Question

The sum of the series formed by the sequence $3,\sqrt{3},\mathrm{1.......}$ upto infinity is :

• A. $\frac{3\sqrt{3}(\sqrt{3}+1)}{2}$
• B. $\frac{3\sqrt{3}(\sqrt{3}-1)}{2}$
• C. $\frac{3(\sqrt{3}+1)}{2}$
• D. $\frac{3(\sqrt{3}-1)}{2}$

Answer 1

The correct option is A $\frac{3\sqrt{3}(\sqrt{3}+1)}{2}$

For the given series

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

So it is a GP with common ratio $\frac{1}{\sqrt{3}}$

Formula for sum of infinite terms of GP $=\frac{a}{1-r}$, where

$a$ is first term and r is the common ratio

$\Rightarrow S_{\infty }=\frac{3}{1-\frac{1}{\sqrt{3}}}=\frac{3\sqrt{3}}{\sqrt{3}-1}$

$=\frac{3\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$ ..... [On rationalizing]

$=\frac{3\sqrt{3}(\sqrt{3}+1)}{2}$

Hence, A is correct.