(A) 27

(B) 24

(C) 23

(D) 21

Correct Answer: 27

Explanation

Let the four terms of the A.P. be

$$ a_1 = a - 3l,\quad a_2 = a - l,\quad a_3 = a + l,\quad a_4 = a + 3l $$

Sum of the four terms:

$$ (a - 3l) + (a - l) + (a + l) + (a + 3l) = 4a $$

Given,

$$ 4a = 48 \Rightarrow a = 12 $$

So the terms are:

$$ 12 - 3l,\; 12 - l,\; 12 + l,\; 12 + 3l $$

Now compute the product:

$$ a_1a_2a_3a_4 = (12 - 3l)(12 - l)(12 + l)(12 + 3l) $$

Group the terms:

$$ = [(12)^2 - (3l)^2][(12)^2 - l^2] $$ $$ = (144 - 9l^2)(144 - l^2) $$

Expand:

$$ = 144^2 - 144l^2 - 1296l^2 + 9l^4 $$ $$ = 20736 - 1440l^2 + 9l^4 $$

Given condition:

$$ a_1a_2a_3a_4 + l^4 = 361 $$

Substitute:

$$ 20736 - 1440l^2 + 9l^4 + l^4 = 361 $$ $$ 10l^4 - 1440l^2 + 20375 = 0 $$

Try integer values of \(l\). For \(l = 5\):

$$ 10(625) - 1440(25) + 20375 $$ $$ = 6250 - 36000 + 20375 = 361 $$

Hence, \(l = 5\).

Largest term of the A.P.:

$$ a_4 = 12 + 3l = 12 + 15 = 27 $$

Therefore, the correct answer is 27.

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