If the coefficients of 2nd, 3rd and the 4th terms in the expansion of

Given coefficient of 2nd, 3rd and 4th terms in the expansion (1 + x)n are A.P

Since coefficient of (r + 1)th terms is nCr

∴ 2nd, 3rd and 4th coefficient are such that 2nCr = nC1 + nC3

i. e. $\quad 2 \times \frac{n(n-1)}{2}=n+\frac{n(n-1)(n-2)}{6}$

i. e. $n(n-1)=n\left[1+\frac{(n-1)(n-2)}{6}\right]$

i.e. $\quad n-1=1+\frac{(n-1)(n-2)}{6}$

i.e. $6(n-1)=6+n^{2}-3 n+2$

i.e. $\quad n^{2}-9 n+14=0$

i.e. $n^{2}-7 n-2 n+14=0$

i.e. $n(n-7)-2(n-7)=0$

i.e. $n=7 \quad(\because n=2$ is not possible since for $n=2,3$ rd and 4 th terms will be 0$)$

Hence, the correct answer is option B.