CAT 2022

Slot 2

QA

Question & Solution

Sum of n terms in an A.P = $\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

$A_n=\frac{n}{2}\left(6+\left(n-1\right)4\right)=n\left(2n+1\right)$

$\Sigma\ A_n=\Sigma\ n\left(2n+1\right)=2\Sigma\ n^2+\Sigma\ n=\ \frac{\ 2n\left(n+1\right)\left(2n+1\right)}{6}+\ \frac{\ n\left(n+1\right)}{2}$

Substituting n = 25, we get

$\frac{1}{25} \sum_{n=1}^{25} A_{n}$ = $\frac{1}{25}\left(\ \frac{\ 2\left(25\right)\left(25+1\right)\left(50+1\right)}{6}+\ \frac{\ 25\left(25+1\right)}{2}\right)$

$\frac{1}{25} \sum_{n=1}^{25} A_{n}$ = $\ 26\left(17\right)+13$ = 455

The answer is option A.