Question

If in the expansion of ${\left(a-2b\right)}^n$, the sum of $5$th and $6$th terms is zero, then the value of $\frac{a}{b}$ is

• A. $\frac{n-4}{5}$
• B. $\frac{2\left(n-4\right)}{5}$
• C. $\frac{5}{n-4}$
• D. $\frac{5}{2\left(n-4\right)}$

Answer 1

The correct option is B

$\frac{2\left(n-4\right)}{5}$

Explanation for the correct option.

Find the value of $\frac{a}{b}$.

In the expansion of ${\left(a+b\right)}^n$ the $r+1$th term is given as $T_{r+1}={}^{n}C_r{a}^{n-r}{b}^r$.

Now, for the expansion ${\left(a-2b\right)}^n$, the fifth term is given as: $T_5={}^{n}C_4{a}^{n-4}{\left(-2b\right)}^4$ and the sixth term is given as: $T_6={}^{n}C_5{a}^{n-5}{\left(-2b\right)}^5$.

As the sum of $5$th and $6$th terms is zero, so $T_5+T_6=0$

$$\begin{gathered} {}^{n}C_4{a}^{n-4}{\left(-2b\right)}^4+{}^{n}C_5{a}^{n-5}{\left(-2b\right)}^5=0 \\ \Rightarrow \frac{n!}{\left(n-4\right)!4!}{a}^{n-4}{\left(-2b\right)}^4+\frac{n!}{\left(n-5\right)!5!}{a}^{n-5}{\left(-2b\right)}^5=0 \\ \Rightarrow \frac{n!}{\left(n-4\right)\left(n-5\right)!4!}a\times a^{n-5}{\left(-2b\right)}^4+\frac{n!}{\left(n-5\right)!·5·4!}{a}^{n-5}\left(-2b\right){\left(-2b\right)}^4=0 \\ \Rightarrow \frac{1}{n-4}a+\frac{1}{5}\left(-2b\right)=0 \\ \Rightarrow \frac{a}{n-4}=\frac{2b}{5} \\ \Rightarrow \frac{a}{b}=\frac{2\left(n-4\right)}{5} \end{gathered}$$

So the value of $\frac{a}{b}$ is $\frac{2\left(n-4\right)}{5}$.

Hence, the correct option is B.