AMC 10 · 2004 · #14

Grade 8 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
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. Adding 2 to the second term and 20 to the third makes the three new numbers 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)
$\ 81$
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

Naming the common difference writes the new trio as 9, 11+d, 29+2d.

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

Use the geometric-progression rule

The middle squared equals the outer product, giving d² + 4d - 140 = 0.

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

Solve for the common difference

Both roots are real and legal: d = 10 or d = -14.

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

Pick the smallest third term

Evaluating the third term at both gives 49 and 1, so the minimum is 1, choice (A).

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

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