Question

In the binomial expansion of $(1+x{)}^n$, where $n$ is a natural number, the co-efficient of 5th, 6th and 7th terms are in A.P., then $n$ is equal to

• A. 7 and 13
• B. 7 and 14
• C. 7 and 17
• D. none of these

Answer 1

Answer: (B)

Since co-eff. of $5th$, $6th$, $7th$ terms are in $A.P$.
$\therefore {}^n{c}_4+{}^n{c}_6=2{\cdot }^n{c}_5$
$\Rightarrow \frac{n!}{n-4!4!}+\frac{n!}{n-6!6!}$
$=\frac{n!}{2n-5!5!}$
$\Rightarrow 6\cdot 5+\left(n-4\right)\left(n-5\right)=2\left(n-4\right)\cdot 6$
$\Rightarrow 30+n^2-9n+60=12n-48$
$\Rightarrow n^2-21n+98=0$
$\Rightarrow n=7$, $14$