Let a 1 , a 2 , a 3 , be in an A.P. such that sum k=1 12 a 2 k-1 =-72/5 a 1 , a 1 neq 0 . If sum k=1 n a k =0 , then n is:

Mathematics · Sequence and Series

JEE Main 2025 — 2 April, Morning Shift — Question 30

Let $a_{1}, a_{2}, a_{3}$ , 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:

Option A:
$17$
Option B:
$18$
Option C:
$11$
Correct
Option D:
$10$

Answer: C

Step-by-step solution

$\sum_{k=1}^{12} a_{2 k-1}=-\frac{72}{5} a_{1}$

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

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

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

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

$\Rightarrow 132 d=-\frac{132}{5} a$

$\Rightarrow a=-5 d …(i)$

Also $\sum_{k=1}^{n} a_{k}=0$

$\Rightarrow S_{n}=0$

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

$\Rightarrow 2 a=-(n-1) d …(ii)$

From equation (i) and (ii)

$(n-1) d=10 d$

$\therefore \quad n=11$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression