Question

The sum of 5th term and 6th term is equal to $0$ of the expansion of the term $(2a-b{)}^n$. The value of a/b is

• A. $\frac{n+4}{10}$
• B. $\frac{10}{(n-4)}$
• C. $\frac{10}{(n+4)}$
• D. $\frac{n-4}{10}$

Answer 1

The correct option is D $\frac{n-4}{10}$

$T_5=nC4{\left(2a\right)}^{(n-4)}{\left(-b\right)}^4{T}_6=nC5{\left(2a\right)}^{(n-5)}{\left(-b\right)}^5=-nC5{\left(2a\right)}^{(n-5)}{b}^5{T}_5+T_6=0nC4{\left(2a\right)}^{(n-4)}{\left(-b\right)}^4=nC5{\left(2a\right)}^{(n-5)}{b}^{5}nC4\left(2a\right)=nC5\left(b\right)\frac{a}{b}=\frac{1}{2}\left(\frac{n}{C}\right)=\frac{n-4}{10}$