Question:
If the coefficient of second, third and fourth terms in the expansion of $(1+x)^{2 n}$ are in A.P. Show that $2 n^{2}-9 n+7=0$.
Solution:
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,
$2 \cdot{ }^{2 n} C_{2}={ }^{2 n} C_{1}+{ }^{2 n} C_{3}$
$2\left[\frac{2 n(2 n-1)(2 n-2) !}{2 \times 1 \times(2 n-2) !}\right]=2 n+\frac{2 n(2 n-1)(2 n-2)(2 n-3) !}{3 !(2 n-3) !}$
$n(2 n-1)=n+\frac{n(2 n-1)(n-1)}{3}$
$3(2 n-1)=3+\left(2 n^{2}-3 n+1\right)$
$6 n-3=2 n^{2}-3 n+4 \Rightarrow 2 n^{2}-9 n+7=0$