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Question: Answered & Verified by Expert
Let a1a2a3,..........., a11 be real numbers satisfying a1=1527-2a2>0 and ak=2ak-1-ak-2 ∀ k=34 .......,11. 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
MathematicsSequences and SeriesJEE Main
Solution:
2503 Upvotes Verified Answer
The correct answer is: 0

a k = 2 a k - 1 - a k - 2


 a1a2,....a11 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, a1+a2+.......a11=1122×15+10(-3)=0

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