KVPY 2019 SA Exam Previous Papers

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Question : 4 of 80

Marks: +1, -0

Let

a_n, n ≥ 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

∑↖{10}↙{n=1}M_n

is :

110
335
770
1100

Solution:

M_n={∑↖{n}↙{r=1}2+(r-1)4}/n

=2(n+1)-2=2n

Hence

∑↖{n}↙{n=1}M_n=110

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