Connect simple interest, arithmetic sequences, and linear growth

Ontario Grade 11 Mathematics

Discrete Functions

Seeing the Same Pattern in Money, Lists of Numbers, and Straight-Line Graphs

Have you ever left money in a savings account and watched the balance grow by the same dollar amount every single year? That steady, equal growth is at the heart of this lesson. We are going to look at three things — simple interest in finance, arithmetic sequences in number patterns, and linear functions in graphing — and discover that they are really three different ways of describing the exact same idea: adding the same amount, over and over, at equal intervals. Once you see that connection, you can move between a bank problem, a number list, and a straight-line graph without missing a beat.

What you will learn

Recognize that simple interest produces a sequence of amounts that grow by equal increments.
Identify the first term and common difference of an arithmetic sequence and write its general term.
Connect the general term of an arithmetic sequence to a linear function and interpret its slope and intercept.
Move fluently among a table of values, a sequence formula, and a graph to describe the same linear growth situation.

Prerequisite Bridge: Linear Functions and Slope

Before connecting three big ideas, it helps to dust off one tool from Grade 10: the equation of a line. Recall that a linear function has the form

f(x) = mx + b

, where

m

is the slope (how steeply the line rises or falls) and

b

is the

y

-intercept (the value when

x = 0

). The slope tells you the constant rate of change — the amount the output increases each time the input goes up by one unit.

For example, if you earn CAD 12 every hour, your total earnings follow

f(x) = 12x

, where

x

is hours worked. The slope is

12

because earnings increase by exactly CAD 12 for each additional hour. This idea of a fixed, repeating increase is exactly what we will see in simple interest and arithmetic sequences.

A linear function $f(x) = mx + b$ has a constant rate of change equal to the slope $m$ .
The $y$ -intercept $b$ is the starting value when the input is zero.
Equal increases in the output for equal increases in the input are the hallmark of linear growth.

Simple Interest: Equal Growth in Your Bank Account

Simple interest is a method of calculating the extra money (interest) earned on a principal — the amount originally deposited or borrowed. Each time period, interest is calculated only on the original principal, never on previously earned interest. Because the same dollar amount is added in every period, the total grows in a perfectly steady way.

The simple interest formula gives the total amount in the account after

t

years as

A = P + Prt

, which can be written as

A = P(1 + rt)

. Here

P

is the principal,

r

is the annual interest rate written as a decimal, and

t

is time in years. Notice that

P

and

r

are both constants for a given situation, so

Pr

is just one fixed number. That means

A

increases by the same amount,

Pr

, every year.

For instance, suppose you deposit CAD 500 at a simple interest rate of 4% per year. Each year, the interest earned is

500 \times 0.04 = 20

dollars. After year 1 the total is CAD 520, after year 2 it is CAD 540, and so on — always CAD 20 more than the year before. This pattern of equal additions is the key link to arithmetic sequences.

A = P(1 + rt)

Simple interest is always calculated on the original principal $P$ , not on accumulated interest.
The interest earned each period, $Pr$ , is constant — it never changes.
Total amount $A = P + Prt$ grows by $Pr$ for every additional unit of time.
The graph of $A$ against $t$ is a straight line with slope $Pr$ and $y$ -intercept $P$ .

Arithmetic Sequences: The Language of Equal Steps

An arithmetic sequence is an ordered list of numbers in which the difference between any two consecutive terms is always the same. That fixed difference is called the common difference, written

d

. The first number in the list is called the first term, written

t_1

.

For example, the list

500, 520, 540, 560, \ldots

has a first term of

t_1 = 500

and a common difference of

d = 20

, because

520 - 500 = 20

,

540 - 520 = 20

, and so on. Compare that to the simple interest example above — they are the exact same numbers!

The general term (also called the

n

th term) of an arithmetic sequence gives you the value of any term without listing every term before it. The formula is

t_n = t_1 + (n-1)d

, where

n

is the term number (a positive integer starting at 1). To see why this works, notice that to reach the

n

th term from the first, you add the common difference exactly

(n - 1)

times.

Using the bank example:

t_n = 500 + (n-1)(20)

. For the 5th year,

t_5 = 500 + 4 \times 20 = 580

. You can verify this by continuing the list: 500, 520, 540, 560, 580. The 5th value is indeed CAD 580.

t_n = t_1 + (n-1)d

An arithmetic sequence has a constant common difference $d$ between consecutive terms.
The first term is $t_1$ ; the $n$ th term is $t_n = t_1 + (n-1)d$ .
The common difference $d$ plays the same role as the slope in a linear function.
The sequence from a simple interest problem is always arithmetic because the interest added each period is constant.

Linear Growth: Connecting the Graph to the Sequence

When you plot the terms of an arithmetic sequence — with the term number

n

on the horizontal axis and the term value

t_n

on the vertical axis — the points always lie on a straight line. This is what we call linear growth: the output increases by a fixed amount for every one-unit increase in the input.

Notice what happens when you expand the general term formula. Starting from

t_n = t_1 + (n-1)d

, distribute

d

to get

t_n = t_1 + dn - d

, and then rearrange to get

t_n = dn + (t_1 - d)

. Written this way,

t_n

is a linear function of

n

. The coefficient of

n

is

d

(the slope), and the constant term is

t_1 - d

(the vertical intercept).

For the bank example, substituting

t_1 = 500

and

d = 20

gives

t_n = 20n + 480

. The slope is 20 (CAD 20 per year), and the intercept is 480. The slope of 20 matches the annual interest

Pr = 500 \times 0.04 = 20

, and the intercept 480 equals

P - Pr = 500 - 20 = 480

.

The connection is complete: simple interest creates arithmetic sequences, and arithmetic sequences, when graphed, produce straight lines. All three representations — the finance formula, the sequence formula, and the linear graph — describe the same equal-step growth.

t_n = dn + (t_1 - d)

Plotting an arithmetic sequence gives points that lie exactly on a straight line.
Expanding $t_n = t_1 + (n-1)d$ gives $t_n = dn + (t_1 - d)$ , matching the form $f(n) = mn + b$ .
The common difference $d$ equals the slope of the corresponding line.
Simple interest, arithmetic sequences, and linear growth are three representations of the same constant-rate pattern.

Putting It All Together: Moving Between Representations

The real power of this connection is flexibility. If you are given a finance problem, you can write it as an arithmetic sequence or read its graph. If you are given a linear equation, you can describe the arithmetic sequence it represents, or imagine the financial situation it might model. The table below shows how the key parts of each representation match up.

When you are asked to find when an account reaches a certain balance, you are really asking: for what value of

n

(or

t

) does the sequence (or linear function) reach that target? You can set up a simple equation and solve it with Grade 10 algebra. No new tools are needed — you already have everything required.

The principal $P$ in simple interest matches the first term $t_1$ in the sequence and the starting value on the graph.
The interest per period $Pr$ matches the common difference $d$ and the slope $m$ .
To find when a target value is reached, set the general term or linear equation equal to that value and solve for $n$ or $t$ .
All three representations — table, formula, graph — carry the same information about the same linear growth.

Three Representations of the Same Linear Growth

Feature	Simple Interest	Arithmetic Sequence	Linear Function
Starting value	Principal $P$	First term $t_1$	$y$ -intercept $b$
Fixed increase per period	Interest per period $Pr$	Common difference $d$	Slope $m$
Value after $n$ periods	$A = P + Prt$	$t_n = t_1 + (n-1)d$	$f(n) = mn + b$
Graph shape	Straight line	Collinear points	Straight line

Worked example

Example 1: From a Simple Interest Account to an Arithmetic Sequence

Aisha deposits CAD 800 in a savings account that earns simple interest at a rate of 3% per year. (a) Write the sequence of account balances at the end of years 1, 2, 3, and 4. (b) Write the general term

t_n

for the balance at the end of year

n

. (c) In which year will the balance first reach or exceed CAD 920?

Find the annual interest amount
The interest earned each year is the principal multiplied by the rate: $800 \times 0.03$ . This amount is added to the account every year and never changes under simple interest, so it will be the common difference $d$ .
$800 \times 0.03 = 24$
List the first four balances
The balance at the end of year 1 is $t_1 = 800 + 24 = 824$ . Each subsequent year, add CAD 24.
$824,\ 848,\ 872,\ 896$
Confirm it is arithmetic
Check that the difference between consecutive terms is constant: $848 - 824 = 24$ , $872 - 848 = 24$ , $896 - 872 = 24$ . The common difference is $d = 24$ , confirming this is an arithmetic sequence.
$d = 24$
Write the general term
Use $t_n = t_1 + (n-1)d$ with $t_1 = 824$ and $d = 24$ . Expand and simplify to get a clean linear expression.
$t_n = 824 + (n-1)(24) = 24n + 800$
Set up an inequality to find the target year
We need the first year $n$ for which $t_n \geq 920$ . Substitute the general term and write the inequality.
$24n + 800 \geq 920$
Solve for n
Subtract 800 from both sides to isolate the term with $n$ , then divide both sides by 24.
$24n \geq 120 \implies n \geq 5$
Interpret the answer
Since $n$ must be a whole number (the sequence records end-of-year balances), the balance first reaches or exceeds CAD 920 at the end of year 5. Verify by substituting $n = 5$ : $t_5 = 24(5) + 800 = 920$ .
$t_5 = 24(5) + 800 = 920$

Answer: The general term is

t_n = 24n + 800

. The balance first reaches CAD 920 at the end of year 5.

Check: Check year 4:

t_4 = 24(4) + 800 = 896 < 920

✓. Check year 5:

t_5 = 24(5) + 800 = 920 \geq 920

✓. The boundary is correct.

Worked example

Example 2: Working Backward from a Linear Graph to a Financial Situation

A graph of account balance

A

(in CAD) against time

t

(in years) shows a straight line passing through the points

(0, 650)

and

(4, 754)

. (a) Find the slope and explain what it represents. (b) Write the linear equation for

A

in terms of

t

. (c) Identify the principal and the annual interest rate. (d) Write the balance as an arithmetic sequence and state

t_1

and

d

.

Calculate the slope
Slope is rise over run. Use the two given points $(0, 650)$ and $(4, 754)$ , where rise is the change in balance and run is the change in time.
$m = \frac{754 - 650}{4 - 0} = \frac{104}{4} = 26$
Interpret the slope
A slope of 26 means the account balance increases by CAD 26 every year. Under simple interest, this fixed annual increase is the interest payment, equal to $Pr$ .
Write the linear equation
The $y$ -intercept is the value of $A$ when $t = 0$ , which the graph shows as 650. Using slope-intercept form $A = mt + b$ , substitute $m = 26$ and $b = 650$ .
$A = 26t + 650$
Identify the principal
The principal $P$ is the starting balance, which is the value of $A$ at $t = 0$ . From the equation, the $y$ -intercept gives the principal directly.
$P = 650$
Find the interest rate
The annual interest is $Pr$ , which equals the slope 26. Divide both sides by $P = 650$ to solve for $r$ .
$r = \frac{26}{650} = 0.04$
Interpret the rate
Since $r = 0.04$ , the annual simple interest rate is 4%. This confirms the slope equals $Pr = 650 \times 0.04 = 26$ , which is consistent with the graph.
Write the arithmetic sequence
The balance at the end of year 1 is found by substituting $t = 1$ into the equation: $A = 26(1) + 650 = 676$ , so $t_1 = 676$ . The common difference equals the slope: $d = 26$ . The general term is $t_n = 676 + (n-1)(26)$ , which simplifies as shown.
$t_n = 26n + 650$

Answer: The linear equation is

A = 26t + 650

. The principal is CAD 650, the annual simple interest rate is 4%, the first term of the sequence is

t_1 = 676

, and the common difference is

d = 26

.

Check: Verify with the given point

(4, 754)

:

A = 26(4) + 650 = 104 + 650 = 754

✓. Also confirm

t_4 = 26(4) + 650 = 754

✓.

Common mistakes and how to avoid them

Confusing the term number

n

with the number of increases. For example, thinking

t_3

requires adding

d

three times instead of two times.

Correction: Remember that

t_n = t_1 + (n-1)d

. To get to the 3rd term from the 1st, you add

d

exactly

3 - 1 = 2

times.

Using the balance at year 0 (the principal) as

t_1

when the sequence is defined as balances at the end of each year.

Correction: Define

t_1

clearly based on the problem. If the sequence lists end-of-year balances, then

t_1 = P + Pr

, not

P

itself.

Treating the

y

-intercept of the linear equation as the first term

t_1

of the sequence without checking.

Correction: The

y

-intercept corresponds to

n = 0

, but the first term is at

n = 1

. These are equal only if

d = 0

, which would not be a useful sequence.

Forgetting to convert the interest rate from a percentage to a decimal before calculating.

Correction: Always divide the percentage by 100 first. A rate of 3% becomes

r = 0.03

, so annual interest on CAD 800 is

800 \times 0.03 = 24

, not

800 \times 3 = 2400

.

Rounding

n

down when asked when a balance will first reach a target amount.

Correction: If

n

must be a whole number and the inequality gives a non-integer, round up to the next whole number, since you need the balance to reach or exceed the target.

Lesson summary

Simple interest adds the same dollar amount, $Pr$ , to an account every period, making balances grow at a constant rate.
This constant-rate growth produces an arithmetic sequence, where $t_1$ is the first balance and the common difference $d$ equals the interest per period.
The general term $t_n = t_1 + (n-1)d$ lets you calculate any term directly, without listing all the terms before it.
Expanding $t_n = t_1 + (n-1)d$ gives $t_n = dn + (t_1 - d)$ , which has the same structure as a linear function $f(n) = mn + b$ .
On a graph, the points of an arithmetic sequence lie on a straight line whose slope equals the common difference and whose structure reflects the starting value.
Simple interest, arithmetic sequences, and linear growth are three representations of the same underlying idea: equal additions at equal intervals.

Check your understanding

Question 1

A simple interest account starts with CAD 1 000 and earns CAD 35 per year. Which arithmetic sequence correctly lists the balance at the end of years 1, 2, and 3?

$1035, 1070.25, 1106.02, \ldots$
$1035, 1070, 1105, \ldots$
$1000, 1035, 1070, \ldots$
$35, 70, 105, \ldots$

Show answer and explanation

1035, 1070, 1105, \ldots

The balance grows by CAD 35 each year:

1000 + 35 = 1035

at year 1,

1035 + 35 = 1070

at year 2,

1070 + 35 = 1105

at year 3. Option A uses compounding (incorrect for simple interest). Option C starts with the principal, not the end-of-year-1 balance. Option D lists only the interest amounts, not the total balance.

Question 2

An arithmetic sequence has

t_1 = 412

and

d = 18

. What is the value of

t_6

?

$502$
$520$
$484$
$430$

Show answer and explanation

502

Use

t_n = t_1 + (n-1)d

. For

n = 6

:

t_6 = 412 + (6-1)(18) = 412 + 5 \times 18 = 412 + 90 = 502

. Option B adds

d

six times instead of five. Option C adds

d

only four times. Option D adds

d

only once.

Question 3

The general term of an arithmetic sequence is

t_n = 15n + 300

. What is the slope of the corresponding linear function, and what does it represent in a simple interest context?

Slope $= 300$ ; it is the principal.
Slope $= 15$ ; it is the interest earned each period.
Slope $= 315$ ; it is the balance after one period.
Slope $= 15$ ; it is the total interest over all periods.

Show answer and explanation

Slope

= 15

; it is the interest earned each period.

In

t_n = 15n + 300

, the coefficient of

n

is 15, which is the slope. In a simple interest context, this constant increase of 15 per period is the interest earned each period (

Pr

). The 300 is the

y

-intercept, related to the starting value, not the slope.

Question 4

A linear graph of account balance against time has a

y

-intercept of CAD 500 and passes through the point

(5, 650)

. What is the annual simple interest rate?

4%
3%
5%
6%

Show answer and explanation

6%

The slope is

\frac{650 - 500}{5 - 0} = \frac{150}{5} = 30

. The slope equals

Pr

, so

r = \frac{30}{500} = 0.06

, which is 6%. A rate of 3% would give slope 15, 4% gives slope 20, and 5% gives slope 25 — none of those match.

Key terms

Principal ( $P$ ): The original amount of money deposited or borrowed, before any interest is added.
Simple interest: A method of calculating interest where the interest is always based on the original principal, so the same dollar amount is added in every time period.
Arithmetic sequence: An ordered list of numbers in which the difference between any two consecutive terms is always the same constant value.
Common difference ( $d$ ): The fixed amount added from one term to the next in an arithmetic sequence. It can be positive (increasing) or negative (decreasing).
General term ( $t_n$ ): A formula that gives the value of the $n$ th term in a sequence without requiring you to list all the previous terms.
Linear growth: A pattern of growth in which the quantity increases by the same fixed amount for every equal interval of the input variable, producing a straight-line graph.
Slope ( $m$ ): In a linear function $f(x) = mx + b$ , the slope is the constant rate of change — how much the output increases for each one-unit increase in the input.
y-intercept ( $b$ ): The value of a linear function when the input is zero; on a graph, it is where the line crosses the vertical axis.

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