In the expansion of $(a-b)^2 , n \leq 5,$ the sum 5th and 6th term is zero, then $\large\frac{a}{b}$ is equal to

Question

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