If the coefficients of 2^nd, 3^rd , 4^th terms of ${(1+x)}^n$ are in A.P then $n$ is

$\begin{array}{l}{T}_2={}^n{\text{C}}_1\\ T_3={}^n{\text{C}}_2\\ T_4={}^n{\text{C}}_3\\ 2T_3=T_2+T_4\\ 2.\frac{n(n-1)}{2}=n+\frac{n(n-1)(n-2)}{6}\end{array}$

$\begin{array}{l}n(n-1)=\frac{n(6+n^2-3n+2)}{6}\\ 6(n-1)=6+n^2+3n+2\\ 6n-6=n^2-3n+8\\ n^2-9n+14=0\\ n^2-7n-2n+14=0\\ (n-7)(n-2)=0\\ n=2\left(\text{or}\right)7\end{array}$

But $n=2$ is not satisfied

$\therefore n=7$