Mathematics - Binomial Theorem Question with Solution | TestHub

$\begin{array}{rl}& (1+x{)}^{n+4}\\ & {}^{n+4}{C}_4,{}^{n+4}{C}_5,{}^{n+4}{C}_6,\to \text{ A.P. }\\ & \Rightarrow 2\times {}^{n+4}{C}_5={}^{n+4}{C}_4+{}^{n+4}{C}_6\\ & \Rightarrow 4\times {}^{n+4}{C}_5=\left({}^{n+4}{C}_4+{}^{n+4}{C}_5\right)+\left({}^{n+4}{C}_5+{}^{n+4}{C}_6\right)\\ & \Rightarrow 4\times {}^{n+4}{C}_5={}^{n+5}{C}_5+{}^{n+5}{C}_6\end{array}$
$\begin{array}{rl}& \Rightarrow 4\times \frac{(\mathrm{n}+4)!}{5!\cdot (\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\cdot \mathrm{n}!}\\ & \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6\mathrm{n}}\\ & \Rightarrow {\mathrm{n}}^2+11\mathrm{n}+30=24\mathrm{n}\\ & \Rightarrow {\mathrm{n}}^2-13\mathrm{n}+30=0\\ & \Rightarrow \mathrm{n}=3,10(\text{ rejected })\\ & \because \mathrm{n}\ne 10\end{array}$
$\therefore$ Largest binomial coefficient in expansion of
$\begin{array}{rl}& (1+\mathrm{x}{)}^7\\ & (\because \mathrm{n}+4=7)\end{array}$
is coeff. of middle term
$\Rightarrow {}^7{\mathrm{C}}_4={}^7{\mathrm{C}}_3=35$