Question

Find the sum of infinite geometric series: $ 1, - \dfrac{3}{2},\dfrac{9}{4}, - \dfrac{{27}}{8},.... $?

Answer 1

Hint: The given problem requires us to find the sum of an infinite geometric progression whose first four terms are given to us. For finding out the sum of an infinite geometric series, we need to know the first term and the common ratio of that particular geometric series. We can find out the common ratio of a geometric progression by dividing any two consecutive terms of the series.

Complete step by step solution:
Given infinite geometric series: $ 1 + \left( { - \dfrac{3}{2}} \right) + \left( {\dfrac{9}{4}} \right) + \left( { - \dfrac{{27}}{8}} \right) + .... $
We have to find the sum of this infinite geometric progression.
Here, first term $ = a = 1 $ .
Now, we can find the common ratio by dividing any two consecutive terms.
So, common ratio \[ = r = - \dfrac{3}{2}\]
So, \[r = - \dfrac{3}{2}\] .
Now, Using formula of sum of infinite geometric when $ r < 1 $ , we get,
Sum of the given infinite geometric progression $ = {S_\infty } = \dfrac{a}{{1 - r}} $
Substituting the values of a and r, we get the sum of the infinite geometric progression as: $ {S_\infty } = \dfrac{1}{{1 - \left( { - \dfrac{3}{2}} \right)}} $
 $ = {S_\infty } = \dfrac{1}{{1 + \dfrac{3}{2}}} $
Simplifying the expression, we get,
 $ = {S_\infty } = \dfrac{1}{{\dfrac{{2 + 3}}{2}}} $
So, we get the sum as
 $ = {S_\infty } = \dfrac{2}{5} $
So, the sum of infinite geometric progression given to us is $ {S_\infty } = \dfrac{2}{5} $ .
So, the correct answer is “$ {S_\infty } = \dfrac{2}{5} $”.

Note: Geometric progression is a series where any two consecutive terms are in the same ratio. The common ratio of a geometric series can be calculated by division of any two consecutive terms of the series. The sum of n terms of a geometric progression can be calculated if we know the first term and common ratio of the geometric series as: $ {S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}} $ if $ r < 1 $ and $ {S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}} $ if $ r > 1 $.