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…

Question

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 ∑n=110M_n is

MARKS Content Team · Accepted Answer

Correct answer: (A) 110. The sum of first n,n≥1 terms of arithmetic progression with first term 2 and common difference 4, is S_n=n/2[4+(n-1)4]=2n^2 So, the average of the first n terms M_n=S_nn=2n Now, ∑n=110 M_n=2∑n=110 n =2×(10×11)/2=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

Solution

More Sequences and Series Questions

Let an,n≥1, be an arithmetic progression with first term 2 and common difference 4 . Let Mn be the average of the first n terms. Then the sum ∑n=110Mn is

Solution

More Sequences and Series Questions

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