Let 1 x 1 , 1 x 2 , … , 1 x n ( x i ≠ 0 for i = 1 , 2 , … . , n ) be in A.P. such that x 1 = 4 and x 21 = 20 . If n is the least positive integer for which x n > 50 , then ∑ i = 1 n 1 x i is equal to

Step-by-step solution

Given 1 x 1 , 1 x 2 , … … … , 1 x n are in A.P. and x 1 = 4 and x 21 = 20 . ⇒ a = 1 4 Given, x 21 = 20 . ⇒ 1 4 + 20 · d = 1 20 ⇒ d = − 1 100 Given, x n > 50 . ⇒ 1 x n < 1 50 ⇒ 1 4 -   n - 1 100 <   1 50 ⇒ n > 24 ∵   n = 25 Now, ∑ i = 1 25 1 x i = 25 2 2 × 1 4 - 1 100 × 24 = 13 4