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

(Solution)
(Solution)
Line 20: Line 20:
  
 
<cmath>
 
<cmath>
\frac{x}{y}=x-5y\\
+
\frac{x}{y}=x-5y</cmath>
\frac{-3}{y-1}=\frac{-3y}{y-1}-5y\\
+
<cmath>\frac{-3}{y-1}=\frac{-3y}{y-1}-5y</cmath>
-3=-3y-5y(y-1)\\
+
<cmath>-3=-3y-5y(y-1)</cmath>
0=5y^2-2y-3\\
+
<cmath>0=5y^2-2y-3</cmath>
0=(5y+3)(y-1)\\
+
<cmath>0=(5y+3)(y-1)</cmath>
y=-\frac35, 1</cmath>
+
<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 00:16, 30 December 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 (ProblemsAnswer KeyResources)
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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