JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

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

A
27
B
24
C
21
D
23

Solution

Let the four terms be $a - 3 d, a - d, a + d, a + 3 d$ where $d = \frac{l}{2}$ .

Sum: $4 a = 48 \Rightarrow a = 12$ .

Product: $(a - 3 d) (a - d) (a + d) (a + 3 d) = (a^{2} - 9 d^{2}) (a^{2} - d^{2})$ . With $l^{4} = 16 d^{4}$ :

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

$20736 - 144 d^{2} - 1296 d^{2} + 9 d^{4} + 16 d^{4} = 361$

$25 d^{4} - 1440 d^{2} + 20375 = 0$

Notice $25 d^{4} - 1440 d^{2} + 20736 = (5 d^{2} - 144)^{2}$ , so

$(5 d^{2} - 144)^{2} = 361 = 1 9^{2} \Rightarrow 5 d^{2} - 144 = \pm 19$ .

$5 d^{2} = 163 \Rightarrow d^{2} = 163/5$ (rejected, gives non-integer $l$ ), or $5 d^{2} = 125 \Rightarrow d^{2} = 25 \Rightarrow d = 5$ ( $l = 10$ , integer).

Largest term $= a + 3 d = 12 + 15 = 27$ .

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