2003 AMC 10B Problems/Problem 24

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Problem

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?

$\textbf{(A)}\ -\frac{15}{8}\qquad\textbf{(B)}\ -\frac{6}{5}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{27}{20}\qquad\textbf{(E)}\ \frac{123}{40}$


Solution

We get the equations:

\begin{align*}x+y+d&=x-y\\ x-y+d&=xy\\ xy+d &=\frac{x}{y}\end{align*}

Simplifying and rearranging, we can keep substituting and find that $(d, x, y)=\left(\frac{6}{5},-\frac{9}{8},-\frac{3}{5}\right)$

Finally, \[\frac{x}{y}+d=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\frac{123}{40}\rightarrow\boxed{\textbf{E}}.\]