Question

If ${}^n{C}_4{,}^n{C}_5$ and ${}^n{C}_6$ are in AP then n is

• A. 8
• B. 9
• C. 14
• D. 7

Answer 1

Answer: (C), (D)

The correct options are
C 7
D 14

If ${}^n{C}_4{,}^n{C}_5$ and ${}^n{C}_6$ are in A.P.
$\Rightarrow {2.}^n{C}_5{=}^n{C}_4{+}^n{C}_6$
$\Rightarrow \frac{2}{5}=\frac{1}{(n-4)}+\frac{n-5}{6}$
$\Rightarrow n^2-21n+98=0\Rightarrow n=7,14$