Question

The sum of n terms of the AP $\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},....\mathrm{is}$
(a) 1
(b) 2n (n + 1)
(c) $\frac{1}{2}n(n+1)$
(d) $\frac{1}{\sqrt{2}}n(n+1)$

Answer 1

(d) $\frac{1}{\sqrt{2}}n\left(n+1\right)$
​
$$\begin{gathered} Here,a=\sqrt{2}\mathrm{and}d=\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2} \\ \therefore Sn=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ \Rightarrow \frac{n}{2}\left[2\times \sqrt{2}+\left(n-1\right)\times \sqrt{2}\right] \\ \Rightarrow \left(\frac{n}{2}\right)\left(\sqrt{2}\right)\left[2+n-1\right] \\ \Rightarrow \left(\frac{n}{\sqrt{2}}\right)\left[n+1\right] \end{gathered}$$