Given :
$nC_1,nC_2,nC_3$ are in A.P
$\Rightarrow 2nC_2=nC_1+nC_3$
$\Rightarrow \large\frac{2n(n-1)}{1.2}=\frac{n}{1}+\frac{n(n-1)(n-2)}{1.2.3}$
$\Rightarrow 6n(n-1)=6n+n(n-1)(n-2)$
$\Rightarrow n^3-9n^2+14n=0$
$\Rightarrow n(n^2-9n+14)=0$
$\Rightarrow n=0$
$n^2-9n+14=0$
$n^2-7n-2n+14=0$
$n(n-7)-2(n-7)=0$
$n-2=0$ or $n-7=0$
$n=2,7$
$\therefore n=7$(There are atleast 4 terms in $(1+x)^n$)
Hence (B) is the correct answer.