Question

If ${}^n{C}_4,{}^n{C}_5$ and ${}^n{C}_6$ are in $A.P$., then find $n$.

Answer 1

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

${\Rightarrow }^n{C}_6{-}^n{C}_5{=}^n{C}_5{-}^n{C}_4$

$\Rightarrow \frac{n!}{6!(n-6)!}-\frac{n!}{5!(n-5)!}=\frac{n!}{5!(n-5)!}-\frac{1}{4!(n-4)!}$

$\Rightarrow \frac{1}{6.5!(n-6)!}-\frac{1}{5!(n-5)(n-6)!}=\frac{1}{5!(n-5)(n-6)!}-\frac{1}{4!(n-4)(n-5)(n-6)}$

$\Rightarrow \frac{1}{\mathrm{6.5.4}!}-\frac{1}{5.4!(n-5)}=\frac{1}{5.4!(n-5)}-\frac{1}{4!(n-4)(n-5)}$

$\Rightarrow \frac{1}{6.5}-\frac{1}{5(n-5)}=\frac{1}{5(n-5)}-\frac{1}{(n-4)(n-5)}$

Solving we get

$n^2-21n+98=0$

or $(n-7)(n-14)=0$

$n=7,14$