If C(n, 4), C(n, 5) and C(n, 6) are in AP, then what is the value...

Given that $-{ }^n C_4,{ }^n C_5 and { }^n C_6$ are in A.P $$\begin{aligned} & \therefore 2^n C_5={ }^n C_4+{ }^n C_6 \\ & \Rightarrow 2 \cdot \frac{n!}{(n-5)!5!}=\frac{n!}{(n-4)!4!}+\frac{n!}{(n-6)!6!} \end{aligned}$$\begin{gathered}\Rightarrow \frac{2 n!}{5(n-5)(n-6)!4!}=\frac{n!}{(n-4)(n-5)(n-6)!4!}+\frac{n!}{(n-6)!6.5!4!} \\ \Rightarrow \frac{2 n!}{5(n-5)(n-6)!4!}=\frac{n!}{(n-6)!4!} \\ {\left[\frac{1}{(n-4)(n-5)}+\frac{1}{6.5}\right]}\end{gathered}$$\begin{aligned} & \Rightarrow \frac{2}{(n-5)(n-6)}=\frac{n^2-9 n+50}{6(n-4)(n-5)} \\ & \Rightarrow 12(n-4)=(n-5)\left(n^2-9 n+50\right) \\ & \Rightarrow n^3-15 n^2+92 n-252=0 \\ & \Rightarrow n^3-7 n^2-8 n^2+56 n+36 n-252=0 \\ & \Rightarrow n^2(n-7)-8 n(n-7)+36(n-7)=0 \\ & \Rightarrow(n-7)\left(n^2-8 n+36\right)=0 \\ & \Rightarrow n=7\left[\therefore \text{ roots\text{ of } } n^2-8 n+36 \text{ is not real }\right]\end{aligned}$$$