AMC 10 · 2003 · #8
Grade 8 algebraThe second and fourth terms of a geometric sequence are 2 and 6. Which of the following is a possible first term?
Pick an answer.
AMC 10 2003 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: In a geometric sequence every term is the one before it multiplied by the same fixed number (the common ratio). The second term is $2$ and the fourth term is $6$. Find which of the five listed values could be the first term.
Givens: The sequence is geometric: each term equals the previous term times a fixed ratio $r$; The second term is $2$; The fourth term is $6$; Answer choices: (A) $-\sqrt{3}$, (B) $-\frac{2\sqrt{3}}{3}$, (C) $-\frac{\sqrt{3}}{3}$, (D) $\sqrt{3}$, (E) $3$
Unknowns: A possible value of the first term
Understand
Restated: In a geometric sequence every term is the one before it multiplied by the same fixed number (the common ratio). The second term is $2$ and the fourth term is $6$. Find which of the five listed values could be the first term.
Givens: The sequence is geometric: each term equals the previous term times a fixed ratio $r$; The second term is $2$; The fourth term is $6$; Answer choices: (A) $-\sqrt{3}$, (B) $-\frac{2\sqrt{3}}{3}$, (C) $-\frac{\sqrt{3}}{3}$, (D) $\sqrt{3}$, (E) $3$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #11 Work Backwards, #3 Eliminate Possibilities
The first term is unknown and so is the ratio, so Tool #4 (Introduce a Variable) is the spine: call the first term $a$ and the ratio $r$, then the two facts become $ar = 2$ and $ar^3 = 6$. Tool #11 (Work Backwards) does the heavy lifting — dividing the fourth-term equation by the second-term equation cancels $a$ and leaves a clean equation for $r$, and once $r$ is known one step back from the second term recovers $a$. Tool #3 (Eliminate Possibilities) finishes it: the two facts allow two ratios (a positive and a negative one), giving two candidate first terms, and only one of those candidates appears among the five choices.
Execute — Answer: B
6.EE.A.2 Step 1 Name the unknowns, write the facts
- Let the first term be $a$ and the common ratio be $r$.
- Because each term is the previous one times $r$, the second term is $a r$ and the fourth term is $a r^3$.
- The two given facts turn into two equations.
💡 Giving the unknown first term and ratio letters lets the two clues become two equations you can push around.
8.EE.A.1 Step 2 Divide to cancel the first term
- Divide the fourth-term equation by the second-term equation.
- On the left the $a$ cancels and $r^3 \div r = r^2$; on the right $6 \div 2 = 3$.
- That isolates the ratio: $r^2 = 3$.
💡 Dividing the two equations wipes out the unknown first term and leaves the ratio all by itself.
8.EE.A.2 Step 3 Take the square root — keep both signs
- If $r^2 = 3$ then $r$ is a number whose square is $3$, and there are two such numbers: $r = \sqrt{3}$ or $r = -\sqrt{3}$.
- Both are legal ratios for a geometric sequence, so both must be carried forward.
💡 A squared value always has two square roots, so a positive and a negative ratio both survive.
7.NS.A.2 Step 4 Step back to the first term
- From $a r = 2$, the first term is $a = \dfrac{2}{r}$.
- Put in each ratio: $a = \dfrac{2}{\sqrt{3}}$ or $a = \dfrac{2}{-\sqrt{3}}$.
- Clear the root from the bottom by multiplying top and bottom by $\sqrt{3}$, using $\sqrt{3}\cdot\sqrt{3} = 3$.
- This gives $a = \dfrac{2\sqrt{3}}{3}$ or $a = -\dfrac{2\sqrt{3}}{3}$.
💡 Undo one multiplication by the ratio to walk back from the second term to the first.
8.NS.A.1 Step 5 Match a candidate to the choices
- The two possible first terms are $\dfrac{2\sqrt{3}}{3}$ and $-\dfrac{2\sqrt{3}}{3}$.
- Scanning the five choices, the positive one is not listed, but the negative one $-\dfrac{2\sqrt{3}}{3}$ is exactly choice (B).
- Choices (A) $-\sqrt{3}$ and (D) $\sqrt{3}$ are the ratio, not the first term, and (C) and (E) match nothing, so the answer is (B).
💡 Only one of the two honest candidates is printed among the options, so that is the possible first term.
6.EE.A.2 Let the first term be $a$ and the common ratio be $r$. Because each term is the 8.EE.A.1 Divide the fourth-term equation by the second-term equation. On the left the $a$ 8.EE.A.2 If $r^2 = 3$ then $r$ is a number whose square is $3$, and there are two such nu 7.NS.A.2 From $a r = 2$, the first term is $a = \dfrac{2}{r}$. Put in each ratio: $a = \d 8.NS.A.1 The two possible first terms are $\dfrac{2\sqrt{3}}{3}$ and $-\dfrac{2\sqrt{3}}{ Review
Reasonableness: Test the winner directly. If $a = -\dfrac{2\sqrt{3}}{3}$, then for the second term to be $2$ the ratio must be $r = \dfrac{2}{a} = \dfrac{2}{-2\sqrt{3}/3} = -\sqrt{3}$. Building the sequence: first $-\dfrac{2\sqrt{3}}{3}$, second $-\dfrac{2\sqrt{3}}{3}\cdot(-\sqrt{3}) = \dfrac{2\cdot 3}{3} = 2$ (correct), third $2\cdot(-\sqrt{3}) = -2\sqrt{3}$, fourth $-2\sqrt{3}\cdot(-\sqrt{3}) = 2\cdot 3 = 6$ (correct). The second and fourth terms come out to $2$ and $6$, so choice (B) genuinely works.
Alternative: Tool #3 (Eliminate Possibilities) can be run more aggressively from the start: since $r^2 = 3$ forces $r = \pm\sqrt{3}$, the first term must be $\dfrac{2}{\pm\sqrt{3}} = \pm\dfrac{2\sqrt{3}}{3}$, a number whose size is about $\pm 1.15$. Only choice (B) has that magnitude, since (A) and (D) are about $\pm 1.73$ and (C) is about $\pm 0.58$ — this rough sizing alone points to (B).
CCSS standards used (min grade 8)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the first term $a$ and ratio $r$ and turning the two given terms into the equations $ar = 2$ and $ar^3 = 6$.)8.EE.A.1Know and apply the properties of integer exponents (Dividing $ar^3$ by $ar$ so that $r^3 \div r = r^2$, cancelling the first term to leave $r^2 = 3$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Solving $r^2 = 3$ and keeping both roots $r = \sqrt{3}$ and $r = -\sqrt{3}$.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Recovering $a = \dfrac{2}{r}$ for each ratio and rationalizing $\dfrac{2}{\sqrt{3}}$ into $\dfrac{2\sqrt{3}}{3}$.)8.NS.A.1Know that numbers that are not rational are called irrational numbers (Recognising that $\pm\dfrac{2\sqrt{3}}{3}$ are irrational and matching the negative one to choice (B).)
⭐ Name the first term and the ratio, divide the fourth-term equation by the second to get the ratio squared, and remember the square root has two signs.
⭐ Name the first term and the ratio, divide the fourth-term equation by the second to get the ratio squared, and remember the square root has two signs.
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