JEE Main 2025 — Binomial Theorem Question with Solution

$$\begin{array}{rl}& (1+\mathrm{x}{)}^{\mathrm{n}+4}\\ & {}^{\mathrm{n}+4}{\mathrm{C}}_4{}^{\mathrm{n}+4}{\mathrm{C}}_5{}^{\mathrm{n}+4}{\mathrm{C}}_6,\to \text{ A.P. }\\ & \Rightarrow 2\times {}^{\mathrm{n}+4}{\mathrm{C}}_5={}^{\mathrm{n}+4}{\mathrm{C}}_4+{}^{\mathrm{n}+4}{\mathrm{C}}_6\\ & \Rightarrow 4\times {}^{\mathrm{n}+4}{\mathrm{C}}_5=\left({}^{n+4}{\mathrm{C}}_4+{}^{\mathrm{n}+4}{\mathrm{C}}_5\right)+\left({}^{\mathrm{n}+4}{\mathrm{C}}_5+{}^{\mathrm{n}+4}{\mathrm{C}}_6\right)\\ & \Rightarrow 4\times {}^{n+4}{\mathrm{C}}_5={}^{\mathrm{n}+5}{\mathrm{C}}_5+{}^{\mathrm{n}+5}{\mathrm{C}}_6\\ & \Rightarrow 4\times \frac{(\mathrm{n}+4)!}{5!.(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\mathrm{n}!}\\ & \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6\mathrm{n}}\\ & \Rightarrow {\mathrm{n}}^2+11\mathrm{n}+30=24\mathrm{n}\\ & \Rightarrow {\mathrm{n}}^2-13\mathrm{n}+30=0\\ & \Rightarrow \mathrm{n}=3,10(\text{ rejected })\\ & \because \mathrm{n}\ne 10\end{array}$$

$\therefore$ Largest binomial coefficient in expansion of

$$\begin{array}{rl}& (1+x{)}^7\\ & (\because \mathrm{n}+4=7)\end{array}$$

is coeff. of middle term

$$\Rightarrow {}^7{\mathrm{C}}_4={}^7{\mathrm{C}}_3=35$$

N.T.A. Ans Option (2)