If n C 4 ,   n C 5 and n C 6 are in A.P., then n can be

Since, ${}^n{C}_4,{ }^n{C}_5,{ }^n{C}_6$ are in A.P.

$\Rightarrow 2{\times }^n{C}_5{=}^n{C}_4{+}^n{C}_6$

$\Rightarrow 2\times \frac{n!}{5!\left(n-5\right)!}=\frac{n!}{4!\left(n-4\right)!}+\frac{n!}{6!\left(n-6\right)!}$

$\Rightarrow \frac{2}{5!\left(n-5\right)!}=\frac{30+\left(n-5\right)\left(n-4\right)}{6!\left(n-4\right)!}$

$\Rightarrow 6\times 2\left(n-4\right)=30+n^2-9n+20$

$\Rightarrow n^2-21n+98=0$

$\Rightarrow \left(n-7\right)\left(n-14\right)=0$

$\Rightarrow n=7$ or $14$