Question

If ${}^n{\text{C}}_4,{}^n{\text{C}}_5$ and ${}^n{\text{C}}_6$ are in A.P., then $n$ can be :

• A. $9$
• B. $11$
• C. $12$
• D. $14$

Answer 1

The correct option is D $14$

${}^n{\text{C}}_4,{}^n{\text{C}}_5,{}^n{\text{C}}_6$ are in A.P.

$\therefore 2\times {}^n{\text{C}}_5={}^n{\text{C}}_4+{}^n{\text{C}}_6$
$\Rightarrow 2\times \frac{n!}{5!\times (n-5)!}=\frac{n!}{4!\times (n-4)!}+\frac{n!}{6!\times (n-6)!}$

$\Rightarrow \frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{30}$

From options, $n=14$ satisfies the above equation.