The first four terms in an arithmetic sequence are $x+y$, $x-y$, $xy$, and $\frac{x}{y}$, in that order. What is the fifth term? 
$\textbf{(A)}\ -\frac{15}{8}\qquad\textbf{(B)}\ -\frac{6}{5}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{27}{20}\qquad\textbf{(E)}\ \frac{123}{40}$
The difference between consecutive terms is $(x-y)-(x+y)=-2y.$ Therefore we can also express the third and fourth terms as $x-3y$ and $x-5y.$ Then we can set them equal to $xy$ and $\frac{x}{y}$ because they are the same thing.
\begin{align*} xy&=x-3y\\ xy-x&=-3y\\ x(y-1)&=-3y\\ x&=\frac{-3y}{y-1} \end{align*}
Substitute into our other equation.
\[\frac{x}{y}=x-5y\]
\[\frac{-3}{y-1}=\frac{-3y}{y-1}-5y\]
\[-3=-3y-5y(y-1)\]
\[0=5y^2-2y-3\]
\[0=(5y+3)(y-1)\]
\[y=-\frac35, 1\]
But $y$ cannot be $1$ because then the first term would be $x+1$ and the second term $x-1$ while the last two terms would be equal to $x.$ Therefore $y=-\frac35.$ Substituting the value for $y$ into any of the equations, we get $x=-\frac98.$ Finally,
\[\frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}\]