## Free Mathematics Tutorials

## 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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## 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 .

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)

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:

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

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.

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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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- 12.3 Geometric Sequences and Series
- Introduction
- 1.1 Use the Language of Algebra
- 1.2 Integers
- 1.3 Fractions
- 1.4 Decimals
- 1.5 Properties of Real Numbers
- Key Concepts
- Review Exercises
- Practice Test
- 2.1 Use a General Strategy to Solve Linear Equations
- 2.2 Use a Problem Solving Strategy
- 2.3 Solve a Formula for a Specific Variable
- 2.4 Solve Mixture and Uniform Motion Applications
- 2.5 Solve Linear Inequalities
- 2.6 Solve Compound Inequalities
- 2.7 Solve Absolute Value Inequalities
- 3.1 Graph Linear Equations in Two Variables
- 3.2 Slope of a Line
- 3.3 Find the Equation of a Line
- 3.4 Graph Linear Inequalities in Two Variables
- 3.5 Relations and Functions
- 3.6 Graphs of Functions
- 4.1 Solve Systems of Linear Equations with Two Variables
- 4.2 Solve Applications with Systems of Equations
- 4.3 Solve Mixture Applications with Systems of Equations
- 4.4 Solve Systems of Equations with Three Variables
- 4.5 Solve Systems of Equations Using Matrices
- 4.6 Solve Systems of Equations Using Determinants
- 4.7 Graphing Systems of Linear Inequalities
- 5.1 Add and Subtract Polynomials
- 5.2 Properties of Exponents and Scientific Notation
- 5.3 Multiply Polynomials
- 5.4 Dividing Polynomials
- Introduction to Factoring
- 6.1 Greatest Common Factor and Factor by Grouping
- 6.2 Factor Trinomials
- 6.3 Factor Special Products
- 6.4 General Strategy for Factoring Polynomials
- 6.5 Polynomial Equations
- 7.1 Multiply and Divide Rational Expressions
- 7.2 Add and Subtract Rational Expressions
- 7.3 Simplify Complex Rational Expressions
- 7.4 Solve Rational Equations
- 7.5 Solve Applications with Rational Equations
- 7.6 Solve Rational Inequalities
- 8.1 Simplify Expressions with Roots
- 8.2 Simplify Radical Expressions
- 8.3 Simplify Rational Exponents
- 8.4 Add, Subtract, and Multiply Radical Expressions
- 8.5 Divide Radical Expressions
- 8.6 Solve Radical Equations
- 8.7 Use Radicals in Functions
- 8.8 Use the Complex Number System
- 9.1 Solve Quadratic Equations Using the Square Root Property
- 9.2 Solve Quadratic Equations by Completing the Square
- 9.3 Solve Quadratic Equations Using the Quadratic Formula
- 9.4 Solve Equations in Quadratic Form
- 9.5 Solve Applications of Quadratic Equations
- 9.6 Graph Quadratic Functions Using Properties
- 9.7 Graph Quadratic Functions Using Transformations
- 9.8 Solve Quadratic Inequalities
- 10.1 Finding Composite and Inverse Functions
- 10.2 Evaluate and Graph Exponential Functions
- 10.3 Evaluate and Graph Logarithmic Functions
- 10.4 Use the Properties of Logarithms
- 10.5 Solve Exponential and Logarithmic Equations
- 11.1 Distance and Midpoint Formulas; Circles
- 11.2 Parabolas
- 11.3 Ellipses
- 11.4 Hyperbolas
- 11.5 Solve Systems of Nonlinear Equations
- 12.1 Sequences
- 12.2 Arithmetic Sequences
- 12.4 Binomial Theorem

## Learning Objectives

By the end of this section, you will be able to:

- Determine if a sequence is geometric
- Find the general term (nth term) of a geometric sequence
- Find the sum of the first n n terms of a geometric sequence
- Find the sum of an infinite geometric series
- Apply geometric sequences and series in the real world

## Be Prepared 12.7

Before you get started, take this readiness quiz.

Simplify: 24 32 . 24 32 . If you missed this problem, review Example 1.24 .

## Be Prepared 12.8

Evaluate: ⓐ 3 4 3 4 ⓑ ( 1 2 ) 4 . ( 1 2 ) 4 . If you missed this problem, review Example 1.19 .

## Be Prepared 12.9

If f ( x ) = 4 · 3 x , f ( x ) = 4 · 3 x , find ⓐ f ( 1 ) f ( 1 ) ⓑ f ( 2 ) f ( 2 ) ⓒ f ( 3 ) . f ( 3 ) . If you missed this problem, review Example 3.49 .

Determine if a Sequence is Geometric

We are now ready to look at the second special type of sequence, the geometric sequence.

A sequence is called a geometric sequence if the ratio between consecutive terms is always the same. The ratio between consecutive terms in a geometric sequence is r , the common ratio , where n is greater than or equal to two.

## Geometric Sequence

A geometric sequence is a sequence where the ratio between consecutive terms is always the same.

The ratio between consecutive terms, a n a n − 1 , a n a n − 1 , is r , the common ratio . n is greater than or equal to two.

Consider these sequences.

## Example 12.21

Determine if each sequence is geometric. If so, indicate the common ratio.

ⓐ 4 , 8 , 16 , 32 , 64 , 128 , … 4 , 8 , 16 , 32 , 64 , 128 , …

ⓑ −2 , 6 , −12 , 36 , −72 , 216 , … −2 , 6 , −12 , 36 , −72 , 216 , …

ⓒ 27 , 9 , 3 , 1 , 1 3 , 1 9 , … 27 , 9 , 3 , 1 , 1 3 , 1 9 , …

To determine if the sequence is geometric, we find the ratio of the consecutive terms shown.

## Try It 12.41

Determine if each sequence is geometric. If so indicate the common ratio.

ⓐ 7 , 21 , 63 , 189 , 567 , 1,701 , … 7 , 21 , 63 , 189 , 567 , 1,701 , …

ⓑ 64 , 16 , 4 , 1 , 1 4 , 1 16 , … 64 , 16 , 4 , 1 , 1 4 , 1 16 , …

ⓒ 2 , 4 , 12 , 48 , 240 , 1,440 , … 2 , 4 , 12 , 48 , 240 , 1,440 , …

## Try It 12.42

ⓐ −150 , −30 , −15 , −5 , − 5 2 , 0 , … −150 , −30 , −15 , −5 , − 5 2 , 0 , …

ⓑ 5 , 10 , 20 , 40 , 80 , 160 , … 5 , 10 , 20 , 40 , 80 , 160 , …

ⓒ 8 , 4 , 2 , 1 , 1 2 , 1 4 , … 8 , 4 , 2 , 1 , 1 2 , 1 4 , …

If we know the first term, a 1 , a 1 , and the common ratio, r , we can list a finite number of terms of the sequence.

## Example 12.22

Write the first five terms of the sequence where the first term is 3 and the common ratio is r = −2 . r = −2 .

We start with the first term and multiply it by the common ratio. Then we multiply that result by the common ratio to get the next term, and so on.

The sequence is 3 , −6 , 12 , −24 , 48 , … 3 , −6 , 12 , −24 , 48 , …

## Try It 12.43

Write the first five terms of the sequence where the first term is 7 and the common ratio is r = −3 . r = −3 .

## Try It 12.44

Write the first five terms of the sequence where the first term is 6 and the common ratio is r = −4 . r = −4 .

Find the General Term ( n th Term) of a Geometric Sequence

Just as we found a formula for the general term of a sequence and an arithmetic sequence, we can also find a formula for the general term of a geometric sequence.

Let’s write the first few terms of the sequence where the first term is a 1 a 1 and the common ratio is r . We will then look for a pattern.

As we look for a pattern in the five terms above, we see that each of the terms starts with a 1 . a 1 .

The first term, a 1 , a 1 , is not multiplied by any r . In the second term, the a 1 a 1 is multiplied by r . In the third term, the a 1 a 1 is multiplied by r two times ( r · r r · r or r 2 r 2 ). In the fourth term, the a 1 a 1 is multiplied by r three times ( r · r · r r · r · r or r 3 r 3 ) and in the fifth term, the a 1 a 1 is multiplied by r four times. In each term, the number of times a 1 a 1 is multiplied by r is one less than the number of the term. This leads us to the following

## General Term ( n th term) of a Geometric Sequence

The general term of a geometric sequence with first term a 1 a 1 and the common ratio r is

We will use this formula in the next example to find the fourteenth term of a sequence.

## Example 12.23

Find the fourteenth term of a sequence where the first term is 64 and the common ratio is r = 1 2 . r = 1 2 .

## Try It 12.45

Find the thirteenth term of a sequence where the first term is 81 and the common ratio is r = 1 3 . r = 1 3 .

## Try It 12.46

Find the twelfth term of a sequence where the first term is 256 and the common ratio is r = 1 4 . r = 1 4 .

Sometimes we do not know the common ratio and we must use the given information to find it before we find the requested term.

## Example 12.24

Find the twelfth term of the sequence 3, 6, 12, 24, 48, 96, … Find the general term for the sequence.

To find the twelfth term, we use the formula, a n = a 1 r n − 1 , a n = a 1 r n − 1 , and so we need to first determine a 1 a 1 and the common ratio r .

## Try It 12.47

Find the ninth term of the sequence 6, 18, 54, 162, 486, 1,458, … Then find the general term for the sequence.

## Try It 12.48

Find the eleventh term of the sequence 7, 14, 28, 56, 112, 224, … Then find the general term for the sequence.

## Find the Sum of the First n Terms of a Geometric Sequence

We found the sum of both general sequences and arithmetic sequence. We will now do the same for geometric sequences. The sum, S n , S n , of the first n terms of a geometric sequence is written as S n = a 1 + a 2 + a 3 + ... + a n . S n = a 1 + a 2 + a 3 + ... + a n . We can write this sum by starting with the first term, a 1 , a 1 , and keep multiplying by r to get the next term as:

Let’s also multiply both sides of the equation by r .

Next, we subtract these equations. We will see that when we subtract, all but the first term of the top equation and the last term of the bottom equation subtract to zero.

## Sum of the First n Terms of a Geometric Series

The sum, S n , S n , of the first n terms of a geometric sequence is

where a 1 a 1 is the first term and r is the common ratio, and r is not equal to one.

We apply this formula in the next example where the first few terms of the sequence are given. Notice the sum of a geometric sequence typically gets very large when the common ratio is greater than one.

## Example 12.25

Find the sum of the first 20 terms of the geometric sequence 7, 14, 28, 56, 112, 224, …

To find the sum, we will use the formula S n = a 1 ( 1 − r n ) 1 − r . S n = a 1 ( 1 − r n ) 1 − r . We know a 1 = 7 , a 1 = 7 , r = 2 , r = 2 , and n = 20 . n = 20 .

## Try It 12.49

Find the sum of the first 20 terms of the geometric sequence 3, 6, 12, 24, 48, 96, …

## Try It 12.50

Find the sum of the first 20 terms of the geometric sequence 6, 18, 54, 162, 486, 1,458, …

In the next example, we are given the sum in summation notation. While adding all the terms might be possible, most often it is easiest to use the formula to find the sum of the first n terms.

To use the formula, we need r . We can find it by writing out the first few terms of the sequence and find their ratio. Another option is to realize that in summation notation, a sequence is written in the form ∑ i = 1 k a ( r ) i , ∑ i = 1 k a ( r ) i , where r is the common ratio.

## Example 12.26

Find the sum: ∑ i = 1 15 2 ( 3 ) i . ∑ i = 1 15 2 ( 3 ) i .

To find the sum, we will use the formula S n = a 1 ( 1 − r n ) 1 − r , S n = a 1 ( 1 − r n ) 1 − r , which requires a 1 a 1 and r . We will write out a few of the terms, so we can get the needed information.

## Try It 12.51

Find the sum: ∑ i = 1 15 6 ( 2 ) i . ∑ i = 1 15 6 ( 2 ) i .

## Try It 12.52

Find the sum: ∑ i = 1 10 5 ( 2 ) i . ∑ i = 1 10 5 ( 2 ) i .

Find the Sum of an Infinite Geometric Series

If we take a geometric sequence and add the terms, we have a sum that is called a geometric series. An infinite geometric series is an infinite sum whose first term is a 1 a 1 and common ratio is r and is written

## Infinite Geometric Series

An infinite geometric series is an infinite sum whose first term is a 1 a 1 and common ratio is r and is written

We know how to find the sum of the first n terms of a geometric series using the formula, S n = a 1 ( 1 − r n ) 1 − r . S n = a 1 ( 1 − r n ) 1 − r . But how do we find the sum of an infinite sum?

Let’s look at the infinite geometric series 3 + 6 + 12 + 24 + 48 + 96 + … . 3 + 6 + 12 + 24 + 48 + 96 + … . Each term gets larger and larger so it makes sense that the sum of the infinite number of terms gets larger. Let’s look at a few partial sums for this series. We see a 1 = 3 a 1 = 3 and r = 2 r = 2

As n gets larger and larger, the sum gets larger and larger. This is true when | r | ≥ 1 | r | ≥ 1 and we call the series divergent. We cannot find a sum of an infinite geometric series when | r | ≥ 1 . | r | ≥ 1 .

Let’s look at an infinite geometric series whose common ratio is a fraction less than one, 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + 1 64 + … 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + 1 64 + … . Here the terms get smaller and smaller as n gets larger. Let’s look at a few finite sums for this series. We see a 1 = 1 2 a 1 = 1 2 and r = 1 2 . r = 1 2 .

Notice the sum gets larger and larger but also gets closer and closer to one. When | r | < 1 , | r | < 1 , the expression r n r n gets smaller and smaller. In this case, we call the series convergent. As n approaches infinity, (gets infinitely large), r n r n gets closer and closer to zero. In our sum formula, we can replace the r n r n with zero and then we get a formula for the sum, S , for an infinite geometric series when | r | < 1 . | r | < 1 .

This formula gives us the sum of the infinite geometric sequence. Notice the S does not have the subscript n as in S n S n as we are not adding a finite number of terms.

## Sum of an Infinite Geometric Series

For an infinite geometric series whose first term is a 1 a 1 and common ratio r ,

If | r | < 1 , the sum is If | r | < 1 , the sum is

If | r | ≥ 1 , the infinite geometric series does not have a sum. We say the series diverges. If | r | ≥ 1 , the infinite geometric series does not have a sum. We say the series diverges.

## Example 12.27

Find the sum of the infinite geometric series 54 + 18 + 6 + 2 + 2 3 + 2 9 + … 54 + 18 + 6 + 2 + 2 3 + 2 9 + …

To find the sum, we first have to verify that the common ratio | r | < 1 | r | < 1 and then we can use the sum formula S = a 1 1 − r . S = a 1 1 − r .

## Try It 12.53

Find the sum of the infinite geometric series 48 + 24 + 12 + 6 + 3 + 3 2 + … 48 + 24 + 12 + 6 + 3 + 3 2 + …

## Try It 12.54

Find the sum of the infinite geometric series 64 + 16 + 4 + 1 + 1 4 + 1 16 + … 64 + 16 + 4 + 1 + 1 4 + 1 16 + …

An interesting use of infinite geometric series is to write a repeating decimal as a fraction.

## Example 12.28

Write the repeating decimal 0. 5 – 0. 5 – as a fraction.

## Try It 12.55

Write the repeating decimal 0. 4 – 0. 4 – as a fraction.

## Try It 12.56

Write the repeating decimal 0. 8 – 0. 8 – as a fraction.

Apply Geometric Sequences and Series in the Real World

One application of geometric sequences has to do with consumer spending. If a tax rebate is given to each household, the effect on the economy is many times the amount of the individual rebate.

## Example 12.29

The government has decided to give a $1,000 tax rebate to each household in order to stimulate the economy. The government statistics say that each household will spend 80% of the rebate in goods and services. The businesses and individuals who benefitted from that 80% will then spend 80% of what they received and so on. The result is called the multiplier effect. What is the total effect of the rebate on the economy?

Every time money goes into the economy, 80% of it is spent and is then in the economy to be spent. Again, 80% of this money is spent in the economy again. This situation continues and so leads us to an infinite geometric series.

Here the first term is 1,000, a 1 = 1000 . a 1 = 1000 . The common ratio is 0.8 , 0.8 , r = 0.8 . r = 0.8 . We can evaluate this sum since 0.8 < 1 . 0.8 < 1 . We use the formula for the sum on an infinite geometric series.

The total effect of the $1,000 received by each household will be a $5,000 growth in the economy.

## Try It 12.57

What is the total effect on the economy of a government tax rebate of $1,000 to each household in order to stimulate the economy if each household will spend 90% of the rebate in goods and services?

## Try It 12.58

What is the total effect on the economy of a government tax rebate of $500 to each household in order to stimulate the economy if each household will spend 85% of the rebate in goods and services?

We have looked at a compound interest formula where a principal, P , is invested at an interest rate, r , for t years. The new balance, A , is A = P ( 1 + r n ) n t A = P ( 1 + r n ) n t when interest is compounded n times a year. This formula applies when a lump sum was invested upfront and tells us the value after a certain time period.

An annuity is an investment that is a sequence of equal periodic deposits. We will be looking at annuities that pay the interest at the time of the deposits. As we develop the formula for the value of an annuity, we are going to let n = 1 . n = 1 . That means there is one deposit per year.

Suppose P dollars is invested at the end of each year. One year later that deposit is worth P ( 1 + r ) 1 P ( 1 + r ) 1 dollars, and another year later it is worth P ( 1 + r ) 2 P ( 1 + r ) 2 dollars. After t years, it will be worth A = P ( 1 + r ) t A = P ( 1 + r ) t dollars.

After three years, the value of the annuity is

This a sum of the terms of a geometric sequence where the first term is P and the common ratio is 1 + r . 1 + r . We substitute these values into the sum formula. Be careful, we have two different uses of r . The r in the sum formula is the common ratio of the sequence. In this case, that is 1 + r 1 + r where r is the interest rate.

Remember our premise was that one deposit was made at the end of each year.

We can adapt this formula for n deposits made per year and the interest is compounded n times a year.

## Value of an Annuity with Interest Compounded n n Times a Year

For a principal, P , invested at the end of a compounding period, with an interest rate, r , which is compounded n times a year, the new balance, A, after t years, is

## Example 12.30

New parents decide to invest $100 per month in an annuity for their baby daughter. The account will pay 5% interest per year which is compounded monthly. How much will be in the child’s account at her eighteenth birthday?

To find the Annuity formula, A t = P ( ( 1 + r n ) n t − 1 ) r n , A t = P ( ( 1 + r n ) n t − 1 ) r n , we need to identify P , r , n , and t .

## Try It 12.59

New grandparents decide to invest $200 per month in an annuity for their grandson. The account will pay 5% interest per year which is compounded monthly. How much will be in the child’s account at his twenty-first birthday?

## Try It 12.60

Arturo just got his first full-time job after graduating from college at age 27. He decided to invest $200 per month in an IRA (an annuity). The interest on the annuity is 8%, which is compounded monthly. How much will be in the Arturo’s account when he retires at his sixty-seventh birthday?

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

- Geometric Sequences
- Geometric Series
- Future Value Annuities and Geometric Series
- Application of a Geometric Series: Tax Rebate

## Section 12.3 Exercises

Practice makes perfect.

In the following exercises, determine if the sequence is geometric, and if so, indicate the common ratio.

3 , 12 , 48 , 192 , 768 , 3072 , … 3 , 12 , 48 , 192 , 768 , 3072 , …

2 , 10 , 50 , 250 , 1250 , 6250 , … 2 , 10 , 50 , 250 , 1250 , 6250 , …

72 , 36 , 18 , 9 , 9 2 , 9 4 , … 72 , 36 , 18 , 9 , 9 2 , 9 4 , …

54 , 18 , 6 , 2 , 2 3 , 2 9 , … 54 , 18 , 6 , 2 , 2 3 , 2 9 , …

−3 , 6 , −12 , 24 , −48 , 96 , … −3 , 6 , −12 , 24 , −48 , 96 , …

2 , −6 , 18 , −54 , 162 , −486 , … 2 , −6 , 18 , −54 , 162 , −486 , …

In the following exercises, determine if each sequence is arithmetic, geometric or neither. If arithmetic, indicate the common difference. If geometric, indicate the common ratio.

48 , 24 , 12 , 6 , 3 , 3 2 , … 48 , 24 , 12 , 6 , 3 , 3 2 , …

12 , 6 , 0 , −6 , −12 , −18 , … 12 , 6 , 0 , −6 , −12 , −18 , …

−7 , −2 , 3 , 8 , 13 , 18 , … −7 , −2 , 3 , 8 , 13 , 18 , …

5 , 9 , 13 , 17 , 21 , 25 , … 5 , 9 , 13 , 17 , 21 , 25 , …

1 2 , 1 4 , 1 8 , 1 16 , 1 32 , 1 64 , … 1 2 , 1 4 , 1 8 , 1 16 , 1 32 , 1 64 , …

4 , 8 , 12 , 24 , 48 , 96 , … 4 , 8 , 12 , 24 , 48 , 96 , …

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

a 1 = 4 a 1 = 4 and r = 3 r = 3

a 1 = 9 a 1 = 9 and r = 2 r = 2

a 1 = −4 a 1 = −4 and r = −2 r = −2

a 1 = −5 a 1 = −5 and r = −3 r = −3

a 1 = 27 a 1 = 27 and r = 1 3 r = 1 3

a 1 = 64 a 1 = 64 and r = 1 4 r = 1 4

In the following exercises, find the indicated term of a sequence where the first term and the common ratio is given.

Find a 11 a 11 given a 1 = 8 a 1 = 8 and r = 3 . r = 3 .

Find a 13 a 13 given a 1 = 7 a 1 = 7 and r = 2 . r = 2 .

Find a 10 a 10 given a 1 = −6 a 1 = −6 and r = −2 . r = −2 .

Find a 15 a 15 given a 1 = −4 a 1 = −4 and r = −3 . r = −3 .

Find a 10 a 10 given a 1 = 100,000 a 1 = 100,000 and r = 0.1 . r = 0.1 .

Find a 8 a 8 given a 1 = 1,000,000 a 1 = 1,000,000 and r = 0.01 . r = 0.01 .

In the following exercises, find the indicated term of the given sequence. Find the general term for the sequence.

Find a 9 a 9 of the sequence, 9 , 18 , 36 , 72 , 144 , 288 , … 9 , 18 , 36 , 72 , 144 , 288 , …

Find a 12 a 12 of the sequence, 5 , 15 , 45 , 135 , 405 , 1215 , … 5 , 15 , 45 , 135 , 405 , 1215 , …

Find a 15 a 15 of the sequence, −486 , 162 , −54 , 18 , −6 , 2 , … −486 , 162 , −54 , 18 , −6 , 2 , …

Find a 16 a 16 of the sequence, 224 , −112 , 56 , −28 , 14 , −7 , … 224 , −112 , 56 , −28 , 14 , −7 , …

Find a 10 a 10 of the sequence, 1 , 0.1 , 0.01 , 0.001 , 0.0001 , 0.00001 , … 1 , 0.1 , 0.01 , 0.001 , 0.0001 , 0.00001 , …

Find a 9 a 9 of the sequence, 1000 , 100 , 10 , 1 , 0.1 , 0.01 , … 1000 , 100 , 10 , 1 , 0.1 , 0.01 , …

Find the Sum of the First n terms of a Geometric Sequence

In the following exercises, find the sum of the first fifteen terms of each geometric sequence.

8 , 24 , 72 , 216 , 648 , 1944 , … 8 , 24 , 72 , 216 , 648 , 1944 , …

7 , 14 , 28 , 56 , 112 , 224 , … 7 , 14 , 28 , 56 , 112 , 224 , …

−6 , 12 , −24 , 48 , −96 , 192 , … −6 , 12 , −24 , 48 , −96 , 192 , …

−4 , 12 , −36 , 108 , −324 , 972 , … −4 , 12 , −36 , 108 , −324 , 972 , …

81 , 27 , 9 , 3 , 1 , 1 3 , … 81 , 27 , 9 , 3 , 1 , 1 3 , …

256 , 64 , 16 , 4 , 1 , 1 4 , 1 16 , … 256 , 64 , 16 , 4 , 1 , 1 4 , 1 16 , …

In the following exercises, find the sum of the geometric sequence.

∑ i = 1 15 ( 2 ) i ∑ i = 1 15 ( 2 ) i

∑ i = 1 10 ( 3 ) i ∑ i = 1 10 ( 3 ) i

∑ i = 1 9 4 ( 2 ) i ∑ i = 1 9 4 ( 2 ) i

∑ i = 1 8 5 ( 3 ) i ∑ i = 1 8 5 ( 3 ) i

∑ i = 1 10 9 ( 1 3 ) i ∑ i = 1 10 9 ( 1 3 ) i

∑ i = 1 15 4 ( 1 2 ) i ∑ i = 1 15 4 ( 1 2 ) i

In the following exercises, find the sum of each infinite geometric series.

1 + 1 3 + 1 9 + 1 27 + 1 81 + 1 243 + 1 729 + … 1 + 1 3 + 1 9 + 1 27 + 1 81 + 1 243 + 1 729 + …

1 + 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + 1 64 + … 1 + 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + 1 64 + …

6 − 2 + 2 3 − 2 9 + 2 27 − 2 81 + … 6 − 2 + 2 3 − 2 9 + 2 27 − 2 81 + …

−4 + 2 − 1 + 1 2 − 1 4 + 1 8 − … −4 + 2 − 1 + 1 2 − 1 4 + 1 8 − …

6 + 12 + 24 + 48 + 96 + 192 + … 6 + 12 + 24 + 48 + 96 + 192 + …

5 + 15 + 45 + 135 + 405 + 1215 + … 5 + 15 + 45 + 135 + 405 + 1215 + …

1,024 + 512 + 256 + 128 + 64 + 32 + … 1,024 + 512 + 256 + 128 + 64 + 32 + …

6,561 + 2187 + 729 + 243 + 81 + 27 + … 6,561 + 2187 + 729 + 243 + 81 + 27 + …

In the following exercises, write each repeating decimal as a fraction.

0. 3 – 0. 3 –

0. 6 – 0. 6 –

0. 7 – 0. 7 –

0. 2 – 0. 2 –

0. 45 — 0. 45 —

0. 27 — 0. 27 —

In the following exercises, solve the problem.

Find the total effect on the economy of each government tax rebate to each household in order to stimulate the economy if each household will spend the indicated percent of the rebate in goods and services.

New grandparents decide to invest $ 100 $ 100 per month in an annuity for their grandchild. The account will pay 6 % 6 % interest per year which is compounded monthly (12 times a year). How much will be in the child’s account at their twenty-first birthday?

Berenice just got her first full-time job after graduating from college at age 30. She decided to invest $ 500 $ 500 per quarter in an IRA (an annuity). The interest on the annuity is 7 % 7 % which is compounded quarterly (4 times a year). How much will be in the Berenice’s account when she retires at age 65?

Alice wants to purchase a home in about five years. She is depositing $ 500 $ 500 a month into an annuity that earns 5 % 5 % per year that is compounded monthly (12 times a year). How much will Alice have for her down payment in five years?

Myra just got her first full-time job after graduating from college. She plans to get a master’s degree, and so is depositing $ 2,500 $ 2,500 a year from her year-end bonus into an annuity. The annuity pays 6.5 % 6.5 % per year and is compounded yearly. How much will she have saved in five years to pursue her master’s degree?

## Writing Exercises

In your own words, explain how to determine whether a sequence is geometric.

In your own words, explain how to find the general term of a geometric sequence.

In your own words, explain the difference between a geometric sequence and a geometric series.

In your own words, explain how to determine if an infinite geometric series has a sum and how to find it.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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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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## Privacy Overview

## Geometric Sequence & Applications

These lessons help High School students to express and interpret geometric sequence applications.

Related Pages Number Sequences Linear Sequences Geometric Sequences: n-th Term Quadratic and Cubic Sequences

Examples, solutions, videos, and lessons to help High School students learn to choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.

A. Factor a quadratic expression to reveal the zeros of the function it defines.

B. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines.

C. Use the properties of exponents to transform expressions for exponential functions. For example, the expression 1.15 t can be rewritten as (1.15 1/12 ) 12t ≈ 1.012 12t to reveal the approximate equivalent monthly interest rate if the annual rate is 15%.

## Suggested Learning Targets

Use properties of exponents (such as power of a power, product of powers, power of a product, and rational exponents, etc.) to write an equivalent form of an exponential function to reveal and explain specific information about its approximate rate of growth or decay.

Common Core: HSA-SSE.B.3c

Geometric Sequences Word Problems

- Bruno has 3 pizza stores and wants to dramatically expand his franchise nationwide. If the number of stores he owns doubles in number each month, what month will he launch 6,144 stores?
- On January 1, Abby’s troop sold three boxes of Girl Scout cookies online. Their daily goal is to sell double the number of boxes as the previous day. At this rate, how many boxes will they sell on day 7? If this pattern continues, on what day will they sell 24,576 boxes of cookies?

Compounding Interest and other Geometric Sequence Word Problems

- Suppose you invest $1,000 in the bank. You leave the money in for 3 years, each year getting 5% interest per annum. How much money do you have in the bank after 3 years?
- You invest $5000 for 20 years at 2% p.a. How much will we end up with? How does this change if the interest is given quarterly? monthly?
- If I can invest at 5% and I want $50,000 in 10 years, how much should I invest now?
- I decide to run a rabbit farm. I have 50 rabbits. The rabbit grows at 7% per week. How many will I have in 15 weeks?

Geometric sequence - salary

Example: You land a job as a police officer. Your salary for the first year is $43,125. You will receive 7% increase every year. How much will your salary be at the start of year six?

Solve Word Problems using Geometric Sequences

Example: Wilma bought a house for $170,000. Each year, it increases 2% of its value. a. Write the equation that represents the house’s value over time. b. What will the house be worth in 10 years?

Application of a Geometric Sequence

Example: Bouncing ball application of a geometric sequence When a ball is dropped onto a flat floor, it bounces to 65% of the height from which it was dropped. If the ball is dropped from 80 cm, find the height of the fifth bounce.

Population Growth and Compound Interest

This video gives examples of population growth and compound interest. Remember these examples are variations on geometric sequence.

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## VIDEO

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Solution to Problem 1: Use the definition of a geometric sequence a2 = a1 × r = 10(−1) = −10 a3 = a2 × r = −10(−1) = 10 a4 = a3 × r = 10(−1) = −10 a5 = a4 × r = −10(−1) = 10 a 2 = a 1 × r = 10 ( − 1) = − 10 a 3 = a 2 × r = − 10 ( − 1) = 10 a 4 = a 3 × r = 10 ( − 1) = − 10 a 5 = a 4 × r = − 10 ( − 1) = 10 Problem 2

Definition 12.4.4. An infinite geometric series is an infinite sum whose first term is a1 and common ratio is r and is written. a1 + a1r + a1r2 + … + a1rn − 1 + …. We know how to find the sum of the first n terms of a geometric series using the formula, Sn = a1(1 − rn) 1 − r.

an = a1rn − 1 GeometricSequence. In fact, any general term that is exponential in n is a geometric sequence. Example 9.3.1: Find an equation for the general term of the given geometric sequence and use it to calculate its 10th term: 3, 6, 12, 24, 48…. Solution. Begin by finding the common ratio, r = 6 3 = 2.

Intro to geometric sequences Google Classroom About Transcript Sal introduces geometric sequences and their main features, the initial term and the common ratio. Created by Sal Khan. Questions Tips & Thanks Sort by: Top Voted Beorn MacRupert 7 years ago

Geometric sequence formulas give a ( n) , the n th term of the sequence. This is the explicit formula for the geometric sequence whose first term is k and common ratio is r : a ( n) = k ⋅ r n − 1. This is the recursive formula of that sequence: { a ( 1) = k a ( n) = a ( n − 1) ⋅ r.

Solving Application Problems with Geometric Sequences. 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.

Level up on all the skills in this unit and collect up to 1400 Mastery points! Start Unit test. Sequences are a special type of function that are useful for describing patterns. In this unit, we'll see how sequences let us jump forwards or backwards in patterns to solve problems.

= {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 The Rule We can also calculate any term using the Rule: xn = ar(n-1) (We use "n-1" because ar0 is for the 1st term) Example: 10, 30, 90, 270, 810, 2430, ...

Learn how to solve geometric sequence problems using a formula and detailed solutions. See examples of finding the common ratio, the first term, and the nth term of a geometric sequence. Download a free worksheet to practice your skills.

Solution. Finding the common ratio is a matter of dividing any term by its previous term: 45 15 = 3 = r 45 15 = 3 = r. Therefore, the general term of the sequence is: an = 15 ⋅3n−1 a n = 15 ⋅ 3 n − 1. The general term gives us a formula to find a10 a 10. Plug n = 10 n = 10 into the general term an a n.

Solving Application Problems with Geometric Sequences. In real-world scenarios involving arithmetic sequences, we may need to use an initial term of \(a_0\) instead of \(a_1\). ... Solving Application Problems with Geometric Sequences. In 2013, the number of students in a small school is \(284\). It is estimated that the student population will ...

1 Properties 2 Sum 2.1 Finite 2.2 Infinite 3 Problems 3.1 Intermediate 4 See also Properties Because each term is a common multiple of the one before it, every term of a geometric sequence can be expressed as the sum of the first term and a multiple of the common ratio.

Problem 1: Find the sum of the first nine (9) terms of the geometric series if [latex] {a_1} = 1 [/latex] and [latex]r=2 [/latex]. Answer Problem 2: Find the sum of the first ten (10) terms of the geometric series if [latex] {a_1} = 4 [/latex] and [latex]r = {\large- {1 \over 2}} [/latex]. Answer

The nth n t h term rule is an = 16(12)n−1 a n = 16 ( 1 2) n − 1. Finally, let's find the nth n t h term rule for the geometric sequence in which a5 = 8 a 5 = 8 and a10 = 14 a 10 = 1 4. Using the same method at the previous problem, we can solve for r r and a1 a 1. Then, write the general rule. Equation 1: a5 = 8 a 5 = 8, so 8 = a1r4 8 = a 1 ...

( 1) If we multiply both sides of our equation by 𝑟, we have 𝑟 𝑆 = 𝑇 𝑟 + 𝑇 𝑟 + 𝑇 𝑟 + ⋯ + 𝑇 𝑟 + 𝑇 𝑟. ( 2) When we subtract the terms in equation (2) from the terms in equation (1), all but the terms 𝑇 and 𝑇 𝑟 cancel out: 𝑆 = 𝑇 + 𝑇 𝑟 + 𝑇 𝑟 + ⋯ + 𝑇 𝑟 + 𝑇 𝑟, 𝑟 𝑆 = 𝑇 𝑟 + 𝑇 𝑟 + 𝑇 𝑟 + ⋯ + 𝑇 𝑟 + 𝑇 𝑟.

Learn how to find any term in a geometric sequence using a formula and a common ratio. See examples with answers and exercises to practice these concepts. Explore the summary of geometric sequences and the difference between increasing and decreasing sequences.

A geometric sequence is a sequence where the ratio between consecutive terms is always the same. The ratio between consecutive terms, a n a n − 1, is r, the common ratio. n is greater than or equal to two. Consider these sequences. Example 12.21 Determine if each sequence is geometric. If so, indicate the common ratio. ⓐ 4, 8, 16, 32, 64, 128, …

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.

Grade Ten students discuss Geometric Sequences through word problem solving, and application. There are also bonus practice problems to fully test if the ski...

Example 1: continuing a geometric sequence. Calculate the next three terms for the geometric progression 1, 2, 4, 8, 16, 1, 2,4,8,16, …. Take two consecutive terms from the sequence. Here we will take the numbers 4 4 and 8 8. 2 Divide the second term by the first term to find the value of the common ratio, r r.

Compounding Interest and other Geometric Sequence Word Problems. Examples: Suppose you invest $1,000 in the bank. You leave the money in for 3 years, each year getting 5% interest per annum. ... Solve Word Problems using Geometric Sequences. Example: Wilma bought a house for $170,000. Each year, it increases 2% of its value. a. Write the ...

Solution. 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. 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.

Solution. 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 = − 5 n + 1500.