MCR3U Arithmetic Sequences: Notes and Practice

An arithmetic sequence is the simplest and most common type: you add the SAME number every single step. Your allowance going up by 5 dollars each week, a phone plan's data cap dropping by 2 GB each overage tier, a staircase rising by 18 cm each step — all arithmetic. That constant add-on is called the common difference, $d$ . Because the change never varies, arithmetic sequences are completely predictable: from just the first term and the common difference you can find any term, work backward to find which position a value sits in, or decide that a value never appears at all.

The common difference

To find $d$ , subtract any term from the term right after it: $d = t_{n} - t_{n - 1}$ . If this comes out the same no matter which consecutive pair you pick, the sequence is arithmetic. If $d > 0$ the sequence increases; if $d < 0$ it decreases. Order matters when you subtract: for $20, 17, 14$ the difference is $17 - 20 = - 3$ , not $+ 3$ . The common difference can be a fraction or decimal too, as in $1.5, 2, 2.5, 3$ with $d = 0.5$ , and it can involve variables: $x, x + 4, x + 8$ has $d = 4$ , while $3 k, 5 k, 7 k$ has $d = 2 k$ . Check at least two or three pairs before calling a sequence arithmetic: $2, 4, 8$ has first differences $2$ and $4$ , so it is not.

The explicit formula

Starting from the first term $a$ , to reach the $n$ th term you add $d$ a total of $(n - 1)$ times (not $n$ times — you don't add anything to get to the very first term). It is the fence-post idea: 10 posts in a row have only 9 gaps between them, and the 10th term is 9 jumps of $d$ past the first. The table below builds the terms of $a = 5$ , $d = 3$ this way, so you can see the coefficient of $d$ always lags one behind the term number. The same reasoning works from any known term, not just the first: to get from $t_{4}$ to $t_{15}$ you make $15 - 4 = 11$ jumps, so $t_{15} = t_{4} + 11 d$ . Thinking in jumps between positions is often faster than finding $a$ first, and it is exactly why the gap in position appears in the denominator when you find $d$ from two terms.

t_{n} = a + (n - 1) d

n	Build from the first term	$t_{n}$
1	$5 + 0 (3)$	5
2	$5 + 1 (3)$	8
3	$5 + 2 (3)$	11
4	$5 + 3 (3)$	14
10	$5 + 9 (3)$	32

The number of jumps of d is always one less than the term number.

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Drag the sliders for the first term and the common difference to watch an arithmetic sequence build itself term by term — simple interest follows exactly this pattern.

This one formula does double duty: given $a$ , $d$ , and $n$ you can find $t_{n}$ ; given $t_{n}$ instead, you can rearrange it to solve for $n$ (which term number produced that value) or for $d$ or $a$ . Expanding it shows why arithmetic sequences are the discrete version of lines: $t_{n} = a + (n - 1) d = d n + (a - d)$ . That is a linear function of $n$ with slope $d$ , so its dots all lie on a straight line. For $a = 5$ , $d = 3$ you get $t_{n} = 3 n + 2$ . This gives a quick way to read an explicit formula backwards: in $t_{n} = 6 n - 4$ , the coefficient of $n$ is the common difference ( $d = 6$ ), and the first term is found by substituting $n = 1$ , giving $a = 2$ . Be careful not to read the constant $- 4$ as the first term; it is the value the line would have at $n = 0$ , which is not a term of the sequence at all.

VideoWorked examples of finding any term from the first term and the common difference. The video writes the formula as a_n = a_1 + (n - 1)d, which is the same formula as t_n = a + (n - 1)d in this lesson.Watch on YouTube

Solving for unknowns

Tip

Whenever a problem gives you two terms that AREN'T consecutive (say

t_{3}

and

t_{8}

), find

d

first by dividing the difference in VALUE by the difference in POSITION:

d = \frac{t _{8} - t _{3}}{8 - 3}

. Then work back to

a

. Equivalently, write

a + 2 d = t_{3}

and

a + 7 d = t_{8}

and solve the linear system by elimination; subtracting the equations gives

5 d = t_{8} - t_{3}

, the same thing.

Key idea

n

must come out to be a positive integer. If solving for

n

gives a decimal or negative number, the value you were given is not actually a term of that sequence.

Identify $a$ (the first term), $d$ (the change per step) and what $n$ counts. In context, decide carefully which quantity is $t_{1}$ : "week 1", "row 1" or the starting value before any change.
Write the explicit formula with your numbers, e.g. $t_{n} = 40 + 15 (n - 1)$ .
To find a term, substitute $n$ . To find a position, set $t_{n}$ equal to the value and solve for $n$ .
For "first term to exceed" questions, solve the inequality and round $n$ UP to the next whole number, then check the terms on both sides.

Counting terms and spotting arithmetic situations

How many terms are in $7, 11, 15, \dots, 203$ ? Treat the last term as $t_{n}$ and solve: $203 = 7 + (n - 1) (4)$ , so $196 = 4 (n - 1)$ , $n - 1 = 49$ and $n = 50$ . (Subtracting $203 - 7$ and dividing by 4 gives 49 gaps, and 49 gaps means 50 terms: the fence-post idea again.) In word problems, look for a fixed amount added or removed each step: a salary rising by a fixed 1500 dollars per year, seats increasing by 2 per row, simple interest adding the same dollar amount every year, or a candle burning down 0.8 cm per hour. A fixed PERCENTAGE change is not arithmetic; that is the geometric pattern in the next lesson.

AArithmetic (fixed amount)

Salary rises by 1500 dollars every year
Each row of a theatre has 2 more seats than the row before
Simple interest: the same interest is added every year
A candle burns down 0.8 cm every hour
Test: first differences are constant

BNot arithmetic

Salary rises by 3% every year (geometric)
A population doubles every decade (geometric)
Compound interest: interest earns interest (geometric)
Square tile patterns 1, 4, 9, 16 (quadratic)
Test: first differences change

Arithmetic Sequences

Key things to remember

The common difference

The explicit formula

Solving for unknowns

Counting terms and spotting arithmetic situations

Worked examples

Where marks get lost

Flashcards 8

Practice