AMC10 2003 10B, Problem 24

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?

Solution: 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, \[\frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\fbox{\frac{123}{40}}\]

Answer: \frac{123}{40}