2016/085)Let n be a positive integer. if $2^{nd}$ , $3^{rd}$ and $4^{th}$ term of expansion of $(1+x)^n$ are in AP find n

we are given
${n \choose 1} +  {n \choose 3}=  2{n \choose 2}$
or
$n + \frac{n(n-1)(n-2)}{6} = 2\frac{n(n-1)}{2}$
dividing both sides by $n$ we get
$1  + \frac{(n-1)(n-2)}{6} = (n-1)$
or $(n-2)(1-\frac{n-1}{6} = 0$ giving n= 2 or 7
but $n > =3$ so n = 4