Question

The sum of 24 terms of the following series : $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+$ ..................

• A. $300$ $\sqrt{2}$
• B. $150$ $\sqrt{2}$
• C. $250$ $\sqrt{2}$
• D. $325$ $\sqrt{2}$

Answer 1

The correct option is A $300$ $\sqrt{2}$

We have $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+$ ..................

We can rewrite the series as $1\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}+........$

Taking $\sqrt{2}$ common, we get:

$\sqrt{2}[1+2+3+4+.......\text{upto}24\text{terms}]$

The ${24}^{th}$ term will be $24$.

Hence, $\sqrt{2}[1+2+3+4+.......+24]$

We know that, the sum of $n$ terms of an AP is given by

$S_n=\frac{n}{2}(a+l)$

Here, $a=1,l=24,n=24$

$\therefore S_{24}=\sqrt{2}\left[\frac{24}{2}\right(1+24\left)\right]$

$\Rightarrow \sqrt{2}\times \frac{24\times 25}{2}=300\sqrt{2}$

Hence, the sum of $24$ terms of the given series is $300\sqrt{2}$.