AMC 10 · 2022 · #15
Grade 8 arithmeticProblem
Let Sn be the sum of the first n terms of an arithmetic sequence that has a common difference of 2. The quotient SnS3n does not depend on n. What is S20?
Pick an answer.
AMC 10 2022 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: An arithmetic sequence has first term $a$ and common difference $2$. Let $S_n$ be the sum of its first $n$ terms. The ratio $S_{3n}/S_n$ comes out to the same value no matter which $n$ you plug in. Find $S_{20}$.
Givens: Arithmetic sequence with common difference $d = 2$; First term $a$ is unknown; $S_n = a_1 + a_2 + \cdots + a_n$; $S_{3n}/S_n$ is constant — does not depend on $n$; Answer choices: (A) 340, (B) 360, (C) 380, (D) 400, (E) 420
Unknowns: First term $a$; $S_{20}$
Understand
Restated: An arithmetic sequence has first term $a$ and common difference $2$. Let $S_n$ be the sum of its first $n$ terms. The ratio $S_{3n}/S_n$ comes out to the same value no matter which $n$ you plug in. Find $S_{20}$.
Givens: Arithmetic sequence with common difference $d = 2$; First term $a$ is unknown; $S_n = a_1 + a_2 + \cdots + a_n$; $S_{3n}/S_n$ is constant — does not depend on $n$; Answer choices: (A) 340, (B) 360, (C) 380, (D) 400, (E) 420
Plan
Primary tool: #9 Easier Related Problem
Secondary: #5 Look for a Pattern, #6 Guess and Check, #13 Convert to Algebra, #3 Eliminate Possibilities
Tool #9 (Easier Problem): try a few simple values of $a$ and watch the ratio $S_{3}/S_{1}, S_{6}/S_{2}, S_{9}/S_{3}$. The ratio should be the same number every time. If $a = 1$ the sequence is the odd numbers $1, 3, 5, \ldots$, and $S_n = n^2$ — a pattern that even a 4th grader recognizes. So $S_{3n}/S_n = (3n)^2 / n^2 = 9$ for every $n$. Tool #5 (Pattern) confirms the odd-number sum law $1 + 3 + 5 + \cdots + (2n - 1) = n^2$. 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 $S_{20} = 400$ to choice (D).
Execute — Answer: D
4.OA.C.5 Step 1 - Try the simplest first term: $a = 1$, common difference 2.
- The sequence is $1, 3, 5, 7, 9, \ldots$ — the odd numbers.
- The classic pattern says the sum of the first $n$ odd numbers equals $n^2$ (you can check $1 = 1, 1+3 = 4, 1+3+5 = 9$, $\ldots$).
- So $S_n = n^2$.
💡 Grade 4 number pattern: odd numbers stack into squares — a pattern you can literally draw with dots.
6.RP.A.1 Step 2 - Check the ratio.
- With $S_n = n^2$, $S_{3n} = (3n)^2 = 9n^2$.
- So $S_{3n}/S_n = 9n^2 / n^2 = 9$, a single number for every $n$.
- The constancy condition is satisfied — so $a = 1$ is a working first term.
💡 Grade 6 ratio: $9n^2$ to $n^2$ simplifies to $9:1$, no leftover $n$ — exactly what "doesn't depend on $n$" needs.
6.RP.A.2 Step 3 - Rule out other choices of $a$.
- Try $a = 2$: sequence $2, 4, 6, 8, \ldots$, $S_n = n(n + 1)$.
- Then $S_{3n}/S_n = 3n(3n + 1)/(n(n+1)) = 3(3n + 1)/(n + 1)$, which still has $n$ inside — not constant.
- So $a = 2$ fails, and similarly $a = 0, 3, 4, \ldots$ all fail.
- Only $a = 1$ keeps the ratio fixed.
💡 Grade 6 unit-rate sense: if $n$ stays in the simplified ratio, it isn't a fixed rate.
8.F.B.4 Step 4 - Back-up algebra check.
- The arithmetic-sum formula with $d = 2$ gives $S_n = \tfrac{n}{2}(2a + 2(n - 1)) = n(a + n - 1)$.
- Then $\tfrac{S_{3n}}{S_n} = \tfrac{3(a + 3n - 1)}{a + n - 1}$.
- For this to be a single number for every $n$, the $n$-coefficients $(9 \text{ over } 1)$ must match the constant-term ratio $(3a - 3)$ over $(a - 1)$: $\tfrac{3a - 3}{a - 1} = 9$, which simplifies (when $a \neq 1$) to $3 = 9$ — impossible.
- So $a - 1 = 0$, i.e.
- $a = 1$.
💡 Grade 8 linear function: a fraction (linear in $n$) / (linear in $n$) is constant iff both linear pieces are proportional — forces $a = 1$.
3.OA.C.7 Step 5 - Compute $S_{20}$.
- With $a = 1$, $S_n = n^2$, so $S_{20} = 20^2 = 400$.
💡 Grade 3 multiplication fact: $20 \times 20 = 400$.
6.EE.B.5 Step 6 Match 400 to the answer choices: that is (D).
💡 Final compare — only (D) lands on 400.
4.OA.C.5 Try the simplest first term: $a = 1$, common difference 2. The sequence is $1, 3 6.RP.A.1 Check the ratio. With $S_n = n^2$, $S_{3n} = (3n)^2 = 9n^2$. So $S_{3n}/S_n = 9n 6.RP.A.2 Rule out other choices of $a$. Try $a = 2$: sequence $2, 4, 6, 8, \ldots$, $S_n 8.F.B.4 Back-up algebra check. The arithmetic-sum formula with $d = 2$ gives $S_n = \tfr 3.OA.C.7 Compute $S_{20}$. With $a = 1$, $S_n = n^2$, so $S_{20} = 20^2 = 400$. 6.EE.B.5 Match 400 to the answer choices: that is (D). Review
Reasonableness: Spot-check the sum directly with the standard arithmetic-sum formula. With $a_1 = 1, a_{20} = 1 + 19 \cdot 2 = 39$, $S_{20} = \tfrac{20}{2} \cdot (1 + 39) = 10 \cdot 40 = 400$. Matches. Also $S_n = n^2$ for the odd-number sequence is well-known: $1, 4, 9, 16, 25, \ldots$ — perfect squares. And $S_{3n}/S_n = 9$ is what you'd expect intuitively, since tripling the index of a perfect-square sequence should multiply by 9.
Alternative: Tool #1 (Diagram): draw the odd-number sum as L-shaped layers around a growing square. Each new odd number adds an L of size $2n + 1$ around an $n \times n$ square to make an $(n+1) \times (n+1)$ square. The picture shows $S_n = n^2$ without any algebra, and then $S_{20} = 20^2 = 400$ is immediate.
CCSS standards used (min grade 8)
3.OA.C.7Fluently multiply and divide within 100 (Computing $20 \times 20 = 400$ once $S_n = n^2$ is known.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing the pattern $1, 1 + 3 = 4, 1 + 3 + 5 = 9, \ldots$ for the sum of the first $n$ odd numbers.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Matching $S_{20} = 400$ to answer choice (D).)6.RP.A.1Understand the concept of a ratio and use ratio language (Reading $S_{3n}/S_n = (3n)^2/n^2$ as the ratio $9:1$ that does not depend on $n$.)6.RP.A.2Understand the concept of a unit rate and use rate language (Testing $a = 2$ and showing $\tfrac{3(3n+1)}{n+1}$ still depends on $n$, so the ratio is not a fixed rate.)8.F.B.4Construct a function to model a linear relationship between two quantities (Writing $S_n$ as the linear-in-$n$ expression $n(a + n - 1)$ and forcing the ratio of two linear-in-$n$ expressions to be constant — pinning down $a = 1$.)
⭐ 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 $1, 3, 5, \ldots$, whose sums follow the famous square pattern $S_n = n^2$, so $S_{20} = 400$.
⭐ 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 $1, 3, 5, \ldots$, whose sums follow the famous square pattern $S_n = n^2$, so $S_{20} = 400$.
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