Let a1, a2, a3, .... is an A. P. and ∑ α2k-1 k ∈ to (k=1, 12) = - 72/5 α1 and ∑ αk ∈ to (k=1, n) = 0.

Question

Let \(a_1, a_2, a_3, ....\) is an A. P. and \(\sum_\limits{k=1}^{12} α_{2k-1} = -\frac{72}{5}α_1\) and \(\sum_\limits{k=1}^{n} α_{k} =0\). Then the value of n is 

(1) 8  

(2) 10

(3) 11

(4) 13

Answer 1

Correct option is: (3) 13