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Let a₁, a₂, a₃, a₄ be an A.P. of four terms such that each term of the A.P. and its common difference l are integers. If a₁+a₂+a₃+a₄=48 and a₁ a₂ a₃ a₄+l⁴=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 + 3k with common difference l = 2k . Sum = 48 is satisfied. Product condition: (144 - 9k^2)(144 - k^2) + 16k^4 = 361 25k^4 - 1440k^2 + 20375 = 0 Let u = k^2 : 25u^2 - 1440u + 20375 = 0 u = 1440 2073600 - 2037500 50 = 1440 190 50 u = 25 (integer) or u = 32.6 (rejected). k = 5 , l = 10 . Terms: -3, 7, 17, 27 . Largest term = 27 .

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