SSC CHSL 6 Mar 2018 Shift 1 Paper

First eleven multiples of 11 will form an arithmetic progression : 11,22,33,.......,121
First term, and common difference, $\mathit{d}=11$
Sum $\mathit{n}$ of terms = $\frac{\mathit{n}}{2}\left(\mathit{a}+\mathit{l}\right)$
⇒ ${\mathit{S}}_{11}=\frac{11}{2}\left(11+121\right)$
= $11\times 66$
Now, average of 11 terms = $=\frac{11\times 66}{11}=66$