Question

If ${{}^{n}C}_4,{{}^{n}C}_5$ and ${{}^{n}C}_6$ are in AP, then $n$ is

• A. $7$ or $14$
• B. $7$
• C. $14$
• D. $14$ or $21$

Answer 1

The correct option is A $7$ or $14$

Since, ${{}^{n}C}_4,{{}^{n}C}_5$ and ${{}^{n}C}_6$ are in AP
$\therefore 2{{}^{n}C}_5={{}^{n}C}_4+{{}^{n}C}_6$
$\Rightarrow 2\times \frac{n!}{5!(n-5)!}=\frac{n!}{4!(n-4)!}+\frac{n!}{6!(n-6)!}$
$\Rightarrow \frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{30}$
$\Rightarrow \frac{2}{5(n-5)}=\frac{30+n^2-9n+50}{30(n-4)(n-5)}$
$\Rightarrow 12(n-4)=n^2-9n+50$
$\Rightarrow n^2-21n+98=0$
$\Rightarrow (n-14)(n-7)=0$
$\Rightarrow n=7,14$