2003 AMC 10B Problems/Problem 24

Revision as of 09:47, 14 August 2011 by Mrdavid445 (talk | contribs) (Created page with "==Problem== The first four terms in an arithmetic sequence are <math>x+y</math>,<math>x-y</math> ,<math>xy</math> , and <math>\frac{x}{y}</math>, in that order. What is the fi...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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