JEE Main 2026 — 24 January, Evening Shift — Question 5

$\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \mathrm{a}_{4}$ as

$\mathrm{a}-3 \mathrm{~d}, \mathrm{a}-\mathrm{d}, \mathrm{a}+\mathrm{d}, \mathrm{a}+3 \mathrm{~d}$ where $\mathrm{d}=\frac{\ell}{2}$

$\because a_{1}+a_{2}+a_{3}+a_{4}=48$

$\Rightarrow 4 a=48 \Rightarrow a=12$

$a_{1} a_{2} a_{3} a_{4}+\ell^{4}=361$

$\Rightarrow\left(a^{2}-9 d^{2}\right)\left(a^{2}-d^{2}\right)+16 d^{4} =361$

$\Rightarrow\left(144-9 d^{2}\right)\left(144-d^{2}\right)+16 d^{4}=361$

$\Rightarrow 25 d^{4}-1440 d^{2}+(144)^{2}=361$

$\left(5 d^{2}-144\right)^{2}=19^{2}$

$\therefore 5 d^{2}-144=19$ or $-19$

$d^{2}=\frac{163}{5}$ or $d^{2}=\frac{125}{5}=25$

$d=\sqrt{\frac{163}{5}}$or $d=5$

$\therefore \ell=2 \sqrt{\frac{163}{5}}$ or $\ell=10$ (rejected)

$\because$ common difference is an integer therefore largest term $=12+15=27$