Let a1, a2, a3,.... is in A.P. and ∑ a2k-1 = -72/5 a1 k ∈ to (k=1, 12) and ∑ ak = 0 k ∈ to (k=1, n) . Then the value of n is

Question

Let \(a_{1}, a_{2}, a_{3}, \ldots.\) is in A.P. and \(\sum_\limits{k=1}^{12} a_{2 k-1}=-\frac{72}{5} a_{1}\) and \(\sum_\limits{k=1}^{n} a_{k}=0.\) Then the value of n is

(1) 8

(2) 10

(3) 11

(4) 13

Answer 1

Correct option is: (3) 11

\(\sum_\limits{k=1}^{12} a_{2 k-1}=-\frac{72}{5} a_{1}\)

\(a_{1}+a_{3}+\cdots+a_{23}=-\frac{72}{5} a_{1}\)

\(a+a+2 d+\cdots+a+22 d=-\frac{72}{5} a\)

\(12 a+2 d(1+2+\cdots 11)=-\frac{72}{5} a\)

\(\Rightarrow 12 a+2 d\left(\frac{11 \times 12}{2}\right)=-\frac{72}{5} a\)

\(\Rightarrow 132 d=-\frac{132}{5} a\)  

\(\Rightarrow a=-5 d \)   .....(i)

Also \(\sum_\limits{k=1}^{n} a_{k}=0\)

\(\Rightarrow S_{n}=0\)

\(\Rightarrow \frac{n}{2}[2 a+(n-1) d]=0\)

\(\Rightarrow 2 a=-(n-1) d \)    ......(ii)

From equation (i) and (ii)

(n - 1)d = 10d

\(\therefore n=11\)