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

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==Solution 2==
 
==Solution 2==
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Because this is an arithmetic sequence, we conclude from the first two terms that the common difference is <math>-2y</math>. Therefore, <math>xy = x-3y</math> and <math>\frac{x}{y}=x-5y</math>. If we multiply <math>xy</math> and <math>\frac{x}{y}</math>, we see:
  
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>xy\cdot\frac{x}{y} = (x-3y)(x-5y) = x^2 - 8xy + 15y^2</math>
<math>x-3y=xy</math> and
 
<math>x-5y=x/y</math>.
 
Then,
 
<math>x=3y+xy \implies y=\frac{x}{x+3}</math>.
 
So,
 
<math>x-\frac{5x}{x+3}=x+3 \implies x(x+3)-5x=(x+3)^2</math>.
 
We rearrange to get
 
<math>x^2+3x-5x=x^2+6x+9 \implies 8x=-9 \implies x=-\frac{9}{8}</math>.
 
Then,
 
<math>y=\frac{-\frac{9}{8}}{\frac{15}{8}}=-\frac{3}{5}</math>.
 
So, the fifth term is
 
<math>x-7y=-\frac{9}{8}+\frac{21}{5}=\boxed{\frac{123}{40}}</math>.
 
  
~peace09
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Because <math>xy\cdot\frac{x}{y}</math>, by basic multiplication, is <math>x^2</math>, we have
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 +
<math>x^2 = x^2 - 8xy + 15y^2</math>
 +
 
 +
<math>8xy = 15y^2</math>
 +
 
 +
<math>8x = 15y</math>
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 +
<math>x = \frac{15y}{8}</math>
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 +
Now that we have <math>x</math> in terms of <math>y</math>, we substitute in <math>\frac{15y}{8}</math> in for <math>x</math> in <math>\frac{x}{y}</math> (the fourth term). This leaves us with <math>\frac{\frac{15y}{8}}{y} = \frac{15}{8}</math>.
 +
 
 +
Recall that <math>\frac{x}{y}</math> can be written as <math>x - 5y</math>. Thus, <math>x - 5y = \frac{15}{8}</math>. Substitute in <math>\frac{15}{8}</math> in for <math>x</math>, and we see:
 +
 
 +
<math>\frac{15y}{8} - 5y = \frac{15}{8}</math>
 +
 
 +
<math>15y - 40y = 15</math>
 +
 
 +
<math>y = \frac{15}{-25} = \frac{-3}{5}</math>
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 +
Aha! This means <math>2y</math>, the common difference, is <math>2\cdot y = 2\cdot \frac{-3}{5} = \frac{-6}{5}</math>. Now, all we need to do is find the fifth term, which is just <math>\frac{x}{y}-2y</math>. We can substitute known values to solve:
 +
 
 +
<math>\frac{x}{y}-2y = \frac{15}{8} - \frac{-6}{5} = \frac{15}{8} + \frac{6}{5} = \frac{75 + 48}{40} = \boxed{\textbf{(E)}\ \frac{123}{40}}</math>.
 +
 
 +
~SXWang
  
 
==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 18:10, 14 July 2022

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

Because this is an arithmetic sequence, we conclude from the first two terms that the common difference is $-2y$. Therefore, $xy = x-3y$ and $\frac{x}{y}=x-5y$. If we multiply $xy$ and $\frac{x}{y}$, we see:

$xy\cdot\frac{x}{y} = (x-3y)(x-5y) = x^2 - 8xy + 15y^2$

Because $xy\cdot\frac{x}{y}$, by basic multiplication, is $x^2$, we have

$x^2 = x^2 - 8xy + 15y^2$

$8xy = 15y^2$

$8x = 15y$

$x = \frac{15y}{8}$

Now that we have $x$ in terms of $y$, we substitute in $\frac{15y}{8}$ in for $x$ in $\frac{x}{y}$ (the fourth term). This leaves us with $\frac{\frac{15y}{8}}{y} = \frac{15}{8}$.

Recall that $\frac{x}{y}$ can be written as $x - 5y$. Thus, $x - 5y = \frac{15}{8}$. Substitute in $\frac{15}{8}$ in for $x$, and we see:

$\frac{15y}{8} - 5y = \frac{15}{8}$

$15y - 40y = 15$

$y = \frac{15}{-25} = \frac{-3}{5}$

Aha! This means $2y$, the common difference, is $2\cdot y = 2\cdot \frac{-3}{5} = \frac{-6}{5}$. Now, all we need to do is find the fifth term, which is just $\frac{x}{y}-2y$. We can substitute known values to solve:

$\frac{x}{y}-2y = \frac{15}{8} - \frac{-6}{5} = \frac{15}{8} + \frac{6}{5} = \frac{75 + 48}{40} = \boxed{\textbf{(E)}\ \frac{123}{40}}$.

~SXWang

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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