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, $T_4={}^n{C}_3{\left(a\right)}^{n-3}{\left(-2b\right)}^3$
and $T_5={}^n{C}_4{\left(a\right)}^{n-4}{\left(-2b\right)}^4$
Given, $T_4+T_5=0$
$\Rightarrow {}^n{C}_3{\left(a\right)}^{n-3}{\left(-2b\right)}^3+{}^n{C}_4{\left(a\right)}^{n-4}{\left(-2b\right)}^4=0$
$\Rightarrow {\left(a\right)}^{n-4}{\left(-2b\right)}^3\left[a{}^n{C}_3+{}^n{C}_4\left(-2b\right)\right]=0$
$\Rightarrow \frac{a}{b}=\frac{2{}^n{C}_4}{{}^n{C}_3}$
$=\frac{2\lfloor n}{\lfloor 4\lfloor n-4}\times \frac{\lfloor 3\lfloor n-3}{\lfloor n}$
$=\frac{n-3}{2}$