1.4.9 Geometric sequences and series (A-level only) - Revision Guide

What you'll learn

How to recognise a geometric sequence and find its common ratio.
How to calculate the nth term and the sum of a finite geometric series.
When an infinite geometric series converges, and how to find its sum to infinity.
How modulus notation expresses the convergence condition precisely.

Before you start

A sequence is an ordered list of terms. For example,

24,…3,\ 6,\ 12,\ 24,\ldots

A series is formed by adding the terms of a sequence:

3+6+12+24+⋯3+6+12+24+\cdots

The position of a term is represented by $n$ , where $n$ is a positive integer. The first term corresponds to $n = 1$ , the second to $n = 2$ , and so on.

Geometric sequences

A geometric sequence is created by repeatedly multiplying by the same number.

Definition

Geometric sequence

A geometric sequence is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed number called the common ratio, denoted by $r$ .

If consecutive terms are $u_n$ and $u_{n+1}$ , then

r=un+1un.r=\frac{u_{n+1}}{u_n}.

For a geometric sequence with first term $a$ and common ratio $r$ , the terms are

ar3,…a,\ ar,\ ar^2,\ ar^3,\ldots

The ratio can be positive, negative or zero. If $r$ is negative, the signs of the terms alternate.

Example

Finding the common ratio

Determine whether the sequence

−10,…80,\ -40,\ 20,\ -10,\ldots

is geometric.

Divide the second term by the first:
$−4080=−12.\frac{-40}{80}=-\frac12.$
Check another pair of consecutive terms:
$20−40=−12.\frac{20}{-40}=-\frac12.$
The ratio is constant, so the sequence is geometric with $a = 80$ and $r=−12r=-\frac12$ .

Common Mistake

Subtracting instead of dividing

In an arithmetic sequence you look for a common difference. In a geometric sequence, divide consecutive terms to find the common ratio.

The nth term

The first term is $a$ . To reach the second term, you multiply by $r$ once; to reach the third term, you multiply by $r$ twice.

Therefore, reaching term number $n$ requires $n - 1$ multiplications by $r$ .

Key Idea

Nth term formula

The nth term of a geometric sequence is

u_n=ar^{n-1}.

Here, $a$ is the first term and $r$ is the common ratio.

Example

Finding a term of a geometric sequence

Find the eighth term of the geometric sequence with first term 5 and common ratio 3.

Use the nth term formula with $a = 5$ , $r = 3$ and $n = 8$ :
$u8=5(3)8−1.u_8=5\left(3\right)^{8-1}.$
Simplify the exponent:
$u8=5(3)7.u_8=5\left(3\right)^7.$
Evaluate:
$u_8=10935.$

Common Mistake

Using the exponent n

The exponent is $n - 1$ , not $n$ . Check that your formula gives $u_1=a$ when $n = 1$ .

Finding an unknown position

You may need to solve $ar^{n-1}=k$ to find when a particular value occurs. When the unknown appears in the exponent, logarithms may be needed.

Example

Finding when a value is first exceeded

The nth term of a geometric sequence is $un=4(1.3)n−1u_n=4\left(1.3\right)^{n-1}$ . Find the first term greater than 50.

Form the inequality
$4(1.3)n−1>50,4\left(1.3\right)^{n-1}>50,$
so
$(1.3)n−1>12.5.\left(1.3\right)^{n-1}>12.5.$
Take logarithms and divide by the positive value $log⁡(1.3)\log(1.3)$ :
$n−1>log⁡(12.5)log⁡(1.3)≈9.63.n-1>\frac{\log(12.5)}{\log(1.3)}\approx 9.63.$
Hence $n > 10.63$ . Since $n$ must be an integer, the first possible value is $n = 11$ .

Finite geometric series

A finite geometric series is the sum of a fixed number of terms from a geometric sequence.

For example, the sum of the first $n$ terms is

Sn=a+ar+ar2+⋯+arn−1.S_n=a+ar+ar^2+\cdots+ar^{n-1}.

To derive the formula, multiply by $r$ :

rSn=ar+ar2+⋯+arn−1+arn.rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n.

Subtracting the second equation from the first causes the middle terms to cancel:

Sn−rSn=a−arn,Sn(1−r)=a(1−rn).\begin{aligned} S_n-rS_n&=a-ar^n,\\ S_n(1-r)&=a(1-r^n). \end{aligned}

Key Idea

Finite sum formula

For $r≠1r\neq1$ , the sum of the first $n$ terms is

Sn=a(1−rn)1−r.S_n=\frac{a(1-r^n)}{1-r}.

An equivalent form is

Sn=a(rn−1)r−1.S_n=\frac{a(r^n-1)}{r-1}.

Example

Summing a finite geometric series

Find the sum of the first six terms of

.7+21+63+\cdots.

Identify $a = 7$ , $r = 3$ and $n = 6$ .
Substitute into the finite sum formula:
$S6=7(1−36)1−3.S_6=\frac{7(1-3^6)}{1-3}.$
Evaluate:
$S6=7(1−729)−2=2548.S_6=\frac{7(1-729)}{-2}=2548.$

Tip

Choosing the form

When $r > 1$ , the form $a(rn−1)r−1\frac{a(r^n-1)}{r-1}$ often keeps both the numerator and denominator positive.

Common Mistake

The case r equals 1

The finite sum formula divides by $1 - r$ , so it cannot be used when $r = 1$ . In that case every term equals $a$ , giving $S_n=na$ .

Sum to infinity

An infinite geometric series has no final term:

a+ar+ar2+ar3+⋯a+ar+ar^2+ar^3+\cdots

Its partial sum $S_n$ is the sum of its first $n$ terms. A series converges if its partial sums approach a finite limiting value as $n→∞n\to\infty$ .

Graphs of positive and alternating geometric partial sums approaching finite limits

Definition

Modulus notation

The modulus $∣r∣\lvert r\rvert$ is the distance of $r$ from zero, so it is always non-negative. The condition $∣r∣<1\lvert r\rvert<1$ is equivalent to

- 1 < r < 1.

When $∣r∣<1\lvert r\rvert<1$ , repeated powers of $r$ approach zero:

rn→0asn→∞.r^n\to0\quad\text{as}\quad n\to\infty.

Applying this to the finite sum formula gives

S∞=a1−r.S_\infty=\frac{a}{1-r}.

Key Idea

Convergence condition

A geometric series has a sum to infinity only when

∣r∣<1.\lvert r\rvert<1.

If this condition holds, then

S∞=a1−r.S_\infty=\frac{a}{1-r}.

Example

Finding a sum to infinity

Find the sum to infinity of

.18-6+2-\frac23+\cdots.

Find the common ratio:
$r=−618=−13.r=\frac{-6}{18}=-\frac13.$
Check convergence:
$∣−13∣=13<1.\left\lvert-\frac13\right\rvert=\frac13<1.$
Use the sum-to-infinity formula:
$S∞=181−(−13)=1843=272.S_\infty=\frac{18}{1-\left(-\frac13\right)} =\frac{18}{\frac43} =\frac{27}{2}.$

Common Mistake

Ignoring convergence

Never use $S∞=a1−rS_\infty=\frac{a}{1-r}$ before checking $∣r∣<1\lvert r\rvert<1$ . For example, $r = - 2$ is negative but its modulus is greater than 1, so the series diverges.

Forming a geometric model

Geometric series can model repeated percentage changes, depreciation and situations in which each new amount is a fixed proportion of the previous one.

Example

Modelling repeated bounces

A ball is dropped from a height of 10 m. After each bounce, it rises to 60% of its previous height. Find the total vertical distance travelled.

The initial downward distance is 10 m. The upward bounce heights form a geometric series with $a = 6$ and $r = 0.6$ .
Since $∣0.6∣<1\lvert0.6\rvert<1$ , the total upward distance is
$m.\frac{6}{1-0.6}=15\text{ m}.$
Each bounce height is travelled once upwards and once downwards, so the total distance is
$m.10+2(15)=40\text{ m}.$

Common Mistake

Missing the return journey

In bounce problems, each bounce height usually contributes an upward and a downward distance. The original drop occurs only once.

Exam technique

In the exam

Identify $a$ , $r$ and, for a finite sum, $n$ before substituting into a formula.
Distinguish carefully between an nth-term question, which uses $u_n=ar^{n-1}$ , and a sum question, which uses $S_n$ .
Before finding a sum to infinity, explicitly state that $∣r∣<1\lvert r\rvert<1$ .
Keep exact fractions where possible, especially when the ratio is fractional or negative.

Self review

Check yourself

Can you find the nth term of the sequence $24,−12,6,−3,…24,-12,6,-3,\ldots$ ?
How would you find the sum of the first eight terms of $5+15+45+⋯5+15+45+\cdots$ ?
Why does a geometric series with $r = - 1.2$ not have a sum to infinity?

What you'll learn

How to recognise a geometric sequence and find its common ratio.
How to calculate the nth term and the sum of a finite geometric series.
When an infinite geometric series converges, and how to find its sum to infinity.
How modulus notation expresses the convergence condition precisely.

Before you start

A sequence is an ordered list of terms. For example,

24,…3,\ 6,\ 12,\ 24,\ldots

A series is formed by adding the terms of a sequence:

3+6+12+24+⋯3+6+12+24+\cdots

The position of a term is represented by $n$ , where $n$ is a positive integer. The first term corresponds to $n = 1$ , the second to $n = 2$ , and so on.

Geometric sequences

A geometric sequence is created by repeatedly multiplying by the same number.

Definition

Geometric sequence

A geometric sequence is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed number called the common ratio, denoted by $r$ .

If consecutive terms are $u_n$ and $u_{n+1}$ , then

r=un+1un.r=\frac{u_{n+1}}{u_n}.

For a geometric sequence with first term $a$ and common ratio $r$ , the terms are

ar3,…a,\ ar,\ ar^2,\ ar^3,\ldots

The ratio can be positive, negative or zero. If $r$ is negative, the signs of the terms alternate.

Example

Finding the common ratio

Determine whether the sequence

−10,…80,\ -40,\ 20,\ -10,\ldots

is geometric.

Divide the second term by the first:
$−4080=−12.\frac{-40}{80}=-\frac12.$
Check another pair of consecutive terms:
$20−40=−12.\frac{20}{-40}=-\frac12.$
The ratio is constant, so the sequence is geometric with $a = 80$ and $r=−12r=-\frac12$ .

Common Mistake

Subtracting instead of dividing

In an arithmetic sequence you look for a common difference. In a geometric sequence, divide consecutive terms to find the common ratio.

The nth term

The first term is $a$ . To reach the second term, you multiply by $r$ once; to reach the third term, you multiply by $r$ twice.

Therefore, reaching term number $n$ requires $n - 1$ multiplications by $r$ .

Key Idea

Nth term formula

The nth term of a geometric sequence is

u_n=ar^{n-1}.

Here, $a$ is the first term and $r$ is the common ratio.

Example

Finding a term of a geometric sequence

Find the eighth term of the geometric sequence with first term 5 and common ratio 3.

Use the nth term formula with $a = 5$ , $r = 3$ and $n = 8$ :
$u8=5(3)8−1.u_8=5\left(3\right)^{8-1}.$
Simplify the exponent:
$u8=5(3)7.u_8=5\left(3\right)^7.$
Evaluate:
$u_8=10935.$

Common Mistake

Using the exponent n

The exponent is $n - 1$ , not $n$ . Check that your formula gives $u_1=a$ when $n = 1$ .

Finding an unknown position

You may need to solve $ar^{n-1}=k$ to find when a particular value occurs. When the unknown appears in the exponent, logarithms may be needed.

Example

Finding when a value is first exceeded

The nth term of a geometric sequence is $un=4(1.3)n−1u_n=4\left(1.3\right)^{n-1}$ . Find the first term greater than 50.

Form the inequality
$4(1.3)n−1>50,4\left(1.3\right)^{n-1}>50,$
so
$(1.3)n−1>12.5.\left(1.3\right)^{n-1}>12.5.$
Take logarithms and divide by the positive value $log⁡(1.3)\log(1.3)$ :
$n−1>log⁡(12.5)log⁡(1.3)≈9.63.n-1>\frac{\log(12.5)}{\log(1.3)}\approx 9.63.$
Hence $n > 10.63$ . Since $n$ must be an integer, the first possible value is $n = 11$ .

Finite geometric series

A finite geometric series is the sum of a fixed number of terms from a geometric sequence.

For example, the sum of the first $n$ terms is

Sn=a+ar+ar2+⋯+arn−1.S_n=a+ar+ar^2+\cdots+ar^{n-1}.

To derive the formula, multiply by $r$ :

rSn=ar+ar2+⋯+arn−1+arn.rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n.

Subtracting the second equation from the first causes the middle terms to cancel:

Sn−rSn=a−arn,Sn(1−r)=a(1−rn).\begin{aligned} S_n-rS_n&=a-ar^n,\\ S_n(1-r)&=a(1-r^n). \end{aligned}

Key Idea

Finite sum formula

For $r≠1r\neq1$ , the sum of the first $n$ terms is

Sn=a(1−rn)1−r.S_n=\frac{a(1-r^n)}{1-r}.

An equivalent form is

Sn=a(rn−1)r−1.S_n=\frac{a(r^n-1)}{r-1}.

Example

Summing a finite geometric series

Find the sum of the first six terms of

.7+21+63+\cdots.

Identify $a = 7$ , $r = 3$ and $n = 6$ .
Substitute into the finite sum formula:
$S6=7(1−36)1−3.S_6=\frac{7(1-3^6)}{1-3}.$
Evaluate:
$S6=7(1−729)−2=2548.S_6=\frac{7(1-729)}{-2}=2548.$

Tip

Choosing the form

When $r > 1$ , the form $a(rn−1)r−1\frac{a(r^n-1)}{r-1}$ often keeps both the numerator and denominator positive.

Common Mistake

The case r equals 1

The finite sum formula divides by $1 - r$ , so it cannot be used when $r = 1$ . In that case every term equals $a$ , giving $S_n=na$ .

Sum to infinity

An infinite geometric series has no final term:

a+ar+ar2+ar3+⋯a+ar+ar^2+ar^3+\cdots

Its partial sum $S_n$ is the sum of its first $n$ terms. A series converges if its partial sums approach a finite limiting value as $n→∞n\to\infty$ .

Graphs of positive and alternating geometric partial sums approaching finite limits

Definition

Modulus notation

The modulus $∣r∣\lvert r\rvert$ is the distance of $r$ from zero, so it is always non-negative. The condition $∣r∣<1\lvert r\rvert<1$ is equivalent to

- 1 < r < 1.

When $∣r∣<1\lvert r\rvert<1$ , repeated powers of $r$ approach zero:

rn→0asn→∞.r^n\to0\quad\text{as}\quad n\to\infty.

Applying this to the finite sum formula gives

S∞=a1−r.S_\infty=\frac{a}{1-r}.

Key Idea

Convergence condition

A geometric series has a sum to infinity only when

∣r∣<1.\lvert r\rvert<1.

If this condition holds, then

S∞=a1−r.S_\infty=\frac{a}{1-r}.

Example

Finding a sum to infinity

Find the sum to infinity of

.18-6+2-\frac23+\cdots.

Find the common ratio:
$r=−618=−13.r=\frac{-6}{18}=-\frac13.$
Check convergence:
$∣−13∣=13<1.\left\lvert-\frac13\right\rvert=\frac13<1.$
Use the sum-to-infinity formula:
$S∞=181−(−13)=1843=272.S_\infty=\frac{18}{1-\left(-\frac13\right)} =\frac{18}{\frac43} =\frac{27}{2}.$

Common Mistake

Ignoring convergence

Never use $S∞=a1−rS_\infty=\frac{a}{1-r}$ before checking $∣r∣<1\lvert r\rvert<1$ . For example, $r = - 2$ is negative but its modulus is greater than 1, so the series diverges.

Forming a geometric model

Geometric series can model repeated percentage changes, depreciation and situations in which each new amount is a fixed proportion of the previous one.

Example

Modelling repeated bounces

A ball is dropped from a height of 10 m. After each bounce, it rises to 60% of its previous height. Find the total vertical distance travelled.

The initial downward distance is 10 m. The upward bounce heights form a geometric series with $a = 6$ and $r = 0.6$ .
Since $∣0.6∣<1\lvert0.6\rvert<1$ , the total upward distance is
$m.\frac{6}{1-0.6}=15\text{ m}.$
Each bounce height is travelled once upwards and once downwards, so the total distance is
$m.10+2(15)=40\text{ m}.$

Common Mistake

Missing the return journey

In bounce problems, each bounce height usually contributes an upward and a downward distance. The original drop occurs only once.

Exam technique

In the exam

Identify $a$ , $r$ and, for a finite sum, $n$ before substituting into a formula.
Distinguish carefully between an nth-term question, which uses $u_n=ar^{n-1}$ , and a sum question, which uses $S_n$ .
Before finding a sum to infinity, explicitly state that $∣r∣<1\lvert r\rvert<1$ .
Keep exact fractions where possible, especially when the ratio is fractional or negative.

Self review

Check yourself

Can you find the nth term of the sequence $24,−12,6,−3,…24,-12,6,-3,\ldots$ ?
How would you find the sum of the first eight terms of $5+15+45+⋯5+15+45+\cdots$ ?
Why does a geometric series with $r = - 1.2$ not have a sum to infinity?

1.7.5 Geometric sequences and series (A-level only)

What you'll learn

Before you start

Geometric sequences

The nth term

Finding an unknown position

Finite geometric series

Sum to infinity

Forming a geometric model

Review 1.4.9 Geometric sequences and series (A-level only) by teaching Genie

1.7.5 Geometric sequences and series (A-level only)

What you'll learn

Before you start

Geometric sequences

The nth term

Finding an unknown position

Finite geometric series

Sum to infinity

Forming a geometric model

Review 1.4.9 Geometric sequences and series (A-level only) by teaching Genie

1.7.5 Geometric sequences and series (A-level only)

What you'll learn

Before you start

Geometric sequences

The nth term

Finding an unknown position

Finite geometric series

Sum to infinity

Forming a geometric model

Review 1.4.9 Geometric sequences and series (A-level only) by teaching Genie

1.4.9 Geometric sequences and series (A-level only) Revision Guide

1.7.5 Geometric sequences and series (A-level only)

What you'll learn

Before you start

Geometric sequences

The nth term

Finding an unknown position

Finite geometric series

Sum to infinity

Forming a geometric model

Review 1.4.9 Geometric sequences and series (A-level only) by teaching Genie

1.4.9 Geometric sequences and series (A-level only) Revision Guide