AMC 10 · 2010 · #10
Grade 8 algebraPick an answer.
The problem names p and q but hides the quantity the question really depends on: the common difference. Introduce it as a third variable d, and the single word arithmetic turns into equations. Two of the terms, 3p - q and 3p + q, are built from the same pair, so their gap collapses to 2q on sight; spotting that pattern hands over d almost for free. Once p, q, and d are known, the sequence gets reorganized from a step-by-step rule into a direct formula for the nth term, which is what a jump to term 2010 needs. Elimination from the answer choices is listed last on purpose: it works as a final check, but as the review shows it quietly assumes something the problem never states.
Name the common difference
Every neighbouring pair shares one common gap.
The word arithmetic is really a statement about gaps, so giving the gap a name turns a description into equations.
6.EE.B.6Introduce A VariableThe last two terms hand you d
The last two terms hand the gap over directly.
A pair written as something minus q and something plus q always differs by exactly 2q, whatever the something is.
A pair written as something minus a step and something plus that step differs by twice the step, whatever the something is.
▸ Why?
The shared middle value cancels in the subtraction, leaving only the two steps.
▸ Why?
An arithmetic sequence is defined by having the same gap between neighbours, so that gap is the whole hypothesis.
Turn the other two gaps into equations
The other gaps become two plain equations.
Each equal gap is worth one equation, and two equations in two unknowns is exactly enough to pin p and q down.
8.EE.C.8Introduce A VariableSolve the two equations
Solving gives a first term of 5 and a gap of 4.
Substitution collapses two unknowns into one equation with a single unknown, and that leaves exactly one possible pair.
8.EE.C.8Introduce A VariableCheck that such a sequence exists
The opening terms really do step by 4.
Solving tells you what the numbers would have to be; substituting back is what proves they actually work.
4.OA.C.5Look For A PatternWrite a direct rule for term n
A direct rule beats listing two thousand terms.
Reaching term n costs n - 1 gaps, not n, because the first term is already there before any gap is crossed.
8.F.B.4Organize Information In More WaysEvaluate the rule at n = 2010
The rule gives 8041, choice (A).
A direct formula lands on term 2010 in one multiplication instead of 2009 separate additions.
5.NBT.B.5Organize Information In More WaysWhen a sequence adds the same amount every time, name that amount first, then count gaps carefully: term 2010 is 2009 gaps past the start, not 2010.
- Name the common difference
- The last two terms hand you d
- Turn the other two gaps into equations
- Solve the two equations
- Check that such a sequence exists
- Write a direct rule for term n
- Evaluate the rule at n = 2010