# Difference between revisions of "2003 AMC 10B Problems/Problem 24"

## 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

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 every term would be equal to $x.$ Therefore $y=-\frac35.$ Substituting the value for $y$ into any of the equations, we get $x=-\frac98.$ Finally,

$$\frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}$$