AMC 10 · 2009 · #7

Grade 8 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation
sequences-arithmeticlinear-equations-one-varsystems-of-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: sequences-arithmetic
📏 Medium solution 💡 2 insights
Problem
Three expressions in one letter are the first three terms of an arithmetic sequence. Find the position at which the sequence produces a given value.

Pick an answer.

(A)
255
(B)
502
(C)
1004
(D)
1506
(E)
8037
How to solve
Strategy Convert to Algebra

The word "arithmetic" is not decoration — it is an equation in disguise. Tool #13 (Convert to Algebra) turns it into a₂ - a₁ = a₃ - a₂, one equation in one unknown x. That equation carries the whole problem, so it is worth being precise about what it proves. Solving it shows only that x = 4 is forced if the sequence is arithmetic; it does not by itself show that x = 4 makes the sequence arithmetic. So Tool #6 (Guess and Check) is used deliberately in the other direction — substitute x = 4 back and confirm the three numbers really do step by a constant. With a₁ and d nailed down, Tool #5 (Look for a Pattern) reads off the explicit rule a_n = 4n + 1, and the final step must also check that 2009 is genuinely reached — that the n solving 4n + 1 = 2009 is a positive whole number and not a fraction. Tool #3 (Eliminate Possibilities) gives a fast independent estimate: terms grow by about 4 each step, so n has to land near 2009 ÷ 4.

1STEP 1

Turn "arithmetic" into an equation

Equal gaps become one equation.

(5x-11) - (2x-3) = (3x+1) - (5x-11) ⟹ 3x - 8 = -2x + 12
2STEP 2

Solve for x, and note it is unique

Solving pins the letter uniquely.

3x - 8 = -2x + 12 ⟹ 5x = 20 ⟹ x = 4 (unique, since 5 ≠ 0)
3STEP 3

Substitute back and confirm it works

Substituting back confirms the sequence really is arithmetic.

a₁ = 2(4)-3 = 5, a₂ = 5(4)-11 = 9, a₃ = 3(4)+1 = 13, 9 - 5 = 13 - 9 = 4 = d
4STEP 4

Build the rule for the nth term

The general term is a simple linear rule.

a_n = a₁ + (n-1)d = 5 + 4(n-1) = 4n + 1, a₁ = 5, a₂ = 9, a₃ = 13
5STEP 5

Find n and check 2009 is really reached

Solving gives the whole number 502, choice (A).

4n + 1 = 2009 ⟹ 4n = 2008 ⟹ n = 502 (4 ∣ 2008, so n ∈ Z^+), a₅₀₂ = 4(502)+1 = 2009
Answer
502
Size check first: the sequence climbs by 4 per step from a start of 5, so reaching about 2009 should take roughly 2009 ÷ 4 ≈ 502 steps. Only (B) 502 sits anywhere near that. The other choices are the predictable slips — (C) 1004 ≈ 2009 ÷ 2 comes from mistaking the common difference for 2, (E) 8037 ≈ 4 × 2009 multiplies where you should divide, and (D) 1506 = 3 × 502 misreads which of the three given expressions is the first term. Direct substitution settles it: a₅₀₂ = 4(502) + 1 = 2009. No other index works either, because d = 4 ＞ 0 makes the sequence strictly increasing, so each value appears at most once.
💡Key takeaway

Equal gaps is an equation — write a₂ - a₁ = a₃ - a₂, get x = 4, check it really works, then count 4s up to 2009 to land on term 502.

  • Turn "arithmetic" into an equation
  • Solve for x, and note it is unique
  • Substitute back and confirm it works
  • Build the rule for the nth term
  • Find n and check 2009 is really reached

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