DoAssignment.ca

C3.2 · Connect compound interest, geometric sequences, and exponential growth

Learn to connect compound interest, geometric sequences, and exponential growth through clear examples and targeted practice.

Ontario Grade 11 Mathematics

Discrete Functions

Seeing the same pattern through three different lenses

Imagine you place CAD 1 000 in a savings account that pays 6% interest per year, compounded annually. After one year you have CAD 1 060. After two years you have CAD 1 123.60. After three years you have CAD 1 191.02. Look closely at those amounts: each one is exactly 1.06 times the amount before it. That is the same pattern you see in a geometric sequence, and it is also the defining feature of exponential growth. This lesson ties those three ideas together so you can move fluently between them.

What you will learn

Prerequisite Bridge: Exponents and Percent Increase

Before connecting the three big ideas, let's make sure two Grade 10 tools are sharp. First, recall that increasing a quantity by a percent means multiplying it by a growth factor. A 6% increase means multiplying by 1+0.06=1.061 + 0.06 = 1.06. A 3.5% increase means multiplying by 1.0351.035. The growth factor is always 1+r1 + r, where rr is the interest rate written as a decimal.
Second, recall the exponent rule: multiplying the same base repeatedly gives a power. If you multiply CAD 1 000 by 1.061.06 three times, you get 1000×1.06×1.06×1.06=1000×1.0631000 \times 1.06 \times 1.06 \times 1.06 = 1000 \times 1.06^3. This shorthand is the bridge from repeated multiplication to exponential notation, and it is the key mechanic of this entire lesson.

Geometric Sequences: Multiplication Step by Step

A sequence is simply an ordered list of numbers. A geometric sequence is one where you move from each term to the next by multiplying by the same fixed number every time. That fixed number is called the common ratio, usually written rr. For example, the sequence 500,600,720,864,…500, 600, 720, 864, \ldots has a common ratio of r=1.2r = 1.2 because 500×1.2=600500 \times 1.2 = 600, 600×1.2=720600 \times 1.2 = 720, and so on.
The first term is written t1t_1 (read 'term one'). After one multiplication you get t2=t1⋅rt_2 = t_1 \cdot r. After two multiplications you get t3=t1⋅r2t_3 = t_1 \cdot r^2. The pattern leads to a general formula for the nn-th term.
Notice what this formula is really saying: start at t1t_1 and apply the growth factor rr exactly n−1n - 1 times. This is identical to what a bank does when it applies the same interest rate once per year for several years.
tn=t1⋅rn−1t_n = t_1 · r^{n-1}

Compound Interest: The Same Pattern with a Financial Name

Compound interest means that interest is calculated on both the original amount and any interest already earned. The original amount deposited or borrowed is called the principal, written PP. The interest rate per compounding period is ii (a decimal). The number of compounding periods is nn. The amount after nn periods is called the future value, written AA.
Each period, the account balance is multiplied by (1+i)(1 + i). After one period the balance is P(1+i)P(1+i). After two periods it is P(1+i)2P(1+i)^2. After nn periods it is P(1+i)nP(1+i)^n. Compare this to the geometric sequence formula tn=t1⋅rn−1t_n = t_1 \cdot r^{n-1}: if you set t1=P(1+i)t_1 = P(1+i) (the balance after the first period) and r=1+ir = 1+i, the two formulas describe exactly the same list of numbers.
The compound interest formula is often written as A=P(1+i)nA = P(1 + i)^n. Here nn counts the total number of times interest is applied. This is why nn appears as an exponent: each compounding period is one more multiplication by the growth factor (1+i)(1+i).
A=P(1+i)nA = P(1 + i)^n

Exponential Growth: The Function View

When the number of compounding periods nn can be any non-negative integer, the formula A=P(1+i)nA = P(1+i)^n defines a function. In function notation you can write A(n)=P(1+i)nA(n) = P(1+i)^n, where nn is the input (time in periods) and A(n)A(n) is the output (account balance). This is an exponential function because the input nn appears as an exponent.
The three representations — the sequence list, the compound interest formula, and the exponential function — all describe the same growth pattern. The sequence list shows individual terms one at a time. The formula lets you jump directly to any term without listing every one before it. The function notation makes it clear that the balance depends on time and that the relationship is exponential.
Exponential growth has a distinctive shape: it starts slow and accelerates. The balance grows by more dollars each period even though the percentage rate stays the same, because the base on which interest is calculated keeps getting larger. That is the real-world meaning of the exponent growing.
A(n)=P(1+i)nA(n) = P(1 + i)^n

Linking the Three Representations Side by Side

To see the connection clearly, consider a single situation: CAD 2 000 is invested at 5% per year, compounded annually. List the first few balances, then write the general term as a sequence, then write the formula, then write the function. Every representation gives the same numbers.
As a sequence: t1=2000×1.05=2100t_1 = 2000 \times 1.05 = 2100, t2=2100×1.05=2205t_2 = 2100 \times 1.05 = 2205, t3=2205×1.05=2315.25t_3 = 2205 \times 1.05 = 2315.25, and so on, with r=1.05r = 1.05. As a compound interest formula after nn years: A=2000(1.05)nA = 2000(1.05)^n. As a function: A(n)=2000(1.05)nA(n) = 2000(1.05)^n. Notice that t1t_1 in the sequence corresponds to A(1)A(1), t2t_2 to A(2)A(2), and so on. The sequence index nn and the function input nn count the same thing: how many times the growth factor has been applied.
The key insight to carry forward is this: whenever you see a situation where a quantity is multiplied by the same factor repeatedly, you are looking at a geometric sequence, compound interest, and exponential growth all at once. The context gives it a name; the mathematics is the same.

Three Representations of the Same Growth Pattern

RepresentationNotationWhat n countsExample (P = 1000, r = 1.05)
Geometric sequencetn=t1⋅rn−1t_n = t_1 \cdot r^{n-1}Term number (starting at 1)t3=1050×1.052=1157.63t_3 = 1050 \times 1.05^2 = 1157.63
Compound interestA=P(1+i)nA = P(1+i)^nNumber of compounding periodsA=1000×1.053=1157.63A = 1000 \times 1.05^3 = 1157.63
Exponential functionA(n)=P(1+i)nA(n) = P(1+i)^nInput variable (time in periods)A(3)=1000×1.053=1157.63A(3) = 1000 \times 1.05^3 = 1157.63

Worked example

From Sequence to Future Value

Priya opens a savings account with a deposit of CAD 3 000. The account earns 4% interest per year, compounded annually. (a) Write the first four annual balances as a geometric sequence and state the common ratio. (b) Use the compound interest formula to find the balance after 10 years. Round to the nearest cent.
  1. Identify the growth factor
    A 4% annual interest rate means each year's balance is multiplied by 1+0.04=1.041 + 0.04 = 1.04. This multiplier is the growth factor, and in a geometric sequence it is the common ratio rr.
    r=1.04r = 1.04
  2. List the first four balances
    Start with t1t_1, the balance after year 1, then multiply by r=1.04r = 1.04 each time to get the next term. Round intermediate values to the nearest cent for clarity.
    t1=3000×1.04=3120.00t_1 = 3000 × 1.04 = 3120.00
  3. Continue the sequence
    Apply the common ratio again to get years 2, 3, and 4.
    t2=3120.00×1.04=3244.80, t3=3244.80×1.04=3374.59, t4=3374.59×1.04=3509.57t_2 = 3120.00 × 1.04 = 3244.80, \ t_3 = 3244.80 × 1.04 = 3374.59, \ t_4 = 3374.59 × 1.04 = 3509.57
  4. Write the general term
    Using the geometric sequence formula tn=t1⋅rn−1t_n = t_1 \cdot r^{n-1} with t1=3120t_1 = 3120 and r=1.04r = 1.04, or equivalently using the compound interest formula A=P(1+i)nA = P(1+i)^n with P=3000P = 3000, i=0.04i = 0.04, both correctly model the situation.
    A(n)=3000×(1.04)nA(n) = 3000 × (1.04)^n
  5. Apply the formula for n = 10
    Substitute n=10n = 10 into A(n)=3000×(1.04)nA(n) = 3000 \times (1.04)^n. Calculate (1.04)10(1.04)^{10} first: (1.04)10≈1.480244(1.04)^{10} \approx 1.480244. Then multiply by 3000.
    A(10)=3000×(1.04)10≈3000×1.480244≈4440.73A(10) = 3000 × (1.04)^{10} \approx 3000 × 1.480244 \approx 4440.73
Answer: The common ratio is r=1.04r = 1.04. The first four annual balances are approximately CAD 3 120.00, CAD 3 244.80, CAD 3 374.59, and CAD 3 509.57. After 10 years the balance is approximately CAD 4 440.73.
Check: Verify the 10-year result by checking that 3000×1.04103000 \times 1.04^{10} is reasonable. Since 1.0410≈1.481.04^{10} \approx 1.48, the balance is about 48% larger than the principal. CAD 3 000 increased by 48% gives roughly CAD 3 000 + CAD 1 440 = CAD 4 440, which matches CAD 4 440.73. The answer is consistent.

Worked example

Identifying Exponential Growth from a Sequence

A colony of bacteria doubles every hour. At time zero there are 500 bacteria. (a) Write the number of bacteria at hours 0, 1, 2, 3, and 4 as a sequence. (b) Identify the common ratio and write the count as a function of time tt in hours. (c) Find the number of bacteria after 8 hours.
  1. List the terms
    At time zero there are 500 bacteria. Each hour the count doubles, so multiply by 2 each time.
    500, 1000, 2000, 4000, 8000500, \ 1000, \ 2000, \ 4000, \ 8000
  2. State the common ratio
    Divide any term by the previous term: 1000÷500=21000 \div 500 = 2. The common ratio is r=2r = 2, confirming this is a geometric sequence.
    r=2r = 2
  3. Write the function
    The starting value at t=0t = 0 is 500. After tt hours the count has been multiplied by 2 exactly tt times. Write this as an exponential function using the same structure as the compound interest formula A=P(1+i)nA = P(1+i)^n, with P=500P = 500 and the growth factor equal to 2.
    B(t)=500×2tB(t) = 500 × 2^t
  4. Evaluate at t = 8
    Substitute t=8t = 8 into B(t)=500×2tB(t) = 500 \times 2^t. Compute 28=2562^8 = 256, then multiply by 500.
    B(8)=500×28=500×256=128 000B(8) = 500 × 2^8 = 500 × 256 = 128\ 000
Answer: The sequence for hours 0 through 4 is 500, 1 000, 2 000, 4 000, 8 000. The common ratio is r=2r = 2. The function is B(t)=500×2tB(t) = 500 \times 2^t. After 8 hours there are 128 000 bacteria.
Check: From t=4t = 4 (which gives 8 000) to t=8t = 8 is four more doublings: 8000×24=8000×16=128 0008000 \times 2^4 = 8000 \times 16 = 128\ 000. This matches B(8)=128 000B(8) = 128\ 000, confirming the answer is correct.

Common mistakes and how to avoid them

Using the interest rate as the growth factor instead of adding 1 first. For example, writing A=P×(0.06)nA = P \times (0.06)^n instead of A=P×(1.06)nA = P \times (1.06)^n.
Correction: Always write the growth factor as (1+i)(1 + i). The 1 preserves the principal; the ii adds the interest earned.
Confusing the number of years with the number of compounding periods when interest is compounded more frequently than annually.
Correction: The exponent nn counts compounding periods, not years. Make sure ii and nn use the same time unit (e.g., both monthly or both annual).
Using tn=t1⋅rnt_n = t_1 \cdot r^n instead of tn=t1⋅rn−1t_n = t_1 \cdot r^{n-1}, which gives an answer one multiplication too large.
Correction: The exponent is n−1n - 1 because t1t_1 already exists before any multiplication. Multiply by rr exactly n−1n - 1 times to reach term nn.
Treating the starting value in a bacteria or population problem as t1t_1 and then applying the formula as if it is the balance after one period.
Correction: When the starting value is at time zero, use A(t)=P⋅rtA(t) = P \cdot r^t where the exponent equals the number of periods elapsed, not the term number minus one.
Assuming that a higher interest rate always means more total interest earned, without accounting for the number of compounding periods.
Correction: Both the rate and the number of periods affect the final amount. Use the full formula A=P(1+i)nA = P(1+i)^n to compare scenarios fairly.

Lesson summary

Check your understanding

Question 1

A geometric sequence starts at t1=400t_1 = 400 and has a common ratio of r=1.05r = 1.05. Which expression gives the 6th term?
  1. 400×1.056400 \times 1.05^6
  2. 400×1.055400 \times 1.05^5
  3. 400×6×1.05400 \times 6 \times 1.05
  4. 400+5×1.05400 + 5 \times 1.05
Show answer and explanation
400×1.055400 \times 1.05^5
The general term is tn=t1⋅rn−1t_n = t_1 \cdot r^{n-1}. For the 6th term, the exponent is 6−1=56 - 1 = 5, giving 400×1.055400 \times 1.05^5. The other options use the wrong exponent or incorrectly add instead of multiply.

Question 2

CAD 5 000 is invested at 3% per year, compounded annually. Which formula correctly models the balance after nn years?
  1. A=5000×0.03nA = 5000 \times 0.03^n
  2. A=5000×3nA = 5000 \times 3^n
  3. A=5000×(1.03)nA = 5000 \times (1.03)^n
  4. A=5000+5000×0.03×nA = 5000 + 5000 \times 0.03 \times n
Show answer and explanation
A=5000×(1.03)nA = 5000 \times (1.03)^n
The growth factor for a 3% annual rate is 1+0.03=1.031 + 0.03 = 1.03. The compound interest formula is A=P(1+i)n=5000×(1.03)nA = P(1+i)^n = 5000 \times (1.03)^n. Option A uses only the rate as the base, option B confuses the percent with the base, and option D is simple interest (not compound).

Question 3

A population of insects triples every week. Starting from 200 insects at week 0, how many are there after 4 weeks?
  1. 2 400
  2. 16 200
  3. 6 400
  4. 1 800
Show answer and explanation
16 200
The function is B(t)=200×3tB(t) = 200 \times 3^t. At t=4t = 4: 34=813^4 = 81, and 200×81=16 200200 \times 81 = 16\ 200. Option A divides incorrectly, option C uses base 2 instead of 3 (doubling, not tripling), and option D multiplies by 3 only once.

Question 4

Which statement best explains why compound interest is an example of exponential growth?
  1. The principal increases by the same dollar amount each period.
  2. The balance is multiplied by the same growth factor every period, so the total grows by larger and larger dollar amounts over time.
  3. The interest rate increases each year as the balance rises.
  4. The balance is added to itself once per year, making it grow in a straight line.
Show answer and explanation
The balance is multiplied by the same growth factor every period, so the total grows by larger and larger dollar amounts over time.
Exponential growth occurs when a quantity is multiplied by a constant factor repeatedly. In compound interest, the balance is multiplied by (1+i)(1+i) each period. Because the base keeps growing, the dollar increase each period gets larger, even though the rate stays constant. The other options describe simple interest, increasing rates, or additive (linear) growth, none of which match compound interest.

Key terms

Geometric sequence
An ordered list of numbers where each term is found by multiplying the previous term by a fixed value called the common ratio.
Common ratio
The fixed number by which each term in a geometric sequence is multiplied to get the next term. Found by dividing any term by the term before it.
Principal
The original amount of money deposited or borrowed before any interest is added.
Compound interest
Interest calculated on both the original principal and any interest that has already been earned, causing the balance to grow faster than simple interest.
Growth factor
The number you multiply the current amount by to get the next amount. For a rate of ii, the growth factor is (1+i)(1 + i).
Future value
The total amount in an account after a given number of compounding periods, including principal and all interest earned.
Exponential function
A function of the form f(n)=a⋅bnf(n) = a \cdot b^n, where the input nn appears as an exponent and b>0b > 0, b≠1b \neq 1.
Compounding period
The length of time between one interest calculation and the next. Common periods include annually (once a year), semi-annually, and monthly.

Continue through MCR3U

View the complete Ontario Grade 11 Mathematics learning path

About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation C3.2. It is a study resource, not an official curriculum publication.

Official curriculum reference

Report a correction or ask a question