JEE Main 2025 — 2 April, Evening Shift — Question 38

Let the number of terms be $2 n$

$$\begin{gathered} T_{1}+T_{3}+T_{5} \ldots T_{2 n-1}=24\\ T_{2}+T_{4}+T_{6} \ldots T_{2 n}=30\\ \end{gathered}$$

Subtract (2) from (1)

$(\left.T_{2}-T_{1}\right)+\left(T_{4}-T_{3}\right)+\ldots\left(T_{2 n}-T_{2 n-1}\right)=6$

$n d=6$

$(a+(2 n+1) d)-a=\frac{21}{2}$

$\Rightarrow 2 n d-d=\frac{21}{2}$

$\Rightarrow 12-\frac{21}{2}=d$

$\Rightarrow d=\frac{3}{2}$

$\therefore n=4$

$\therefore$ Total terms $=8$.