8. Sequences and Series
medium

Let $a_1, a_2, a_3 \ldots$ be in an $A.P.$ such that $\sum_{ k =1}^{12} a _{2 k -1}=-\frac{72}{5} a _1, a _1 \neq 0$. If $\sum_{ k =1}^{ n } a _{ k }=0$, then $n$ is:

A$11$
B$10$
C$18$
D$17$
(JEE MAIN-2025)

Solution

 Let $a _1= a$, common difference $= d$
$a_1+a_3+a_5+\ldots \ldots+a_{23}=-\frac{72}{5} a$
$\frac{12}{2}[2 a+11 \times 2 d]=-\frac{72}{5} a$
$12 a+132 d=-\frac{72}{5} a$
$132 a+132 \times 5 d=0$
$a=-5 d$
$\frac{n}{2}(2 a+(n-1) d)=0 \Rightarrow-10 d+n d-d=0$
$n=11$
 
Standard 11
Mathematics

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