C. Discrete Functions · Financial Applications: Interest, Present Value and AnnuitiesSimple Interest, Arithmetic Sequences and Linear Growth
C3.1About 70 minutes
Learning goals- I can use I = Prt and A = P(1 + rt) to find interest, amount, principal, rate or time.
- I can show that the year-end amounts of a simple-interest investment form an arithmetic sequence.
- I can explain why simple interest is linear growth, using first differences, a table, a graph and an equation.
- I can use a spreadsheet to calculate simple interest, find first differences and graph amount against time.
Words to know
principal (P)
The amount of money first invested or borrowed.
interest (I)
The money paid for the use of money. An investor earns it; a borrower pays it.
simple interest
Interest calculated only on the original principal: I = Prt, where r is the annual rate as a decimal and t is the time in years.
amount (A)
Principal plus interest: A = P + I. Also called the future value.
arithmetic sequence
A sequence in which each term is the previous term plus the same number, the common difference d.
linear growth
Growth by the same amount in each equal time step. Its graph is a straight line and its first differences are constant.
Welcome to Financial Applications
You are 16. A relative gives you $2 000 for your birthday and says, "Don't spend it. Put it somewhere it will grow." The bank offers a 5-year GIC (Guaranteed Investment Certificate) that pays 4% simple interest per year. How much will you have at the end of each year? What does that growth look like on a graph? And why does a mathematician get excited that the yearly amounts form a sequence?
This unit connects the money ideas you met in Grades 8 and 9 to the sequences and series you studied in the unit Sequences and Series, and to the exponential functions you studied in the unit Exponential Functions. By the end you will be able to plan a savings goal, compare loans, and work out what a car payment really costs.
Remember from Grades 8 and 9
In Grade 8 you calculated simple interest with I = Prt and compound interest with A = P(1 + r)ⁿ. In Grade 9 you compared compounding periods, used spreadsheets to compare borrowing options, and saw that money earning simple interest grows in a straight-line (linear) way. In this unit you will prove why those patterns happen, using sequences, series and functions.
The simple interest formula
I = Prt
- P = principal (dollars)
- r = annual interest rate, written as a decimal (4% = 0.04)
- t = time in years (6 months = 6/12 = 0.5 years; 146 days = 146/365 years)
- I = interest (dollars)
The amount at the end is A = P + I = P + Prt, which factors to A = P(1 + rt).
The key fact about simple interest: interest is earned only on the principal. The interest from earlier years never earns interest of its own. So the account grows by exactly the same amount, Pr, every year.
Worked exampleExample 1: A simple-interest GIC, year by year
You invest $2 000 at 4% simple interest per year. (a) Find the amount at the end of each of the first 5 years. (b) Show that these amounts form an arithmetic sequence and write its general term.
- Interest for one year: I = Prt = 2000 × 0.04 × 1 = $80.
- (a) Each year adds another $80: Year 1: $2 080; Year 2: $2 160; Year 3: $2 240; Year 4: $2 320; Year 5: $2 400.
- (b) The first differences are 2160 − 2080 = 80, 2240 − 2160 = 80, and so on. Every difference is 80, so the amounts form an arithmetic sequence with first term a = 2080 and common difference d = 80.
- Using the formula from Sequences and Series: tₙ = a + (n − 1)d = 2080 + (n − 1)(80) = 2000 + 80n, where n is the number of years.
- Check: t₅ = 2000 + 80(5) = 2400. ✓ Notice that d = Pr: the common difference is one year's interest.
$2 080, $2 160, $2 240, $2 320, $2 400. This is arithmetic with d = 80, and tₙ = 2000 + 80n.
The bars rise by the same $80 every year. Equal steps mean the year-end amounts form an arithmetic sequence (Year 0 is the $2 000 you start with).Three ways to see linear growth
The amount after t years is A(t) = P + (Pr)t. Compare it with the equation of a line, y = mx + b:
- the slope m is Pr, the interest earned per year;
- the A-intercept b is P, the amount at time 0.
So simple interest is linear growth. You can recognise it three ways:
- Table: the first differences are constant (all equal to Pr).
- Graph: the points lie on a straight line.
- Equation: A is a linear (degree 1) function of t.
Discrete or continuous? If the bank pays interest only at the end of each full year, the amounts exist only at t = 0, 1, 2, … . That is a discrete function (a sequence), graphed as separate dots. If you could cash out at any moment and receive interest for the part of the year, A(t) = P + Prt makes sense for every t ≥ 0, and you would join the dots with a line. You studied this difference in Sequences and Series.
Worked exampleExample 2: Finding the rate or the time
(a) $750 invested at simple interest grows to $862.50 in 3 years. What is the annual rate?
(b) How long will it take $4 000 at 3.5% simple interest to earn $560 of interest?
- (a) The interest is I = A − P = 862.50 − 750 = $112.50.
- Rearrange I = Prt to get r = I ÷ (Pt) = 112.50 ÷ (750 × 3) = 112.50 ÷ 2250 = 0.05.
- So the rate is 5% per year.
- (b) Rearrange I = Prt to get t = I ÷ (Pr) = 560 ÷ (4000 × 0.035) = 560 ÷ 140 = 4 years.
(a) 5% per year. (b) 4 years.
Worked exampleExample 3: Working backward from a sequence
A simple-interest account shows these year-end amounts: $1 545, $1 590, $1 635. Find the principal and the annual rate.
- The common difference is 1590 − 1545 = 45 (and 1635 − 1590 = 45). So one year's interest is Pr = $45.
- The first amount, $1 545, is after 1 year. Go back one step to time 0: P = 1545 − 45 = $1 500.
- Rate: r = 45 ÷ 1500 = 0.03 = 3%.
- Check with A = P(1 + rt): 1500(1 + 0.03 × 3) = 1500(1.09) = 1635 ✓ (the third amount).
P = $1 500 and r = 3% per year.
Worked exampleExample 4: Time in days
Banks often count simple interest by the day. Find the interest on $5 000 at 3.65% per year for 146 days.
- Write the time in years: t = 146 ÷ 365 = 0.4.
- I = Prt = 5000 × 0.0365 × 0.4 = $73.00.
Using technology: a simple-interest spreadsheet
The curriculum asks you to use a spreadsheet (or a graphing calculator) to calculate simple interest, find first differences and graph the amount. Any free spreadsheet works (Google Sheets, LibreOffice Calc, Excel, Numbers). Set it up like this:
| Cell | Type this | What it does |
|---|
| A1, B1, C1 | Year, Amount, First difference | headings |
| A2 | 0 | start time |
| B2 | 2000 | principal |
| A3 | =A2+1 | next year |
| B3 | =B2+2000*0.04 | adds one year of simple interest (always on the original $2 000) |
| C3 | =B3-B2 | first difference |
Just for funWhy was the simple-interest account so calm?
Because every year was exactly like the last one: +$80, +$80, +$80. Nothing ever compounds its stress.
Hands-on activityBuild it and graph it
You need: a computer, tablet or phone with a free spreadsheet app, or grid paper, a ruler and a calculator.
- Choose a principal between $500 and $5 000 and a rate between 2% and 6%. (Example: $1 200 at 5%.)
- Build the spreadsheet above with your numbers, for Years 0 to 10. (On paper: make a table with columns Year, Amount and First difference.)
- Make a scatter graph of Amount against Year. On paper, use 1 square = 1 year across and choose a sensible scale up the side.
- Find the slope of the line from two points. Compare it with P × r.
- Write A(t) = P + Prt for your numbers and check it gives your Year 10 amount.
What to look for: for $1 200 at 5%, the first differences are all $60, the points lie on a straight line with slope 60, A(t) = 1200 + 60t, and A(10) = $1 800. Your numbers will be different, but the first differences must all be equal to P × r.
Your turn
1Find the simple interest and the amount for $3 500 at 2.8% per year for 4 years.
I = 3500 × 0.028 × 4 = $392. A = 3500 + 392 = $3 892.
2$1 200 is invested at 5% simple interest. List the amounts at the end of years 1 to 4 and state the common difference.
$1 260, $1 320, $1 380, $1 440; d = 1200 × 0.05 = $60.
3Write the general term tₙ for the sequence in the last question, where n is the number of years.
tₙ = 1260 + (n − 1)(60) = 1200 + 60n.
4$1 500 earned $270 of simple interest in 3 years. What was the rate?
r = 270 ÷ (1500 × 3) = 270 ÷ 4500 = 0.06 = 6%.
5How many years will it take $4 000 at 3.5% simple interest to earn $700?
t = 700 ÷ (4000 × 0.035) = 700 ÷ 140 = 5 years.
6Find the simple interest on $2 400 at 4.5% for 9 months.
t = 9/12 = 0.75. I = 2400 × 0.045 × 0.75 = $81.00.
7A table of year-end amounts reads: Year 0: $900; Year 1: $954; Year 2: $1 008; Year 3: $1 062. Is this simple interest? If so, find the rate.
The first differences are all $54, so yes, it is linear growth. r = 54 ÷ 900 = 0.06 = 6%.
More practice: warm-up
1In I = Prt, what do P, r and t stand for, and what units must r and t use?
P = principal, r = annual rate as a decimal, t = time in years.
2Find the simple interest on $500 at 4% for 1 year.
3Find the first differences of 1000, 1030, 1060, 1090. What kind of growth is this?
30, 30, 30: constant, so linear growth (an arithmetic sequence).
4The year-end amounts of a simple-interest investment form what kind of sequence?
5Write 8 months as a fraction of a year.
8/12 = 2/3 of a year (about 0.667).
More practice: on your own
1$6 000 is invested at 3.25% simple interest. (a) Write A(t). (b) What does the slope mean? (c) Find A(7).
(a) Pr = 6000 × 0.0325 = 195, so A(t) = 6000 + 195t. (b) The account gains $195 per year. (c) A(7) = 6000 + 1365 = $7 365.
2A simple-interest sequence of year-end amounts is $2 120, $2 240, $2 360, … . Find the principal, the rate and t₁₀.
d = 120, so P = 2120 − 120 = $2 000 and r = 120 ÷ 2000 = 6%. t₁₀ = 2000 + 120(10) = $3 200.
3Find the simple interest on $1 850 at 4.1% for 120 days.
I = 1850 × 0.041 × (120/365) ≈ $24.94.
4Account A: A = 3000 + 150t. Account B: $2 500 at 7% simple interest. Which grows faster? After how many years are they equal?
B gains 2500 × 0.07 = $175 a year, more than A's $150, so B grows faster (steeper slope). 3000 + 150t = 2500 + 175t gives 25t = 500, so t = 20 years (both are worth $6 000).
5Explain, using first differences, why simple interest is linear growth and not exponential growth.
Each year's interest is Pr, calculated on the same principal, so the amount always rises by the same number. Constant first differences mean linear growth. Exponential growth would need a constant ratio, which simple interest does not have (for example 2080/2000 = 1.04 but 2160/2080 ≈ 1.038).
Challenge
1Account A has $5 000 at 3% simple interest. Account B has $4 000 at 4.5% simple interest. When will they hold equal amounts, and how much will each hold?
A(t) = 5000 + 150t, B(t) = 4000 + 180t. 5000 + 150t = 4000 + 180t → 30t = 1000 → t = 33⅓ years (33 years 4 months). Each holds 5000 + 150(100/3) = $10 000.
2A simple-interest investment is worth $2 740 after 3 years and $3 060 after 7 years. Find the principal and the rate.
In 4 years it gained 3060 − 2740 = 320, so d = 320 ÷ 4 = $80 per year. P = 2740 − 3(80) = $2 500. r = 80 ÷ 2500 = 0.032 = 3.2%.
3Prove that the year-end amounts tₙ = P + nPr form an arithmetic sequence. Then explain when it is better to graph simple interest as dots and when as a line.
tₙ₊₁ − tₙ = [P + (n + 1)Pr] − [P + nPr] = Pr, a constant, so the sequence is arithmetic with d = Pr. Use dots (a discrete function) when interest is only paid at whole years, so only those amounts exist. Use a line (continuous) when interest is earned for any part of a year, for example when it is counted by the day.
Quiz
1. Find the simple interest on $800 at 5% per year for 3 years.
2. The year-end amounts of a simple-interest investment form:
3. A simple-interest investment is modelled by A(t) = 4000 + 140t. What is the annual rate?
4. What is the common difference of the year-end amounts for $2 500 at 4% simple interest?
5. The graph of A against t for simple interest is:
6. Find the simple interest on $1 000 at 6% for 6 months.
Big idea: Expectation C3.1 asks students to connect simple interest, arithmetic sequences and linear growth, and to use technology (a spreadsheet or graphing calculator) to calculate interest, find first differences and graph amount against time.
How to help: Ask your teen to explain what the slope and the intercept of A = P + Prt mean in dollars. If you have a savings account statement or a GIC certificate, look together at the rate and how often interest is paid (no need to share balances).
Common difficulties:
- Using the percent instead of the decimal (4 instead of 0.04).
- Using months or days for t without dividing by 12 or 365.
- Mixing up interest (I) and amount (A). Ask, "Does the question want the interest earned, or the total?"
- Counting the terms: the amount after 1 year is the first term; the principal is the amount at year 0.