Geometric sequences and series

A geometric sequence is formed by multiplying each term by the same constant, called the common ratio $r$.

For example,

$3,12,48,192,\ldots$

is geometric because each term is $4$ times the preceding term, so $r=4$.

A geometric series is the result of adding terms of a geometric sequence:

$3+12+48+192+\cdots.$

A sequence is a list. A series is a sum.

Prerequisites

You should be able to:

• use nth terms and recurrence rules;
• apply the laws of indices;
• rearrange formulae;
• solve equations involving powers, using logarithms when necessary.

Recognising a geometric sequence

For consecutive non-zero terms, divide a term by the term before it. A sequence $(u_n)$ is geometric if

$\frac{u_{n+1}}{u_n}=r$

is constant. Equivalently,

$u_{n+1}=ru_n.$

Sequence	Consecutive ratios	Geometric?
$5,15,45,135,\ldots$	$3,3,3,\ldots$	Yes, $r=3$
$80,40,20,10,\ldots$	$\frac12,\frac12,\frac12,\ldots$	Yes, $r=\frac12$
$6,-12,24,-48,\ldots$	$-2,-2,-2,\ldots$	Yes, $r=-2$
$2,6,12,20,\ldots$	$3,2,\frac53,\ldots$	No

A negative common ratio makes the signs alternate. A ratio between $0$ and $1$ makes positive terms decrease towards zero. A common ratio of $1$ gives a constant sequence.

Equal differences or equal ratios?

Arithmetic sequences have a constant difference. Geometric sequences have a constant ratio. The sequence $2,4,8,16,\ldots$ has unequal differences but equal ratios.

The nth term

Let the first term be $a$ and the common ratio be $r$. Repeated multiplication gives

$$\begin{aligned} u_1&=a,\\ u_2&=ar,\\ u_3&=ar^2,\\ u_4&=ar^3. \end{aligned}$$

The $n$th term contains $n-1$ multiplications by $r$, so

$\boxed{u_n=ar^{n-1}}.$

The exponent is $n-1$, not $n$. Substituting $n=1$ must give the first term $a$.

Worked example 1: find a term

Find the eighth term of $160,80,40,20,\ldots$.

Here $a=160$ and

$r=\frac{80}{160}=\frac12.$

Therefore

$$\begin{aligned} u_8&=ar^{8-1}\\ &=160\left(\frac12\right)^7\\ &=\frac54. \end{aligned}$$

Worked example 2: a negative common ratio

For the geometric sequence with $a=7$ and $r=-3$, find $u_6$.

$u_6=7(-3)^5=-1701.$

The fifth power is odd, so the answer is negative. Keep a negative ratio in brackets when using a calculator.

Worked example 3: recover the first term and ratio

In a geometric sequence, $u_3=18$ and $u_6=486$. Find $a$ and $r$.

Using $u_n=ar^{n-1}$:

$ar^2=18$

and

$ar^5=486.$

Divide the second equation by the first. This removes $a$:

$r^3=\frac{486}{18}=27.$

Hence $r=3$. Substituting into $ar^2=18$ gives

$9a=18,$

so $a=2$.

The gap between term numbers is $6-3=3$, which explains the power $r^3$ after division.

Even powers can give two ratios

If the same method gives $r^2=9$, then $r=3$ or $r=-3$. Both possibilities must be checked unless the question supplies more information, such as all terms being positive.

Worked example 4: is a value a term?

Determine whether $1458$ is a term of $2,6,18,54,\ldots$.

Here $a=2$ and $r=3$. Set the nth term equal to $1458$:

$2(3^{n-1})=1458.$

Then

$3^{n-1}=729=3^6,$

so $n-1=6$ and $n=7$. Therefore $1458$ is the seventh term.

If the powers do not match exactly, logarithms can find $n$. The result must be a positive integer for the value to be a term.

Self-check 1

A geometric sequence has $u_2=12$ and $u_5=324$.

1. Find $r$ and $a$.
2. Find $u_8$.
3. State a recurrence relation for the sequence.

Answers

Dividing $ar^4=324$ by $ar=12$ gives

$r^3=27,$

so $r=3$ and $a=12/3=4$.

$u_8=4(3^7)=8748.$

A suitable recurrence relation is

$u_1=4,\qquad u_{n+1}=3u_n.$

The sum of the first n terms

Let

$S_n=a+ar+ar^2+\cdots+ar^{n-1}.$

Multiply every term by $r$:

$rS_n=ar+ar^2+ar^3+\cdots+ar^n.$

Subtract the second equation from the first. All middle terms cancel:

$S_n-rS_n=a-ar^n.$

Therefore

$S_n(1-r)=a(1-r^n),$

and, for $r\ne1$,

$\boxed{S_n=\frac{a(1-r^n)}{1-r}}.$

Multiplying the numerator and denominator by $-1$ gives the equivalent form

$\boxed{S_n=\frac{a(r^n-1)}{r-1}}.$

The first form is often convenient when $|r|<1$; the second often keeps positive quantities when $r>1$. If $r=1$, every term equals $a$, so $S_n=na$.

Worked example 5: sum a finite geometric series

Find the sum of the first $10$ terms of

$5+15+45+135+\cdots.$

Here $a=5$, $r=3$ and $n=10$. Using the second form,

$$\begin{aligned} S_{10}&=\frac{5(3^{10}-1)}{3-1}\\ &=\frac{5(59049-1)}2\\ &=147620. \end{aligned}$$

The power is $r^n$, although the final term is $ar^{n-1}$. This difference follows from the subtraction used in the derivation.

Worked example 6: sum with a negative ratio

Find

$64-32+16-8+\cdots$

up to and including the eighth term.

Here $a=64$, $r=-\frac12$ and $n=8$:

$$\begin{aligned} S_8 &=\frac{64\left(1-\left(-\frac12\right)^8\right)}{1-\left(-\frac12\right)}\\ &=\frac{64\left(1-\frac1{256}\right)}{\frac32}\\ &=\frac{85}{2}. \end{aligned}$$

The sum is positive because the positive terms outweigh the following negative terms.

Worked example 7: sum between two positions

For $u_n=3(2^{n-1})$, find

$u_5+u_6+\cdots+u_{12}.$

There are

$12-5+1=8$

terms. Their first term is

$u_5=3(2^4)=48,$

and their common ratio is still $2$. Hence

$\frac{48(2^8-1)}{2-1}=12240.$

Alternatively, calculate $S_{12}-S_4$. Subtracting $S_5$ would also remove $u_5$.

Self-check 2

1. Find $S_7$ for $81+27+9+\cdots$.
2. Find the sum from the fourth to the ninth term inclusive of $5,10,20,\ldots$.

Answers

For the first series, $a=81$, $r=\frac13$ and $n=7$:

$S_7=\frac{81\left(1-(\frac13)^7\right)}{1-\frac13}=\frac{1093}{9}.$

For the second, the fourth term is $40$ and there are $9-4+1=6$ terms:

$\frac{40(2^6-1)}{2-1}=2520.$

Solving equations involving finite sums

When $n$ is unknown, the equation usually contains $r^n$. Rearrange first, then express both sides with a common base or use logarithms.

Worked example 8: find the number of terms

The sum of the first $n$ terms of $3+6+12+\cdots$ is $3069$. Find $n$.

Here $a=3$ and $r=2$:

$$\begin{aligned} 3069&=\frac{3(2^n-1)}{2-1},\\ 1023&=2^n-1,\\ 1024&=2^n. \end{aligned}$$

Since $1024=2^{10}$, $n=10$.

Worked example 9: a threshold problem

An athlete runs $2$ km in the first week of a programme and increases the weekly distance by $8\%$ each week. Find the least number of weeks for the total distance to exceed $30$ km.

The weekly distances form a geometric sequence with $a=2$ and $r=1.08$. We need

$\frac{2(1.08^n-1)}{1.08-1}>30.$

Therefore

$25(1.08^n-1)>30,$

so

$1.08^n>2.2.$

Taking logarithms gives

$n>\frac{\log 2.2}{\log 1.08}\approx10.25.$

Since $n$ counts complete weeks, the least possible integer is

$\boxed{n=11}.$

Check the boundary: $S_{10}\approx28.97<30$, while $S_{11}\approx33.29>30$.

Infinite geometric series

For a finite sum,

$S_n=\frac{a(1-r^n)}{1-r}.$

If $|r|<1$, repeated multiplication makes $r^n$ approach zero as $n$ grows. The partial sums therefore approach

$\boxed{S_\infty=\frac{a}{1-r}},\qquad |r|<1.$

This is the sum to infinity. It is a limit of finite partial sums, not the result of reaching a final term.

The condition $|r|<1$ is essential:

• if $r\geq1$, the terms do not approach zero, except in the trivial case $a=0$;
• if $r\leq-1$, the terms either alternate without settling or grow in magnitude;
• if $-1<r<0$, the terms alternate, but shrink enough for the partial sums to settle.

A necessary test for any infinite series to converge is that its terms approach zero. For a geometric series, this condition is also sufficient.

Worked example 10: find a sum to infinity

Find the sum to infinity of

$18+12+8+\frac{16}{3}+\cdots.$

The ratio is

$r=\frac{12}{18}=\frac23.$

Since $|r|<1$, the series converges:

$S_\infty=\frac{18}{1-\frac23}=54.$

Worked example 11: an alternating infinite series

Determine whether

$20-8+3.2-1.28+\cdots$

converges, and find its sum if it does.

The common ratio is

$r=\frac{-8}{20}=-0.4.$

Since $|-0.4|<1$, the series converges. Its sum is

$S_\infty=\frac{20}{1-(-0.4)}=\frac{100}{7}.$

Do not replace $r$ by $|r|$ in the formula. Absolute value is used only to test convergence.

Worked example 12: find an unknown ratio

A convergent geometric series has first term $12$ and sum to infinity $20$. Find the common ratio.

$20=\frac{12}{1-r}.$

Thus

$20(1-r)=12,$

so

$1-r=\frac35$

and

$r=\frac25.$

The result satisfies $|r|<1$, as required.

Worked example 13: recurring decimals

Express $0.272727\ldots$ as a fraction.

Write the decimal as a geometric series:

$0.272727\ldots=0.27+0.0027+0.000027+\cdots.$

Here $a=0.27=\frac{27}{100}$ and $r=0.01=\frac1{100}$. Therefore

$$\begin{aligned} 0.272727\ldots &=\frac{\frac{27}{100}}{1-\frac1{100}}\\ &=\frac{27}{99}\\ &=\frac3{11}. \end{aligned}$$

Self-check 3

1. Decide whether $7+8.4+10.08+\cdots$ has a sum to infinity.
2. Find the sum to infinity of $45-15+5-\cdots$.
3. A convergent geometric series has first term $9$ and sum to infinity $6$. Find $r$.

Answers

1. Here $r=1.2$. Since $|r|>1$, there is no sum to infinity.

2. Here $r=-\frac13$, so

   $S_\infty=\frac{45}{1+\frac13}=\frac{135}{4}.$

3. From

   $6=\frac9{1-r},$

   we get $1-r=\frac32$, so $r=-\frac12$. This satisfies $|r|<1$.

Common misconceptions

• Using $ar^n$ for the nth term: there are only $n-1$ multiplications after the first term, so $u_n=ar^{n-1}$.
• Using the nth term formula for a sum: $ar^{n-1}$ is one term; $S_n$ adds all terms from the first through the nth.
• Losing a negative ratio: for an alternating sequence, substitute the negative value of $r$ into every formula.
• Using $S_\infty$ without checking: the formula $a/(1-r)$ is valid only when $|r|<1$.
• Rounding too early: retain exact powers or full calculator values until the final line.
• Accepting a non-integer value of $n$: a term number or number of terms must be a positive integer. In threshold problems, check which neighbouring integer first satisfies the inequality.

Mixed exam-style check

A geometric sequence has positive terms. Its third term is $20$ and its sixth term is $2.5$.

1. Find $a$ and $r$.
2. Find the sum of the first eight terms.
3. Find the sum to infinity.

Answer

Since $ar^2=20$ and $ar^5=2.5$,

$r^3=\frac{2.5}{20}=\frac18.$

The terms are positive, so $r=\frac12$. Then

$a\left(\frac12\right)^2=20,$

which gives $a=80$.

For the first eight terms,

$$S_8=\frac{80\left(1-(\frac12)^8\right)}{1-\frac12} =\frac{1275}{8}.$$

Since $|r|<1$,

$S_\infty=\frac{80}{1-\frac12}=160.$

The finite sum $159.375$ is slightly below its limiting value $160$, which is consistent with all terms being positive.

What to learn next

Continue with sigma notation to write sums compactly, then study sequences and series in modelling to apply geometric methods to repeated growth, decay and payment problems. For the continuous counterpart, see exponential growth and decay.