Tip:A–D to answerE for explanationV for videoS to reveal answer
Let a1,a2,a3,... be in an AP such that ∑k=1122a2k−1=−572. If ∑k=1nak=0, then n is equal to:
- A.
11
(Correct Answer) - B.
10
- C.
18
- D.
17
Explanation
Let first term a1=a
Common difference = d
∑k=112a2k−1=−572a1,a1=0
212[2a+11×2d]=−572a1
12a+132d=−572a1
132a+132×5d=0
a=−5d
∑k=1nak=0,
2n(2a+(n−1)d)=0
⇒
−10d+nd−d=0
n=11
Explanation
Let first term a1=a
Common difference = d
∑k=112a2k−1=−572a1,a1=0
212[2a+11×2d]=−572a1
12a+132d=−572a1
132a+132×5d=0
a=−5d
∑k=1nak=0,
2n(2a+(n−1)d)=0
⇒
−10d+nd−d=0
n=11
