Question

If the coefficients of the $r^{th},{(r+1)}^{th}\&{(r+2)}^{th}$ terms in expansion of ${(1+x)}^{14}$ are in AP, find $r$.

Answer 1

As we know that, the general term in the expansion ${(a+b)}^n$ is given as-

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

Therefore,

In the expansion ${(1+x)}^{14}$,

$T_r={}^{14}{C}_{r-1}{1}^{14-(r-1)}{x}^{r-1}$

Coefficient of $T_r={}^{14}{C}_{r-1}$

$T_{r+1}={}^{14}{C}_r{1}^{14-r}{x}^r$

Coefficient of $T_{r+1}={}^{14}{C}_r$

$T_{r+2}={}^{14}{C}_{r+1}{1}^{14-(r+1)}{x}^{r+1}$

Coefficient of $T_{r+2}={}^{14}{C}_{r+1}$

Given that $T_r,T_{r+1}$ and $T_{r+2}$ are in A.P.

Therefore,

$2T_{r+1}=T_r+T_{r+2}$

$2{}^{14}{C}_r={}^{14}{C}_{r-1}+{}^{14}{C}_{r+1}$

$2\times \frac{14!}{r!(14-r)!}=\frac{14!}{(r-1)!(14-(r-1))!}+\frac{14!}{(r+1)!(14-(r+1))!}$

$\Rightarrow \frac{2}{r!(14-r)!}=\frac{1}{(r-1)!(15-r)!}+\frac{1}{(r+1)!(13-r)!}$

$\Rightarrow \frac{2}{r(14-r)}=\frac{1}{(15-r)(14-r)}+\frac{1}{(r+1)r}$

$\Rightarrow \frac{2}{r(14-r)}=\frac{(r^2+r)+(r^2-29r+210)}{(15-r)(14-r)(r+1)r}$

$\Rightarrow 2(15+14r-r^2)=2r^2-28r+210$

$\Rightarrow 4r^2-56r+180=0$

$\Rightarrow r^2-14r+45=0$

$\Rightarrow (r-9)(r-5)=0$

$\Rightarrow r=9\text{or}r=5$

Hence the value of $r$ can be $5$ or $9$.