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 $110$

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

$\sum _{n=1}^{10}{M}_n=\frac{1}{2}\sum _{n=1}^{10}(2a+(n-1\left)d\right)=2\sum _{n=1}^{10}n[\because a=2,d=4]=110$