AMC 10 · 2002 · #19
Grade 6 arithmeticSuppose that {an} is an arithmetic sequence with
a1+a2+⋯+a100=100 and a101+a102+⋯+a200=200.
What is the value of a_2 - a_1 ?
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: An arithmetic sequence $\{a_n\}$ has a fixed step between neighboring terms. The first $100$ terms add up to $100$, and the next $100$ terms (terms $101$ through $200$) add up to $200$. Find the size of one step, $a_2 - a_1$.
Givens: $\{a_n\}$ is arithmetic: each term is the previous term plus the same fixed step; $a_1 + a_2 + \cdots + a_{100} = 100$; $a_{101} + a_{102} + \cdots + a_{200} = 200$; Answer choices: (A) $0.0001$, (B) $0.001$, (C) $0.01$, (D) $0.1$, (E) $1$
Unknowns: The common step $a_2 - a_1$ (the common difference $d$)
Understand
Restated: An arithmetic sequence $\{a_n\}$ has a fixed step between neighboring terms. The first $100$ terms add up to $100$, and the next $100$ terms (terms $101$ through $200$) add up to $200$. Find the size of one step, $a_2 - a_1$.
Givens: $\{a_n\}$ is arithmetic: each term is the previous term plus the same fixed step; $a_1 + a_2 + \cdots + a_{100} = 100$; $a_{101} + a_{102} + \cdots + a_{200} = 200$; Answer choices: (A) $0.0001$, (B) $0.001$, (C) $0.01$, (D) $0.1$, (E) $1$
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #4 Introduce a Variable, #5 Look for a Pattern
The obvious route — find $a_1$ and the step $d$ separately from the two sum equations — is more work than needed. Tool #15 (Organize Information in More Ways) reorganizes the two totals: instead of adding each block up, subtract the second total from the first block's total term by term. Tool #4 (Introduce a Variable) names the step $d = a_2 - a_1$, which is exactly the unknown, and writes any term $100$ places ahead as $a_{k+100} = a_k + 100d$. Tool #5 (Look for a Pattern) is the payoff: every matched pair in the subtraction differs by the same $100d$, so the messy difference of sums collapses into a single count times $d$, leaving one clean equation to solve.
Execute — Answer: C
6.EE.B.6 Step 1 Name the common step d
- In an arithmetic sequence every term is the one before it plus a single fixed number.
- Call that step $d$, so $d = a_2 - a_1$ is exactly what the problem asks for.
- Because the step never changes, moving ahead by $100$ terms adds the step $100$ times over: $a_{k+100} = a_k + 100d$ for every position $k$.
- For instance $a_{101} = a_1 + 100d$, $a_{102} = a_2 + 100d$, and so on up to $a_{200} = a_{100} + 100d$.
💡 One fixed step repeated $100$ times means each term in the second block is just its partner in the first block, lifted by the same amount $100d$.
6.EE.A.3 Step 2 Subtract the two sums term by term
- Rather than compute each block on its own, subtract the first block's total from the second block's total, matching terms as partners: $a_{101}$ with $a_1$, $a_{102}$ with $a_2$, all the way to $a_{200}$ with $a_{100}$.
- Each partner difference is $a_{k+100} - a_k = 100d$ — the very same value every single time.
- There are $100$ such pairs, so the whole subtraction is $100$ copies of $100d$, giving $100 \times 100d = 10000d$.
- But that same subtraction, read off the given totals, is $200 - 100 = 100$.
- So $10000d = 100$.
💡 Lining up matched partners cancels all the unknown starting values at once and leaves only the step, repeated a countable number of times.
6.EE.B.7 Step 3 Solve for the step
- Now it is one clean equation: $10000d = 100$.
- Divide both sides by $10000$ to get $d = \frac{100}{10000} = \frac{1}{100} = 0.01$.
- Since $a_2 - a_1 = d$, the step is $0.01$, which is choice (C).
💡 Once every unknown but the step has been cancelled away, a single division uncovers it.
6.EE.B.6 In an arithmetic sequence every term is the one before it plus a single fixed nu 6.EE.A.3 Rather than compute each block on its own, subtract the first block's total from 6.EE.B.7 Now it is one clean equation: $10000d = 100$. Divide both sides by $10000$ to ge Review
Reasonableness: Check the scale. The two blocks differ by only $100$ (that is $200 - 100$), yet that gap is built from $10000$ tiny copies of the step, so the step must be a small fraction — about $\frac{100}{10000} = 0.01$. This rules out the big choices (D) $0.1$ and (E) $1$, which would make the blocks differ by $1000$ or $10000$, not $100$. It also rules out (A) $0.0001$ and (B) $0.001$, which would leave a gap of only $1$ or $10$. Only (C) $0.01$ produces the required gap of exactly $100$.
Alternative: Use the arithmetic-sum formula directly. The sum of $n$ terms is $\frac{n}{2}\left(2a_1 + (n-1)d\right)$. The first block gives $\frac{100}{2}\left(2a_1 + 99d\right) = 100$, so $2a_1 + 99d = 2$. Both blocks together are all $200$ terms summing to $100 + 200 = 300$, giving $\frac{200}{2}\left(2a_1 + 199d\right) = 300$, so $2a_1 + 199d = 3$. Subtracting the two equations kills $a_1$: $100d = 1$, hence $d = 0.01$ — the same answer.
CCSS standards used (min grade 6)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the common step $d = a_2 - a_1$ and writing any term $100$ places ahead as $a_{k+100} = a_k + 100d$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Recognizing the term-by-term subtraction as $100$ equal copies of $100d$ and combining them into $10000d$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $10000d = 100$ by dividing both sides by $10000$ to get $d = 0.01$.)
⭐ When two equal-length blocks of an arithmetic sequence sit a fixed distance apart, subtract them partner by partner — the unknown starting values cancel and only the repeated step is left.
⭐ When two equal-length blocks of an arithmetic sequence sit a fixed distance apart, subtract them partner by partner — the unknown starting values cancel and only the repeated step is left.
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