AMC 10 · 2022 · #20
Grade 8 arithmeticProblem
A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are 57, 60, and 91. What is the fourth term of this sequence?
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: A new 4-term sequence is the term-by-term sum of a 4-term arithmetic sequence of positive integers and a 4-term geometric sequence of positive integers. The first three sums are $57$, $60$, $91$. Find the fourth sum.
Givens: Arithmetic: $a, a+d, a+2d, a+3d$ (positive integers); Geometric: $b, br, br^2, br^3$ (positive integers); $a + b = 57$, $(a + d) + br = 60$, $(a + 2d) + br^2 = 91$; Choices: (A) $190$, (B) $194$, (C) $198$, (D) $202$, (E) $206$
Unknowns: $S_4 = (a + 3d) + br^3$
Understand
Restated: A new 4-term sequence is the term-by-term sum of a 4-term arithmetic sequence of positive integers and a 4-term geometric sequence of positive integers. The first three sums are $57$, $60$, $91$. Find the fourth sum.
Givens: Arithmetic: $a, a+d, a+2d, a+3d$ (positive integers); Geometric: $b, br, br^2, br^3$ (positive integers); $a + b = 57$, $(a + d) + br = 60$, $(a + 2d) + br^2 = 91$; Choices: (A) $190$, (B) $194$, (C) $198$, (D) $202$, (E) $206$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #7 Identify Subproblems, #6 Guess and Check, #3 Eliminate Possibilities
Four unknowns ($a, d, b, r$) and three given sums — underdetermined unless we use the positive-integer constraint. Tool #13 (Convert to Algebra) writes the three sums as equations. Tool #7 (Identify Subproblems): take successive differences to kill $a$, then take another difference to kill $d$, leaving a single Diophantine equation in $b$ and $r$. Tool #6 (Guess and Check) over the few divisor cases of $28 = b(r-1)^2$ narrows the candidates. Tool #3 (Eliminate Possibilities) discards the case that fails the positive-integer requirement on the arithmetic sequence.
Execute — Answer: E
8.EE.C.8 Step 1 Write the three given sums as equations.
💡 Translate the three numeric facts into a clean algebraic system.
8.EE.C.8 Step 2 - Subproblem A: kill $a$.
- Subtract equation (1) from (2), and (2) from (3).
💡 Consecutive differences erase the constant first-term $a$.
7.EE.A.1 Step 3 - Subproblem B: kill $d$.
- Subtract the first equation in the new system from the second.
💡 Factoring $b(r-1)$ out leaves a clean $b(r-1)^2 = 28$ — a single equation in $b, r$.
6.NS.B.4 Step 4 - Use the positive-integer constraint to limit cases.
- Since all geometric terms $b, br, br^2, br^3$ must be positive integers and the sequence has $\ge 2$ distinct values (otherwise $b(r-1)^2 = 0$, not $28$), $r$ is a positive integer $\ne 1$.
- So $(r-1)^2$ is a positive perfect square dividing $28$.
- Divisors of $28$ that are perfect squares: $1$ and $4$.
💡 Only two perfect-square divisors of $28$ — exactly two cases to check.
6.EE.B.5 Step 5 - Case I: $r = 2$, so $(r-1)^2 = 1$ and $b = 28$.
- From $d + b(r-1) = 3$: $d + 28 = 3$, so $d = -25$.
- From $a + b = 57$: $a = 29$.
- Check arithmetic sequence: $29, 4, -21, -46$ — third term is negative.
- Violates the positive-integer requirement.
💡 Verify the positivity of every term, not just the visible ones.
6.EE.B.5 Step 6 - Case II: $r = 3$, so $(r-1)^2 = 4$ and $b = 7$.
- From $d + b(r-1) = 3$: $d + 14 = 3$, so $d = -11$.
- From $a + b = 57$: $a = 50$.
- Check both sequences.
- Arithmetic: $50, 39, 28, 17$ — all positive integers.
- Geometric: $7, 21, 63, 189$ — all positive integers.
- Valid.
💡 Both sequences must consist of positive integers — Case II passes.
4.NBT.B.4 Step 7 Compute the fourth sum: $S_4 = (a + 3d) + br^3 = 17 + 189 = 206$.
💡 Add the fourth term of each parent sequence.
8.EE.C.8 Write the three given sums as equations. 8.EE.C.8 Subproblem A: kill $a$. Subtract equation (1) from (2), and (2) from (3). 7.EE.A.1 Subproblem B: kill $d$. Subtract the first equation in the new system from the s 6.NS.B.4 Use the positive-integer constraint to limit cases. Since all geometric terms $b 6.EE.B.5 Case I: $r = 2$, so $(r-1)^2 = 1$ and $b = 28$. From $d + b(r-1) = 3$: $d + 28 = 6.EE.B.5 Case II: $r = 3$, so $(r-1)^2 = 4$ and $b = 7$. From $d + b(r-1) = 3$: $d + 14 = 4.NBT.B.4 Compute the fourth sum: $S_4 = (a + 3d) + br^3 = 17 + 189 = 206$. Review
Reasonableness: Cross-verify all three given sums with $a = 50, d = -11, b = 7, r = 3$. $S_1 = 50 + 7 = 57$ ✓. $S_2 = 39 + 21 = 60$ ✓. $S_3 = 28 + 63 = 91$ ✓. Differences of the four sums: $S_2 - S_1 = 3$, $S_3 - S_2 = 31$, $S_4 - S_3 = 206 - 91 = 115$. Second differences: $31 - 3 = 28$, $115 - 31 = 84 = 3 \cdot 28$. The factor of $3$ matches $r$, confirming the geometric ratio.
Alternative: Tool #6 (Guess and Check) using small $r$: since the geometric values explode quickly, try $r = 2$ first (geometric $b, 2b, 4b, 8b$) — fails the positive-arithmetic check above. Then $r = 3$ (geometric $b, 3b, 9b, 27b$) — works with $b = 7$. Larger $r$ blows up: $r = 4$ would need $b(3)^2 = 28$, i.e. $b = 28/9$, non-integer. So $r = 3$ is forced without ever solving the Diophantine equation directly.
CCSS standards used (min grade 8)
8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Setting up the three-equation system in $a, d, b, r$ and reducing it by consecutive subtraction.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Factoring $b(r-1)(r-1) = 28$ out of the difference of the two reduced equations.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Listing divisors of $28$ that are perfect squares (only $1$ and $4$) to pin down $(r-1)^2$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Checking each candidate $(a, d, b, r)$ against the positivity requirement on every arithmetic and geometric term.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing the fourth sum $17 + 189 = 206$.)
⭐ Three sums and four unknowns — but the positive-integer rule does the rest. Subtract consecutive equations twice and you get $b(r-1)^2 = 28$, so $(r-1)^2 \in \{1, 4\}$. Only $r = 3, b = 7$ keeps both sequences positive ($50, 39, 28, 17$ and $7, 21, 63, 189$), giving $S_4 = 17 + 189 = \textbf{(E) }206$.
⭐ Three sums and four unknowns — but the positive-integer rule does the rest. Subtract consecutive equations twice and you get $b(r-1)^2 = 28$, so $(r-1)^2 \in \{1, 4\}$. Only $r = 3, b = 7$ keeps both sequences positive ($50, 39, 28, 17$ and $7, 21, 63, 189$), giving $S_4 = 17 + 189 = \textbf{(E) }206$.