Question

The sum of $11$ terms of an A.P. whose middle term is $30,$ is

• A. $320$
• B. $330$
• C. $340$
• D. $350$

Answer 1

Answer: (B)

$330$

Given middle term or the sixth term is $30$

As we know, $a_n=a+(n-1)d$, where a & d are the first term and common difference of an AP respectively.
Then, $a+5d=30$
Sum = $\frac{n}{2}(2a+(n-1\left)d\right)$
=$\frac{11}{2}(2a+10d)$
Substituting,
=$\frac{11}{2}(2\times 30)$
=$330$