Question

If the coefficients of $r^{th}$, $(r+1{)}^{th}$ and $(r+2{)}^{th}$ terms in the expansion of $(1+x{)}^{14}$ are in A.P. then $r=$

• A. $5$
• B. $9$
• C. $7$
• D. $5or9$

Answer 1

The correct option is D $5or9$

The coefficients of $r^{th},(r+1{)}^{th},(r+2{)}^{th}$ terms are in A.P.
Therefore, we have ${}^n{C}_{r-1}+{}^n{C}_{r+1}=2{}^n{C}_r$
Then,
$\frac{n!}{(r-1)!(n-r+1)!}+\frac{n!}{(r+1)!(n-r-1)!}=2\times \frac{n!}{r!(n-r)!}$
$\frac{1}{(r-1)!(n-r+1)(n-r)(n-r-1)!}+\frac{1}{(r+1)\left(r\right)(r-1)!(n-r-1)!}=2\times \frac{1}{r(r-1)!(n-r)(n-r-1)!}$

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

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

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

Cross multiplying, we get
$r(r+1)+(n-r)(n-r+1)=2(r+1)(n-r+1)$
$r^2+r+n^2-nr+n-nr+r^2-r=2(nr-r^2+r+n-r+1)$
$n^2-4nr-n+4r^2-2=0$
$n^2-n(4r+1)+4r^2-2=0$
We know that $n=14$
Substituting the value of $n,$ we get
${14}^2-14(4r+1)+4r^2-2=0$
$196-56r-14+4r^2-2=0$
$4r^2-56r+180=0$
$4(r^2-14r+45)=0$
$r^2-9r-5r+45=0$
$r(r-9)-5(r-9)=0$
$(r-9)(r-5)=0$
Hence, $r=9$ or $5$.