Question

In the expansion of $(a-b{)}^n,n\ge 5$, the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $\frac{a}{b}$ equals:

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

Answer 1

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

The genral term of it can be written as

$T_{r+1}{=}^n{C}_r.a^{n-r}.b^r.{(-1)}^r$

$T_5$ $=$ $T_{4+1}{=}^n{C}_4.a^{n-4}.b^4$

$T_6$ $=$ $T_{5+1}={-}^n{C}_5.a^{n-5}.b^5$

As sum of $5^{th}$ and $6^{th}$ term is $0$.

$\Rightarrow$$T_5+T_6=0$

$\Rightarrow$ ${}^n{C}_4.a^{n-4}.b^4{=}^n{C}_5.a^{n-5}.b^5$

$\Rightarrow$ $\frac{a^{n-4}.b^5}{a^{n-5}.b^5}=\frac{{}^n{C}_5}{{}^n{C}_4}$

$\Rightarrow$ $\frac{a}{b}=\frac{n!.(n-4)!.4!}{n!.(n-5)!.5!}$

$\Rightarrow$$\frac{a}{b}=\frac{(n-4)}{5}$