Question

If $^nC_4,\, ^nC_5$ and $^nC_6$ are in $A.P.$, then $n$ is

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

Answer 1

Answer: (A)

7 or 14

Answer 2

Since, ${ }^{n} C_{4},{ }^{n} C_{5}$ and ${ }^{n} C_{6}$ are in AP
$\therefore 2{ }^{n} C_{5}={ }^{n} C_{4}+{ }^{n} C_{6}$
$\Rightarrow 2 \times \frac{n !}{5 !(n-5) !}=\frac{n !}{4 !(n-4) !}+\frac{n !}{(n-6) ! 6 !}$
$\Rightarrow \frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{30}$
$\Rightarrow \frac{2}{5(n-5)}=\frac{30+n^{2}-9 n+20}{30(n-4)(n-5)}$
$\Rightarrow 12(n-4)=n^{2}-9 n+50$
$\Rightarrow n^{2}-21 n+98=0$
$\Rightarrow(n-14)(n-7)=0$
$\Rightarrow n=7,14$