jmc

Note that $(x-y)-(x+y)=xy-(x-y),$ which simplifies to $xy-x+3y=0.$ Solving for $x,$ we find $$x=\frac{3y}{1-y}.$$Also, $(x-y)-(x+y)=\frac{x}{y}-xy,$ which simplifies to $$\frac{x}{y}-xy+2y=0.$$Substituting $x=\frac{3y}{1-y},$ we get $$\frac{3}{1-y}-\frac{3y^2}{1-y}+2y=0.$$This simplifies to $5y^2-2y-3=0,$ which factors as $(y-1)(5y+3)=0,$ so $y=1$ or $y=-\frac{3}{5}.$

If $y=1,$ then $x=\frac{3y}{1-y}$ is not defined, so $y=-\frac{3}{5}.$ Then $$x=\frac{3y}{1-y}=\frac{3(-3/5)}{1+3/5}=-\frac{9}{8}.$$Then the common difference of the arithmetic sequence is $(x-y)-(x+y)=-2y=\frac{6}{5},$ so the fifth term is $$\frac{x}{y}+\frac{6}{5}=\frac{15}{8}+\frac{6}{5}=\boxed{\frac{123}{40}}.$$