratio is [latex]2[/latex].

b) The first term is [latex]-2[/latex]. Find the seventh term if the common ratio is [latex]-3[/latex].

Solution: These two (2) problems are very similar. The first term and the common ratio are both given in the problem. The only thing we have to do is to plug these values into the geometric sequence formula then use it to find the nth term of the sequence.

a) The first term is [latex]\large{{a_1} = 3}[/latex] while its common ratio is [latex]r = 2[/latex].

This gives us

To find the sixth term, we let [latex]n=6[/latex] then simplify.

[latex]\large{a_n} = 3{\left( 2 \right)^{n – 1}}[/latex]

[latex]\large{a_6} = 3{\left( 2 \right)^{6 – 1}}[/latex]

[latex]\large = 3{\left( 2 \right)^5}[/latex]

[latex]\large = 3\left( {32} \right)[/latex]

[latex]\large{a_6} = 96[/latex]

b) The first term is [latex]\large{{a_1} = -2}[/latex] and the common ratio is [latex]r = -3[/latex].

The nth term formula becomes

To find the seventh term, we will set [latex]n=7[/latex] then simplify.

[latex]\large{a_n} = – 2{\left( { – 3} \right)^{n – 1}}[/latex]

[latex]\large{a_7} = – 2{\left( { – 3} \right)^{7 – 1}}[/latex]

[latex]\large = – 2{\left( { – 3} \right)^6}[/latex]

[latex]\large = – 2\left( {729} \right)[/latex]

[latex]\large {a_7} = – 1,458[/latex]

Example 7: Find the tenth term of the geometric sequence below.

[latex]\Large{1 \over 3}\,\,,\,\,\,1\,\,,\,\,\,3\,\,,\,\,\,9\,\,,\,\,…[/latex]

This problem is similar to example 6. The only difference is that the values of the first term and common ratio are not given upfront. However, they can easily be found and computed from the geometric sequence.

The very first term of the sequence is very obvious to identify.

Calculate the common ratio by dividing each term by the previous term. If the quotients are the same, then it is our common ratio.

Since the first term is [latex]\large{a_1} = {1 \over 3}[/latex] and the common ratio is [latex]r = 3[/latex], we write the general formula as

[latex]\Large{a_n} = {1 \over 3}{\left( 3 \right)^{n – 1}}[/latex]

The 10th term is calculated by letting [latex]n=10[/latex].

[latex]\Large{a_{10}} = {1 \over 3}{\left( 3 \right)^{10 – 1}}[/latex]

[latex]\Large = {1 \over 3}{\left( 3 \right)^9}[/latex]

[latex]\large = {1 \over 3}\left( {19,683} \right)[/latex]

[latex]\large {a_{10}} = 6,561[/latex]

Example 8: The second term of a geometric sequence is [latex]2[/latex], and the fifth term is [latex]\Large{1 \over {32}}[/latex]. Find the ninth term.

We will use the given two terms to create a system of equations that we can solve to find the common ratio [latex]r[/latex] and the first term [latex]{a_1}[/latex]. After doing so, it is possible to write the general formula that can find any term in the geometric sequence. In particular, we want to find the ninth term.

Since [latex]\large{a_2} = 2[/latex] and [latex]\large{a_5} = {1 \over {32}}[/latex], we substitute them in the nth term formula [latex]\large{a_n} = {a_1}{\left( r \right)^{n – 1}}[/latex] to get

Equation 1:

[latex]\large2 = {a_1}r[/latex]

Equation 2:

[latex]\Large{{1 \over {32}} = {a_1}{r^4}}[/latex]

Let’s divide equation 2 by equation 1 to cancel the variable [latex]\large{a_1}[/latex]. What’s left is an equation that we can solve for [latex]\large{r}[/latex].

We can now solve for the value of the first term by substituting the value of [latex]r[/latex] to either equation 1 or 2 then solve for [latex]\large{a_1}[/latex]. We will use equation 1 because it is much simpler.

This gives us the nth term formula of the geometric sequence.

Finally, we can solve for the 9th term of the geometric sequence by letting [latex]n=9[/latex] and simplifying it.

Example 9: The third term of a geometric sequence is [latex]5[/latex], and the seventh term is [latex]\Large{5 \over {16}}[/latex]. Find the thirteenth term.

I’m sure that you already know the drill. We will use the information of the given two terms from the problem to create a system of equations with [latex]\color{red}\large{a_1}[/latex] and [latex]\color{red}r[/latex] as the unknown variables.

For [latex]\large{a_3=5}[/latex], our equation 1 is

For [latex]\large{a_7} = {5 \over {16}}[/latex], our equation 2 is

We divide equation 2 by equation 1 to cancel out the [latex]\large{a_1}[/latex] term then solve for [latex]\large{r}[/latex].

Now, we solve for [latex]\large{a_1}[/latex] by substituting the value of [latex]r=1/2[/latex] to any of the equations. We will use equation 1 because it is the simpler one of the two.

We can now put together the nth term formula of the geometric sequence.

Finally, we solve the value of the missing term. Since we are looking for the 13th term of the geometric sequence, we let [latex]n=13[/latex] then simplify.

You may also be interested in these related math lessons or tutorials:

Geometric Series Formula

Geometric Series Practice Problems

Arithmetic Sequence Formula

Arithmetic Sequence Practice Problems with Answers

Arithmetic Series Formula

Arithmetic Series Formula Practice Problems

9.3 Geometric Sequences

Learning objectives.

In this section, you will:

• Find the common ratio for a geometric sequence.
• List the terms of a geometric sequence.
• Use a recursive formula for a geometric sequence.
• Use an explicit formula for a geometric sequence.

Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be $26,520 after one year; $27,050.40 after two years; $27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way.

Finding Common Ratios

The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio . The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.

Definition of a Geometric Sequence

A geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If a 1 a 1 is the initial term of a geometric sequence and r r is the common ratio, the sequence will be

Given a set of numbers, determine if they represent a geometric sequence.

• Divide each term by the previous term.
• Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.

Is the sequence geometric? If so, find the common ratio.

• ⓐ 1 , 2 , 4 , 8 , 16 , ... 1 , 2 , 4 , 8 , 16 , ...
• ⓑ 48 , 12 , 4 ,  2 , ... 48 , 12 , 4 ,  2 , ...

Divide each term by the previous term to determine whether a common ratio exists.

The sequence is geometric because there is a common ratio. The common ratio is 2.

The sequence is not geometric because there is not a common ratio.

The graph of each sequence is shown in Figure 1 . It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.

If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio?

No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio.

Writing Terms of Geometric Sequences

Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is a 1 = − 2 a 1 = − 2 and the common ratio is r = 4, r = 4, we can find subsequent terms by multiplying − 2 ⋅ 4 − 2 ⋅ 4 to get − 8 − 8 then multiplying the result − 8 ⋅ 4 − 8 ⋅ 4 to get − 32 − 32 and so on.

The first four terms are { –2 ,  –8 ,  –32 ,  –128 } . { –2 ,  –8 ,  –32 ,  –128 } .

Given the first term and the common factor, find the first four terms of a geometric sequence.

• Multiply the initial term, a 1 , a 1 , by the common ratio to find the next term, a 2 . a 2 .
• Repeat the process, using a n = a 2 a n = a 2 to find a 3 a 3 and then a 3 a 3 to find a 4, a 4, until all four terms have been identified.
• Write the terms separated by commons within brackets.

Writing the Terms of a Geometric Sequence

List the first four terms of the geometric sequence with a 1 = 5 a 1 = 5 and r = –2. r = –2.

Multiply a 1 a 1 by − 2 − 2 to find a 2 . a 2 . Repeat the process, using a 2 a 2 to find a 3 , a 3 , and so on.

The first four terms are { 5 , –10 , 20 , –40 } . { 5 , –10 , 20 , –40 } .

List the first five terms of the geometric sequence with a 1 = 18 a 1 = 18 and r = 1 3 . r = 1 3 .

Using Recursive Formulas for Geometric Sequences

A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.

Recursive Formula for a Geometric Sequence

The recursive formula for a geometric sequence with common ratio r r and first term a 1 a 1 is

Given the first several terms of a geometric sequence, write its recursive formula.

• State the initial term.
• Find the common ratio by dividing any term by the preceding term.
• Substitute the common ratio into the recursive formula for a geometric sequence.

Write a recursive formula for the following geometric sequence.

The first term is given as 6. The common ratio can be found by dividing the second term by the first term.

Substitute the common ratio into the recursive formula for geometric sequences and define a 1 . a 1 .

The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown in Figure 2

Do we have to divide the second term by the first term to find the common ratio?

No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio.

Using Explicit Formulas for Geometric Sequences

Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.

Let’s take a look at the sequence { 18 ,  36 ,  72 ,  144 ,  288 ,  ... } . { 18 ,  36 ,  72 ,  144 ,  288 ,  ... } . This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is

The graph of the sequence is shown in Figure 3 .

Explicit Formula for a Geometric Sequence

The n n th term of a geometric sequence is given by the explicit formula :

Writing Terms of Geometric Sequences Using the Explicit Formula

Given a geometric sequence with a 1 = 3 a 1 = 3 and a 4 = 24 , a 4 = 24 , find a 2 . a 2 .

The sequence can be written in terms of the initial term and the common ratio r . r .

Find the common ratio using the given fourth term.

Find the second term by multiplying the first term by the common ratio.

The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power.

Given a geometric sequence with a 2 = 4 a 2 = 4 and a 3 = 32 a 3 = 32 , find a 6 . a 6 .

Writing an Explicit Formula for the n n th Term of a Geometric Sequence

Write an explicit formula for the n th n th term of the following geometric sequence.

The first term is 2. The common ratio can be found by dividing the second term by the first term.

The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.

The graph of this sequence in Figure 4 shows an exponential pattern.

Write an explicit formula for the following geometric sequence.

Solving Application Problems with Geometric Sequences

In real-world scenarios involving geometric sequences, we may need to use an initial term of a 0 a 0 instead of a 1 . a 1 . In these problems, we can alter the explicit formula slightly by using the following formula:

In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.

• ⓐ Write a formula for the student population.
• ⓑ Estimate the student population in 2020.

The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.

Let P P be the student population and n n be the number of years after 2013. Using the explicit formula for a geometric sequence we get

We can find the number of years since 2013 by subtracting.

We are looking for the population after 7 years. We can substitute 7 for n n to estimate the population in 2020.

The student population will be about 374 in 2020.

A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.

• ⓐ Write a formula for the number of hits.
• ⓑ Estimate the number of hits in 5 weeks.

Access these online resources for additional instruction and practice with geometric sequences.

• Geometric Sequences
• Determine the Type of Sequence
• Find the Formula for a Sequence

9.3 Section Exercises

What is a geometric sequence?

How is the common ratio of a geometric sequence found?

What is the procedure for determining whether a sequence is geometric?

What is the difference between an arithmetic sequence and a geometric sequence?

Describe how exponential functions and geometric sequences are similar. How are they different?

For the following exercises, find the common ratio for the geometric sequence.

1 , 3 , 9 , 27 , 81 , ... 1 , 3 , 9 , 27 , 81 , ...

− 0.125 , 0.25 , − 0.5 , 1 , − 2 , ... − 0.125 , 0.25 , − 0.5 , 1 , − 2 , ...

− 2 , − 1 2 , − 1 8 , − 1 32 , − 1 128 , ... − 2 , − 1 2 , − 1 8 , − 1 32 , − 1 128 , ...

For the following exercises, determine whether the sequence is geometric. If so, find the common ratio.

− 6 , − 12 , − 24 , − 48 , − 96 , ... − 6 , − 12 , − 24 , − 48 , − 96 , ...

5 , 5.2 , 5.4 , 5.6 , 5.8 , ... 5 , 5.2 , 5.4 , 5.6 , 5.8 , ...

− 1 , 1 2 , − 1 4 , 1 8 , − 1 16 , ... − 1 , 1 2 , − 1 4 , 1 8 , − 1 16 , ...

6 , 8 , 11 , 15 , 20 , ... 6 , 8 , 11 , 15 , 20 , ...

0.8 , 4 , 20 , 100 , 500 , ... 0.8 , 4 , 20 , 100 , 500 , ...

For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio.

a 1 = 8 , r = 0.3 a 1 = 8 , r = 0.3

a 1 = 5 , r = 1 5 a 1 = 5 , r = 1 5

For the following exercises, write the first five terms of the geometric sequence, given any two terms.

a 7 = 64 , a 10 = 512 a 7 = 64 , a 10 = 512

a 6 = 25 , a 8 = 6.25 a 6 = 25 , a 8 = 6.25

For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio.

The first term is 2, 2, and the common ratio is 3. 3. Find the 5 th term.

The first term is 16 and the common ratio is − 1 3 . − 1 3 . Find the 4 th term.

For the following exercises, find the specified term for the geometric sequence, given the first four terms.

a n = { − 1 , 2 , − 4 , 8 , ... } . a n = { − 1 , 2 , − 4 , 8 , ... } . Find a 12 . a 12 .

a n = { − 2 , 2 3 , − 2 9 , 2 27 , ... } . a n = { − 2 , 2 3 , − 2 9 , 2 27 , ... } . Find a 7 . a 7 .

For the following exercises, write the first five terms of the geometric sequence.

a 1 = − 486 , a n = − 1 3 a n − 1 a 1 = − 486 , a n = − 1 3 a n − 1

a 1 = 7 , a n = 0.2 a n − 1 a 1 = 7 , a n = 0.2 a n − 1

For the following exercises, write a recursive formula for each geometric sequence.

a n = { − 1 , 5 , − 25 , 125 , ... } a n = { − 1 , 5 , − 25 , 125 , ... }

a n = { − 32 , − 16 , − 8 , − 4 , ... } a n = { − 32 , − 16 , − 8 , − 4 , ... }

a n = { 14 , 56 , 224 , 896 , ... } a n = { 14 , 56 , 224 , 896 , ... }

a n = { 10 , − 3 , 0.9 , − 0.27 , ... } a n = { 10 , − 3 , 0.9 , − 0.27 , ... }

a n = { 0.61 , 1.83 , 5.49 , 16.47 , ... } a n = { 0.61 , 1.83 , 5.49 , 16.47 , ... }

a n = { 3 5 , 1 10 , 1 60 , 1 360 , ... } a n = { 3 5 , 1 10 , 1 60 , 1 360 , ... }

a n = { − 2 , 4 3 , − 8 9 , 16 27 , ... } a n = { − 2 , 4 3 , − 8 9 , 16 27 , ... }

a n = { 1 512 , − 1 128 , 1 32 , − 1 8 , ... } a n = { 1 512 , − 1 128 , 1 32 , − 1 8 , ... }

a n = − 4 ⋅ 5 n − 1 a n = − 4 ⋅ 5 n − 1

a n = 12 ⋅ ( − 1 2 ) n − 1 a n = 12 ⋅ ( − 1 2 ) n − 1

For the following exercises, write an explicit formula for each geometric sequence.

a n = { − 2 , − 4 , − 8 , − 16 , ... } a n = { − 2 , − 4 , − 8 , − 16 , ... }

a n = { 1 , 3 , 9 , 27 , ... } a n = { 1 , 3 , 9 , 27 , ... }

a n = { − 4 , − 12 , − 36 , − 108 , ... } a n = { − 4 , − 12 , − 36 , − 108 , ... }

a n = { 0.8 , − 4 , 20 , − 100 , ... } a n = { 0.8 , − 4 , 20 , − 100 , ... }

a n = { − 1.25 , − 5 , − 20 , − 80 , ... } a n = { − 1.25 , − 5 , − 20 , − 80 , ... }

a n = { − 1 , − 4 5 , − 16 25 , − 64 125 , ... } a n = { − 1 , − 4 5 , − 16 25 , − 64 125 , ... }

a n = { 2 , 1 3 , 1 18 , 1 108 , ... } a n = { 2 , 1 3 , 1 18 , 1 108 , ... }

a n = { 3 , − 1 , 1 3 , − 1 9 , ... } a n = { 3 , − 1 , 1 3 , − 1 9 , ... }

For the following exercises, find the specified term for the geometric sequence given.

Let a 1 = 4 , a 1 = 4 , a n = − 3 a n − 1 . a n = − 3 a n − 1 . Find a 8 . a 8 .

Let a n = − ( − 1 3 ) n − 1 . a n = − ( − 1 3 ) n − 1 . Find a 12 . a 12 .

For the following exercises, find the number of terms in the given finite geometric sequence.

a n = { − 1 , 3 , − 9 , ... , 2187 } a n = { − 1 , 3 , − 9 , ... , 2187 }

a n = { 2 , 1 , 1 2 , ... , 1 1024 } a n = { 2 , 1 , 1 2 , ... , 1 1024 }

For the following exercises, determine whether the graph shown represents a geometric sequence.

For the following exercises, use the information provided to graph the first five terms of the geometric sequence.

a 1 = 1 , r = 1 2 a 1 = 1 , r = 1 2

a 1 = 3 , a n = 2 a n − 1 a 1 = 3 , a n = 2 a n − 1

a n = 27 ⋅ 0.3 n − 1 a n = 27 ⋅ 0.3 n − 1

Use recursive formulas to give two examples of geometric sequences whose 3 rd terms are 200. 200.

Use explicit formulas to give two examples of geometric sequences whose 7 th terms are 1024. 1024.

Find the 5 th term of the geometric sequence { b , 4 b , 16 b , ... } . { b , 4 b , 16 b , ... } .

Find the 7 th term of the geometric sequence { 64 a ( − b ) , 32 a ( − 3 b ) , 16 a ( − 9 b ) , ... } . { 64 a ( − b ) , 32 a ( − 3 b ) , 16 a ( − 9 b ) , ... } .

At which term does the sequence { 10 , 12 , 14.4 , 17.28 , ... } { 10 , 12 , 14.4 , 17.28 , ... } exceed 100 ? 100 ?

At which term does the sequence { 1 2187 , 1 729 , 1 243 , 1 81 ... } { 1 2187 , 1 729 , 1 243 , 1 81 ... } begin to have integer values?

For which term does the geometric sequence a n = − 36 ( 2 3 ) n − 1 a n = − 36 ( 2 3 ) n − 1 first have a non-integer value?

Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10 th term.

Use the explicit formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8 th term.

Is it possible for a sequence to be both arithmetic and geometric? If so, give an example.

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• Authors: Jay Abramson
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• Book title: College Algebra
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Geometric Sequences and Sums

A Sequence is a set of things (usually numbers) that are in order.

Geometric Sequences

In a Geometric Sequence each term is found by multiplying the previous term by a constant .

Example: 1, 2, 4, 8, 16, 32, 64, 128, 256, ...

This sequence has a factor of 2 between each number.

Each term (except the first term) is found by multiplying the previous term by 2 .

In General we write a Geometric Sequence like this:

{a, ar, ar 2 , ar 3 , ... }

• a is the first term, and
• r is the factor between the terms (called the "common ratio" )

Example: {1,2,4,8,...}

The sequence starts at 1 and doubles each time, so

• a=1 (the first term)
• r=2 (the "common ratio" between terms is a doubling)

And we get:

= {1, 1×2, 1×2 2 , 1×2 3 , ... }

= {1, 2, 4, 8, ... }

But be careful, r should not be 0:

• When r=0 , we get the sequence {a,0,0,...} which is not geometric

We can also calculate any term using the Rule:

x n = ar (n-1)

(We use "n-1" because ar 0 is for the 1st term)

Example: 10, 30, 90, 270, 810, 2430, ...

This sequence has a factor of 3 between each number.

The values of a and r are:

• a = 10 (the first term)
• r = 3 (the "common ratio")

The Rule for any term is:

x n = 10 × 3 (n-1)

So, the 4th term is:

x 4 = 10 × 3 (4-1) = 10 × 3 3 = 10 × 27 = 270

And the 10th term is:

x 10 = 10 × 3 (10-1) = 10 × 3 9 = 10 × 19683 = 196830

A Geometric Sequence can also have smaller and smaller values:

Example: 4, 2, 1, 0.5, 0.25, ...

This sequence has a factor of 0.5 (a half) between each number.

Its Rule is x n = 4 × (0.5) n-1

Why "Geometric" Sequence?

Because it is like increasing the dimensions in geometry :

Geometric Sequences are sometimes called Geometric Progressions (G.P.’s)

Summing a Geometric Series

To sum these:

a + ar + ar 2 + ... + ar (n-1)

(Each term is ar k , where k starts at 0 and goes up to n-1)

We can use this handy formula:

What is that funny Σ symbol? It is called Sigma Notation

And below and above it are shown the starting and ending values:

It says "Sum up n where n goes from 1 to 4. Answer= 10

The formula is easy to use ... just "plug in" the values of a , r and n

Example: Sum the first 4 terms of 10, 30, 90, 270, 810, 2430, ...

The values of a , r and n are:

• n = 4 (we want to sum the first 4 terms)

You can check it yourself:

10 + 30 + 90 + 270 = 400

And, yes, it is easier to just add them in this example , as there are only 4 terms.

But imagine adding 50 terms ... then the formula is much easier.

Using the Formula

Let's see the formula in action:

Example: Grains of Rice on a Chess Board

On the page Binary Digits we give an example of grains of rice on a chess board. The question is asked:

When we place rice on a chess board:

• 1 grain on the first square,
• 2 grains on the second square,
• 4 grains on the third and so on,

... doubling the grains of rice on each square ...

... how many grains of rice in total?

So we have:

• a = 1 (the first term)
• r = 2 (doubles each time)
• n = 64 (64 squares on a chess board)

= 1−2 64 −1 = 2 64 − 1

= 18,446,744,073,709,551,615

Which was exactly the result we got on the Binary Digits page (thank goodness!)

And another example, this time with r less than 1:

Example: Add up the first 10 terms of the Geometric Sequence that halves each time:

{ 1/2, 1/4, 1/8, 1/16, ... }.

• a = ½ (the first term)
• r = ½ (halves each time)
• n = 10 (10 terms to add)

Very close to 1.

(Question: if we continue to increase n , what happens?)

Why Does the Formula Work?

Let's see why the formula works, because we get to use an interesting "trick" which is worth knowing.

Notice that S and S·r are similar?

Now subtract them!

Wow! All the terms in the middle neatly cancel out. (Which is a neat trick)

By subtracting S·r from S we get a simple result:

S − S·r = a − ar n

Let's rearrange it to find S :

Which is our formula (ta-da!):

Infinite Geometric Series

So what happens when n goes to infinity ?

We can use this formula:

But be careful :

r must be between (but not including) −1 and 1

and r should not be 0 because the sequence {a,0,0,...} is not geometric

So our infnite geometric series has a finite sum when the ratio is less than 1 (and greater than −1)

Let's bring back our previous example, and see what happens:

Example: Add up ALL the terms of the Geometric Sequence that halves each time:

{ 1 2 , 1 4 , 1 8 , 1 16 , ... }.

= ½×1 ½ = 1

Yes, adding 1 2 + 1 4 + 1 8 + ... etc equals exactly 1 .

Recurring Decimal

On another page we asked "Does 0.999... equal 1?" , well, let us see if we can calculate it:

Example: Calculate 0.999...

We can write a recurring decimal as a sum like this:

And now we can use the formula:

Yes! 0.999... does equal 1.

So there we have it ... Geometric Sequences (and their sums) can do all sorts of amazing and powerful things.

• Mathematicians
• Math Lessons
• Square Roots
• Math Calculators
• Geometric sequence – Pattern, Formula, and Explanation

JUMP TO TOPIC

What is a geometric sequence?                

Geometric sequence formula, practice questions, geometric sequence – pattern, formula, and explanation.

Geometric sequences are sequences of numbers where two consecutive terms of the sequence will always share a common ratio.

We’ll learn how to identify geometric sequences in this article. We’ll also learn how to apply the geometric sequence’s formulas for finding the next terms and the sum of the sequence.

We’ll also learn how to identify geometric sequences from word problems and apply what we’ve learned to solve and address these problems. So, let’s begin by understanding the definition and conditions of geometric sequences.

Geometric sequences are sequences where the term of the sequence can be determined by multiplying the previous term with a fixed factor we call the common ratio .

The sequence above shows a geometric sequence where we multiply the previous term by $2$ to find the next term.

That’s why we have the following terms:

\begin{aligned} 1 \times 2 &= 2\\2 \times 2 &= 4\\4 \times 2 &= 8\\8 \times 2 &= 16\\.\\.\\. \end{aligned}

This shows that this sequence has a common ratio of $2$.

Let’s look at another way of understanding geometric sequences and common ratios by the figure shown below.

This figure is a visual representation of terms from a geometric sequence with a common ratio of $\dfrac{1}{2}$. We can find the smaller square dimensions by taking half of the length of the previous dimensions.

\begin{aligned} 1 \times \color{green}\dfrac{1}{2} &= \dfrac{1}{2}\\\dfrac{1}{2} \times \color{green}\color{green}\dfrac{1}{2} &= \dfrac{1}{4}\\\dfrac{1}{4} \times \color{green}\color{green}\dfrac{1}{2} &= \dfrac{1}{8}\\.\\.\\. \end{aligned}

This means that when you’re having a memory lapse and need a quick refresher of what a geometric sequence is – it helps to remember how it’s like when we fold papers into a half fold over and over again.

Geometric sequence definition            

Now that we have an idea of what a geometric sequence is, we should define it formally so that we have its general form and we can use it to work on all types of geometric sequences.

Let’s say the first term of the sequence is $a$ and the sequence has a common ratio of $r$, and the second term can be determined by multiplying $a$ by $r$. This process continues throughout the entire process.

Keep in mind as well that $r$ can either be an integer or a rational factor.

• When $|r| < 1$, the next term becomes smaller (disregarding the sign of the term).
• When $|r| > 1$, the next term becomes bigger (disregarding the sign of the term).

Here are four more examples of geometric sequences. We’ll observe what happens when $a$ is positive and when $r$ is a whole number or a rational one.

These examples can help us better understand how the signs of the first term and the value of the common ratio affect the next terms. From this, we can also see that the terms of a geometric sequence drastically increase or decrease because we’re working with the common ratio’s powers, $r$.

How to solve geometric sequences?                

There are different approaches to finding the unknown elements of a geometric sequence. The most important step is finding the common ratio shared by the sequence since, in most formulas, $r$ is essential.

We’ll slowly dive right into these different formulas and understand when they are most useful in the next sections. Why don’t we begin with learning how we can find $a_n$.

In this section, we’ll learn how to find the nth term, $a_n$, of a geometric sequence by expressing $a_n$ using the explicit and recursive rules.

Recursive Rule

We can find any term from the sequence using the recursive rule. For this, we’ll need the term before $a_n$ and the common ratio to find the value of $a_n$.

\begin{aligned}a_n = a_{n – 1} \cdot r\end{aligned}

This makes sense since we need to multiply the previous term, $a_{n-1}$, by the common ratio, $r$, to find the value of the next term, $a_n$.

We can also use this rule to find an expression for $r$ in terms of $a_{n – 1}$ and $a_n$.

\begin{aligned}r = \dfrac{a_n}{a_{n – 1}}\end{aligned}

However, this form has its limitations, especially when we want to find, let’s say, the $40$th term of a sequence. This is also why we need to learn about the explicit rule.

Explicit Rule

Let’s begin by observing the general expressions for the terms of a geometric sequence.

\begin{aligned} a_1 &= a\\a_2 &= a \cdot r\\a_3 &= ar \cdot r\\&= ar^2\\a_4&= ar^2 \cdot r\\&= ar^3\\.\\.\\.\end{aligned}

Notice anything? To find the nth term, we multiply the first term by the ratio raised to the $(n – 1)$th term.

This means that if we continue the equations show above and include $a_n$, we have the explicit rule shown below.

$\begin{aligned}a_n = a_1 r^{n -1}\end{aligned}

This rule is more flexible than the recursive rule, and it would be important for us to remember this rule.

How to find the sum of the geometric sequence?

We’ve just learned how to find the nth term of the geometric sequence, so it’s time for us to learn how to find the sum of a geometric series.

Remember the difference between the arithmetic series and arithmetic sequence ? The same reasoning applies concerning the difference between the geometric series and sequence.

Given the general form of a geometric sequence, $\{a_1, a_2, a_3, …, a_n\}$, the general form of a geometric series is simply $ a_1 + a_2 + a_3 + … + a_n$.

To find this series’s sum, we need the first term and the series’s common ratio.

\begin{aligned}S_n = \dfrac{a_1(1 – r^n)}{1 – r}\end{aligned}

We can also find the sum of an infinite geometric series, $a_1 + a_2 + a_3 + …$, using the first term and the common ratio.

\begin{aligned}S_{\infty} = \dfrac{a_1}{1 – r}\end{aligned}

 We can only use this if we want to find the sum of an infinite geometric series with a common ratio between $-1$ and $1$.

Now that we’ve learned everything we need to know about geometric sequences and series, why don’t we apply this in the problems below?

Calculate the common ratios of the following geometric sequence and find the next two terms of the sequence.

a. $2, 6, 18, …$ b. $-1, -4, -16, …$ c. $\dfrac{1}{3}, -\dfrac{1}{9}, \dfrac{1}{27},…$

Let’s observe the common ratios for each sequence by dividing the next term by the previous term. Once we have this ratio, we can find the next two terms by multiplying the last term with this ratio.

Beginning with $2, 6, 18, …$, we have:

\begin{aligned} r &= \dfrac{6}{2}\\&= 3\\r&= \dfrac{18}{6}\\&= 3\end{aligned}

This means that the common ratio of this geometric sequence is $3$. To find the next two terms, we simply multiply $18$ by $3$ and do the same for the next term.

\begin{aligned}18 \times 3 &=\boldsymbol{54 }\\ 54\times 3 &= \boldsymbol{162}\end{aligned}

Now, let’s work on the second geometric sequence, $-1, -4, -16, …$.

\begin{aligned} r &= \dfrac{-4}{-1}\\&= 4\\r&= \dfrac{-16}{-4}\\&= 4\end{aligned}

This shows that this geometric sequence will have a common ratio of $4$. We can use this to find the next two terms as shown below.

\begin{aligned}-16\times 4 &=\boldsymbol{-64 }\\ -64\times 4 &= \boldsymbol{-256}\end{aligned}

Let’s move on to the last geometric sequence, $\dfrac{1}{3}, -\dfrac{1}{9}, \dfrac{1}{27},…$.

\begin{aligned} r &= \dfrac{-\dfrac{1}{9}}{\dfrac{1}{3}}\\&= -\dfrac{1}{3}\\r &= \dfrac{\dfrac{1}{27}}{-\dfrac{1}{9}}\\&= -\dfrac{1}{3}\end{aligned}

Hence, the third geometric sequence has a common ratio of $-\dfrac{1}{3}$, so we can use this to find the next two terms.

\begin{aligned}\dfrac{1}{27} \times -\dfrac{1}{3} &=\boldsymbol{-\dfrac{1}{81} }\\ -\dfrac{1}{81} \times -\dfrac{1}{3} &=\boldsymbol{\dfrac{1}{243} }\end{aligned}

What is the tenth term of the sequence $\dfrac{1}{4}, \dfrac{1}{2}, 1, 2, …$?

Our approach in our previous example might be tedious if we want to find the tenth term of the sequence that we have now, so instead, let’s use the explicit rule to the 10 th term.

Recall that $a_n = a_1 r^{n-1}$, where for our problem, $a_1 = \dfrac{1}{4}$ and $n = 10$.

We still need to find the value of the common ratio, $r$, by inspecting the terms and dividing the next term by the previous one, as shown below.

\begin{aligned} r &= \dfrac{\dfrac{1}{2}}{\dfrac{1}{4}}\\&= 2\\r &= \dfrac{1}{\dfrac{1}{2}}\\&= 2\end{aligned}

Now that we have $r = 2$, we can now use the explicit rule to find the tenth term’s value.

\begin{aligned}a_{10} &= \dfrac{1}{4} \cdot 2^{10 – 1}\\&= \dfrac{1}{4} \cdot 2^{9}\end{aligned}

You can continue by expanding $2^9 = 512$, but it’ll be easier if we express $4$ as $2^2$. This way, we can use the exponent property, $\dfrac{b^n}{b^m} = b^{n – m}$ to simplify the expression.

\begin{aligned}a_{10} &= \dfrac{1}{2^2} \cdot 2^{9}\\&= \dfrac{2^9}{2^2}\\&= 2^7\\&= 128\end{aligned}

Hence, the tenth term of the sequence, $\dfrac{1}{4}, \dfrac{1}{2}, 1, 2, …$, is equal to $128$.

Find the second and third terms of a geometric sequence if its first and the fourth terms are $-5$ and $-135$?

We’re only given the first and the fourth terms of the sequence – $a_1$ and $a_4$. We need to find $a_2$ and $a_3$.

Why don’t we use the explicit rule, $a_n = a_1 r^{n – 1}$, to express $a_4$ in terms of $a_1$. We can then solve for $r$ by simplifying the resulting equation.

\begin{aligned}a_4 &= a_1 r^{4 – 1}\\a_4 &= a_1 r^3\\-135&= -5r^3\\r^3 &= \dfrac{-135}{-5}\\r^3&= 27\\ r&=3\end{aligned}

Now that we have the common ratio, we can now find the second and third terms by multiplying $-5$ by $3$ and do the same for the third term.

\begin{aligned}a_2 &= a_1 \cdot 3\\&= -5(3)\\&=-15\\a_3 &= a_2 \cdot 3\\&= -15(3)\\&=-45\end{aligned}

This means that the second and third terms are $-15$ and $-45$, respectively.

Find the sum of the following infinite and finite geometric series.

a. $64 + 16 + 4 + 1 + …$ b. $-12 + -6 + -3 + -\dfrac{3}{2} + …$ c. $2 + 4 + 8 + 16 + … + 1024$ d. $\dfrac{1}{27} + \dfrac{1}{9} + \dfrac{1}{3} + 1 + … + 81$

The first two series are infinite geometric series, so we’ll only need the first and common ratios for each series. Recall that the formula for the sum of infinite series is $S_{\infty} = \dfrac{a_1}{1 – r}$, where $-1 < r < 1$.

Let’s work on the first series, we have $64 + 16 + 4 + 1 + …$. Divide the next term by the previous one to find the common ratio, $r$.

\begin{aligned}r &= \dfrac{16}{64}\\&= \dfrac{1}{4}\\r &= \dfrac{4}{16}\\&= \dfrac{1}{4}  \end{aligned}

Since $r = \dfrac{1}{4}$ and is within the allowed range; we can apply the formula for the sum of infinite series.

\begin{aligned}S_{\infty} &= \dfrac{a_1}{1 – r}\\&= \dfrac{64}{1 – \dfrac{1}{4}}\\&= \dfrac{64}{\dfrac{3}{4}}\\&= 64 \cdot \dfrac{4}{3}\\&= \dfrac{256}{3}\end{aligned}

We can apply a similar process for the second infinite series, so let’s go ahead and observe its common ratio.

\begin{aligned}r &= \dfrac{-6}{-12}\\&= \dfrac{1}{2}\\r &= \dfrac{-3}{-6}\\&= \dfrac{1}{2}  \end{aligned}

The common ratio, $r = \dfrac{1}{2}$, is also within the allowed range for $r$, so we can use the same formula to find the sum of $-12 + -6 + -3 + -\dfrac{3}{2} + …$.

\begin{aligned}S_{\infty} &= \dfrac{a_1}{1 – r}\\&= \dfrac{-12}{1 – \dfrac{1}{2}}\\&= \dfrac{-12}{\dfrac{1}{2}}\\&= -12 \cdot \dfrac{2}{1}\\&= -24\end{aligned}

The third series is a finite geometric series. We know this because the series has an end and meaning, has a value for $a_n$ and $n$. For this kind of series, we’ll use the formula $S_n =\dfrac{a_1(1 – r^n)}{1- r}$.

Using the explicit rule for the nth term, $a_n$, we’ll be able to confirm the number of terms present in $2 + 4 + 8 + 16 + … + 1024$.

First, let’s determine the common ratio shared by the series.

\begin{aligned}r &= \dfrac{4}{2}\\&= 2\\r &= \dfrac{8}{4}\\&= 2\end{aligned}

We have $r=2$, $a_1 = 2$, and $a_n = 1024$, so let’s substitute these values and find the value of $n$ using $a_n = a_1 \cdot r^{n – 1}$.

\begin{aligned}a_n &= a_1 \cdot r^{n – 1}\\1024 &= 2(2)^{n-1}\end{aligned}

We can express $1024$ as a power of $2$ so that we can equate the exponents and solve for $n$.

\begin{aligned}2^{10} &= 2(2)^{n-1}\\2^{10} &= 2^{1 + n – 1}\\2^{10} &= 2^n\\n&=10\end{aligned}

Now that we all have the values that we need let’s go ahead and find the sum of the series.

\begin{aligned}S_n &=\dfrac{a_1(1 – r^n)}{1- r}\\&= \dfrac{2(1 – 2^{10})}{1- 2}\\&=\dfrac{2(1  -1024)}{-1}\\&=-2(-1023)\\&= 2046\end{aligned}

We’ll apply a similar process when finding the sum of $\dfrac{1}{27} + \dfrac{1}{9} + \dfrac{1}{3} + 1 + … + 81$. Let’s begin by determining the value of the common ratio, $r$.

\begin{aligned}r &= \dfrac{\dfrac{1}{9}}{\dfrac{1}{27}}\\&= 3\\r &= \dfrac{ \dfrac{1}{3}}{ \dfrac{1}{9}}\\&= 3\end{aligned}

We’ll now use the explicit rule to find the value of $n$ or the number of terms present in the finite geometric series.

\begin{aligned}81 &= \dfrac{1}{27}(3)^{n-1}\\3^{4} &= (3^{-3})3^{n – 1}\\3^{4} &= 3^{-3 + n – 1}\\3^4 &= 3^{n – 4}\\4 &= n – 4\\n&=8\end{aligned}

Now that we have all the values that we need, $a_1 = \dfrac{1}{27}$, $r = 3$, and $n = 8$, so let’s apply the formula for the sum.

\begin{aligned}S_n &=\dfrac{a_1 (1- r^n)}{1 – r}\\&=\dfrac{\dfrac{1}{27}(1- 3^8)}{1- 3}\\&= \dfrac{ \dfrac{1 – 3^8}{27}}{-2}\\&= -\dfrac{1}{2} \cdot \dfrac{6560}{27}\\&= \dfrac{3280}{27}\end{aligned}

Hence, we have the sum for the following geometric series.

a. $64 + 16 + 4 + 1 + … = \dfrac{256}{3}$ b. $-12 + -6 + -3 + -\dfrac{3}{2} + … = -24$ c. $2 + 4 + 8 + 16 + … + 1024 = 2046$ d. $\dfrac{1}{27} + \dfrac{1}{9} + \dfrac{1}{3} + 1 + … + 81 = \dfrac{3280}{27}$

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Arithmetic and Geometric Progressions Problem Solving

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Algebra through puzzles.

Supercharge your algebraic intuition and problem solving skills!

• Vilakshan Gupta
• Andres Gonzalez

To solve problems on this page, you should be familiar with

• arithmetic progressions
• geometric progressions
• arithmetic-geometric progressions .

You can boost up your problem solving on arithmetic and geometric progressions through this wiki. Make sure you hit all the problems listed in this page.

Problem Solving - Basic

Problem solving - intermediate, problem solving - advanced.

This section contains basic problems based on the notions of arithmetic and geometric progressions. Starting with an example, we will head into the problems to solve.

I have an arithmetic progression such that the initial term is 5 and the common difference is 10. What is minimum value of \(n\) such that the \(n^\text{th}\) term is larger than 100? We can just start by listing out the numbers: \[ 5,15,25,35,45,55,65,75,85, 95,105. \] We can clearly see that the \(11^\text{th}\) number is larger than 100, and thus \(n=11.\) However, note that this will become impractical if the common difference becomes smaller and/or the number we are looking for becomes larger. A practical way to solve it is via applying the \(n^\text{th}\) term formula. With \(a = 5, d = 10\), we have \( T_n = a + (n-1) d > 100 \). Then \(5 + (n-1) \cdot 10 > 100 \). Solving for \(n\) yields \(n > 10.5\). So the \(11^\text{th}\) term is the smallest term that satisfies the condition. \(_\square\)

Here comes the problems for you to solve.

The average of the first 100 positive integers is \(\text{__________}.\)

If \(A, B, C, D \) are consecutive terms in an arithmetic progression, what is the value of

\[ \frac{ D^2 - A^2 } { C^2 - B^2} ? \]

Assume \( C^2 - B^2 \neq 0.\)

\[ 54+51+48+45+ \cdots\]

You are given the sum of an arithmetic progression of a finite number of terms, as shown above.

What is the minimum number of terms used to make a total value of 513?

One side of an equilateral triangle is 24 cm. The midpoints of its sides are joined to form another triangle whose midpoints, in turn, are joined to form still another triangle. This process continues indefinitely.

Find the sum of the perimeters of all these triangles that are defined above.

Once a man did a favor to a king that made the king very happy. Out of joy the king told the man to wish for anything and he would be granted. The man wanted to ask for the whole kingdom which was worth 1500 trillion dollars, but obviously that would make the king mad and he would never be granted that wish.

The man who happened to be a mathematician thought a little bit and said the following:

"Bring in a big piece of rug with an \(8\times 8\) grid in it. Starting from the top left square, put one dollar in that square. Put two dollars in the square next to it and then double of that, four dollars, in the next square and so on. When you reach the end of the first row, continue on to the next row, doubling the amount every time as you move to the next square, all the way until the \(64^\text{th}\) square at the bottom right."

The king thought for a second. The first square will take one dollar, the second two dollars, the third, four dollars, and next 8 dollars, and then 16 dollars, and then 32 dollars, 64 dollars, 128 dollars, 256 dollars, and so on. That's not too bad. I can do it.

The king agreed. What happened next?

\[1+2 \cdot 2+ 3 \cdot 2^2 + 4 \cdot 2^3 + \cdots+ 100 \cdot 2^{99}= \, ?\]

This section contains a bit harder problems than the previous section. But all these can be solved using arithmetic and geometric problems. Here we go:

Real numbers \(a_1,a_2,\ldots,a_{99}\) form an arithmetic progression.

Suppose that \[ a_2+a_5+a_8+\cdots+a_{98}=205.\] Find the value of \( \displaystyle \sum_{k=1}^{99} a_k\).

The value of \(\displaystyle \sum_{n=1}^ \infty \frac{ 2n}{ 3^n } \) can be expressed in the form \( \frac{a}{b} \), where \(a\) and \(b\) are coprime positive integers. Find \( a - b \).

Let \(a,b,c\) be positive integers such that \(\frac{b}{a}\) is an integer. If \(a,b,c\) are in geometric progression and the arithmetic mean of \(a,b,c\) is \(b+2,\) find the value of

\[\dfrac{a^2+a-14}{a+1}.\]

If an infinite GP of real numbers has second term \(x\) and sum \(4,\) where does \(x\) belong?

4 positive integers form an arithmetic progression.

If we subtract \(2,6,7\) and \(2,\) respectively, from the 4 numbers, it forms a geometric progression.

What is the sum of these 4 numbers?

Let \(a, b \in R^{+}.\)

\(a, A_{1}, A_{2}, b\) is an arithmetic progression. \(a, G_{1}, G_{2}, b\) is a geometric progression.

Which of the following must be true?

We have three numbers in an arithmetic progression, and another three numbers in a geometric progression. Adding the corresponding terms of the two series, we get \( 120 , 116 , 130 \). If the sum of all the terms in the geometric progression is \( 342 \), what is the largest term in the geometric progression?

\(\) Details and Assumptions:

• If the terms of the AP are A, B, C, and the terms of the GP are X, Y, Z, then adding the corresponding terms will give us A+X, B+Y, C+Z.

This section has problems which need advanced understanding of the notions and generally get solved on using multiple notions at a time. Let's give these problems an attempt.

\[(1-x)(1-2x)(1-4x)\cdots \left(1-2^{101}x\right)\]

What is the coefficient of \(x^{101} \) in the expansion of the above?

\[ \frac {2+6}{4^{100}} + \frac {2+2 \times 6}{4^{99}} + \frac {2+ 3 \times 6}{4^{98}} + \cdots + \frac {2+ 100 \times 6}{4} \]

Evaluate the above expression.

In JEE examination the paper consists of 90 questions. The marks are awarded in such a way that if a person gets a question correct, he gets \(+4\) marks; if he does it wrong, he gets \(-2\) marks; if he leaves the question unanswered, he gets \(0\) marks (as per 2015). Find the sum of all possible marks that a student can get in JEE.

Let \(A=\{a_1, a_2, \ldots, a_n\}\) be a set of the first \(n\) terms of an arithmetic progression. Similarly, let \(B=\{b_1, b_2, \ldots, b_n\}\) be a set of the first \(n\) terms of a geometric progression.

If a new set \(C=A+B=\{a_1+b_1, a_2+b_2, \ldots, a_n+b_n\}\) and the first four terms of \(C\) are \(\{0, 0, 1, 0\},\) what is the \(11^\text{th}\) term of \(C?\)

A linear function \(f(x)=bx+a\) has the property that \(f\big(f(x)\big)=dx+c\) is another linear function such that \(a,b,c,d\) are integers that are consecutive terms in an arithmetic sequence. Find the last three digits of the sum of all possible values of \(f(2013)\).

\[ \begin{eqnarray} &\displaystyle\sum_{n=0}^{7}\log_{3}(x_{n}) &= 308 \\ 56 \leq & \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right )& \leq 57 \\ \end{eqnarray} \]

The increasing geometric sequence \(x_{0},x_{1},x_{2},\ldots\) consists entirely of integral powers of 3. If they satisfy the two conditions above, find \(\log_{3}(x_{14}).\)

Suppose \(2015\) people of different heights are arranged in a straight line from shortest to tallest such that

(i) the tops of their heads are collinear, and

(ii) for any two successive people, the horizontal distance between them is equal to the height of the shorter of the two people.

If the shortest person is \(49\) inches tall and the tallest is \(81\) inches tall, then how tall is the person at the middle of the line, (in inches)?

Given that \(a_1,a_2,a_3\) is an arithmetic progression in that order so that \(a_1+a_2+a_3=15\) and \(b_1,b_2,b_3\) is a geometric progression in that order so that \(b_1b_2b_3=27\).

If \(a_1+b_1, a_2+b_2, a_3+b_3\) are positive integers and form a geometric progression in that order, determine the maximum possible value of \(a_3\).

The answer is of the form \(\dfrac{a+b\sqrt{c}}{d}\), where \(a\), \(b\), \(c\), and \(d\) are positive integers and the fraction is in its simplest form and \(c\) is square free. Submit the value of \( a + b + c + d \).

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8.2: Problem Solving with Arithmetic Sequences

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• Page ID 83159

• Jennifer Freidenreich
• Diablo Valley College

Arithmetic sequences, introduced in Section 8.1, have many applications in mathematics and everyday life. This section explores those applications.

Example 8.2.1

A water tank develops a leak. Each week, the tank loses \(5\) gallons of water due to the leak. Initially, the tank is full and contains \(1500\) gallons.

• How many gallons are in the tank \(20\) weeks later?
• How many weeks until the tank is half-full?
• How many weeks until the tank is empty?

This problem can be viewed as either a linear function or as an arithmetic sequence. The table of values give us a few clues towards a formula.

The problem allows us to begin the sequence at whatever \(n\)−value we wish. It’s most convenient to begin at \(n = 0\) and set \(a_0 = 1500\).

Therefore, \(a_n = −5n + 1500\)

Since the leak is first noticed in week one, \(20\) weeks after the initial week corresponds with \(n = 20\). Use the formula where \(\textcolor{red}{n = 20}\):

\(a_{20} = −5(\textcolor{red}{20}) + 1500 = −100 + 1500 = 1400\)

Therefore, \(20\) weeks later, the tank contains \(1400\) gallons of water.

• How many weeks until the tank is half-full? A half-full tank would be \(750\) gallons. We need to find \(n\) when \(\textcolor{red}{a_n = 750}\).

\(\begin{array} &750 &= −5n + 1500 &\text{Substitute \(a_n = 750\) into the general term.} \\ 750 − 1500 &= −5n + 1505 − 1500 &\text{Subtract \(1500\) from each side of the equation.} \\ −750 &= −5n &\text{Simplify each side of the equation.} \\ \dfrac{−750}{−5} &= \dfrac{−5n}{−5} &\text{Divide both sides by \(−5\).} \\ 150 &= n & \end{array}\)

Since \(n\) is the week-number, this answer tells us that on week \(150\), the tank is half full. However, most people would better understand the answer if stated in the following way, “The tank is half full after 150 weeks.” This answer sounds more natural and is preferred.

• How many weeks until the tank is empty? The tank is empty when \(a_n = 0\) gallons. Find \(n\) such that \(\textcolor{red}{a_n = 0}\).

\(\begin{array}& 0 &= −5n + 1500 &\text{Substitute \(a_n=0\) into the general term.} \\ 0 − 1500 &= −5n + 1500 − 1500 &\text{Subtract \(1500\) from each side of the equation.} \\ −1500 &= −5n &\text{Simplify.} \\ \dfrac{−1500}{−5} &= \dfrac{−5n}{−5} &\text{Divide both sides by \(−5\).} \\ 300 &= n & \end{array}\)

Since \(n\) is the week-number, this answer tells us that on week \(300\), the tank is empty. However, most people would better understand the answer if stated in the following way, “ The tank is empty after 300 weeks. ” This answer sounds more natural and is preferred.

Example 8.2.2

Three stages of a pattern are shown below, using matchsticks. Each stage requires a certain number of matchsticks. If we keep up the pattern…

• How many matchsticks are required to make the figure in stage \(34\)?
• What stage would require \(220\) matchsticks?

Let’s create a table of values. Let \(n =\) stage number, and let \(a_n =\) the number of matchsticks used in that stage. Then note the common difference.

Find the value \(a_0\):

\(\begin{array} &a_0 + 3 &= 4 \\ a_0 + 3 − 3 &= 4 − 3 \\ a_0 &= 1 \end{array}\)

The general term of the sequence is:

\(a_n = 3n + 1\)

• Compute \(a_{34}\) to find the number of matchsticks in stage \(34\):

\(a_{34} = 3(\textcolor{red}{34}) + 1 = 103\).

There are \(103\) matchsticks in stage \(34\).

• What stage would require \(220\) matchsticks? We are looking for the stage-number, given the number of matchsticks. Find \(n\) if \(a_n = 220\).

\(\begin{array} &220 &= 3n + 1 \\ 219 &= 3n \\ 73 &= n \end{array}\)

Answer Stage \(73\) would require \(220\) matchsticks.

Example 8.2.3

Cory buys \(5\) items at the grocery store with prices \(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\) which is an arithmetic sequence. The least expensive item is \($1.89\), while the total cost of the \(5\) items is \($12.95\). What is the cost of each item?

Put the \(5\) items in order of expense: least to most and left to right. Because it is an arithmetic sequence, each item is \(d\) more dollars than the previous item. Each item’s price can be written in terms of the price of the least expensive item, \(a_1\), and \(a_1 = $1.89\).

The diagram above gives \(5\) expressions for the costs of the \(5\) items in terms of \(a_1\) and the common difference is \(d\).

\(\begin{array} &a_1 + a_2 + a_3 + a_4 + a_5 &= 12.95 &\text{Total cost of \(5\) items is \($12.95\).} \\ a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + (a_1 + 4d) &= 12.95 &\text{See diagram for substitutions.} \\ 5s_1 + 10d &= 12.95 &\text{Gather like terms.} \\ 5(1.89) + 10d &= 12.95 &a_1 = 1.89. \\ 9.45 + 10d &= 12.95 &\text{Simplify.} \\ 9.45 + 10d − 9.45 &= 12.95 − 9.45 &\text{Subtract \(9.45\) from each side of equation.} \\ 10d &= 3.50 &\text{Simplify. Then divide both sides by \(10\).} \\ d &= 0.35 &\text{The common difference is \($0.35\).} \end{array}\)

Now that we know the common difference, \(d = $0.35\), we can answer the question.

The price of each item is as follows: \($1.89, $2.24, $2.59, $2.94, $3.29\).

Try It! (Exercises)

1. ZKonnect cable company requires customers sign a \(2\)-year contract to use their services. The following describes the penalty for breaking contract: Your services are subject to a minimum term agreement of \(24\) months. If the contract is terminated before the end of the \(24\)-month contract, an early termination fee is assessed in the following manner: \($230\) termination fee is assessed if contract is terminated in the first \(30\) days of service. Thereafter, the termination fee decreases by \($10\) per month of contract.

• If Jack enters contract with ZKonnect on April 1 st of \(2021\), but terminates the service on January 10 th of \(2022\), what are Jack’s early termination fees?
• The general term \(a_n\) describes the termination fees for the stated contract. Describe the meaning of the variable \(n\) in the context of this problem. Find the general term \(a_n\).
• Is the early termination fee a finite sequence or an infinite sequence? Explain.
• Find the value of \(a_{13}\) and interpret its meaning in words.

2. A drug company has manufactured \(4\) million doses of a vaccine to date. They promise additional production at a rate of \(1.2\) million doses/month over the next year.

• How many doses of the vaccine, in total, will have been produced after a year?
• The general term \(a_n\) describes the total number of doses of the vaccine produced. Describe the meaning of the variable \(n\) in the context of this problem. Find the general term\(a_n\).
• Find the value of \(a_8\) and interpret its meaning in words.

3. The theater shown at right has \(22\) seats in the first row of the “A Center” section. Each row behind the first row gains two additional seats.

• Let \(a_n = 22 + 2n\), starting with \(n = 0\). Give the first \(10\) values of this sequence.
• Using \(a_n = 22 + 2n\), Find the value of \(a_{10}\) and interpret its meaning in words in the context of this problem. Careful! Does \(n=\) row number?
• How many seats, in total, are in “A Center” section if there are \(12\) rows in the section?

4) Logs are stacked in a pile with \(48\) logs on the bottom row and \(24\) on the top row. Each row decreases by three logs.

• The stack, as described, has how many rows of logs?
• Write the general term \(a_n\) to describe the number of logs in a row in two different ways. Each general term should produce the same sequence, regardless of its starting \(n\)-value.

i. Start with \(n = 0\).

ii. Start with \(n = 1\).

5) The radii of the target circle are an arithmetic sequence. If the area of the innermost circle is \(\pi \text{un}^2\) and the area of the entire target is \(49 \pi \text{un}^2\), what is the area of the blue ring? [The formula for area of a circle is \(A = \pi r^2\)].

6) Three stages of a pattern are shown below, using matchsticks. Each stage adds another triangle and requires a certain number of matchsticks. If we keep up the pattern…

• What stage would require \(325\) matchsticks?

7) Three stages of a pattern are shown below, using matchsticks. Each stage requires a certain number of matchsticks. If we keep up the pattern…

• How many matchsticks are required to make the figure in stage \(22\)?
• What stage would require \(424\) matchsticks?