If the coefficient of second,

Given $(1+x)^{2 n}$

Now, coefficient of $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms are ${ }^{2 n} C_{1},{ }^{2 n} C_{2}$ and ${ }^{2 n} C_{3}$, respectively.

Given that, ${ }^{2 n} C_{1},{ }^{2 n} C_{2}$ and ${ }^{2 n} C_{3}$ are in A.P.

Then,

$2 \cdot{ }^{2 n} C_{2}={ }^{2 n} C_{1}+{ }^{2 n} C_{3}$

$2\left[\frac{2 n(2 n-1)(2 n-2) !}{2 \times 1 \times(2 n-2) !}\right]=2 n+\frac{2 n(2 n-1)(2 n-2)(2 n-3) !}{3 !(2 n-3) !}$

$n(2 n-1)=n+\frac{n(2 n-1)(n-1)}{3}$

$3(2 n-1)=3+\left(2 n^{2}-3 n+1\right)$

$6 n-3=2 n^{2}-3 n+4 \Rightarrow 2 n^{2}-9 n+7=0$