JEE Main 2025 Jan 24 Shift 1, Mathematics Q16: Properties Of Binomial Coefficient

JEE Main2025Jan 24, Shift 1Mathematics

For some $n \neq = 10$ , let the coefficients of the $5^{t h}$ , $6^{t h}$ and $7^{t h}$ terms in the binomial expansion of $(1 + x)^{n + 4}$ be in A.P. Then the largest coefficient in the expansion of $(1 + x)^{n + 4}$ is :

A
$70$
B
$35$
C
$20$
D
$10$

Solution

The $5^{t h}$ , $6^{t h}$ , $7^{t h}$ term coefficients of $(1 + x)^{n + 4}$ are $(4 n + 4), (5 n + 4), (6 n + 4)$ . Being in A.P.:

$2 (5 n + 4) = (4 n + 4) + (6 n + 4)$

Use the ratio identity $\frac{( k + 1 N )}{( k N )} = \frac{N - k}{k + 1}$ with $N = n + 4$ to get a clean polynomial in $n$ . Equivalently, divide everything by $(5 n + 4)$ :

$2 = \frac{5}{n} \cdot \frac{1}{1} + \frac{n - 1}{6}$ (using consecutive ratios $(4 N) / (5 N) = \frac{5}{N - 4} = \frac{5}{n}$ , $(6 N) / (5 N) = \frac{N - 5}{6} = \frac{n - 1}{6}$ )

Multiply by $6 n$ : $12 n = 30 + n (n - 1) \Rightarrow n^{2} - 13 n + 30 = 0 \Rightarrow n = 3$ or $n = 10$ .

Given $n \neq = 10$ , take $n = 3$ , so the exponent is $n + 4 = 7$ .

The largest binomial coefficient in $(1 + x)^{7}$ is the central one: $(3 7) = (4 7) = 35$ .

$∴$ Option (2).

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JEE Main 2025 Jan 24 Shift 1, Mathematics Q16: Properties Of Binomial Coefficient

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