When simplified, the third term in the expansion of $(\frac{a}{\sqrt{x}}-\frac{\sqrt{x}}{a^2})^6$ is:
$\textbf{(A)}\ \frac{15}{x}\qquad \textbf{(B)}\ -\frac{15}{x}\qquad \textbf{(C)}\ -\frac{6x^2}{a^9} \qquad \textbf{(D)}\ \frac{20}{a^3}\qquad \textbf{(E)}\ -\frac{20}{a^3}$
By the [Binomial Theorem](https://artofproblemsolving.com/wiki/index.php/Binomial_Theorem), the third term in the expansion is
\[\binom{6}{2}(\frac{a}{\sqrt{x}})^{4}(\frac{-\sqrt{x}}{a^2})^2\]
\[15 \cdot \frac{a^4}{x^2} \cdot \frac{x}{a^4}\]
\[\frac{15}{x}\]
The answer is $\boxed{\textbf{(A)}}$.