AMC 10 · 2004 · #18
Grade 8 algebraA sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?
Pick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The original three numbers are in arithmetic progression (a constant is added to get from each term to the next).; The first term of the arithmetic progression is $9$.; Adding $2$ to the second term and $20$ to the third term produces three numbers in geometric progression (each term is a constant multiple of the previous one).; Answer choices: (A) $1$, (B) $4$, (C) $36$, (D) $49$, (E) $81$
Unknowns: The smallest possible value of the third term of the geometric progression.
Understand
Restated: 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.
Givens: The original three numbers are in arithmetic progression (a constant is added to get from each term to the next).; The first term of the arithmetic progression is $9$.; Adding $2$ to the second term and $20$ to the third term produces three numbers in geometric progression (each term is a constant multiple of the previous one).; Answer choices: (A) $1$, (B) $4$, (C) $36$, (D) $49$, (E) $81$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #14 Extreme Principle
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.
Execute — Answer: A
7.EE.B.4 Step 1 Name the common difference
- Let $d$ be the common difference of the arithmetic progression.
- Starting from $9$, the three terms are $9$, $9+d$, and $9+2d$.
- Now apply the changes: nothing is added to the first term, $2$ is added to the second, and $20$ is added to the third.
- So the three numbers of the geometric progression are $9$, $(9+d)+2 = 11+d$, and $(9+2d)+20 = 29+2d$.
💡 One unknown, the common difference, controls all three terms, so naming it captures the entire problem.
7.EE.A.1 Step 2 Use the geometric-progression rule
- In a geometric progression the ratio between consecutive terms is the same, so $\frac{11+d}{9} = \frac{29+2d}{11+d}$.
- Cross-multiplying gives the standard fact that the middle term squared equals the product of the outer terms: $(11+d)^2 = 9(29+2d)$.
- Expand both sides: $121 + 22d + d^2 = 261 + 18d$.
- Move everything to one side: $d^2 + 22d - 18d + 121 - 261 = 0$, which simplifies to $d^2 + 4d - 140 = 0$.
💡 Equal ratios cross-multiply into 'middle squared equals ends multiplied,' which converts the pattern into one solvable equation.
8.EE.A.2 Step 3 Solve for the common difference
- Solve $d^2 + 4d - 140 = 0$ with the quadratic formula: $d = \dfrac{-4 \pm \sqrt{4^2 - 4(1)(-140)}}{2} = \dfrac{-4 \pm \sqrt{16 + 560}}{2} = \dfrac{-4 \pm \sqrt{576}}{2}$.
- Since $\sqrt{576} = 24$, this gives $d = \dfrac{-4 + 24}{2} = 10$ or $d = \dfrac{-4 - 24}{2} = -14$.
- Both are real, so both are valid common differences.
💡 A quadratic can hold two answers, and both must be kept because each gives a genuinely different progression.
7.EE.B.4 Step 4 Pick the smallest third term
- The third term of the geometric progression is $29 + 2d$.
- For $d = 10$: $29 + 2(10) = 49$.
- For $d = -14$: $29 + 2(-14) = 29 - 28 = 1$.
- Both $49$ and $1$ are achievable, so the smallest possible value is $1$.
- That is choice (A).
- Stopping at the positive difference $d=10$ would wrongly give $49$ (choice (D)); the negative difference is what produces the true minimum.
💡 'Smallest possible' means compare every valid case, and a negative common difference can beat a positive one.
7.EE.B.4 Let $d$ be the common difference of the arithmetic progression. Starting from $9 7.EE.A.1 In a geometric progression the ratio between consecutive terms is the same, so $ 8.EE.A.2 Solve $d^2 + 4d - 140 = 0$ with the quadratic formula: $d = \dfrac{-4 \pm \sqrt{ 7.EE.B.4 The third term of the geometric progression is $29 + 2d$. For $d = 10$: $29 + 2( Review
Reasonableness: 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 $\frac{-3}{9} = -\frac{1}{3}$ and $\frac{1}{-3} = -\frac{1}{3}$ 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 $\frac{21}{9} = \frac{7}{3}$ and $\frac{49}{21} = \frac{7}{3}$, also valid, third term $49$. Both $1$ and $49$ appear among the choices, and $1$ is smaller, confirming (A).
Alternative: Introduce the common ratio $r$ instead (Tool #4). The geometric progression is $9, 9r, 9r^2$, so the original arithmetic terms were $9,\, 9r-2,\, 9r^2-20$. Being arithmetic means equal gaps: $(9r-2)-9 = (9r^2-20)-(9r-2)$, which simplifies to $9r^2 - 18r - 7 = 0$. The quadratic formula gives $r = \frac{18 \pm \sqrt{324+252}}{18} = \frac{18 \pm 24}{18}$, so $r = \frac{7}{3}$ or $r = -\frac{1}{3}$. The third term $9r^2$ is $9\cdot\frac{49}{9}=49$ or $9\cdot\frac{1}{9}=1$, giving the same minimum of $1$.
CCSS standards used (min grade 8)
7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Naming the common difference $d$, writing all three terms in terms of it, and evaluating the third term $29+2d$ for each solution.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Expanding $(11+d)^2$ and $9(29+2d)$ and collecting like terms to reach $d^2+4d-140=0$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Evaluating $\sqrt{576}=24$ in the quadratic formula to find $d=10$ and $d=-14$.)
⭐ 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.
⭐ 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.
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