AMC 10 · 2022 · #15

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum

sequences-arithmeticratio-proportionconvert-to-algebrapattern-recognition convert-to-algebraeasier-related-problempattern-recognition ↑ Prerequisites: sequences-arithmeticratio-proportion

📏 Medium solution 💡 2 insights

Problem

An arithmetic sequence has first term a and common difference 2. Let S_n be the sum of its first n terms. The ratio

\frac{S_{3n}}{S_n}

comes out to the same value no matter which n you plug in. Find S₂0.

Pick an answer.

(A)

340

(B)

360

(C)

380

(D)

400

(E)

420

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Easier Related Problem

Tool #9 (Easier Problem): try a few simple values of a and watch the ratio $\frac{S_3}{S_1}$ , $\frac{S_6}{S_2}$ , $\frac{S_9}{S_3}$ . The ratio should be the same number every time. If a = 1 the sequence is the odd numbers 1, 3, 5, …, and S_n = n² — a pattern that even a 4th grader recognizes. So $\frac{S_{3n}}{S_n}$ = $\frac{(3n)²}{n²}$ = 9 for every n. Tool #5 (Pattern) confirms the odd-number sum law 1 + 3 + 5 + … + (2n - 1) = n². Tool #6 (Guess and Check) tests a = 2, 3 etc. and shows the ratio depends on n unless a = 1. Tool #13 (Algebra) gives a clean back-up: S_n = n(a + n - 1), and forcing the ratio to be constant forces a - 1 = 0. Tool #3 (Eliminate) matches S₂0 = 400 to choice (D).

1STEP 1

Simplest first term a = 1 gives the odd numbers 1, 3, 5, …, whose running sums are the perfect squares, so S_n = n².

S_n = 1 + 3 + 5 + … + (2n - 1) = n²

2STEP 2

With S_n = n², S₃n = 9n², so $\frac{S_{3n}}{S_n}$ = 9 for every n — the constant-ratio condition holds, confirming a = 1.

\frac{S_{3n}}{S_n}

\frac{(3n)²}{n²}

= 9 (for every n)

3STEP 3

Test a = 2: then $\frac{S_{3n}}{S_n}$ = $\frac{3(3n + 1)}{(n + 1)}$ still contains n, so a = 2 fails — only a = 1 keeps the ratio fixed.

a = 2 →

\frac{S_{3n}}{S_n}

\frac{3(3n+1)}{n+1}

(depends on n)

4STEP 4

Algebra back-up: S_n = n(a + n - 1) gives a ratio constant for all n only when a - 1 = 0, i.e. a = 1.

\frac{S_{3n}}{S_n}

\frac{9n + (3a - 3)}{n + (a - 1)}

constant → a = 1

5STEP 5

Compute S₂0: with a = 1, S_n = n², so S₂0 = 20² = 400.

S₂0 = 20² = 400

6STEP 6

Match 400 to the answer choices: that is (D).

400 → (D)

Answer

400

Spot-check the sum directly with the standard arithmetic-sum formula. With a₁ = 1, a₂0 = 1 + 19 · 2 = 39, S₂0 =

\frac{20}{2}

· (1 + 39) = 10 · 40 = 400. Matches. Also S_n = n² for the odd-number sequence is well-known: 1, 4, 9, 16, 25, … — perfect squares. And

\frac{S_{3n}}{S_n}

= 9 is what you'd expect intuitively, since tripling the index of a perfect-square sequence should multiply by 9.

💡Key takeaway

This AMC 10 problem only needs Grade 8 linear-function reasoning you already know — the constant-ratio condition forces the sequence to be the odd numbers 1, 3, 5, …, whose sums follow the famous square pattern S_n = n², so S₂0 = 400.

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