Geometric Progression

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

(i) The common ratio

(ii) The $n^{\mathrm{th}}$ 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 = $\frac{\mathrm{Term2}}{\mathrm{Term1}}$

Term2 = 5

Term1 = 20

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

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

(ii)

How to find nth term of a geometric progression

Formula

$n^{\mathrm{th}}=a{r}^{\mathrm{n\; -\; 1}}$

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

$=20\times {\frac{1}{4}}^{\mathrm{-\; 1}}\times {\frac{1}{4}}^{\mathrm{n}}$

$=20\times 4\times {\frac{1}{4}}^{\mathrm{n}}$

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

(iii)

How to find the sum of a geometric progression

Formula

$sn=a\left(\frac{1\; -\; <msup><mi>\; r\; </mi>\; <mn>n</mn>\; </msup>}{1\; -\; r}\right)$

$sn=20\left(\frac{1\; -\; <msup><mi>\; 1/4\; </mi>\; <mn>8</mn>\; </msup>}{1\; -\; 1/4}\right)$

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

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

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

$\frac{8}{4}$

Answer: Common Ratio = $2$

$n^{\mathrm{th}}=a{r}^{\mathrm{n\; -\; 1}}$

second term

second term = 4 / r

= 4 / 2

Answer: second term = 2

first term

first term = 2 / r

= 2 / 2

Answer: first term = 1

(ii)

$sn=a\left(\frac{1\; -\; <msup><mi>\; r\; </mi>\; <mn>n</mn>\; </msup>}{1\; -\; r}\right)$

a = 1, r = 2, n = 10

$sn=1\left(\frac{1\; -\; <msup><mi>\; 2\; </mi>\; <mn>10</mn>\; </msup>}{1\; -\; 2}\right)$

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

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

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

Answer: $sn10=1023$

(iii)

How to solve Sum to infinity

Formula

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

a = 1, r = 2

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

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

$s=\mathrm{-1}$

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

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

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

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

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

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

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

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

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

$\mathrm{-6x\; +\; 10}=0$

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

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

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

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

(ii)

first term

x + 1

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

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

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

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

$\frac{8}{3}$

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

(iii)

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

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 = $\frac{\mathrm{-4}}{3}\times \frac{3}{8}$

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

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

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

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

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

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

s = $\frac{8}{3}÷\frac{3}{2}$

s = $\frac{8}{3}\times \frac{2}{3}$

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

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