Arithmetic Sequence: Calculate Total Collection After 35 Contributions

Arithmetic Sequences 9th-10th Grade
Problem
Apply your knowledge of arithmetic sequences to this real-world question. A collection is taken up to support a family in need. The initial amount in the collection is $5. Everyone places $2 in the collection. When the 35th person puts $2 in the collection, how much is present in the collection?

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:

aₙ = a₁ + (n - 1)d

Substituting our values:

a₃₆ = 5 + (36 - 1) × 2
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

Total contributions = 35 × $2 = $70

Step 3 — Add initial amount

Final total = Initial amount + Total contributions
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."

After the 35th person contributes $2, there is $75 in the collection.

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

✗ Off-by-one error

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.

Wrong: a₃₅ = 5 + (35-1) × 2 = 5 + 68 = $73
✗ Ignoring the initial amount

Calculating only 35 × $2 = $70 and forgetting to add the initial $5. This treats the problem as if the collection started empty.

✗ Including an extra contribution

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:

Total = Initial Value + (Number of Additions) × (Common Difference)

Or in sequence notation:

aₙ₊₁ = a₁ + n × d

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?

1
Reverse the Unknown
The collection starts with $5. Each person contributes $2. After a certain number of people have contributed, the total is $95. How many people have contributed?
Step 1 — Set up the equation

Using a₃₆ = a₁ + (n-1)d, we know the total after n contributions is the (n+1)th term: aₙ₊₁ = 5 + n × 2

Step 2 — Solve for n

95 = 5 + n × 2
90 = n × 2
n = 45

Step 3 — Verify

After 45 people contribute: $5 + 45 × $2 = $5 + $90 = $95

Answer: 45 people have contributed.

2
Different Initial Amount
The collection starts with $12. Each person contributes $3. How much money is in the collection after 28 people have contributed?
Step 1 — Identify the sequence

First term a₁ = $12, common difference d = $3

Step 2 — Find the correct term

After 28 contributions, we need the 29th term: a₂₉

Step 3 — Apply the formula

a₂₉ = 12 + (29-1) × 3 = 12 + 28 × 3 = 12 + 84 = 96

Step 4 — Verify

Direct method: $12 + 28 × $3 = $12 + $84 = $96

Answer: $96 is in the collection.

3
Target Amount
The collection starts with $8. Each person contributes $4. What is the minimum number of people needed so that the total exceeds $200?
Step 1 — Set up the inequality

We need: 8 + n × 4 > 200 where n is the number of contributors

Step 2 — Solve the inequality

n × 4 > 192
n > 48

Step 3 — Find minimum integer

Since n must be a whole number of people, the minimum is n = 49

Step 4 — Verify

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.

4
Variable Contributions
The collection starts with $10. The first person contributes $3, the second contributes $6, the third contributes $9, and so on, with each person contributing $3 more than the previous person. How much is in the collection after 15 people have contributed?
Step 1 — Recognize this is different

The contributions themselves form an arithmetic sequence: 3, 6, 9, 12, ... with a₁ = 3 and d = 3

Step 2 — Sum the contributions

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

Step 3 — Add initial amount

Total = Initial + Sum of contributions = $10 + $360 = $370

Step 4 — Verify with smaller case

After 2 people: $10 + $3 + $6 = $19
Using formula: $10 + 2/2×(6+3) = $10 + 9 = $19

Answer: $370 is in the collection.

Frequently Asked Questions

How do you find the nth term in an arithmetic sequence word problem?
Use the formula aₙ = a₁ + (n-1)d, where a₁ is the first term, d is the common difference, and n is the term number. In this collection problem, the initial $5 is a₁, each $2 contribution is d, and after 35 people contribute, we need a₃₆ = 5 + (36-1)×2 = $75.
What's the difference between the number of people and the term number in sequence problems?
The term number is often one more than the number of contributions. Here, after 35 people contribute, we have the 36th term of the sequence because the initial amount counts as the first term. This indexing confusion is the most common error in real-world arithmetic sequence problems.
When should you use arithmetic sequences instead of simple multiplication?
Use arithmetic sequences when there's an initial value that's different from the repeating pattern. In this problem, the collection starts with $5, then each person adds $2. Simple multiplication (35 × $2) would ignore the initial $5, giving the wrong answer of $70 instead of $75.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-06-06