# Geometric Series

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Geometric Series

series is a sequence where the goal is to add all the terms together. We will study arithmetic series and geometric series.

Recall: Notation from Sequences:

$a$ is first term.

$r$ is the ratio, the amount we multiply by each time.

$n$ is the number of terms in the series.

As for arithmetic series, we can add together all the terms in a geometric series.

Make sure you are happy with the following topics before continuing.

A Level

The sum of a geometric series with $n$ terms is:

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

We can prove this result:

$S_{n}=a+ar+ar^{2}+...+ar^{n-1}$

$rS_{n}=ar+ar^{2}+ar^{3}+...+ar^{n}$

$S_{n}-rS_{n}=a-ar^{n}$

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

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

A Level

## Sum to Infinity

If $|r|<1$ in a geometric progression, then we can sum up an infinite number of terms and still end with a finite answer. This is called the sum to infinity. The sum to infinity is given by:

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

Series that have a sum to infinity are convergent.

Series that do not have a sum to infinity are divergent.

A Level
A Level

## Example 1: Sum of a Geometric Series

What is the sum of the first ten terms of the series that begins $1,3,9,27...$?

[2 marks]

$a=1$

$r=3$

$n=10$

Substitute into formula:

\begin{aligned}S_{10}&=\dfrac{1(1-3^{10})}{1-3}\\[1.2em]&=\dfrac{1-3^{10}}{-2}\\[1.2em]&=\dfrac{1-59049}{-2}\\[1.2em]&=\dfrac{-59048}{-2}\\[1.2em]&=29524\end{aligned}

A Level

## Example 2: Sum to Infinity

What is the sum to infinity of the geometric series that begins $1,\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8}...$?

[2 marks]

$a=1$

$r=\dfrac{1}{2}$

Substitute into formula:

\begin{aligned}S_{\infty}&=\dfrac{a}{1-r}\\[1.2em]&=\dfrac{1}{1-\dfrac{1}{2}}\\[1.2em]&=\dfrac{1}{\dfrac{1}{2}}\\[1.2em]&=2\end{aligned}

A Level

## Example Questions

i) $a=729$

$r=\dfrac{2}{3}$

$n=7$

Substitute into formula:

\begin{aligned}S_{n}&=\dfrac{a(1-r^{n})}{1-r}\\[1.2em]&=\dfrac{729\left(1-\left(\dfrac{2}{3}\right)^{7}\right)}{1-\dfrac{2}{3}}\\[1.2em]&=\dfrac{729\left(1-\dfrac{128}{2187}\right)}{\dfrac{1}{3}}\\[1.2em]&=3\times 729\left(\dfrac{2059}{2187}\right)\\[1.2em]&=2187\times\dfrac{2059}{2187}\\[1.2em]&=2059\end{aligned}

ii) \begin{aligned}S_{\infty}&=\dfrac{a}{1-r}\\[1.2em]&=\dfrac{729}{1-\dfrac{2}{3}}\\[1.2em]&=\dfrac{729}{\dfrac{1}{3}}\\[1.2em]&=3\times 729\\[1.2em]&=2187\end{aligned}

i) A sum to infinity is not sensible for two reasons:

The tree will not live forever.

$r=1.01$ so $|r|>1$ so there is no finite sum to infinity.

ii) $a=0.2$

$r=1.01$

$n=22$

Substitute into formula:

\begin{aligned}S_{n}&=\dfrac{a(1-r^{n})}{1-r}\\[1.2em]&=\dfrac{0.2(1-1.01^{21})}{1-1.01}\\[1.2em]&=4.65\end{aligned}m

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

$a=2$

$390=\dfrac{2}{1-r}$

$1-r=\dfrac{2}{390}$

$r=1-\dfrac{2}{390}$

$r=\dfrac{388}{390}$

$r=\dfrac{194}{195}$

The first option is an arithmetic series:

$a=1$

$d=1$

$n=64$

Substitute into formula:

\begin{aligned}S_{n}&=\dfrac{n}{2}(2a+(n-1)d)\\[1.2em]&=\dfrac{64}{2}(2\times 1+(64-1)1)\\[1.2em]&=32(2+63)\\[1.2em]&=32\times 65\\[1.2em]&=£2080\end{aligned}

The second option is a geometric series:

$a=0.01$

$r=2$

$n=64$

Substitute into formula:

\begin{aligned}S_{n}&=\dfrac{a(1-r^{n})}{1-r}\\[1.2em]&=\dfrac{0.01(1-2^{64})}{1-2}\\[1.2em]&=£1.84\times 10^{17}\end{aligned}

So the second chessboard is far more valuable.

A Level

A Level

A Level

A Level

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