jmc

Since the difference of the first two terms is $-2y$, the third and fourth terms of the sequence must be $x-3y$ and $x-5y$. Thus $$x-3y=xy\text{and}x-5y=\frac{x}{y},$$so $xy-5y^2=x.$ Combining these equations we obtain $$(x-3y)-5y^2=x\text{and, therefore, }-3y-5y^2=0.$$Since $y$ cannot be 0, we have $y=-\frac{3}{5}$, and it follows that $x=-\frac{9}{8}$. The fifth term in the sequence is $x-7y=\boxed{\frac{123}{40}}$.