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

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<math> \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} </math>
 
<math> \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} </math>
  
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==Solution==
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==Solution 1==
  
 
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.
 
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.
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<cmath> \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}</cmath>
 
<cmath> \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}</cmath>
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 +
==Solution 2==
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 +
The common difference of the sequence is <math>(x-y)-(x+y)=-2y</math>. Then, the next two terms are equal to <math>x-y-2y=x-3y</math> and <math>x-3y-2y=x-5y</math>. We are also given that the next two terms are equal to <math>xy</math> and <math>x/y</math>, respectively, so we set up two equations:
 +
<math>x-3y=xy</math>
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<math>x-5y=x/y</math>.
 +
Then,
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<math>x=3y+xy \implies y=\frac{x}{x+3}</math>.
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So,
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<math>x-\frac{5x}{x+3}=x+3 \implies x(x+3)-5x=(x+3)^2</math>
 +
We rearrange to get
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<math>x^2+3x-5x=x^2+6x+9 \implies 8x=-9 \implies x=-\frac{9}{8}</math>
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Then,
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<math>y=\frac{-\frac{9}{8}}{\frac{15}{8}}=-\frac{3}{5}</math>
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So, the fifth term is
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<math>x-7y=-\frac{9}{8}+\frac{21}{5}=\boxed{\frac{123}{40}}</math>.
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 +
~peace09
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2003|ab=B|num-b=23|num-a=25}}
 
{{AMC10 box|year=2003|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:59, 2 September 2020

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 1

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 the first term would be $x+1$ and the second term $x-1$ while the last two terms would be equal to $x.$ Therefore $y=-\frac35.$ Substituting the value for $y$ into any of the equations, we get $x=-\frac98.$ Finally,

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

Solution 2

The common difference of the sequence is $(x-y)-(x+y)=-2y$. Then, the next two terms are equal to $x-y-2y=x-3y$ and $x-3y-2y=x-5y$. We are also given that the next two terms are equal to $xy$ and $x/y$, respectively, so we set up two equations: $x-3y=xy$ $x-5y=x/y$. Then, $x=3y+xy \implies y=\frac{x}{x+3}$. So, $x-\frac{5x}{x+3}=x+3 \implies x(x+3)-5x=(x+3)^2$ We rearrange to get $x^2+3x-5x=x^2+6x+9 \implies 8x=-9 \implies x=-\frac{9}{8}$ Then, $y=\frac{-\frac{9}{8}}{\frac{15}{8}}=-\frac{3}{5}$ So, the fifth term is $x-7y=-\frac{9}{8}+\frac{21}{5}=\boxed{\frac{123}{40}}$.

~peace09

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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