AMC 10 · 2003 · #8

Grade 8 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

sequences-geometricexponentssigned-square-root work-backwards ↑ Prerequisites: sequences-geometric

📏 Medium solution 💡 2 insights

Problem

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.

Pick an answer.

(A)

$-\sqrt{3}$

(B)

$-\frac{2\sqrt{3}}{3}$

(C)

$-\frac{\sqrt{3}}{3}$

(D)

$\sqrt{3}$

(E)

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

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 ar³ = 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.

1STEP 1

Name the unknowns, write the facts

Let the first term be $a$ and the ratio be $r$ . Then the second term is $ar$ and the fourth is $ar^3$ , so the two facts become two equations.

a r = 2 a r³ = 6

2STEP 2

Divide to cancel the first term

Divide the fourth-term equation by the second-term one: $a$ cancels, the right side is $6/2$ , and what is left is $r^2 = 3$ .

(a r³)/(a r) = 6/2 → r² = 3

3STEP 3

Take the square root — keep both signs

Two numbers square to $3$ , so $r = \pm\sqrt{3}$ ; both are legal ratios for a geometric sequence, so both travel on.

r² = 3 → r = √(3) or r = -√(3)

4STEP 4

Step back to the first term

From $ar = 2$ the first term is $a = 2/r$ , and rationalizing $2/(\pm\sqrt{3})$ gives $a = \pm\frac{2\sqrt{3}}{3}$ .

a = 2/±√(3) = ±2/√(3)·√(3)/√(3) = ±2√(3)/3

5STEP 5

Match a candidate to the choices

Only the negative candidate is printed, choice (B); $\pm\sqrt{3}$ is the ratio, not the first term, and (C) and (E) match neither.

a ∈ { 2√(3)/3, -2√(3)/3 } → -2√(3)/3 = (B)

Answer

-2√(3)/3

Test the winner directly. If a = -2√(3)/3, then for the second term to be 2 the ratio must be r = 2/a = 2/(-2√(3)/3) = -√(3). Building the sequence: first -2√(3)/3, second -2√(3)/3·(-√(3)) = (2 · 3)/3 = 2 (correct), third 2·(-√(3)) = -2√(3), fourth -2√(3)·(-√(3)) = 2 · 3 = 6 (correct). The second and fourth terms come out to 2 and 6, so choice (B) genuinely works.

💡Key takeaway

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 unknowns, write the facts
Divide to cancel the first term
Take the square root — keep both signs
Step back to the first term
Match a candidate to the choices

Name the unknowns, write the facts

Divide to cancel the first term

Take the square root — keep both signs

Step back to the first term

Match a candidate to the choices

More like this