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

(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...")
 
Line 8: Line 8:
 
==Solution==
 
==Solution==
  
−
We get the equations:
+
The difference between consecutive terms is <math>(x-y)-(x+y)=-2y.</math> Therefore we can also express the third and fourth terms as <math>x-3y</math> and <math>x-5y.</math> Then we can set them equal to <math>xy</math> and <math>\frac{x}{y}</math> because they are the same thing.
  
−
<cmath> \begin{align*}x+y+d&=x-y\\ x-y+d&=xy\\ xy+d &=\frac{x}{y}\end{align*} </cmath>
+
<cmath>\begin{align*}
 +
xy&=x-3y\\
 +
xy-x&=-3y\\
 +
x(y-1)&=-3y\\
 +
x&=\frac{-3y}{y-1}
 +
\end{align*}</cmath>
  
−
Simplifying and rearranging, we can keep substituting and find that <math> (d, x, y)=\left(\frac{6}{5},-\frac{9}{8},-\frac{3}{5}\right) </math>
+
Substitute into our other equation.
  
−
Finally, <cmath> \frac{x}{y}+d=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\frac{123}{40}\rightarrow\boxed{\textbf{E}}. </cmath>
+
<cmath>\begin{align*}
 +
\frac{x}{y}&=x-5y\\
 +
\frac{-3}{y-1}&=\frac{-3y}{y-1}-5y\\
 +
-3&=-3y-5y(y-1)\\
 +
0&=5y^2-2y-2\\
 +
0&=(5y+3)(y-1)\\
 +
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,
 +
 
 +
<cmath> \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}</cmath>
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2003|ab=B|num-b=23|num-a=25}}

Revision as of 21:15, 26 November 2011

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. 

\begin{align*}
\frac{x}{y}&=x-5y\\
\frac{-3}{y-1}&=\frac{-3y}{y-1}-5y\\
-3&=-3y-5y(y-1)\\
0&=5y^2-2y-2\\
0&=(5y+3)(y-1)\\
y&=-\frac35, 1 (Error compiling LaTeX. Unknown error_msg)

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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