Question

In the binomial expansion of $(a-b{)}^n,n>5,$ the sum of 5${}^{th}$ and 6${}^{th}$ terms is zero, then $\frac{a}{b}$ equals-

• A. $\frac{5}{n-4}$
• B. $\frac{6}{n-5}$
• C. $\frac{n-5}{6}$
• D. $\frac{n-4}{5}$

Answer 1

The correct option is D $\frac{n-4}{5}$

Since, in binomial expansion of ${(a-b)}^n,n\ge 5$,

the sum of fifth and sixth term is equal to zero
${\therefore }^n{C}_4{a}^{n-4}{b}^4{+}^n{C}_5{a}^{n-5}{b}^5=0$
$\Rightarrow \frac{n!}{(n-4)!4!}{a}^{n-4}{b}^4-\frac{n!}{(n-5)!5!}{a}^{n-5}{b}^5=0$
$\Rightarrow \frac{a}{b}=\frac{n-4}{5}$