Question

Let $a_n,n\ge 1$, be an arithmetic progression with first term $2$ and common difference $4$. Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum _{n=1}^{10}{M}_n$ is

• A. $110$
• B. $335$
• C. $770$
• D. $1100$

Answer 1

The correct option is A $110$

$A.P$ $a_n$ $n\ge 1$

First term $a_1=2$

Common difference $d=4$

$M_n=$ Average of first $n$ terms.

$\Rightarrow$ $M_n=\frac{S_n}{n}=\frac{\frac{n}{2}[2a+(n-1\left)d\right]}{n}$

$=\frac{1}{2}[2\times 2+(n-1\left)4\right]$

$=2+2(n-1)$

$\Rightarrow$ $M_n=2+2n-2$

$\Rightarrow$ $M_n=2n$

$\Rightarrow$ $\sum _{n=1}^{10}{M}_n=\sum _{n=1}^{10}\left(2n\right)=\frac{2\left(10\right)\times \left(11\right)}{2}=110$