Arithmetic Sequences Using Recursion Revision Notes for VCE SSCE General Mathematics

Arithmetic Sequences Using Recursion

What is a recurrence relation?

A recurrence relation (also called a recursion relation) is a mathematical way to describe how to generate the terms of a sequence. It consists of two parts:

A starting value (also called the initial value)
A rule that tells you how to find each new term from the previous term

This method is particularly useful for arithmetic sequences, where each term differs from the previous term by a constant amount.

Understanding recurrence relations through an example

Let's look at the arithmetic sequence:

$10, 15, 20, 25, 30, \ldots$

We can describe this sequence using words:

Start the sequence with $10$
To find the next term, add $5$ to the current term, and keep repeating this process

infoNote

Using mathematical notation with subscripts, where $t_0$ represents the first term, $t_1$ the second term, and so on, we can show how this works step by step. This subscript notation is a standard way to label terms in sequences, where the number in the subscript indicates the term's position.

The pattern shows that after $n$ applications of the rule, we have $t_{n+1} = t_n + 5$ .

The general form of a recurrence relation

The recurrence relation provides a precise and compact way to express both the starting value and the rule for generating an arithmetic sequence.

For an arithmetic sequence with:

First term $t_0 = a$ (the starting value)
Common difference $D$ (the amount added each time)

The recurrence relation is:

$t_0 = a, \quad t_{n+1} = t_n + D$

This notation means: start with $a$ , then to get each new term, add $D$ to the previous term.

chatImportant

The common difference $D$ can be positive (sequence increases) or negative (sequence decreases). This determines whether your sequence grows or shrinks with each term.

Generating arithmetic sequences using recurrence relations

lightbulbExample

Worked Example: Generating and Graphing Terms

Question: Generate and graph the first five terms of the arithmetic sequence defined by the recurrence relation:

$t_0 = 24, \quad t_{n+1} = t_n - 2$

Solution:

Step 1: Identify the starting term

$t_0 = 24$

Step 2: Apply the rule repeatedly

The rule tells us to subtract 2 from each term to get the next term:

$t_1 = t_0 - 2 = 24 - 2 = 22$

$t_2 = t_1 - 2 = 22 - 2 = 20$

$t_3 = t_2 - 2 = 20 - 2 = 18$

$t_4 = t_3 - 2 = 18 - 2 = 16$

Step 3: Graph the sequence

To graph the terms, we plot $t_n$ (the term value) on the vertical axis against $n$ (the term number) on the horizontal axis. The points are: $(0, 24)$ , $(1, 22)$ , $(2, 20)$ , $(3, 18)$ , $(4, 16)$ .

The graph shows a linear decreasing pattern, which is characteristic of an arithmetic sequence with a negative common difference.

Finding the nth term directly

While we can use repeated addition (or subtraction) to find each term in an arithmetic sequence, this becomes very tedious when finding terms like $t_{50}$ or $t_{100}$ . Instead, we can develop a formula to calculate any term directly.

infoNote

Imagine having to calculate $t_{100}$ by hand using the recurrence relation—you would need to perform the same operation 100 times! The direct formula solves this problem by letting you jump straight to any term you want.

Developing the direct formula

Consider this arithmetic sequence: $8, 13, 18, 23, 28, 33, \ldots$

This sequence is defined by the recurrence relation:

$t_0 = 8, \quad t_{n+1} = t_n + 5$

We can visualise this sequence and see how the common difference applies:

Looking at the pattern more carefully, we can write each term in terms of the starting value and the number of times we've added $5$ .

Notice the pattern that emerges:

$t_0 = 8 + 0 \times 5$ (no applications of the rule)
$t_1 = 8 + 1 \times 5$ (one application of the rule)
$t_2 = 8 + 2 \times 5$ (two applications of the rule)
$t_3 = 8 + 3 \times 5$ (three applications of the rule)

After $n$ applications of the rule:

$t_n = 8 + n \times 5$

Using this formula, we can find any term directly. For example:

$t_{50} = 8 + 50 \times 5 = 8 + 250 = 258$

This is much faster than calculating all 51 terms from $t_0$ to $t_{50}$ !

The general rule for the nth term

We can generalise this pattern for any arithmetic sequence:

For a sequence defined by the recurrence relation:

$t_0 = a, \quad t_{n+1} = t_n + D$

chatImportant

The direct formula for the nth term is:

$t_n = a + nD$

where:

$a$ is the starting value (initial term)
$D$ is the common difference
$n$ is the term number ( $n = 0, 1, 2, 3, \ldots$ )

This formula tells us: to find the $n$ th term, start with $a$ and add the common difference $D$ exactly $n$ times.

lightbulbExample

Worked Example: Finding a Specific Term

Question: Consider the recurrence relation:

$t_0 = 21, \quad t_{n+1} = t_n - 3$

Find $t_{20}$ .

Solution:

Step 1: Identify the values

From the recurrence relation:

Starting value: $a = 21$
Common difference: $D = -3$ (we subtract 3 each time)

Step 2: Determine the term number

We need to find $t_{20}$ , so $n = 20$

Step 3: Substitute into the formula

Using the formula $t_n = a + nD$ :

$t_{20} = 21 + 20(-3)$

Step 4: Calculate

$t_{20} = 21 + (-60)$

$t_{20} = -39$

Answer: $t_{20} = -39$

bookmarkSummary

Key Points to Remember:

A recurrence relation describes an arithmetic sequence using a starting value and a rule for finding each next term: $t_0 = a$ , $t_{n+1} = t_n + D$
The common difference $D$ is the amount added to each term (it can be negative for decreasing sequences)
The direct formula $t_n = a + nD$ allows you to find any term without calculating all previous terms
In the formula $t_n = a + nD$ , remember that $n$ represents the number of times you apply the rule (so $t_0$ is the first term, $t_1$ is the second term, etc.)
Always identify $a$ (starting value) and $D$ (common difference) before using the formula

Arithmetic Sequences Using Recursion (VCE SSCE General Mathematics): Revision Notes