Solve the following

Solution:

Since nC4 , nC5 and nC6 are in AP.

$\therefore$ 2. ${ }^{n} C_{5}={ }^{n} C_{4}+{ }^{n} C_{6}$

$\Rightarrow 2 \times \frac{n !}{5 !(n-5) !}=\frac{n !}{4 !(n-4) !}+\frac{n !}{6 !(n-6) !}$

$\Rightarrow \frac{2}{5 \times 4 !(n-5)(n-6) !}=\frac{1}{4 !(n-5)(n-4)(n-6) !}+\frac{1}{6 \times 5 \times 4 !(n-6) !}$

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

 

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

$\Rightarrow \frac{2 n-8-5}{5(n-5)(n-4)}=\frac{1}{30}$

 

$\Rightarrow \frac{2 n-13}{(n-5)(n-4)}=\frac{1}{6}$

$\Rightarrow 12 n-78=n^{2}-9 n+20$

 

$\Rightarrow n^{2}-21 n+98=0$

$\Rightarrow(n-7)(n-14)=0$

 

$\therefore n=7$ and 14