# 2003 AMC 10B Problems/Problem 24

## 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 ~