AMC 10 · 2007 · #7
Grade 7 algebraPick an answer.
Two unknowns really run this problem: where the sequence sits and how fast it climbs. Tool #4 (Introduce a Variable) names the second one f, the common difference. The choice of anchor matters — anchoring at a makes the sum 5a+10f, which mixes both unknowns, while anchoring at the middle term c makes the sequence symmetric and lets Tool #15 (Organize Information in More Ways) regroup the sum into mirror pairs so that f cancels. That settles one term. But the question asks which term can be found, and a sum collapsing to 5c only proves c is forced; it proves nothing about the other four. Claiming "we can't find a because we don't know f" describes our ignorance, not the mathematics. So the second half of the work is Tool #3 (Eliminate Possibilities): write down the complete family of legal sequences, then exhibit two of them that disagree on a, b, d, and e. Disagreement is proof of undetermined; failure to find is not.
Anchor the sequence at its middle term
Anchoring at the middle term writes all five symmetrically.
Measuring from the center makes the list symmetric, so the steps taken left and right are set up to cancel.
6.EE.B.6Introduce A VariableAdd the five terms and watch f cancel
Adding makes the common difference cancel.
Two terms equally far from the center overshoot and undershoot by the same amount, so their gaps cancel and only the center survives.
Two terms equally far from the centre overshoot and undershoot by the same amount, so only the centre survives.
▸ Why?
The list climbs by the same fixed step, so a step out to the right is matched by an equal step to the left.
▸ Why?
Terms placed symmetrically about the middle always pair to the same total, namely twice the middle.
Solve for the middle term
So the middle term is forced to be 6.
The whole sum is exactly five copies of the middle term, so dividing by 5 hands it straight back.
6.EE.B.7Introduce A VariableWrite down every sequence that works
Every difference still works, so the family is infinite.
The conditions pin one number and leave one dial free, so a term is determined only if turning the dial does not move it.
6.EE.B.5Introduce A VariableTwo legal sequences settle the rest
Two legal sequences agree only in the middle, so the answer is the middle term, choice (C).
Showing two legal sequences that disagree on a term is what proves the term is genuinely undetermined.
4.OA.C.5Eliminate PossibilitiesFive terms in an arithmetic sequence always add up to five times the middle one, so the total pins down the middle term — and two different sequences with the same total, like 6,6,6,6,6 and 4,5,6,7,8, show nothing else is pinned down.
- Anchor the sequence at its middle term
- Add the five terms and watch f cancel
- Solve for the middle term
- Write down every sequence that works
- Two legal sequences settle the rest