Question

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

Accepted Answer

C. AP has first term $a = 3$, common difference $d = 4$. $A_n = \frac{n}{2} [2(3) + (n-1)4] = \frac{n}{2} [4n + 2] = n(2n + 1) = 2n^2 + n$. We need $\frac{1}{25} \sum_{n=1}^{25} (2n^2 + n) = \frac{1}{25} \left( 2 \sum_{n=1}^{25} n^2 + \sum_{n=1}^{25} n ight)$. $\sum_{n=1}^{25} n = \frac{25 	imes 26}{2} = 325$. $\sum_{n=1}^{25} n^2 = \frac{25 	imes 26 	imes 51}{6} = 5525$. Sum = $2(5525) + 325 = 11050 + 325 = 11375$. Value = $\frac{11375}{25} = 455$.

Sum of First N Terms of AP

Explanation

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