Question

If the coefficients of $5^{th},6^{th},7^{th}$ terms of $(1+x{)}^n$ are in A.P., then $n=$

• A. 10
• B. 12
• C. 14
• D. 15

Answer 1

The correct option is C 14

In the expansion $(1+x{)}^n$
5th term ${=}^n{C}_4\cdot x^4$
6th term ${=}^n{C}_5{x}^5$
7th term, ${}^n{C}_6{x}^6$

Given coefficient of these terms are in A.P.

$\Rightarrow 2{\cdot }^n{C}_5{=}^n{C}_4{+}^n{C}_6$

$\Rightarrow \frac{2n!}{(n-5)!5!}=\frac{n!}{(n-4)!4!}+\frac{n!}{(n-6)!6!}$

$\Rightarrow \frac{2}{(n-5)5}=\frac{1}{(n-4)(n-5)}+\frac{1}{6.5}$

$\Rightarrow \frac{2}{(n-5)5}=\frac{30+(n-4)(n-5)}{6.5(n-4)(n-5)}$

$\Rightarrow 12(n-4)=30+(n-4)(n-5)$

$\Rightarrow (12-n+5)(n-4)=30$

$\Rightarrow (17-n)(n-4)=30$

$\Rightarrow -n^2+21n-68=30$

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

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

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

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