Question

The sum of first $10$ terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+...$ is

• A. $121(\sqrt{6}+\sqrt{2})$
• B. $243(\sqrt{3}+1)$
• C. $\frac{121}{\sqrt{3}-1}$
• D. $242(\sqrt{3}-1)$

Answer 1

The correct option is A $121(\sqrt{6}+\sqrt{2})$

$s=\sqrt{2}+\sqrt{6}+\sqrt{18}+....$

$\Rightarrow s=\sqrt{2}+\sqrt{2}\times \sqrt{3}+\sqrt{2}(\sqrt{3}{)}^2+.....$

This is a G.P with $1$st term $\left(a\right)=\sqrt{2}$ and common ratio$\left(r\right)=\sqrt{3}$

Sum of $10$ terms of this G.P., $S=\frac{a(r^{10}-1)}{r-1}$

$=\frac{\sqrt{2}\left(\right(\sqrt{3}{)}^{10}-1)}{\sqrt{3}-1}$

$=\frac{\sqrt{2}\left(242\right)}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$=121\times \sqrt{2}(\sqrt{3}+1)$

$=121(\sqrt{6}+\sqrt{2})$

$\Rightarrow \left(A\right)$ Option.