Let a_n, n geq 1,be an arithmetic progression | vedclass

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(A) The sum of the first $n$ terms of an arithmetic progression with first term $a = 2$ and common difference $d = 4$ is given by:$S_n = \frac{n}{2}[2

Vedclass · Accepted Answer

$110$

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(A) The sum of the first $n$ terms of an arithmetic progression with first term $a = 2$ and common difference $d = 4$ is given by:$S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[2(2) + (n-1)4] = \frac{n}{2}[4 + 4n - 4] = \frac{n}{2}[4n] = 2n^2$.The average of the first $n$ terms is $M_n = \frac{S_n}{n} = \frac{2n^2}{n} = 2n$.We need to calculate the sum $\sum_{n=1}^{10} M_n = \sum_{n=1}^{10} 2n$.$= 2 \sum_{n=1}^{10} n = 2 	imes \frac{10(10+1)}{2} = 10 	imes 11 = 110$.

Let $a_n, n \geq 1$,be an arithmetic progression with first term $2$ and common difference $4$. Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum_{n=1}^{10} M_n$ is

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