If ${ }^n{C}_4{,}^n{\mathrm{C}}_5$ and ${ }^n{C}_6$ are in A.P., then a value of $n$can be 

As ${ }^n{C}_4{,}^n{C}_5{\text{ and }}^n{C}_6$ are in A.P.

$2\left({ }^n{C}_5\right){=}^n{C}_4{+}^n{C}_6$

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

$=\frac{n!}{4!(n-4)!}\frac{5!(n-5)!}{n!}+\frac{n!}{6!(n-6)!}\frac{5!(n-5)!}{n!}$

$\begin{array}{l}\Rightarrow 2=\frac{5}{n-4}+\frac{n-5}{6}\Rightarrow n^2-21n+98=0\\ \Rightarrow n=7,14.\end{array}$