Arithmetic Sequence: Calculate Total Collection After 35 Contributions
What This Problem Teaches
- Recognizing arithmetic sequences in real-world scenarios with different starting values
- Distinguishing between the number of contributions and the sequence term number
- Applying the explicit formula for arithmetic sequences: aₙ = a₁ + (n-1)d
- Understanding how initial conditions affect sequence indexing
- Connecting sequence notation to practical counting situations
Solution: The Sequence Model Approach
This problem is asking us to model a growing collection as an arithmetic sequence. The key insight is recognizing that the total amount in the collection after each person contributes forms a sequence.
Step 1 — Set up the sequence
Let's define our sequence where each term represents the total amount in the collection:
- a₁ = $5 (initial amount, before anyone contributes)
- a₂ = $5 + $2 = $7 (after 1st person contributes)
- a₃ = $7 + $2 = $9 (after 2nd person contributes)
- a₄ = $9 + $2 = $11 (after 3rd person contributes)
This is an arithmetic sequence with first term a₁ = 5 and common difference d = 2.
Step 2 — Determine which term we need
Here's where students often get confused: after the 35th person contributes, we need the 36th term of our sequence. Why? Because our first term represents the initial state (before any contributions), so the nth contribution corresponds to the (n+1)th term.
After 35 people contribute, we want a₃₆.
Step 3 — Apply the arithmetic sequence formula
The formula for the nth term of an arithmetic sequence is:
Substituting our values:
a₃₆ = 5 + 35 × 2
a₃₆ = 5 + 70
a₃₆ = 75
Solution: Method 2 — Direct Counting Approach
Instead of thinking about sequence terms, we can approach this by separating the initial amount from the total contributions.
Step 1 — Identify the components
The final total consists of:
- Initial amount: $5
- Total contributions from 35 people: 35 × $2
Step 2 — Calculate total contributions
Step 3 — Add initial amount
Final total = $5 + $70 = $75
This method is more intuitive for many students because it directly mirrors the problem's language: "start with $5, then 35 people each add $2."
Verification
Let's verify our answer using both perspectives:
Sequence verification
Using our sequence formula: a₃₆ = 5 + (36-1) × 2 = 5 + 70 = 75 ✓
Direct counting verification
Initial amount + contributions: $5 + 35 × $2 = $5 + $70 = $75 ✓
Pattern check
Let's check a smaller case. After the 3rd person contributes:
- Sequence method: a₄ = 5 + (4-1) × 2 = 5 + 6 = $11
- Direct method: $5 + 3 × $2 = $5 + $6 = $11
- Manual count: $5 → $7 → $9 → $11 ✓
Watch Out For These
Using a₃₅ instead of a₃₆. Students often think "35th person means 35th term," but after 35 contributions we need the 36th term because the initial amount is the first term.
Calculating only 35 × $2 = $70 and forgetting to add the initial $5. This treats the problem as if the collection started empty.
Thinking the initial $5 is also a $2 contribution, leading to 36 × $2 = $72. The initial amount is separate from the pattern of contributions.
The Pattern Behind This
This problem illustrates a general pattern for arithmetic sequences that start with an "offset" value:
Or in sequence notation:
where n is the number of times the pattern repeats. This shows up frequently in real-world problems:
- Bank account: initial balance + deposits
- Height growth: birth height + yearly increases
- Temperature change: starting temp + hourly changes
The key insight is that real-world sequences often don't start at zero, so the indexing requires careful attention to what counts as the "first term."
How to Spot This Problem Type
Look for these telltale signs that signal an arithmetic sequence approach:
- "Initial amount" + "each person adds the same amount" — classic arithmetic sequence setup
- "After the nth person..." — this phrasing often creates the off-by-one indexing situation
- Any scenario with a starting value followed by regular, equal increments
- "How much total after..." rather than "how much did the nth person add" — asking for cumulative rather than individual terms
This same mathematical structure appears in:
- Savings account problems (initial deposit + regular additions)
- Height tracking (starting height + monthly growth)
- Inventory problems (initial stock + daily deliveries)
- Distance problems (head start + regular intervals)
What If?
Using a₃₆ = a₁ + (n-1)d, we know the total after n contributions is the (n+1)th term: aₙ₊₁ = 5 + n × 2
95 = 5 + n × 290 = n × 2n = 45
After 45 people contribute: $5 + 45 × $2 = $5 + $90 = $95 ✓
Answer: 45 people have contributed.
First term a₁ = $12, common difference d = $3
After 28 contributions, we need the 29th term: a₂₉
a₂₉ = 12 + (29-1) × 3 = 12 + 28 × 3 = 12 + 84 = 96
Direct method: $12 + 28 × $3 = $12 + $84 = $96 ✓
Answer: $96 is in the collection.
We need: 8 + n × 4 > 200 where n is the number of contributors
n × 4 > 192n > 48
Since n must be a whole number of people, the minimum is n = 49
With 48 people: $8 + 48 × $4 = $200 (equals, doesn't exceed)
With 49 people: $8 + 49 × $4 = $204 (exceeds) ✓
Answer: At least 49 people are needed.
The contributions themselves form an arithmetic sequence: 3, 6, 9, 12, ... with a₁ = 3 and d = 3
We need the sum of the first 15 terms: S₁₅ = n/2 × (2a₁ + (n-1)d)S₁₅ = 15/2 × (2×3 + 14×3) = 15/2 × (6 + 42) = 15/2 × 48 = 360
Total = Initial + Sum of contributions = $10 + $360 = $370
After 2 people: $10 + $3 + $6 = $19
Using formula: $10 + 2/2×(6+3) = $10 + 9 = $19 ✓
Answer: $370 is in the collection.
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2026-06-06