Question

If in the expansion of ${(a-2b)}^n$, the sum of $5$th and $6$th term is $0$, then value of $\frac{a}{b}$ is

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

Answer 1

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

$T_5={{}^{n}C}_4{a}^{n-4}{(-2b)}^4$
$T_6={{}^{n}C}_5{a}^{n-5}{(-2b)}^5$
$T_5+T_6=0{{}^{n}C}_4{2}^4{a}^{n-4}{b}^4={{}^{n}C}_5{2}^5{a}^{n-5}{b}^5$
$\Rightarrow \frac{a^{n-4}{b}^4}{a^{n-5}{b}^5}=\frac{n!2^5}{5!(n-5)!}\cdot \frac{4!(n-4)!}{n!2^4}$
$\Rightarrow \frac{a}{b}=\frac{2(n-4)}{5}$