2003 AMC 10B Problem 24

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

Once the common difference is read off the first two terms, every later term is a linear expression in $x$ and $y$, and the two remaining conditions become a small system. Isolating the same combination $x(1 - y)$ from both equations eliminates $x$ instantly and avoids the extraneous root that appears if you substitute blindly. Always verify by writing out the sequence.

Divide $xy = x - 3y$ by $y$: $x = \frac{x}{y} - 3$, so $\frac{x}{y} = x + 3$. Since also $\frac{x}{y} = x - 5y$, we get $-5y = 3$ and $y = -\frac{3}{5}$ at once. Then $x(1 - y) = 3y$ gives $x = -\frac{9}{8}$, and the fifth term is $\frac{x}{y} - 2y = \frac{15}{8} + \frac{6}{5} = \frac{123}{40}$.

• Solving the resulting quadratic in $y$ and keeping the extraneous root $y = 1$, which makes the terms $x+1, x-1, x, x$ (not arithmetic) and leads to a wrong fifth term.
• Taking the common difference as $2y$ instead of $-2y$, or computing the fifth term as $x - 6y$.