Question

If the $5^{th},6^{th}\&7^{th}$ term in the expansion of ${(1+y)}^n$ are in AP then find n.

Answer 1

In the expansion of $(1+y{)}^n$ we get, 5th, 6th and 7th terms as, ${}^n{C}_4{,}^n{C}_5{,}^n{C}_6$

Given,

${}^n{C}_4{,}^n{C}_5{,}^n{C}_6$ are in A.P.

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

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

$2\times \frac{1}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{6\times 5}$

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

$2\times 6(n-4)=30+n^2-9n+20$

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

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

$\therefore n=7,14$