JEE Main 2025 — Sequences And Series Question with Solution

Given:

$\sum _{k=1}^{12}{a}_{2k-1}=-\frac{72}{5}{a}_1$

This means:

$a_1+a_3+\cdots +a_{23}=-\frac{72}{5}{a}_1$

Express each term in A.P.:

The general term of the A.P. is given by $a_k=a+(k-1)d$. Thus, for the odd indices:

$a_1+(a+2d)+(a+4d)+\cdots +(a+22d)$

Simplify the equation:

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

Use the formula for sum of an arithmetic series (first 11 terms):

$1+2+\cdots +11=\frac{11\times 12}{2}=66$

So, the equation becomes:

$12a+2d(66)=-\frac{72}{5}a$

$12a+132d=-\frac{72}{5}a$

Solve for the relationship between $a$ and $d$:

$132d=-\frac{72}{5}a-12a$

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

So,

$a=-5d\text{(i)}$

Second condition:

${\sum }_{k=1}^n{a}_k=0\Rightarrow S_n=0$

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

$2a=-(n-1)d\text{(ii)}$

Combine equation (i) and (ii):

Substitute $a=-5d$ into equation (ii):

$2(-5d)=-(n-1)d$

$-10d=-(n-1)d$

Therefore:

$n-1=10$

$n=11$

Thus, $n=11$.