Question

If ${}^n{C}_4,{}^n{C}_{5}and{}^n{C}_6$ are in A.P, then the value of n is

• A. 14
• B. 11
• C. 7
• D. 8

Answer 1

Answer: (A), (C)

The correct options are
A 14
C 7

Given ${}^n{C}_4,{}^n{C}_5$ and ${}^n{C}_6$ are in A.P

$\Rightarrow {}^n{C}_5-{}^n{C}_4={}^n{C}_6-{}^n{C}_5$

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

$\Rightarrow \frac{1}{5.4!(n-5)(n-6)!}-\frac{1}{4!(n-4)(n-5)(n-6)!}=\frac{1}{\mathrm{6.5.4}!(n-6)!}-\frac{1}{5.4!(n-5)(n-6)!}$

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

multiply on both sides by $30(n-4)(n-5)$

$\Rightarrow \frac{2}{5n-25}=\frac{1}{30}+\frac{1}{n^2-9n+20}$

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

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

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

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

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

$\Rightarrow n=7,14$.