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8. Sequences and Series
hard
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 $\sum \limits_{n=1}^{10} M_n$ is
A
$110$
B
$335$
C
$770$
D
$1100$
(KVPY-2019)
Solution
(a)
The sum of first $n, n \geq 1$ terms of arithmetic progression with first term $2$ and common difference $4$, is
$S_n=\frac{n}{2}[4+(n-1) 4]=2 n^2$
So, the average of the first $n$ terms
$M_n=\frac{S_n}{n}=2 n$
Now, $\sum\limits_{n=1}^{10} M_n=2 \sum\limits_{n=1}^{10} n$
$=2 \times\left(\frac{10 \times 11}{2}\right)=110$
Standard 11
Mathematics
