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 = 1 10 M n is

Step-by-step solution

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 ] = 2 n 2 So, the average of the first n terms M n = S n n = 2 n  Now,  ∑ n = 1 10   M n = 2 ∑ n = 1 10   n = 2 × 10 × 11 2 = 110