JEE Main 12 Jan 2019 Evening Question 11
Question: If $ ^{n}C _4,{{,}^{n}}C _5 $ and $ {{,}^{n}}C _6 $ are in A.R, then n can be
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 12
B) 14
C) 9
D) 11
Show Answer
Answer:
Correct Answer: B
Solution:
- Since, $ ^{n}C _4,{{,}^{n}}C _5 $ and $ {{,}^{n}}C _6 $ are in A.P.
$ \therefore $ $ 2{{\times }^{n}}C _5{{=}^{n}}C _4{{+}^{n}}C _6 $
$ \Rightarrow 2\times \frac{n}{5(n-5)}=\frac{n}{4(n-4)}+\frac{n}{6(n-6)} $
$ \Rightarrow $ $ \frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{6\times 5} $
$ \Rightarrow $ $ \frac{2}{5}(n-4)=1+\frac{(n-4)(n-5)}{6\times 5} $
$ \Rightarrow $ $ 12(n-4)=30+n^{2}-9n+20 $
$ \Rightarrow $ $ 21n-n^{2}-98=0\Rightarrow n^{2}-21n+98=0 $
$ \Rightarrow $ $ (n-14)(n-7)=0\Rightarrow n=7,14 $ But only n = 14 lies in the options.
