NTA Abhyas JEE Main2020MathematicsSequences and SeriesPractice
Let a 1 , a 2 , a 3 ,..........., a 1 1 be real numbers satisfying a 1 = 1 5 , 2 7 - 2 a 2 > 0 and a k = 2 a k - 1 - a k - 2 ∀ k = 3 , 4 , ......., 1 1 . If a 1 2 + a 2 2 + .... + a 1 1 2 1 1 = 9 0 , then the value of a 1 + a 2 + .... + a 1 1 1 1 is equal to
Correct answer
0
Step-by-step solution
a k = 2 a k - 1 - a k - 2 ⇒ a 1 , a 2 , .... , a 1 1 are in AP with let common difference be d ∴ a 1 2 + a 2 2 + … + a 1 1 2 1 1 = 1 1 a 2 + 3 5 × 1 1 d 2 + 11 0 a d 1 1 = 9 0 ⇒ 2 2 5 + 3 5 d 2 + 1 5 0 d = 9 0 3 5 d 2 + 1 5 0 d + 1 3 5 = 0 ⇒ d = - 3 , - 9 7 Given, a 2 < 2 7 2 ∴ d = - 3 and d ≠ - 9 7 Hence, a 1 + a 2 + . . . . . . . a 11 = 11 2 2 × 15 + 10 ( - 3 ) = 0