Year 2 Pure · Chapter 6

Geometric Series

Specification: Edexcel Pure 4.5, 4.6 — Sequences and series · AQA section D5, D6

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In this chapter

1. Geometric sequences
2. Geometric series
3. The sum to infinity
4. Modelling with geometric sequences
Practice

Key facts — this chapter

A geometric sequence multiplies by the same common ratio $r$ each time: $u_n = ar^{n-1}$ .
The sum of the first $n$ terms is $S_n = \frac{a(1 - r^n)}{1 - r}$ , for $r \neq 1$ .
If $|r| < 1$ the series converges, and its sum to infinity is $S_\infty = \frac{a}{1 - r}$ .
If $|r| \geq 1$ the partial sums never settle, so there is no sum to infinity.
To find when a term or a sum passes a value, take logs. Dividing by $\log r$ when $0 < r < 1$ reverses the inequality.
Growth or decay by a fixed percentage each time step is geometric: multiply by $1.05$ for 5% growth, by $0.85$ for a 15% fall.

Where an arithmetic sequence adds the same amount each time, a geometric sequence multiplies by the same amount. That is how compound interest, depreciation and population growth behave. It also leads to something surprising: when the ratio is small enough, adding up infinitely many terms gives a finite answer.

1. Geometric sequences

Key result

With first term $a$ and common ratio $r$ , the $n$ th term is

u_n = ar^{n-1}

The ratio of any term to the one before is always $r$ : $\dfrac{u_{n+1}}{u_n} = r$ .

Worked example 6.1 · Two terms fix the sequence

The 3rd term of a geometric sequence is $18$ and the 6th term is $486$ . Find the first term and the common ratio.

Write both terms using the formula:
$\begin{aligned} ar^2 &= 18 \\ ar^5 &= 486 \end{aligned}$
Divide the second by the first: $r^3 = 27$ , so $r = 3$ .
Then $a = \dfrac{18}{9} = 2$ .

Worked example 6.2 · When does a term pass a value?

The sequence $3, 6, 12, 24, \dots$ is geometric. Find the first term that is greater than $10\,000$ .

$a = 3$ and $r = 2$ , so we need $3 \times 2^{n-1} > 10\,000$ .
Take logs of both sides:
$\begin{aligned} (n - 1)\log 2 &> \log \tfrac{10\,000}{3} \\ n - 1 &> 11.7 \text{ (3 s.f.)} \end{aligned}$
So $n = 13$ . Check: $u_{12} = 6144$ and $u_{13} = 12\,288$ .

2. Geometric series

The sum of a geometric series

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

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

The two forms are equal. The second avoids negative signs when $r > 1$ .

Proof

Write out $S_n$ , then multiply it by $r$ :

\begin{aligned} S_n &= a + ar + ar^2 + \dots + ar^{n-1} \\ rS_n &= ar + ar^2 + \dots + ar^{n-1} + ar^n \end{aligned}

Subtracting, everything cancels except the first term of one line and the last term of the other: $S_n - rS_n = a - ar^n$ . So $S_n(1 - r) = a(1 - r^n)$ , and dividing by $1 - r$ gives the formula.

Worked example 6.3 · The sum of ten terms

Find the sum of the first 10 terms of $2 + 6 + 18 + \dots$

Here $a = 2$ and $r = 3$ , so

\begin{aligned} S_{10} &= \frac{2(3^{10} - 1)}{3 - 1} \\ &= 3^{10} - 1 = 59\,048 \end{aligned}

Worked example 6.4 · A sum in sigma notation

Evaluate $\displaystyle\sum_{k=1}^{6} 3 \times 2^k$ .

The terms are $6, 12, 24, \dots$ : a geometric series with $a = 6$ , $r = 2$ and $6$ terms. So

\sum_{k=1}^{6} 3 \times 2^k = \frac{6(2^6 - 1)}{2 - 1} = 378

Common mistake

Read the first term off the series, not the formula. In $\sum_{k=1}^{6} 3 \times 2^k$ the first term is $3 \times 2^1 = 6$ , not $3$ , because the counter starts at $k = 1$ .

3. The sum to infinity

When $|r| < 1$ the terms shrink towards $0$ , and the sums $S_1, S_2, S_3, \dots$ close in on a fixed value.

The sum to infinity

A geometric series converges if and only if $|r| < 1$ , and then

S_\infty = \frac{a}{1 - r}

Why it works

In $S_n = \dfrac{a(1 - r^n)}{1 - r}$ , the only part that changes with $n$ is $r^n$ . When $|r| < 1$ , $r^n$ gets as close to $0$ as you like as $n$ grows, so $S_n$ gets as close as you like to $\dfrac{a}{1 - r}$ . When $|r| \geq 1$ , $r^n$ doesn’t shrink, and the sums never settle.

A unit square divided into a half, a quarter, an eighth and so on, each piece half the size of the last, together filling the square — ½ + ¼ + ⅛ + … fills the square exactly: its sum to infinity is 1.

Each dot is a partial sum S_n of 1 + r + r² + …. Change r: while −1 < r < 1 the dots close in on the dashed line at 1/(1 − r); outside that range they never settle.

Worked example 6.5 · A sum to infinity

Find the sum to infinity of $24 + 12 + 6 + \dots$

$a = 24$ and $r = \tfrac{1}{2}$ , and $|r| < 1$ , so

S_\infty = \frac{24}{1 - \frac{1}{2}} = 48

Worked example 6.6 · A recurring decimal as a fraction

Write $0.272727\ldots$ as a fraction in its simplest form.

The decimal is a geometric series:
$0.2727\ldots = \frac{27}{100} + \frac{27}{10\,000} + \dots$
with $a = \tfrac{27}{100}$ and $r = \tfrac{1}{100}$ .
So
$\begin{aligned} S_\infty &= \frac{27/100}{1 - 1/100} \\ &= \frac{27}{99} = \frac{3}{11} \end{aligned}$

Worked example 6.7 · Finding the ratio

A geometric series has first term $5$ and sum to infinity $20$ . Find the common ratio.

$\dfrac{5}{1 - r} = 20$ , so $1 - r = \tfrac{1}{4}$ and $r = \tfrac{3}{4}$ . This has $|r| < 1$ , as it must for the series to converge.

4. Modelling with geometric sequences

A quantity that rises or falls by the same percentage each year is multiplied by the same number each year, so its values form a geometric sequence.

Worked example 6.8 · Depreciation

A car is worth $£18\,000$ when new, and loses $15\%$ of its value each year. After how many whole years is it first worth less than $£5000$ ?

Each year the value is multiplied by $0.85$ , so after $n$ years it is $18\,000 \times 0.85^n$ . We need $18\,000 \times 0.85^n < 5000$ .
Take logs. $\log 0.85$ is negative, so dividing by it reverses the inequality:
$\begin{aligned} n \log 0.85 &< \log \tfrac{5000}{18\,000} \\ n &> \frac{\log (5/18)}{\log 0.85} \\ &= 7.88 \text{ (3 s.f.)} \end{aligned}$
So after $8$ years. Check: $18\,000 \times 0.85^7 = £5770$ and $18\,000 \times 0.85^8 = £4905$ (to the nearest pound).

Common mistake

Dividing an inequality by a negative number flips it. For $0 < r < 1$ , $\log r$ is negative, so from $n\log r < \log c$ you get $n > \dfrac{\log c}{\log r}$ . Forgetting the flip gives " $n < 7.88$ ", which is the opposite of the truth.

The model's limits

A fixed percentage each year is a simplification. Real cars lose value fastest in their first year or two, and the model predicts a value that never quite reaches zero, which a scrapped car does.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [3 marks]

A geometric sequence begins $5, 15, 45, \dots$

State the common ratio. [1]
Find the 8th term. [1]
Find the sum of the first 8 terms. [1]

Show solution

(a) $r = \dfrac{15}{5} = 3$ .

(b) $u_8 = 5 \times 3^7 = 10\,935$ .

(c) $S_8 = \dfrac{5(3^8 - 1)}{3 - 1} = \dfrac{5 \times 6560}{2} = 16\,400$ .

2 [3 marks]

For the series $12 - 6 + 3 - \dots$

Explain why the series has a sum to infinity. [1]
Find the sum to infinity. [2]

Show solution

(a) The common ratio is $r = -\tfrac{1}{2}$ , and $|r| < 1$ , so the series converges.

(b)

S_\infty = \frac{12}{1 - \left(-\frac{1}{2}\right)} = \frac{12}{\frac{3}{2}} = 8

3 [2 marks]

Evaluate $\displaystyle\sum_{k=1}^{5} 2^k$ .

Show solution

The terms are $2, 4, 8, 16, 32$ : a geometric series with $a = 2$ , $r = 2$ and $5$ terms.

\sum_{k=1}^{5} 2^k = \frac{2(2^5 - 1)}{2 - 1} = 62

4 [3 marks]

Use a geometric series to write $0.454545\ldots$ as a fraction in its simplest form.

Show solution

$0.4545\ldots = \dfrac{45}{100} + \dfrac{45}{10\,000} + \dots$ , with $a = \tfrac{45}{100}$ and $r = \tfrac{1}{100}$ :

S_\infty = \frac{45/100}{99/100} = \frac{45}{99} = \frac{5}{11}

Standard

5 [4 marks]

The 2nd term of a geometric sequence is $6$ and the 5th term is $162$ . Find the first term, the common ratio, and the sum of the first 6 terms.

Show solution

$ar = 6$ and $ar^4 = 162$ . Dividing, $r^3 = 27$ , so $r = 3$ and $a = 2$ .

S_6 = \frac{2(3^6 - 1)}{3 - 1} = 728

6 [4 marks]

A geometric series has second term $4$ and sum to infinity $16$ . Find the first term and the common ratio.

Show solution

$ar = 4$ and $\dfrac{a}{1 - r} = 16$ , so $a = 16(1 - r)$ . Substituting,

\begin{aligned} 16r(1 - r) &= 4 \\ 4r^2 - 4r + 1 &= 0 \\ (2r - 1)^2 &= 0 \end{aligned}

So $r = \tfrac{1}{2}$ and $a = 8$ .

7 [4 marks]

Find the smallest number of terms of the series $4 + 6 + 9 + \dots$ whose sum is greater than $1000$ .

Show solution

$a = 4$ and $r = \tfrac{3}{2}$ , so $S_n = \dfrac{4(1.5^n - 1)}{0.5} = 8(1.5^n - 1)$ . We need $8(1.5^n - 1) > 1000$ , so $1.5^n > 126$ :

n > \frac{\log 126}{\log 1.5} = 11.9 \text{ (3 s.f.)}

So $n = 12$ . Check: $S_{11} = 684$ and $S_{12} = 1030$ , to 3 s.f.

8 [4 marks]

Prove that the sum of the first $n$ terms of a geometric series with first term $a$ and common ratio $r$ , where $r \neq 1$ , is $\dfrac{a(1 - r^n)}{1 - r}$ .

Show solution

\begin{aligned} S_n &= a + ar + \dots + ar^{n-1} \\ rS_n &= ar + ar^2 + \dots + ar^n \end{aligned}

Subtracting the second line from the first, all but two terms cancel: $S_n - rS_n = a - ar^n$ . So $S_n(1 - r) = a(1 - r^n)$ , and since $r \neq 1$ we can divide by $1 - r$ :

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

9 [4 marks]

The series $1 + 2x + 4x^2 + 8x^3 + \dots$ is geometric.

Find the values of $x$ for which the series converges. [2]
Find the sum to infinity when $x = \tfrac{1}{3}$ . [2]

Show solution

(a) The common ratio is $2x$ . It converges when $|2x| < 1$ , that is $-\tfrac{1}{2} < x < \tfrac{1}{2}$ .

(b) $r = \tfrac{2}{3}$ , so $S_\infty = \dfrac{1}{1 - \frac{2}{3}} = 3$ .

Exam style

10 [6 marks]

A car costs $£22\,000$ new. Its value falls by $12\%$ each year.

Find its value after 4 years, to the nearest pound. [2]
Find the number of whole years after which it is first worth less than $£10\,000$ . [3]
Give one reason why the model may not be accurate. [1]

Show solution

(a) $22\,000 \times 0.88^4 = £13\,193$ .

(b) We need $22\,000 \times 0.88^n < 10\,000$ , so $0.88^n < \tfrac{5}{11}$ . $\log 0.88$ is negative, so dividing by it reverses the inequality:

n > \frac{\log (5/11)}{\log 0.88} = 6.17 \text{ (3 s.f.)}

After $7$ years. Check: the value is $£10\,217$ after 6 years and $£8991$ after 7.

(c) For example, cars usually lose value faster in the first year or two, so a single fixed percentage won’t match every year.

11 [6 marks]

A runner runs $5$ km in the first week of training, and increases the distance by $10\%$ each week after that.

Find the distance she runs in the 8th week. [2]
Find the total distance run in the first 12 weeks. [2]
In which week does she first run more than $20$ km? [2]

Show solution

(a) $5 \times 1.1^7 = 9.74$ km (3 s.f.).

(b)

S_{12} = \frac{5(1.1^{12} - 1)}{1.1 - 1} = 107 \text{ km (3 s.f.)}

(c) We need $5 \times 1.1^{n-1} > 20$ , so $1.1^{n-1} > 4$ and

n - 1 > \frac{\log 4}{\log 1.1} = 14.5 \text{ (3 s.f.)}

So $n = 16$ : she first runs more than $20$ km in the 16th week.

12 [8 marks]

The first three terms of a geometric series are $k$ , $k - 6$ and $k - 9$ .

Show that $k = 12$ . [3]
Find the sum to infinity. [2]
Find the smallest value of $n$ for which $S_\infty - S_n < 0.01$ . [3]

Show solution

(a) The ratios of consecutive terms are equal, so $\dfrac{k - 6}{k} = \dfrac{k - 9}{k - 6}$ . Then

\begin{aligned} (k - 6)^2 &= k(k - 9) \\ k^2 - 12k + 36 &= k^2 - 9k \\ k &= 12 \end{aligned}

(b) The series is $12 + 6 + 3 + \dots$ , with $r = \tfrac{1}{2}$ , so $S_\infty = \dfrac{12}{1 - \frac{1}{2}} = 24$ .

(c) $S_n = 24\left(1 - \left(\tfrac{1}{2}\right)^n\right)$ , so $S_\infty - S_n = \dfrac{24}{2^n}$ . We need $2^n > 2400$ :

n > \frac{\log 2400}{\log 2} = 11.2 \text{ (3 s.f.)}

So $n = 12$ .