Question

The $\operatorname{sum}$ of $3-1+\frac{1}{3}-\frac{1}{9}+\ldots$ is equal to

(A) $\frac{20}{9}$

(B) $\frac{9}{20}$

(C) $\frac{9}{4}$

(D) $\frac{4}{9}$

Answer 1

Given series is $3-1+\frac{1}{3}-\frac{1}{9}+\ldots \infty$

Here, $\quad \frac{-1}{3}=\frac{1 / 3}{-1}=\frac{-1 / 9}{1 / 3}$ so it is in GP.

Here,

$r=-\frac{1}{3}, a=3 $

$\therefore \quad S_{\infty}=\frac{a}{1-r}=\frac{3}{1+\frac{1}{3}}=\frac{9}{4} $

So, The correct option will be (C).