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JEE Mains · Maths · STD 11 - 8. sequence and series

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

  1. A 27
  2. B 24
  3. C 21
  4. D 23
Verified Solution

Answer & Solution

Correct Answer

(A) 27

Step-by-step Solution

Detailed explanation

\(a_1, a_2, a_3, a_4\) as \(a-3 d, a-d, a+d, a+3 d\) where \(d =\frac{\ell}{2}\) \(\because a_1+a_2+a_3+a_4=48 \Rightarrow 4 a=48 \Rightarrow a=12\) & \(a_1 a_2 a_3 a_4+\ell^4=361 \Rightarrow\left(a^2-9 d^2\right)\left(a^2-d^2\right)+16 d^4\) = 361…
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