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 \(\displaystyle\sum^{10}_{n=1}\) M_n is

∑¹⁰_n=1 M_n

(a) 110

(b) 335

(c) 770

(d) 1100

Answer 1

Answer is : (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] = 2n²

So, the average of the first n terms