For some n ≠ 10, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of (1+ x)^ n+4 be in A.P. Then the largest…

$\begin{aligned} & (1+x)^{n+4} \\ & { }^{n+4} C_4,{ }^{n+4} C_5,{ }^{n+4} C_6, \rightarrow \text { A.P. } \\ & \Rightarrow 2 \times { }^{n+4} C_5={ }^{n+4} C_4+{ }^{n+4} C_6 \\ & \Rightarrow 4 \times{ }^{n+4} C_5=\left({ }^{n+4} C_4+{ }^{n+4} C_5\right)+\left({ }^{n+4} C_5+{ }^{n+4} C_6\right) \\ & \Rightarrow 4 \times{ }^{n+4} C_5={ }^{n+5} C_5+{ }^{n+5} C_6\end{aligned}$​(1+x)n+4n+4C4​,n+4C5​,n+4C6​,→ A.P. ⇒2×n+4C5​=n+4C4​+n+4C6​⇒4×n+4C5​=(n+4C4​+n+4C5​)+(n+4C5​+n+4C6​)⇒4×n+4C5​=n+5C5​+n+5C6​​ $\begin{aligned} & \Rightarrow 4 \times \frac{(\mathrm{n}+4)!}{5!\cdot(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\cdot \mathrm{n}!} \\ & \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6 \mathrm{n}} \\ & \Rightarrow \mathrm{n}^2+11 \mathrm{n}+30=24 \mathrm{n} \\ & \Rightarrow \mathrm{n}^2-13 \mathrm{n}+30=0 \\ & \Rightarrow \mathrm{n}=3,10(\text { rejected }) \\ & \because \mathrm{n} \neq 10\end{aligned}$​⇒4×5!⋅(n−1)!(n+4)!​=6!⋅n!(n+6)!​⇒4=6n(n+6)(n+5)​⇒n2+11n+30=24n⇒n2−13n+30=0⇒n=3,10( rejected )∵n=10​ $\therefore$∴ Largest binomial coefficient in expansion of $\begin{aligned}
& (1+\mathrm{x})^7 \
& (\because \mathrm{n}+4=7)
\end{aligned}$is coeff. of middle term$iscoeff.ofmiddleterm\Rightarrow{ }^7 \mathrm{C}_4={ }^7 \mathrm{C}_3=35$