Question

If ${}^{n}C_4$, ${}^{n}C_5$and ${}^{n}C_6$ are in A.P. then the value of $n$ can be?

• A. $14$
• B. $11$
• C. $9$
• D. $5$

Answer 1

The correct option is A

$14$

Explanation for the correct option

If $x,y,z$ are in an A.P then $2y=x+z$

Similarly, as ${}^{n}C_4$, ${}^{n}C_5$and ${}^{n}C_6$ are in an A.P

$\Rightarrow$ $2·{}^{n}C_5={}^{n}C_4+{}^{n}C_6$

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

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

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

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

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

$\Rightarrow$ $\frac{2n-13}{\left(n^2-9n+20\right)}=\frac{1}{6}$

$\Rightarrow$ $12n-78=\left(n^2-9n+20\right)$

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

$\Rightarrow$ $n^2-7n-14n+98=0$

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

$\Rightarrow$ $n=7$ or $n=14$

Hence option(A) is the correct answer.