AMC 10 · 2004 · #18

Grade 8 algebra
sequences-arithmeticsequences-geometricquadratic-equations convert-to-algebracasework ↑ Prerequisites: quadratic-equations
📏 Medium solution 💡 2 insights
Problem
Three real numbers form an arithmetic progression whose first term is 9. If you add 2 to the second term and 20 to the third term, the three new numbers form a geometric progression. Find the smallest possible value of the third term of that geometric progression.

Pick an answer.

(A)
$\ 1$
(B)
$\ 4$
(C)
$\ 36$
(D)
$\ 49$
(E)
$\ 81$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Name the common difference d (Tool #4): then the arithmetic terms are 9, 9+d, 9+2d, and after the additions the geometric terms are 9, 11+d, 29+2d. The geometric condition — middle term squared equals the product of the ends — turns the whole situation into a single equation in d (Tool #13), which is a quadratic. A quadratic can have two solutions, so there can be two valid common differences and therefore two possible third terms. The word "smallest" is the signal to solve completely and then take the minimum (Tool #14, Extreme Principle) — the trap is stopping at the first (positive-difference) answer 49 and missing the smaller one that a negative difference produces.

1STEP 1

Name the common difference

Let d be the common difference: the arithmetic terms are 9, 9+d, 9+2d, so the geometric terms become 9, 11+d, and 29+2d.

AP: 9, 9+d, 9+2d → GP: 9, 11+d, 29+2d
2STEP 2

Use the geometric-progression rule

Equal ratios mean middle squared equals the ends multiplied: (11+d)² = 9(29+2d), which expands to d² + 4d - 140 = 0.

(11+d)² = 9(29+2d) → d² + 4d - 140 = 0
3STEP 3

Solve for the common difference

Solving d² + 4d - 140 = 0 gives d = 10 or d = -14; both are real, so both common differences are allowed.

d = (-4 ± √(576))/2 = (-4 ± 24)/2 → d = 10 or d = -14
4STEP 4

Pick the smallest third term

The third term is 29 + 2d: d = 10 gives 49, while d = -14 gives 1, so the smallest possible value is 1.

29 + 2(10) = 49, 29 + 2(-14) = 1 → min = 1 = (A)
Answer
1
Check both progressions directly. For d=-14 the arithmetic terms are 9, -5, -19; adding 2 and 20 gives 9, -3, 1, and the ratios -3/9 = -1/3 and 1/-3 = -1/3 match, so it is a genuine geometric progression with third term 1. For d=10 the terms 9, 19, 29 become 9, 21, 49, with ratios 21/9 = 7/3 and 49/21 = 7/3, also valid, third term 49. Both 1 and 49 appear among the choices, and 1 is smaller, confirming (A).
💡Key takeaway

When a problem asks for the smallest possible value, solve for every case first — a negative common difference can give a smaller answer than the obvious positive one.

  • Name the common difference
  • Use the geometric-progression rule
  • Solve for the common difference
  • Pick the smallest third term