Question

If ${}^n{C}_4{,}^n{C}_5$ and ${}^n{C}_6$ are in A.P, then possible value of n is

• A. $6$
• B. $12$
• C. $14$
• D. $21$

Answer 1

The correct option is C $14$

If a,b,c are in A.P, then $2b=a+c$

Applying the same condition, we get

$2\times {}^n{C}_5={}^n{C}_6+{}^n{C}_4$

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

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

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

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

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

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

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

$12n-78=n^2-9n+20$

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

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

$n=7,14$