Geometric Sequences: Savings Account Growth Formula

Sequences & Series 9th-10th Grade
PROBLEM
Bailee deposits $500 into a savings account. Her balance increases by a factor of 1.0025 per month. Write an explicit formula describing the amount of money in her account, aₙ, after n months, assuming she makes no deposits or withdrawals.

What This Problem Teaches

  • Geometric sequence recognition — Identifying when a sequence follows multiplicative rather than additive growth
  • Explicit formula construction — Converting from recursive patterns to closed-form expressions
  • Financial literacy foundations — Understanding compound growth and how small monthly rates accumulate over time
  • Mathematical modeling — Translating real-world scenarios into algebraic expressions
  • Exponential notation mastery — Working with bases and exponents in practical contexts

Solution: Method 1 — The Geometric Sequence Approach

This problem is asking us to find an explicit formula for a geometric sequence. Let's work through this step-by-step by identifying the pattern and applying the standard formula.

Step 1 — Identify the initial term and growth factor

Bailee starts with $500 as her initial deposit. Each month, her balance is multiplied by 1.0025. This means:

  • Initial amount (when n = 0): a₀ = 500
  • Growth factor: r = 1.0025

Step 2 — Trace the first few terms to confirm the pattern

Let's see what happens month by month:

  • After 0 months: a₀ = 500
  • After 1 month: a₁ = 500 × 1.0025 = 501.25
  • After 2 months: a₂ = 501.25 × 1.0025 = 502.50
  • After 3 months: a₃ = 502.50 × 1.0025 = 503.76

This confirms we have a geometric sequence where each term equals the previous term multiplied by 1.0025.

Step 3 — Apply the geometric sequence explicit formula

For any geometric sequence, the explicit formula is:

aₙ = a₀ × rⁿ

where a₀ is the initial term and r is the common ratio.

Step 4 — Substitute our specific values

Substituting a₀ = 500 and r = 1.0025:

aₙ = 500 × (1.0025)ⁿ

This is our explicit formula describing Bailee's account balance after n months.

Solution: Method 2 — Percentage Growth Model

We can also approach this by recognizing that a multiplication factor of 1.0025 represents percentage growth. This perspective helps connect the mathematics to financial concepts.

Step 1 — Convert the factor to a growth rate

A factor of 1.0025 means:

  • 1 represents keeping 100% of the original balance
  • 0.0025 represents gaining an additional 0.25% each month

So Bailee's account grows by 0.25% monthly.

Step 2 — Apply compound growth formula

For compound growth, we use the formula:

Final Amount = Initial Amount × (1 + growth rate)ⁿ

Step 3 — Substitute the values

With initial amount = $500, growth rate = 0.0025, and time period = n months:

aₙ = 500 × (1 + 0.0025)ⁿ = 500 × (1.0025)ⁿ

This matches our result from Method 1, confirming our formula is correct.

The explicit formula is: aₙ = 500 × (1.0025)ⁿ

Verification

Let's verify our formula by checking it against our manually calculated values from earlier.

For n = 1:

a₁ = 500 × (1.0025)¹ = 500 × 1.0025 = 501.25

For n = 2:

a₂ = 500 × (1.0025)² = 500 × 1.00500625 = 502.503125 ≈ 502.50

For n = 0:

a₀ = 500 × (1.0025)⁰ = 500 × 1 = 500

Our formula correctly produces the expected values, confirming that aₙ = 500 × (1.0025)ⁿ is the correct explicit formula.

Common Pitfalls

✗ Mistake 1: Using addition instead of multiplication

Writing aₙ = 500 + 0.0025n treats this as an arithmetic sequence. This would mean Bailee gains a fixed $0.0025 each month, which contradicts the problem statement that her balance "increases by a factor of 1.0025."

✗ Mistake 2: Confusing the growth rate with the factor

Writing aₙ = 500 × (0.0025)ⁿ uses the growth rate (0.0025) instead of the full factor (1.0025). This would mean her balance shrinks dramatically each month, which makes no sense for a savings account.

✗ Mistake 3: Wrong initial term indexing

Writing aₙ = 500 × (1.0025)ⁿ⁻¹ assumes the first term is a₁ instead of a₀. This formula would give a₀ = 500 × (1.0025)⁻¹ ≈ 498.75, suggesting Bailee started with less than her $500 deposit.

The Underlying Pattern

This problem illustrates the general structure of geometric sequences in financial contexts. The explicit formula for any geometric sequence is:

aₙ = a₀ × rⁿ

Where:

  • a₀ = initial value (principal amount)
  • r = common ratio (growth factor per period)
  • n = number of periods elapsed

In financial applications, r > 1 represents growth (like savings account interest), while 0 < r < 1 represents decay (like depreciation). The key insight is that when a quantity changes by a constant percentage each period, we get exponential growth or decay, not linear change.

Important Note: This formula assumes the growth factor is applied at discrete intervals (monthly, in this case). For continuous compounding, we would use A = Pe^(rt) instead.

How to Spot This Problem Type

Geometric sequence problems often contain these key phrases:

  • "increases by a factor of" or "multiplied by" — signals multiplication, not addition
  • "growth rate of" or "percentage increase" — when combined with compounding
  • "each month/year/period" — indicates regular, repeating process
  • "explicit formula" — asks for closed-form expression, not recursive formula

Watch out for problems that mention "compound interest," "population growth," "radioactive decay," or "appreciation/depreciation." These contexts almost always involve geometric sequences.

The key test: if each term equals the previous term times a constant, you have a geometric sequence. If each term equals the previous term plus a constant, you have an arithmetic sequence.

Real Applications

  • Investment accounts: Mutual funds and retirement accounts often grow by a relatively consistent monthly factor, making this formula essential for financial planning.
  • Population modeling: When birth rates are stable, populations grow geometrically. Epidemiologists use similar models to track disease spread.
  • Technology scaling: Moore's Law (computer processing power doubling every two years) follows geometric growth, as do many technology adoption rates.

Four "What-If?" Problems

1
Different Initial Amount
Marcus deposits $750 into the same type of savings account that grows by a factor of 1.0025 per month. Write the explicit formula and find his balance after 18 months.
Step 1 — Identify the parameters

Initial amount: a₀ = 750, Growth factor: r = 1.0025

Step 2 — Apply the formula

aₙ = 750 × (1.0025)ⁿ

Step 3 — Calculate for n = 18

a₁₈ = 750 × (1.0025)¹⁸ = 750 × 1.04603 ≈ 784.52

Verification

Check: 750 × 1.0460$784.52

2
Find the Time Period
Bailee's account starts at $500 and grows by a factor of 1.0025 per month. How many months will it take for her balance to exceed $520? Round to the nearest month.
Step 1 — Set up the inequality

500 × (1.0025)ⁿ > 520

Step 2 — Isolate the exponential

(1.0025)ⁿ > 520/500 = 1.04

Step 3 — Take logarithms

n × ln(1.0025) > ln(1.04)

n > ln(1.04)/ln(1.0025) ≈ 0.0392/0.00250 ≈ 15.7

Step 4 — Round up

Since we need the balance to exceed $520, it takes 16 months.

Verification

a₁₆ = 500 × (1.0025)¹⁶ ≈ 520.51 > 520

3
Different Growth Factor
Sarah's savings account starts with $500 but grows by a factor of 1.003 per month (higher interest rate). Write the explicit formula and compare her balance to Bailee's after 24 months.
Step 1 — Write Sarah's formula

aₙ = 500 × (1.003)ⁿ

Step 2 — Calculate Sarah's balance at n = 24

Sarah: 500 × (1.003)²⁴ = 500 × 1.0744 ≈ 537.20

Step 3 — Calculate Bailee's balance at n = 24

Bailee: 500 × (1.0025)²⁴ = 500 × 1.0618 ≈ 530.90

Step 4 — Compare

Sarah has $537.20 - $530.90 = $6.30 more than Bailee after 24 months.

Verification

The higher growth factor (1.003 vs 1.0025) compounds over time, creating a meaningful difference.

4
Piecewise Growth Rates
Kai deposits $500. For the first 6 months, the balance grows by a factor of 1.0025 per month. Starting month 7, the factor changes to 1.004 per month (promotional rate ends). Find his balance after 12 months total.
Step 1 — Calculate balance after first 6 months

a₆ = 500 × (1.0025)⁶ = 500 × 1.01508 ≈ 507.54

Step 2 — Apply new rate for months 7-12

Starting balance for second phase: $507.54

Growth for 6 more months at 1.004 factor: 507.54 × (1.004)⁶

Step 3 — Calculate final balance

a₁₂ = 507.54 × (1.004)⁶ = 507.54 × 1.02429 ≈ 519.87

Step 4 — Write piecewise formula

For 0 ≤ n ≤ 6: aₙ = 500 × (1.0025)ⁿ

For n > 6: aₙ = 500 × (1.0025)⁶ × (1.004)ⁿ⁻⁶

Verification

Check: 500 × 1.01508 × 1.02429 ≈ 519.87

DN

Dr. Neven Jurkovic

Mathematics Professor • Sequence & Series Specialist • 15+ years teaching experience

Frequently Asked Questions

How do you write an explicit formula for a geometric sequence? +
For a geometric sequence, use the formula aₙ = a₁ × rⁿ⁻¹, where a₁ is the first term and r is the common ratio. In this problem, Bailee starts with $500 and multiplies by 1.0025 each month, giving us aₙ = 500 × (1.0025)ⁿ.
What does a growth factor of 1.0025 mean in finance? +
A growth factor of 1.0025 means the balance increases by 0.25% each period. This represents compound growth where 100% of the original amount (the 1) plus 0.25% additional growth (the 0.0025) is retained. Here, Bailee's account grows by 0.25% monthly.
How is geometric sequence different from arithmetic sequence in savings? +
Geometric sequences involve multiplication by a constant factor (like compound interest), while arithmetic sequences involve adding a constant amount. Bailee's account uses geometric growth because each month's balance is 1.0025 times the previous month, not a fixed dollar addition.
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.

2026-07-11