Competition · AMC preparation · step 4 of 4

AMC 10 · 2022A · #20

Grade 8 arithmetic
sequences-arithmeticsequences-geometricsystems-of-equations convert-to-algebraguess-and-check ↑ Prerequisites: sequences-arithmeticsequences-geometric
📏 Long solution 💡 3 insights
Problem
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.

Pick an answer.

(A)
190
(B)
194
(C)
198
(D)
202
(E)
206

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

How to solve
Strategy Convert to Algebra

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.

1STEP 1

Write the three sums

Write the three given sums as equations.

a + b = 57 ; (a+d) + br = 60 ; (a+2d) + br² = 91
2STEP 2

Subtract to remove a

Subproblem A: kill a. Subtract equation (1) from (2), and (2) from (3).

d + b(r-1) = 3 ; d + br(r-1) = 31
3STEP 3

Subtract to remove d

Subproblem B: kill d. Subtract the first equation in the new system from the second.

br(r-1) - b(r-1) = 28 ⟹ b(r-1)² = 28
4STEP 4

Limit the cases by integers

Positive integers force r to be an integer ≠ 1, so (r-1)² is a perfect-square divisor of 28 — only 1 and 4, giving r = 2 or r = 3.

(r-1)² ∈ {1, 4} ⟹ r ∈ {2, 3}
5STEP 5

Test r equal to 2

Case I (r = 2): b = 28, d = -25, a = 29. The arithmetic sequence 29, 4, -21, … turns negative — rejected.

r=2: b=28, d=-25, a=29; arithmetic 29, 4, -21, … — invalid.
6STEP 6

Test r equal to 3

Case II (r = 3): b = 7, d = -11, a = 50. Arithmetic 50, 39, 28, 17 and geometric 7, 21, 63, 189 are all positive integers — valid.

r=3: b=7, d=-11, a=50; arithmetic 50, 39, 28, 17; geometric 7, 21, 63, 189 — valid.
7STEP 7

Compute the fourth sum

Add the fourth terms: S₄ = (a + 3d) + br³ = 17 + 189 = 206.

S₄ = 17 + 189 = 206 → (E)
Answer
206
Cross-verify all three given sums with a = 50, d = -11, b = 7, r = 3. S₁ = 50 + 7 = 57 ✓. S₂ = 39 + 21 = 60 ✓. S₃ = 28 + 63 = 91 ✓. Differences of the four sums: S₂ - S₁ = 3, S₃ - S₂ = 31, S₄ - S₃ = 206 - 91 = 115. Second differences: 31 - 3 = 28, 115 - 31 = 84 = 3 · 28. The factor of 3 matches r, confirming the geometric ratio.
💡Key takeaway

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 ∈ {1, 4}. Only r = 3, b = 7 keeps both sequences positive (50, 39, 28, 17 and 7, 21, 63, 189), giving S₄ = 17 + 189 = (E) 206.

  • Write the three sums
  • Subtract to remove a
  • Subtract to remove d
  • Limit the cases by integers
  • Test r equal to 2
  • Test r equal to 3
  • Compute the fourth sum

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