Geometric Sequences and Series

In short: A geometric sequence multiplies by the same ratio $r$ each time, like 2, 6, 18, 54. Its $n$ th term is $T_{n} = a r^{n - 1}$ , and when $- 1 < r < 1$ the infinite sum is $S_{\infty} = \frac{a}{1 - r}$ .

Starting point: Years 9 and 10

Multiplying patterns

In 3, 6, 12, 24, ..., each term is double the one before. To find the multiplier, divide a term by the one before it: $\frac{6}{3} = 2$ .

In 80, 40, 20, 10, ..., the multiplier is $\frac{40}{80} = \frac{1}{2}$ , so the terms halve.

Adding or multiplying?

Sequence	Same difference?	Same ratio?	Type
5, 9, 13, 17	yes, $+ 4$	no	arithmetic
5, 10, 20, 40	no	yes, $\times 2$	geometric
1, 4, 9, 16	no	no	neither (square numbers)

Repeated percentage change

Growing by 10% means multiplying by 1.1 each step. A 1000 dollar investment growing 10% a year gives 1000, 1100, 1210, 1331, ..., a geometric sequence with multiplier 1.1.

Geometric sequences

Common ratio r

r = T₂ / T₁
Multiply each step

General term

Tₙ = arⁿ⁻¹
Exponential pattern

Sum

Sₙ = a(rⁿ - 1)/(r - 1)

Infinite sum

Needs -1 < r < 1
S = a / (1 - r)

Uses

Interest, decay, bouncing

Year 12Geometric series and the limiting sum

The sum of the first $n$ terms is

$S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1} (r \neq = 1)$

or $\frac{a ( 1 - r ^{n} )}{1 - r}$ , which is easier when $r < 1$ .

Example. $3 + 6 + 12 + \dots$ for 8 terms: $S_{8} = \frac{3 ( 2 ^{8} - 1 )}{1} = 765$ .

Limiting sum. If $- 1 < r < 1$ , the terms shrink towards 0 and $r^{n} \to 0$ . The sum of infinitely many terms approaches

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

Kai's ball. Down 2 m first. Then each bounce goes up and down: $2 (1.2 + 0.72 + \dots)$ . The bounces form a series with $a = 1.2$ and $r = 0.6$ :

$total = 2 + 2 \times \frac{1.2}{1 - 0.6} = 2 + 6 = 8 m$

Recurring decimals. $0. 3 = 0.3 + 0.03 + 0.003 + \dots = \frac{0.3}{1 - 0.1} = \frac{1}{3}$ .

If $∣ r ∣ \geq 1$ , the sum grows without limit and there is no limiting sum.

When does a limiting sum exist? If the ratio contains $x$ , solve $- 1 < r < 1$ . For $1 + 2 x + 4 x^{2} + \dots$ , $r = 2 x$ , so $- \frac{1}{2} < x < \frac{1}{2}$ . For $5 - 5 x + 5 x^{2} - \dots = 12$ : $\frac{5}{1 + x} = 12$ , so $x = - \frac{7}{12}$ , which is allowed because $∣ x ∣ < 1$ .

Exponential terms. $e^{- x} + e^{- 2 x} + \dots$ has $a = r = e^{- x}$ , which is less than 1 for $x > 0$ , so the sum is $\frac{e ^{- x}}{1 - e ^{- x}} = \frac{1}{e ^{x} - 1}$ .

Long-run problems. A dose $D$ is taken every day and 80% of the drug remains a day later. Just after a dose, the amount is $D + 0.8 D + 0. 8^{2} D + \dots$ , which approaches $\frac{D}{1 - 0.8} = 5 D$ .

Taking turns. Ann and Ben take turns, Ann first; each try succeeds with probability $\frac{1}{3}$ for Ann and $\frac{1}{2}$ for Ben. Ann wins on her first try, or after both miss once, and so on: $\frac{1}{3} + (\frac{2}{3} \cdot \frac{1}{2}) \frac{1}{3} + \dots = \frac{1/3}{1 - 1/3} = \frac{1}{2}$ .

Quick summary

A geometric sequence multiplies by the same ratio $r$ each step: $r = \frac{T _{2}}{T _{1}}$ .
$T_{n} = a r^{n - 1}$ .
$S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ .
If $- 1 < r < 1$ , the infinite sum is $S_{\infty} = \frac{a}{1 - r}$ .
Repeated percentage change gives a geometric sequence.

Quiz

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Try 3 questions now

What is the next term of 2, 6, 18, 54, ...?
1. A 108
2. B 162
3. C 72
4. D 216
Show answer
B: 162
Multiply by 3.
What is the common ratio of 100, 50, 25, ...?
1. A 2
2. B $\frac{1}{2}$
3. C $- 50$
4. D 25
Show answer
B: $\frac{1}{2}$
$\frac{50}{100} = \frac{1}{2}$ .
Which sequence is geometric?
1. A 1, 3, 5, 7
2. B 1, 3, 9, 27
3. C 1, 4, 9, 16
4. D 2, 4, 7, 11
Show answer
B: 1, 3, 9, 27
Each term is 3 times the last.