QJEEM25Ma0763: JEE Main 2025 Mathematics - Class 11 JEEM25Ma0763 Solution: For some , let the coefficients of the 5 th, 6 th and 7 th…

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

$\therefore$ Largest binomial coefficient in expansion of

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

is coeff. of middle term

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

N.T.A. Ans Option (2)