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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Sequences and Series

Let a_ 1 , a_ 2 , a_ 3 , a_ 4 be an A.P. of four terms such that...

Let a1,a2,a3,a4a_{1}, a_{2}, a_{3}, a_{4} be an A.P. of four terms such that each term of the A.P. and its common difference ll are integers. If a1+a2+a3+a4=48a_{1}+a_{2}+a_{3}+a_{4}=48 and a1a2a3a4+l4=361a_{1} a_{2} a_{3} a_{4}+l^{4}=361, then the largest term of the A.P. is equal to

Options

  1. A23
  2. B21
  3. C27
  4. D24

Correct answer

C. 27

Step-by-step solution

Let the four terms be 12−3k, 12−k, 12+k, 12+3k12 - 3k,\ 12 - k,\ 12 + k,\ 12 + 3k with common difference l=2kl = 2k.
Sum =48= 48 is satisfied. Product condition:
(144−9k2)(144−k2)+16k4=361(144 - 9k^2)(144 - k^2) + 16k^4 = 361
25k4−1440k2+20375=025k^4 - 1440k^2 + 20375 = 0
Let u=k2u = k^2: 25u2−1440u+20375=025u^2 - 1440u + 20375 = 0
u=1440±2073600−203750050=1440±19050u = \frac{1440 \pm \sqrt{2073600 - 2037500}}{50} = \frac{1440 \pm 190}{50}
u=25u = 25 (integer) or u=32.6u = 32.6 (rejected).
k=5k = 5, l=10l = 10. Terms: −3,7,17,27-3, 7, 17, 27.
Largest term =27= 27.
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