Question

If ${}^{n}C_4,{}^{n}C_5$and ${}^{n}C_6$ are in A.P. then $n$ is

• A. $7$ or $14$
• B. $7$
• C. $14$
• D. $14$ or $21$

Answer 1

The correct option is A

$7$ or $14$

Explanation for the correct option

Given that ${}^{n}C_4,{}^{n}C_5$and ${}^{n}C_6$ are in A.P.

We know that if $a,b,c$ are in A.P. then $2b=a+c$

$\begin{array}{crclc}\therefore & 2{}^{n}C_5& =& {}^{n}C_4+{}^{n}C_6& \\ \Rightarrow & 2\frac{n!}{5!\left(n-5\right)!}& =& \frac{n!}{4!\left(n-4\right)!}+\frac{n!}{6!\left(n-6\right)!}& \left[\because {}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right]\\ \Rightarrow & \frac{2}{5\left(n-5\right)!}& =& \frac{1}{\left(n-4\right)!}+\frac{1}{6\times 5\times \left(n-6\right)!}& \\ \Rightarrow & \frac{2}{5\left(n-5\right)}& =& \frac{1}{\left(n-4\right)\left(n-5\right)}+\frac{1}{30}& \\ \Rightarrow & \frac{2}{5\left(n-5\right)}-\frac{1}{\left(n-4\right)\left(n-5\right)}& =& \frac{1}{30}& \\ \Rightarrow & \frac{2\left(n-4\right)-5}{\left(n-4\right)\left(n-5\right)}& =& \frac{1}{6}& \\ \Rightarrow & \frac{2n-8-5}{\left(n-4\right)\left(n-5\right)}& =& \frac{1}{6}& \\ \Rightarrow & \frac{2n-13}{n^2-9n+20}& =& \frac{1}{6}& \\ \Rightarrow & 12n-78& =& n^2-9n+20& \\ \Rightarrow & n^2-21n+98& =& 0& \\ \Rightarrow & n^2-7n-14n+98& =& 0& \\ \Rightarrow & \left(n-7\right)\left(n-14\right)& =& 0& \\ \Rightarrow & n& =& 7,14& \end{array}$

Hence, the correct option is option (A) i.e. $7$ or $14$