Question

Find the sum of the following geometric series $1+\frac{1}{3}+\frac{1}{9}+...........$ upto $\infty$

Answer 1

It is given that the first term of G.P is $a_1=1$ and the second term is $a_2=\frac{1}{3}$, therefore,

$r=\frac{\frac{1}{3}}{1}=\frac{1}{3}$

We know that the formula for sum of infinite terms of G.P is $S_{\infty }=\frac{a}{1-r}$, where $a$ is the first term, $r$ is the common ratio.

In the given geometric series, the first term is $a=1$, the common ratio is $r=\frac{1}{3}$, therefore,

$S_{\infty }=\frac{a}{1-r}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{3-1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$

Hence, $S_{\infty }=\frac{3}{2}$.