SSC CGL Tier II 25 Oct 2015 Paper 1

(2) Sum of new numbers = na + (2 + 4 + 8 + 16 ..... to n terms)
Now] S = 2 + 4 + 8 + 16 + ..... to n terms
Here, a = first term = 2
r = common ratio = $\frac{4}{2}=2$
It is a geometric series.
$\therefore \mathit{S}=\frac{\mathit{a}\left({\mathit{r}}^{\mathit{n}}-1\right)}{\mathit{r}-1}=\frac{2\left(2^{\mathit{n}}-1\right)}{2-1}$ $=2\left(2^{\mathit{n}}-1\right)$
∴ Required average $=\frac{\mathit{n}\mathit{a}+2\left(2^{\mathit{n}}-1\right)}{\mathit{n}}$$=\mathit{a}+\frac{2\left(2^{\mathit{n}}-1\right)}{\mathit{n}}$