Question

If the coefficients of $r^{th},(r+1{)}^{th}$ and $(r+2{)}^{th}$ terms in the binomial expansion of $(1+y{)}^m$are in $A.P.$, then $m$ and $r$ will satisfy the equation

• A. $m^2-m(4r+1)+4r^2+2=0$
• B. $m^2-m(4r-1)+4r^2-2=0$
• C. $m^2-m(4r-1)+4r^2+2=0$
• D. $m^2-m(4r+1)+4r^2-2=0$

Answer 1

The correct option is D $m^2-m(4r+1)+4r^2-2=0$

Given the coefficients of $r^{th},(r+1{)}^{th}$ and $(r+2{)}^{th}$ terms in the binomial expression of $(1+y{)}^m$ are in A.P.
Since, $(k+1{)}^{th}$ term of $(1+y{)}^m$ is ${}^m{C}_r{y}^r$
Thus, the coefficient of $(k+1{)}^{th}$ term of $(1+y{)}^m$ is ${}^m{C}_r$
So, the coefficients of $r^{th},(r+1{)}^{th}$ and $(r+2{)}^{th}$ are ${}^m{C}_{r-1}{,}^m{C}_r$ and ${}^m{C}_{r+1}$ respectively. And they are in A.P
$\Rightarrow 2{}^m{C}_r={}^m{C}_{r-1}+{}^m{C}_{r+1}$ [Since, if $a_1,a_2$ and $a_3$ are in A.P., then $2a_2=a_1+a_3$]

$\Rightarrow 2=\frac{{}^m{C}_{r-1}}{{}^m{C}_r}+\frac{{}^m{C}_{r+1}}{{}^m{C}_r}$

$\Rightarrow 2=\frac{r}{m-r+1}+\frac{m-r}{r+1}$ [$\because \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}$]

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

$\Rightarrow 2=\frac{r^2+r+m^2-mr+m-mr+r^2-r}{mr+m-r^2-r+r+1}$

$\Rightarrow 2=\frac{2r^2+m^2-2mr+m}{mr+m-r^2+1}$

$\Rightarrow 2(mr+m-r^2+1)=2r^2+m^2-2mr+m$

$\Rightarrow 2mr+2m-2r^2+2=2r^2+m^2-2mr+m$

$\Rightarrow 2mr+2m-2r^2+2-2r^2-m^2+2mr-m=0$

$\Rightarrow 4mr+2m-4r^2+2-m^2-m=0$

$\Rightarrow -4mr-m+4r^2-2+m^2=0$

$\therefore m^2-m(4r+1)+4r^2-2=0$