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

Solution:

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

$a=1, b=x$ and $N=2 n$

We have, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{N} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{N}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

For the $2^{\text {nd }}$ term, $r=1$

$\therefore \mathrm{t}_{2}=\mathrm{t}_{1+1}$

$=\left(\begin{array}{c}2 \mathrm{n} \\ 1\end{array}\right)(1)^{2 \mathrm{n}-1}(\mathrm{x})^{1}$

$=(2 \mathrm{n}) \times \ldots\left[\because\left(\begin{array}{l}\mathrm{n} \\ 1\end{array}\right)=\mathrm{n}\right]$

Therefore, the coefficient of $2^{\text {nd }}$ term $=(2 n)$

For the $3^{\text {rd }}$ term, $r=2$

$\therefore t_{3}=t_{2+1}$

$=\left(\begin{array}{c}2 n \\ 2\end{array}\right)(1)^{2 n-2}(x)^{2}$

$=\frac{(2 \mathrm{n}) !}{(2 \mathrm{n}-2) ! \times 2 !} \mathrm{x}^{2}$

$=\frac{(2 \mathrm{n})(2 \mathrm{n}-1)(2 \mathrm{n}-2) !}{(2 \mathrm{n}-2) ! \times 2} \mathrm{x}^{2} \ldots \ldots \ldots(\mathrm{n} !=\mathrm{n} .(\mathrm{n}-1) !)$

$=(n)(2 n-1) x^{2}$

Therefore, the coefficient of $3^{\text {rd }}$ term $=(n)(2 n-1)$

For the $4^{\text {th }}$ term, $r=3$

$\therefore t_{4}=t_{3+1}$

$=\left(\begin{array}{c}2 \mathrm{n} \\ 3\end{array}\right)(1)^{2 \mathrm{n}-3}(\mathrm{x})^{3}$

$=\frac{(2 n) !}{(2 n-3) ! \times 3 !} \mathrm{x}^{3}$

$=\frac{(2 \mathrm{n})(2 \mathrm{n}-1)(2 \mathrm{n}-2)(2 \mathrm{n}-3) !}{(2 \mathrm{n}-3) ! \times 6} \mathrm{X}^{3}$ ..........$(n !=n \cdot(n-1) !)$

$=\frac{(n)(2 n-1) \cdot 2(n-1)}{3} \mathrm{x}^{3}$

$=\frac{2(n)(2 n-1) \cdot(n-1)}{3} \mathrm{X}^{3}$

Therefore, the coefficient of $3^{\text {rd }}$ term $=\frac{2(\mathrm{n})(2 \mathrm{n}-1) \cdot(\mathrm{n}-1)}{3}$

As the coefficients of $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms are in A.P.

Therefore,

$2 \times$ coefficient of $3^{\text {rd }}$ term $=$ coefficient of $2^{\text {nd }}$ term $+$ coefficient of the $4^{\text {th }}$ term

$\therefore 2 \times(\mathrm{n})(2 \mathrm{n}-1)=(2 \mathrm{n})+\frac{2(\mathrm{n})(2 \mathrm{n}-1) \cdot(\mathrm{n}-1)}{3}$

Dividing throughout by (2n),

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

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

- $3(2 n-1)=3+(2 n-1)(n-1)$

- $6 n-3=3+\left(2 n^{2}-2 n-n+1\right)$

- $6 n-3=3+2 n^{2}-3 n+1$

- $3+2 n^{2}-3 n+1-6 n+3=0$

- $2 n^{2}-9 n+7=0$

$\frac{\text { Conclusion }}{9 n+7=0}$ : If the coefficients of $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms of $(1+x)^{2 n}$ are in A.P. then $2 n^{2}-$