2003 AMC 10B Problem 24

1. In an arithmetic sequence, you add the same amount each step. From the first term to the second: $(x - y) - (x + y) = -2y$. So each step adds $-2y$.
2. Keep adding $-2y$ to write every term in a simple form:
- 3rd term: $x - 3y$, and it is also $xy$.
- 4th term: $x - 5y$, and it is also $\frac{x}{y}$.
- 5th term: $x - 7y$. This is what we want.
3. So there are two equations:
$$xy = x - 3y, \qquad \frac{x}{y} = x - 5y.$$
4. The first equation has a $y$ in every term, and dividing it by $y$ will produce $\frac{x}{y}$, which also appears in the second equation. ($y$ is not $0$, because the fourth term divides by $y$.) Divide both sides of $xy = x - 3y$ by $y$:
$$x = \frac{x}{y} - 3, \quad\text{so}\quad \frac{x}{y} = x + 3.$$
5. Now the fourth term equals both $x + 3$ and $x - 5y$. So $x + 3 = x - 5y$, which gives $3 = -5y$ and $y = -\frac{3}{5}$.
6. Find $x$ from $xy = x - 3y$ with $y = -\frac{3}{5}$:
$$-\frac{3}{5}x = x + \frac{9}{5}.$$
Multiply by $5$: $-3x = 5x + 9$, so $-8x = 9$ and $x = -\frac{9}{8}$.
7. The fifth term:
$$x - 7y = -\frac{9}{8} + \frac{21}{5} = \frac{-45}{40} + \frac{168}{40} = \frac{123}{40}.$$
8. Check by writing out the terms (all over $40$): $x + y = -\frac{69}{40}$, $x - y = -\frac{21}{40}$, $xy = \frac{27}{40}$, $\frac{x}{y} = \frac{15}{8} = \frac{75}{40}$, then $\frac{123}{40}$. Each is $\frac{48}{40}$ more than the one before.

The answer is $\boxed{\textbf{(E)}\ \frac{123}{40}}$.

Once the first two terms give you the step size, every later term can be written simply in terms of $x$ and $y$, and the problem becomes two equations. Look for a move that makes the same expression appear in both equations (here $\frac{x}{y}$); then one unknown drops out right away. Writing out the finished sequence is a quick, reliable check.

Rewrite both equations so the same expression appears. From $xy = x - 3y$: $x - xy = 3y$, which is $x(1 - y) = 3y$. From $\frac{x}{y} = x - 5y$, multiply by $y$: $x = xy - 5y^2$, so $x - xy = -5y^2$, which is $x(1 - y) = -5y^2$. So $3y = -5y^2$. Since $y$ is not $0$, divide by $y$: $3 = -5y$, and again $y = -\frac{3}{5}$.

The first two terms give the common difference $d = (x - y) - (x + y) = -2y$. So the third and fourth terms are $x - 3y$ and $x - 5y$, and the fifth is $x - 7y$. Matching with the given forms:
$$xy = x - 3y, \qquad \frac{x}{y} = x - 5y.$$

Rearrange the first as $x - xy = 3y$, i.e. $x(1 - y) = 3y$. Rearrange the second as $\frac{x}{y} - x = -5y$, i.e. $x(1 - y) = -5y^2$. Comparing the two expressions for $x(1-y)$:
$$3y = -5y^2 \quad\Longrightarrow\quad y(5y + 3) = 0.$$
$y = 0$ is impossible (the fourth term divides by $y$), so $y = -\frac{3}{5}$.

Then $x\left(1 + \frac{3}{5}\right) = -\frac{9}{5}$ gives $x = -\frac{9}{5} \cdot \frac{5}{8} = -\frac{9}{8}$.

Fifth term:
$$x - 7y = -\frac{9}{8} + \frac{21}{5} = \frac{-45 + 168}{40} = \frac{123}{40}.$$

Check: the terms are $-\frac{69}{40}, -\frac{21}{40}, \frac{27}{40}, \frac{75}{40}, \frac{123}{40}$, each $\frac{48}{40}$ more than the last, and indeed $xy = \frac{27}{40}$ and $\frac{x}{y} = \frac{15}{8} = \frac{75}{40}$.

The answer is $\boxed{\textbf{(E)}\ \frac{123}{40}}$.