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

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==Solution 2==
 
==Solution 2==
 +
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>.
+
Because <math>xy\cdot\frac{x}{y}</math>, by basic multiplication, is <math>x^2</math>, we have
−
Then,
+
 
−
<math>x=3y+xy \implies y=\frac{x}{x+3}</math>.
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<math>x^2 = x^2 - 8xy + 15y^2</math>
−
So,
+
 
−
<math>x-\frac{5x}{x+3}=x+3 \implies x(x+3)-5x=(x+3)^2</math>.
+
<math>8xy = 15y^2</math>
−
We rearrange to get
+
 
−
<math>x^2+3x-5x=x^2+6x+9 \implies 8x=-9 \implies x=-\frac{9}{8}</math>.
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<math>8x = 15y</math>
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Then,
+
 
−
<math>y=\frac{-\frac{9}{8}}{\frac{15}{8}}=-\frac{3}{5}</math>.
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<math>x = \frac{15y}{8}</math>
−
So, the fifth term is
+
 
−
<math>x-7y=-\frac{9}{8}+\frac{21}{5}=\boxed{\frac{123}{40}}</math>.
+
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>
 +
 
 +
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
 +
 
 +
 
 +
==Solution 3==
 +
From the first two terms, we can figure out the common difference, <math>x-y-(x+y)=-2y</math>.
 +
This means that the third, fourth and fifth terms are <math>x-3y</math>, <math>x-5y</math> and <math>x-7y</math> respectively.
 +
 
 +
The third term is also equal to <math>xy</math>, so  
 +
<math>xy=x-3y</math>, which can be rearranged to <math>xy+3y=x</math>.
 +
 
 +
Dividing by y, we have
 +
<math>x+3=\frac{x}{y}</math>.
 +
 
 +
<math>\frac{x}{y}</math> happens to be the fourth term. Therefore,
 +
<math>\frac{x}{y}=x-5y=x+3</math>.
 +
 
 +
<math>x-5y=x+3</math>
 +
 
 +
<math>-5y=3</math>
 +
 
 +
<math>y=-\frac{3}{5}</math>
 +
 
 +
Now that we have <math>y</math>, we can substitute it into an equation above, like <math>\frac{x}{y}=x+3</math>.
 +
 
 +
<math>\frac{x}{\frac{-3}{5}}=x+3</math>
 +
 
 +
<math>-\frac{5x}{3}=x+3</math>
 +
 
 +
<math>-\frac{5x+3x}{3}=3</math>
 +
 
 +
<math>-\frac{8x}{3}=3</math>
 +
 
 +
<math>8x=-9</math>
 +
 
 +
<math>x=\frac{-9}{8}</math>
 +
 
 +
As mentioned earlier, the fifth term we want in the end is equal to <math>x-7y</math>. Substitute some more and...
 +
 
 +
<math>\frac{-9}{8}-7(\frac{-3}{5})=\frac{-9\cdot5}{8\cdot5}+\frac{-3\cdot-7\cdot8}{5\cdot8}=\frac{-45}{40}+\frac{168}{40}=\boxed{\textbf{(E)}\ \frac{123}{40}}</math>
 +
 
 +
(Alternatively, like the above solutions suggest, you could also use the admittedly easier <math>\frac{x}{y}-2y</math> as the final step.)
 +
 
 +
~ a seesaw named owlly81 ~
  
−
~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 15:16, 12 March 2024

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


Solution 3

From the first two terms, we can figure out the common difference, $x-y-(x+y)=-2y$. This means that the third, fourth and fifth terms are $x-3y$, $x-5y$ and $x-7y$ respectively.

The third term is also equal to $xy$, so $xy=x-3y$, which can be rearranged to $xy+3y=x$.

Dividing by y, we have $x+3=\frac{x}{y}$.

$\frac{x}{y}$ happens to be the fourth term. Therefore, $\frac{x}{y}=x-5y=x+3$.

$x-5y=x+3$

$-5y=3$

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

Now that we have $y$, we can substitute it into an equation above, like $\frac{x}{y}=x+3$.

$\frac{x}{\frac{-3}{5}}=x+3$

$-\frac{5x}{3}=x+3$

$-\frac{5x+3x}{3}=3$

$-\frac{8x}{3}=3$

$8x=-9$

$x=\frac{-9}{8}$

As mentioned earlier, the fifth term we want in the end is equal to $x-7y$. Substitute some more and...

$\frac{-9}{8}-7(\frac{-3}{5})=\frac{-9\cdot5}{8\cdot5}+\frac{-3\cdot-7\cdot8}{5\cdot8}=\frac{-45}{40}+\frac{168}{40}=\boxed{\textbf{(E)}\ \frac{123}{40}}$

(Alternatively, like the above solutions suggest, you could also use the admittedly easier $\frac{x}{y}-2y$ as the final step.)

~ a seesaw named owlly81 ~


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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