Let a 1 , a 2 , … , a 21 be an A . P . such that ∑ n = 1 20 1 a n a n + 1 = 4 9 . If the sum of this A . P . is 189 , then a 6 a 16 is equal to :

Step-by-step solution

∑ n = 1 20 1 a n a n + 1 = 4 9  [let common difference of  AP   is   d = a n + 1 - a n = 1 d ∑ n = 1 20 1 a n - 1 a n + 1 ⇒ 1 d 1 a 1 - 1 a 2 + 1 a 2 - 1 a 3 + … . 1 a 20 - 1 a 21 = 4 9 ⇒ 1 d 1 a 1 - 1 a 21 = 4 9 ⇒ 1 d a 21 - a 1 a 1 · a 21 = 4 9 ⇒ 20 d d a 1 a 21 = 4 9 ⇒ a 1 a 21 = 45 a 1 a 1 + 20 d = 45       . . . ( 1 ) Now sum of first 21 terms = 21 2 2 a 1 + 20 d = 189 ⇒ a 1 + 10 d = 9     . . . ( 2 ) by usi