Question

If the $5^{th},6^{th},$ and $7^{th}$ term in the expansion of ${(1+y)}^n$ are in A.P. then find the value of $n$.

Answer 1

The coefficients of the $5^{th},6^{th}$ and $7^{th}$ terms are ${}^n{C}_4{,}^n{C}_5{,}^n{C}_6$

Since the coefficients of the $5^{th},6^{th}$ and $7^{th}$ terms are in $AP$. we have,

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

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

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

$\Rightarrow \frac{1}{24(n-5)!}[\frac{1}{5}-\frac{1}{n-4}]=\frac{1}{120(n-6)!}[\frac{1}{6}-\frac{1}{n-5}]$

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

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

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

$\Rightarrow n=7,14$