Question

If in the expansion of $(a - 2b)^n$, the sum of $4^{th}$ and $5^{th}$ term is zero, then the value of $\frac{a}{b}$ is

• A. $\frac{n-4}{5}$
• B. $\frac{n-3}{2}$
• C. $\frac{5}{n-4}$
• D. $\frac{5}{2\left(n-4\right)}$

Answer 1

Answer: (B)

Here, $T_{4} = \,{}^{n}C_{3}\,\left(a\right)^{n-3}\,\left(-2\,b\right)^{3}$
and $T_{5} = \,{}^{n}C_{4} \,\left(a\right)^{n-4}\,\left(-2\,b\right)^{4}$
Given, $T_{4}+T_{5} = 0$
$\Rightarrow \,{}^{n}C_{3}\,\left(a\right)^{n-3}\,\left(-2\,b\right)^{3}+\,{}^{n}C_{4} \,\left(a\right)^{n-4}\,\left(-2\,b\right)^{4} = 0$
$\Rightarrow \left(a\right)^{n-4} \left(-2\,b\right)^{3}\,\left[a\,{}^{n}C_{3}+\,{}^{n}C_{4}\left(-2\,b\right) \right]=0$
$\Rightarrow \frac{a}{b} = \frac{2\,{}^{n}C_{4}}{^{n}C_{3}}$
$= \frac{2\lfloor n}{\lfloor4\lfloor n-4}\times\frac{\lfloor3\lfloor n-3}{\lfloor n}$
$= \frac{n-3}{2}$