AMC 10 · 2013 · #7

Grade 4 patternalgebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps
recursive-sequencepattern-recognitionmulti-digit-arithmetic work-backwardspattern-recognition ↑ Prerequisites: recursive-sequence
📏 Medium solution 💡 2 insights
📘 View easy version →
Problem
Each term adds the two before it, and two later terms are known. Find an earlier term.

Pick an answer.

(A)
4
(B)
6
(C)
10
(D)
12
(E)
16
How to solve
Strategy Work Backwards

The rule points forward, but the known terms sit above the wanted term, so run the rule in reverse. The rule says that in any block of three consecutive terms, the last is the sum of the first two; addition and subtraction are inverses, so knowing any two terms of such a block forces the third. Starting from the block (S₇, S₈, S₉), which already has two known members, each new block picks up a freshly filled term and hands back one more, walking down the list to S₄. Because every filled value is forced, the chain proves S₄ can only be one number; a forward run then confirms such a sequence really exists.

1STEP 1

Turn the rule into a filling tool

The rule read backwards becomes a subtraction.

S_n = S_n-2 + S_n-1 ⇔ S_n-1 = S_n - S_n-2 ⇔ S_n-2 = S_n - S_n-1
2STEP 2

Fill the gap between the givens

The gap between the givens fills in first.

S₈ = S₉ - S₇ = 110 - 42 = 68
3STEP 3

Step down to S₆

Stepping down gives the next term, 26.

S₆ = S₈ - S₇ = 68 - 42 = 26
4STEP 4

Step down to S₅, then S₄

Two more steps reach 10.

S₅ = S₇ - S₆ = 42 - 26 = 16, S₄ = S₆ - S₅ = 26 - 16 = 10
5STEP 5

Check the sequence really exists

Rebuilding the list confirms it, choice (A).

10, 16, 26, 42, 68, 110 → S₇ = 42 ✓, S₉ = 110 ✓
Answer
10
A size estimate made without any of the arithmetic above agrees. A sum-of-the-previous-two sequence multiplies by roughly 1.618 at each step once it settles, so over the three steps from S₄ to S₇ it grows by about 1.618³ ≈ 4.24. That predicts S₄ ≈ 42/4.24 ≈ 9.9, which singles out 10 and rules out 4, 6, and 16 outright while leaving 12 clearly too big. The exact values also behave: pushing the backward rule past S₄ gives S₃ = 16 - 10 = 6, S₂ = 10 - 6 = 4, S₁ = 6 - 4 = 2, so the whole list 2, 4, 6, 10, 16, 26, 42, 68, 110, 178 stays in positive whole numbers with no contradiction anywhere.
💡Key takeaway

When three terms in a row add up in a fixed way, knowing any two of them forces the third — so you can walk backwards down the list just as easily as forwards.

  • Turn the rule into a filling tool
  • Fill the gap between the givens
  • Step down to S₆
  • Step down to S₅, then S₄
  • Check the sequence really exists

More like this

Same archetype — closest grade level first.