Question

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

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

Answer 1

The correct option is A $14$

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

So applying the same condition, we have

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

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

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

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

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

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

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

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

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

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

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

$\therefore$ $n=14$ or $n=7$