CAT 2022 Slot 2 QA Question 2: Consider the arithmetic progression $3, 7, 11, \ldots$ and let $A_n$ denote the sum of the first $n$ terms of this progression. Then the value of $\dfrac{1}{25}\sum_{n=1}^{25} A_n$ is

Question

Consider the arithmetic progression $3, 7, 11, \ldots$ and let $A_n$ denote the sum of the first $n$ terms of this progression. Then the value of $\dfrac{1}{25}\sum_{n=1}^{25} A_n$ is

Accepted Answer

The first term is 3 and the common difference is 4.

$A_n = \dfrac{n}{2} (2 	imes 3 + (n - 1) 	imes 4)$
$A_n = \dfrac{n}{2} (4n + 2)$
$A_n = n(2n + 1) = 2n^2 + n$

$\dfrac{1}{25} \sum_{n=1}^{25} A_n = \dfrac{1}{25} (2 \sum_{n=1}^{25} n^2 + \sum_{n=1}^{25} n)$
$= \dfrac{1}{25} (2 	imes \dfrac{25 \

Consider the arithmetic progression 3,7,11,…3, 7, 11, \ldots3,7,11,… and let AnA_nAn​ denote the sum of the first nnn terms of this progression. Then the value of 125∑n=125An\dfrac{1}{25}\sum_{n=1}^{25} A_n251​∑n=125​An​ is

Consider the arithmetic progression $3, 7, 11, \ldots$ and let $A_n$ denote the sum of the first $n$ terms of this progression. Then the value of $\dfrac{1}{25}\sum_{n=1}^{25} A_n$ is