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

<p>$$\begin{aligned} & (1+\mathrm{x})^{\mathrm{n+4}} \\ & { }^{\mathrm{n}+4} \mathrm{C}_4{ }^{\mathrm{n+4}} \mathrm{C}_5{ }^{\mathrm{n}+4} \mathrm{C}_6, \rightarrow \text { A.P. } \\ & \Rightarrow 2 \times{ }^{\mathrm{n}+4} \mathrm{C}_5={ }^{\mathrm{n+4}} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_6 \\ & \Rightarrow 4 \times{ }^{\mathrm{n+4}} \mathrm{C}_5=\left({ }^{n+4} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_5\right)+\left({ }^{\mathrm{n+4}} \mathrm{C}_5+{ }^{\mathrm{n}+4} \mathrm{C}_6\right) \\ & \Rightarrow 4 \times{ }^{n+4} \mathrm{C}_5={ }^{\mathrm{n}+5} \mathrm{C}_5+{ }^{\mathrm{n}+5} \mathrm{C}_6 \\ & \Rightarrow 4 \times \frac{(\mathrm{n}+4)!}{5!.(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\mathrm{n}!} \\ & \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6 \mathrm{n}} \\ & \Rightarrow \mathrm{n}^2+11 \mathrm{n}+30=24 \mathrm{n} \\ & \Rightarrow \mathrm{n}^2-13 \mathrm{n}+30=0 \\ & \Rightarrow \mathrm{n}=3,10(\text { rejected }) \\ & \because \mathrm{n} \neq 10 \end{aligned}$$</p> <p>$\therefore$ Largest binomial coefficient in expansion of</p> <p>$$\begin{aligned} & (1+x)^7 \\ & (\because \mathrm{n}+4=7) \end{aligned}$$</p> <p>is coeff. of middle term</p> <p>$\Rightarrow{ }^7 \mathrm{C}_4={ }^7 \mathrm{C}_3=35$</p> <p>N.T.A. Ans Option (2)</p>