CAT 2022 Slot 2 Paper

Question : 62 of 66

Marks: +1, -0

Consider the arithmetic progression $3, 7, 11, ...$ and let $\An$ denote the sum of the first $n$ terms of this progression. Then the value
of $1/25 ∑^{25}↙{n=1} \An$ is

Validate

Solution:

Sum of

n

terms in an

\A . \P

={n}/{2}(2 a+(n-1) d)

\A_n={n}/{2}(6+(n-1) 4)=n(2 n+1)

Σ \A_n=Σ n(2 n+1)=2 Σ n^2+Σ n= {2 n(n+1)(2 n+1)}/{6}+{n(n+1)}/{2}

Substituting

n=25

, we get

{1}/{25} ∑↙{{n=1}}↖{{25}} \A_n={1}/{25} ({2(25)(25+1)(50+1)}/{6}+{25(25+1)}/{2})

{1}/{25} ∑↙{{n=1}}↖{{25}} \A_n=26(17)+13=455

The answer is option A.

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