2004 AMC 10A Problem 18

1. Call the common difference $d$. The arithmetic sequence is $9,\ 9 + d,\ 9 + 2d$.
2. Add $2$ to the second term and $20$ to the third. The new numbers are
$$9,\qquad 11 + d,\qquad 29 + 2d.$$
3. In a geometric sequence $a, b, c$ each term is the previous one times the same ratio, so $\tfrac{b}{a} = \tfrac{c}{b}$, which means $b^2 = ac$. Here:
$$(11 + d)^2 = 9(29 + 2d).$$
4. Multiply out both sides: $d^2 + 22d + 121 = 261 + 18d$. Move everything to one side:
$$d^2 + 4d - 140 = 0.$$
5. Look for two numbers that multiply to $-140$ and add to $4$: they are $14$ and $-10$. So $(d + 14)(d - 10) = 0$, giving $d = 10$ or $d = -14$.
6. Check both.
- $d = 10$: the new numbers are $9, 21, 49$. Each is $\tfrac73$ times the one before. Third term $49$.
- $d = -14$: the new numbers are $9, -3, 1$. Each is $-\tfrac13$ times the one before. Third term $1$.
7. Both work, and the smaller third term is $1$.

The answer is $\boxed{\textbf{(A)}\ 1}$.

Write the arithmetic sequence using its common difference, then turn "geometric" into the single equation (middle term) squared equals (first term) times (third term). That always gives a quadratic, so expect two answers. Nothing says the ratio must be positive, and the words "smallest possible" tell you to keep both.

Test the smallest answer choice. A third term of $1$ means $29 + 2d = 1$, so $d = -14$. The new numbers are $9, -3, 1$, and $(-3)^2 = 9 = 9 \times 1$, so they really are geometric. Choice (A) is the smallest choice and it works, so it is the answer.