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Difference between revisions of "2003 AMC 10B Problems/Problem 24"
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Substitute into our other equation.
 
Substitute into our other equation.
  
−
<cmath>\begin{align*}
+
<cmath>
−
\frac{x}{y}&=x-5y\\
+
\frac{x}{y}=x-5y\\
−
\frac{-3}{y-1}&=\frac{-3y}{y-1}-5y\\
+
\frac{-3}{y-1}=\frac{-3y}{y-1}-5y\\
−
-3&=-3y-5y(y-1)\\
+
-3=-3y-5y(y-1)\\
−
0&=5y^2-2y-3\\
+
0=5y^2-2y-3\\
−
0&=(5y+3)(y-1)\\
+
0=(5y+3)(y-1)\\
−
y&=-\frac35, 1</cmath>
+
y=-\frac35, 1</cmath>
  
 
But <math>y</math> cannot be <math>1</math> because then every term would be equal to <math>x.</math> Therefore <math>y=-\frac35.</math> Substituting the value for <math>y</math> into any of the equations, we get <math>x=-\frac98.</math> Finally,
 
But <math>y</math> cannot be <math>1</math> because then every term would be equal to <math>x.</math> Therefore <math>y=-\frac35.</math> Substituting the value for <math>y</math> into any of the equations, we get <math>x=-\frac98.</math> Finally,

Revision as of 19:19, 26 May 2015

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,

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

See Also

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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