AMC 10 · 2007 · #15

Grade 11 algebra

Archetype Geometric Series Share └ Sub-archetype Cyclic Participant Share

sequences-geometricgeometric-series-infinitesystems-of-equations identify-subproblemsconvert-to-algebra ↑ Prerequisites: geometric-series-infinite

📏 Medium solution 💡 3 insights

Problem

A geometric series has a known total, and the terms in odd positions have their own known total. Find the sum of the first term and the ratio.

Pick an answer.

(A)

$\frac{4}{3}$

(B)

$\frac{12}{7}$

(C)

$\frac{3}{2}$

(D)

$\frac{7}{3}$

(E)

$\frac{5}{2}$

How to solve

Strategy Organize Information in More Ways

Two facts are given about one series, so the series has to be looked at in two different arrangements (Tool #15). Sorting the terms into an even-power pile and an odd-power pile (Tool #7) is the move that pays: the piles must add back to 7, so the even pile is 7 - 3 = 4 for free, and lining the piles up term by term shows the odd pile is exactly r times the even pile. That single structural fact turns the problem into the linear equation 3 = 4r (Tool #13), no series formula required. Only afterwards is the closed form a/(1-r) = 7 needed, to work backwards from the total to the first term (Tool #11). The last step is not decoration: everything before it shows only what a and r must be, so the candidate pair gets substituted back to confirm such a series actually exists (Tool #6).

1STEP 1

Unpack what having a sum means

Having a sum at all restricts the ratio.

|r| ＜ 1, a/(1-r) = 7

2STEP 2

Sort the terms by parity

Sorting by position splits the total into two piles.

E + O = 7, O = 3 → E = 4

3STEP 3

Odd pile equals r times even

One pile is the ratio times the other, giving one equation.

O = rE → 3 = 4r

4STEP 4

Solve for the ratio

Solving gives the ratio as 3/4.

4r = 3 → r = 3/4

5STEP 5

Work back to the first term

The total then gives the first term as 7/4.

a = 7(1-r) = 7 · 1/4 = 7/4

6STEP 6

Verify, then add

Adding gives 5/2, choice (D).

a + r = 7/4 + 3/4 = 10/4 = 5/2

Answer

5/2

The odd pile (3) came out smaller than the even pile (4), which is exactly what r = 3/4 ＜ 1 predicts, since each odd term sits right after a larger even term. Listing the terms 7/4, 21/16, 63/64, … ≈ 1.75, 1.31, 0.98, 0.74, 0.55, 0.42 and adding the first six gives about 5.75, still climbing toward 7 — consistent. The tempting slip is to read the ratio as E/O = 4/3 instead of O/E = 3/4, and that gets killed immediately: r = 4/3 ＞ 1 makes the terms grow, so the series would have no sum and could not have been given as 7. Finally a + r = 5/2 = 2.5, matching choice (E).

💡Key takeaway

Sort the terms into an odd pile and an even pile: the piles must add back to the whole, and the odd pile is exactly r times the even one, so the two given sums hand you the ratio.

Unpack what having a sum means
Sort the terms by parity
Odd pile equals r times even
Solve for the ratio
Work back to the first term
Verify, then add