Question

Find $n$ if the coefficient of $5^{th},6^{th}$ & $7^{th}$ terms in the expansion of $(1+x{)}^n$ are in $A.P.$

• A. $7,14$
• B. $3,11$
• C. $7,11$
• D. $Noneofthese$

Answer 1

The correct option is A $7,14$

The coefficient of 5th,6th,7th terms are

${}^n{C}_4{,}^n{C}_5{,}^n{C}_6$

since the coefficient are in A.P

${}^n{C}_5{-}^n{C}_4{=}^n{C}_6{-}^n{C}_5$

$\Rightarrow \frac{n!}{5!\left(n-5\right)!}-\frac{n!}{4!\left(n-4\right)!}=\frac{n!}{6!\left(n-6\right)!}-\frac{n!}{5!\left(n-5\right)!}$

$\Rightarrow \frac{1}{120\left(n-5\right)!}-\frac{1}{24\left(n-4\right)\left(n-5\right)!}=\frac{1}{720\left(n-6\right)!}-\frac{1}{120\left(n-5\right)\left(n-6\right)!}$

$\Rightarrow \frac{1}{120\left(n-5\right)!}\left[\frac{1}{5}-\frac{1}{n-4}\right]=\frac{1}{120\left(n-6\right)!}\left[\frac{1}{6}-\frac{1}{n-5}\right]$

$\Rightarrow \frac{1}{\left(n-5\right)}\left[\frac{n-4-5}{5\left(n-4\right)}\right]=\frac{1}{5}\left[\frac{n-5-6}{6\left(n-5\right)}\right]$

$\Rightarrow \frac{n-9}{n-4}=\frac{n-11}{6}$

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

$\Rightarrow n=7,14$