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

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<math>y = \frac{15}{-25} = \frac{-3}{5}</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>\frac{-6}{5}</math>! Therefore, subtracting <math>2y</math> from the fourth term, <math>\frac{x}{y}</math> or <math>\frac{15}{8}</math>, we see:  
+
Aha! This means <math>2y</math>, the common difference, is <math>\frac{-6}{5}</math>! Therefore, subtracting <math>2y</math> from the fourth term, <math>\frac{x}{y}</math> or <math>\frac{15}{8}</math>, we see the fifth term is:  
  
 
<math>\frac{15}{8} - \frac{-6}{5} = \frac{15}{8} + \frac{6}{5} = \frac{75 + 48}{40} = \boxed{\textbf{(E)}\ \frac{123}{40}}</math>
 
<math>\frac{15}{8} - \frac{-6}{5} = \frac{15}{8} + \frac{6}{5} = \frac{75 + 48}{40} = \boxed{\textbf{(E)}\ \frac{123}{40}}</math>

Revision as of 17:43, 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 notice that $\frac{x}{y}$ (the fourth term) is simply $\frac{\frac{15y}{8}}{y}$, which is just $\frac{15}{8}$. Additionally, we notice that the fourth term can be represented as $x - 5y = \frac{15}{8}$; rewritten as $\frac{15y}{8} - 5y = \frac{15}{8}$, we can simplify as follows:

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

$15y - 40y = 15$

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

Aha! This means $2y$, the common difference, is $\frac{-6}{5}$! Therefore, subtracting $2y$ from the fourth term, $\frac{x}{y}$ or $\frac{15}{8}$, we see the fifth term is:

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

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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