Q37 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
Let $a_{1}, a_{2}, a_{3}, a_{4}$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $a_{1}+a_{2}+a_{3}+a_{4}=48$ and $a_{1} a_{2} a_{3} a_{4}+l^{4}=361$, then the largest term of the A.P. is equal to
- A23
- B21
- Correct answerC27
- D24
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 $u = k^2$: $25u^2 - 1440u + 20375 = 0$
$u = \frac{1440 \pm \sqrt{2073600 - 2037500}}{50} = \frac{1440 \pm 190}{50}$
$u = 25$ (integer) or $u = 32.6$ (rejected).
$k = 5$, $l = 10$. Terms: $-3, 7, 17, 27$.
Largest term $= 27$.
Sum $= 48$ is satisfied. Product condition:
$(144 - 9k^2)(144 - k^2) + 16k^4 = 361$
$25k^4 - 1440k^2 + 20375 = 0$
Let $u = k^2$: $25u^2 - 1440u + 20375 = 0$
$u = \frac{1440 \pm \sqrt{2073600 - 2037500}}{50} = \frac{1440 \pm 190}{50}$
$u = 25$ (integer) or $u = 32.6$ (rejected).
$k = 5$, $l = 10$. Terms: $-3, 7, 17, 27$.
Largest term $= 27$.
Hence, the correct answer is option: 'C'.