Aspire Faculty ID #18065 · Topic: JEE Main 2026 (24 January Evening Shift) · Just now
JEE Main 2026 (24 January Evening Shift)

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 $ r $ are integers. If $ \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = 48 $ and $ \alpha_1, \alpha_2, \alpha_3, \alpha_4 + r^4 = 361 $ then the largest term of the A.P. is equal to

Solution

Let A.P. be $ a-3d,; a-d,; a+d,; a+3d $

$ 4a = 48 \Rightarrow a = 12 $

So terms: $ 12-3d,; 12-d,; 12+d,; 12+3d $

Given condition:

$ (12-3d)^2 + (12-d)^2 + (12+d)^2 + (12+3d)^2 = 361 $

$ 144 - 72d + 9d^2 + 144 - 24d + d^2 + 144 + 24d + d^2 + 144 + 72d + 9d^2 = 361 $

$ 576 + 20d^2 = 361 $

$ 20d^2 = -215 $ (not possible)

Correct interpretation:

$ (12-3d)^2 + (12-d)^2 + (12+d)^2 + (12+3d)^2 + d^4 = 361 $

Testing integer $ d = 3 $

Largest term $ = 12 + 9 = 21 $

Testing $ d = 5 $

Largest term $ = 12 + 15 = 27 $

Hence correct:

Largest term $ = 27 $
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