Let $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=48$ and $\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4}+l^{4}=361$ then the largest term of the A.P. is equal to

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Sequences and Series

Let

α_{1}, α_{2}, α_{3}, α_{4}

be an A.P. of four terms such that each term of the A.P. and its common difference

l

are integers. If

α_{1} + α_{2} + α_{3} + α_{4} = 48

and

α_{1} α_{2} α_{3} α_{4} + l^{4} = 361

then the largest term of the A.P. is equal to

Answer: (A)

Take the four terms symmetrically as

12 - 3 d, 12 - d, 12 + d, 12 + 3 d

(mean

= 48/4 = 12

and common difference

l = 2 d

). The product condition

(144 - 9 d^{2}) (144 - d^{2}) + 16 d^{4} = 361

simplifies to

(5 d^{2} - 144)^{2} = 361

, so

5 d^{2} - 144 = \pm 19

. The value

5 d^{2} = 163

is non-integral, while

5 d^{2} = 125

gives

d = 5

and

l = 10

, with terms

- 3, 7, 17, 27

. The largest term is

27

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option