Question

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

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

Answer 1

The correct option is B $\frac{2(n-4)}{5}$

We know, $(a-2b{)}^n={\sum }_{r=0}^n{}^n{C}_r(a{)}^{n-r}(-2b{)}^r$
The $(r+1)$th term $=t_{r+1}={}^n{C}_r(a{)}^{n-r}.(-2b{)}^r$
$\therefore t_5+t_6=0$
$\Rightarrow {}^n{C}_4\left(a{)}^{n-4}\right(-2b{)}^4+{}^n{C}_5\left(a{)}^{n-5}\right(-2b{)}^5=0$
$\Rightarrow \frac{n!}{4!(n-4)!}{a}^{n-4}(-2b{)}^4$$=-\frac{n!}{5!(n-5)!}\left(a{)}^{n-5}\right(-2b{)}^5$
$\Rightarrow \frac{1}{(n-4)}\times a=\frac{-1}{5}(-2b)$
$\Rightarrow \frac{a}{b}=\frac{2(n-4)}{5}$