Full of Math Examples
FULL OF MATH EXAMPLES

# Geometric Series Examples

Here are the all important examples on Geometric Series.

## What is geometric series ?

Geometric series is a series in which ratio of two successive terms is always constant.

Lets take a example.

Example : 1/2,1/4,1/8,1/16,....

Here the ratio of any two terms is 1/2 , and the series terms values get increased by factor of 1/2.

Example : 2,4,8,16,32,64..... is also an example of geometric series.

or in a general way geometric series can represented as $a,ar,ar^{2},ar^{3},ar^{4}.......$

## Sum of geometric series

The formula for sum of n terms of geometric progression is

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

Formula for sum of infinite terms of geometric series

Geometric series has sum if and only if $|r|<1$ and this case sum is,

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

Then the series is called convergent.

## Examples

### Example #1

#### Find the 8th and nth terms of the G.P. 5,25,125, ...

Here $a=5$ and $r=5 .$ Thus, $a_{8}=5(5)^{8-1}=5(5)^{7}=5^{8}$
and $\quad a_{n}=a r^{n-1}=5(5)^{n-1}=5^{n}$

### Example #2

#### Find the sum of first n terms and the sum of first 5 terms of the geometricseries $1+\frac{2}{3}+\frac{4}{9}+\ldots$

Here $a=1$ and $r=\frac{2}{3} .$ Therefore
$S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$$=\frac{\left[1-\left(\frac{2}{3}\right)^{n}\right]}{1-\frac{2}{3}}=3\left[1-\left(\frac{2}{3}\right)^{n}\right] In particular, \mathrm{S}_{5}=3\left[1-\left(\frac{2}{3}\right)^{5}\right]$$=3 \times \frac{211}{243}=\frac{211}{81}$

### Example #3

#### Find the sum of the sequence 7,77,777,7777, ... to n terms.

This is not a G.P., however, we can relate it to a G.P. by writing the terms as
$\mathrm{S}_{n}=7+77+777+7777+\ldots$ to $n$ terms
$=\frac{7}{9}[9+99+999+9999+\ldots .$ to $n$ term $]$
$=\frac{7}{9}[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\left(10^{4}-1\right)+\ldots n$ terms]
$=\frac{7}{9}\left[\left(10+10^{2}+10^{3}+\right.\right.$ + ... n $terms$ )-(1+1+1+ ... n $terms$ ) ] 
$=\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]$$=\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]$