Let a1, a2, a3, be in an A.P. such that k=112 a2k-1 = - 725a1, a1 0. If k=1n ak = 0, then n is :
- 18
- 17
- 11
- 10
Answer
(3) 11
Solution
Let d be the common difference. Then a1, a3, , a23 form an A.P. of 12 terms with common difference 2d, so
k=112 a2k-1 = 122[2a1 + 11(2d)] = 12a1 + 132d
12a1 + 132d = - 725a1 132d = - 1325a1 d = - a15
Now the sum of the first n terms is zero:
k=1n ak = n2[2a1 + (n - 1)d] = 0
2a1 - (n - 1)a15 = 0, and as a1 0, n - 1 = 10.
n = 11
Answer: (3)