# If the sum of $5^{th}\:and\:6^{th}$ terms in the expansion of $(a-b)^n$, $n\geq5$, is 0, then $\large\frac{a}{b}$ = ?

$\begin{array}{1 1} \frac{n-5}{6} \\ \frac{n-4 }{5} \\ \frac{5}{n-4} \\ \frac{6}{n-5} \end{array}$

Given: $T_5+T_6=0$ in the expansion of $(a-b)^n$
$\Rightarrow\:^nC_4.a^{n-4}.(-b)^4+^nC_5.a^{n-5}.(-b)^5=0$
$\Rightarrow\:^nC_4.a^{n-4}.b^4-^nC_5.a^{n-5}.b^5=0$
$\Rightarrow\:^nC_4.a^{n-4}.b^4=^nC_5.a^{n-5}.b^5$
$\Rightarrow\:\large\frac{a}{b}=\frac{^nC_5}{^nC_4}=\frac{n-4}{5}$