Let the sum of the first n terms of a non-constant A . P . ,   a 1 ,   a 2 ,   a 3 ,   . . . . ,   a n be 50 n + n ( n - 7 ) 2 A , where A is a constant. If d is the common difference of this A . P . , then the ordered pair d ,   a 50 is equal to

Step-by-step solution

Given sum of n terms is S n = 50 n + n n - 7 2 A       … ( 1 ) ⇒ S n - 1 = 50 ( n - 1 ) + ( n - 1 ) n - 8 2 A       … 2 Subtracting ( 1 ) and ( 2 ) , we get S n - S n - 1 = 50 n - 50 n - 1 + n n - 7 2 A - n - 1 n - 8 2 A ⇒ S n - S n - 1 = 50 n - n + 1 + A 2 n n - 7 - n - 1 n - 8 ⇒ S n - S n - 1 = 50 + A 2 n 2 - 7 n - n 2 + 9 n - 8 ⇒ T n = S n - S n - 1 = 50 + A ( n - 4 ) Hence, T 1 = 50 - 3 A and T 2 = 50 - 2 A ⇒ d = T 2 - T 1 = A and T 50 = 50