# Geometric Progression

A Geometric Progression is a sequence of numbers where the ratio between the numbers in the same.

## Example in geometric progression

For the Geometric Progression 20, 5, $1\frac{1}{4}$, ...., find

(i) The common ratio

(ii) The ${n}^{th}$ term

(iii) The sum of the first 8 terms

Solution

(i)

## How to find the ratio of a geometric progression

### Point to note

The ratio can be find by dividing a term which is next in the sequence by a previous one

Ratio = $\frac{\mathrm{Term2}}{Term1}$

Term2 = 5

Term1 = 20

Ratio = $\frac{5}{20}$

Ratio = $\frac{1}{4}$

(ii)

## How to find nth term of a geometric progression

Formula

${n}^{th}=a{r}^{n - 1}$

${n}^{th}=20{\frac{1}{4}}^{n - 1}$

$=20×{\frac{1}{4}}^{- 1}×{\frac{1}{4}}^{n}$

$=20×4×{\frac{1}{4}}^{n}$

${n}^{th}=80\left({\frac{1}{4}}^{n}\right)$

(iii)

## How to find the sum of a geometric progression

Formula

$sn=a\left(\frac{1 -{r}^{n}}{1 - r}\right)$

$sn=20\left(\frac{1 -{\mathrm{1/4}}^{8}}{1 - 1/4}\right)$

$sn=20\frac{\mathrm{1 - 1/65536}}{3/4}$

$sn=20\frac{\mathrm{65535/65536}}{3/4}$

sn = 26.7

## Example of how to find the nth term in geometric progression

The 3rd and 4th terms of a geometric progression are 4 and 8 respectively. Find

(i) The common ratio, first term and second term

(ii) The sum of the first 10 terms

(iii) The sum infinity of this geometric progression

Solutions

(i)

Common Ratio = T2 / T1

T4 = 8, T3 = 4

Common Ratio = $\frac{\mathrm{T2}}{T1}$

$\frac{8}{4}$

Answer: Common Ratio = $2$

${n}^{th}=a{r}^{n - 1}$

second term

second term = 4 / r

= 4 / 2

first term

first term = 2 / r

= 2 / 2

(ii)

$sn=a\left(\frac{1 -{r}^{n}}{1 - r}\right)$

a = 1, r = 2, n = 10

$sn=1\left(\frac{1 -{2}^{10}}{1 - 2}\right)$

$sn=1\left(\frac{1 - 1024}{-1}\right)$

$sn=1\left(\frac{-1023}{-1}\right)$

$sn=\left(\frac{-1023}{-1}\right)$

Answer: $sn10=1023$

(iii)

## How to solve Sum to infinity

Formula

$s=\frac{a}{1 - r}$

a = 1, r = 2

$s=\frac{1}{1 - 2}$

$s=\frac{1}{-1}$

$s=\mathrm{-1}$

Answer: $s=\mathrm{-1}$

## how to solve the first time in geometric progression

The first three terms of a geometric progression are x + 1, x - 3 and x - 1, find;

(i) The value of x

(ii) The first term

(iii) The sum to infinite

Solutions

(i)

Formula

$\frac{\mathrm{T2}}{T1}=\frac{\mathrm{T3}}{T2}$

T1 = x + 1, T2 = x - 3, T3 = x - 1

$\frac{\mathrm{x - 3}}{x + 1}=\frac{\mathrm{x - 1}}{x - 3}$

(x - 3 )(x - 3 ) = (x + 1)(x - 1)

${x}^{2}- 3x - 3x + 9={x}^{2}-x + x - 1$

${x}^{2}-6x + 9={x}^{2}- 1$

${x}^{2}-6x + 9{\mathrm{- x}}^{2}+ 1=0$

${x}^{2}{\mathrm{- x}}^{2}-6x + 9+ 1=0$

$-6x + 10=0$

$-6x=-10$

$\frac{\mathrm{-6x}}{-6}=\frac{\mathrm{-10}}{-6}$

$x=\frac{5}{3}$

Answer: $x=1\frac{2}{3}$

(ii)

first term

x + 1

$\frac{5}{3}+1$

$\frac{5}{3}+\frac{1}{1}$

$\frac{\mathrm{1\left(5\right) + 3\left(1\right)}}{3}$

$\frac{\mathrm{5 + 3}}{3}$

$\frac{8}{3}$

Answer $2\frac{2}{3}$

(iii)

$s=\frac{a}{1 - r}$

r = T2/T1

T2 = x - 3

T2 = 5/3 - 3

T2 = $\frac{\mathrm{5 - 9}}{3}$

T2 = $\frac{\mathrm{-4}}{3}$

T1 = 8/3 , T2 = -4/3

r = $\frac{\mathrm{-4}}{3}÷\frac{8}{3}$

r = $\frac{\mathrm{-4}}{3}×\frac{3}{8}$

r = $\frac{\mathrm{-4}}{8}$

r = $\frac{\mathrm{-1}}{2}$

a = 8/3 , r = -1/2

$s=\frac{\mathrm{8/3}}{1 - -1/2}$

$s=\frac{\mathrm{8/3}}{1 + 1/2}$

$s=\frac{\mathrm{8/3}}{3/2}$

s = $\frac{8}{3}÷\frac{3}{2}$

s = $\frac{8}{3}×\frac{2}{3}$

s = $\frac{\mathrm{16}}{9}$

Answer: s = $1\frac{7}{9}$