Finding Rules for Number Patterns
A number pattern such as \(4, 7, 10, 13, \ldots\) changes by the same step each time, and a rule describes every term at once. In NSW Year 8 maths, students turn patterns into tables and algebraic rules and use them to predict terms (NESA outcome MA4‑LIN‑C‑01).
You will continue increasing and decreasing patterns, find the rule linking \(x\) and \(y\) in a table of values, and write a rule for the \(n\)th term, such as \(T = 2n + 5\). You will use a rule to find a term far along the pattern, test whether a number belongs to it, and solve real problems about chairs around tables, stacked cups and weekly savings. A few patterns, like \(1, 1, 2, 3, 5, \ldots\), grow in other ways. It is part of Year 8 Graphing Linear Equations.
Every question with a fully worked solution.
- Finding Rules for Number Patterns - Video - Making an algebraic rule from a simple pattern Watch
Theory
A pattern that goes up or down by the same step has a rule: multiply the term number by the step, then add or take away a number so the rule fits. For \(10, 7, 4, 1, \ldots\) the step is \(-3\) and the rule is \(T_n = -3n + 13\), so \(T_{10} = -17\). This Year 8 maths page shows how.
Find the step. Take each term from the one after it. The same answer every time means the pattern has a constant step. A pattern that goes down has a negative step.
Describe it in words with the start number and the step: “start at \(20\) and take away \(3\) each time”.
Build the rule. With step \(s\), the rule starts \(T = sn\), where \(n\) is the term number. Put \(n = 1\) into \(sn\) and compare with the 1st term; the difference is the number to add. For \(20, 17, 14, \ldots\): \(-3 \times 1 = -3\) but \(T_1 = 20\), so add \(23\): \(T = -3n + 23\).
From a table, \(y = (\text{step}) \times x + (\text{number})\). When the table starts at \(x = 0\), the number to add is the first \(y\)-value.
Use the rule. Substitute a term number to reach a far term. To go the other way, make the rule equal the value and solve. A term number must be a whole number.
Not every pattern has a constant step. In \(2, 6, 18, 54, \ldots\) each term is multiplied by \(3\); in \(3, 4, 7, 11, 18, \ldots\) each term is the sum of the two before it. Look at how each term is made; if a rule is given, substitute into it.
Building the rule for the nth term:
| Pattern | Working | Rule |
|---|---|---|
| \(5, 7, 9, \ldots\) | step \(2\); \(2 \times 1 = 2\), add \(3\) | \(T = 2n + 3\) |
| \(10, 7, 4, \ldots\) | step \(-3\); \(-3 \times 1 = -3\), add \(13\) | \(T = -3n + 13\) |
Finding the rule for the nth term
- Find the step \(s\) and check it is the same between every pair.
- Write \(T = sn\) and work out \(s \times 1\).
- Compare with the 1st term: add (or take away) the difference.
- Check the rule in another term.
| \(a\) | \(=\) | \(7 + 2 = 9\) |
| \(b\) | \(=\) | \(9 + 2 = 11\) |
The terms go up by \(2\): \(a = 9\) and \(b = 11\).
The terms go down by \(3\), so \(T_n = -3n + b\). With \(n = 1\): \(10 = -3 + b\), so \(b = 13\).
| \(T_n\) | \(=\) | \(-3n + 13\) |
| \(T_{10}\) | \(=\) | \(-3 \times 10 + 13 = -17\) |
\(T_n = -3n + 13\) and \(T_{10} = -17\).
(i) What is the step?
(ii) Write the next two terms.
(iii) Describe the pattern in words.
\(11 - 4 = 7\), \(18 - 11 = 7\), \(25 - 18 = 7\).
(i) \(7\)
(ii) \(32\) and \(39\)
(iii) Start at \(4\) and add \(7\) each time.
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
| \(y\) | \(9\) | \(5\) | \(1\) | \(-3\) | \(-7\) |
The step is \(5 - 9 = -4\). The table starts at \(x = 0\), so the number to add is \(9\).
| \(y\) | \(=\) | \(-4x + 9\) |
| \(-4 \times 12 + 9\) | \(=\) | \(-39\) |
\(y = -4x + 9\), and \(y = -39\) when \(x = 12\). Check \(x = 4\): \(-16 + 9 = -7\).
(i) Find a rule for \(T\).
(ii) Find the \(20\)th term.
(iii) Is \(100\) a term?
The step is \(5\). At \(n = 1\), \(5 \times 1 = 5\) but \(T = 7\), so add \(2\).
| \(T\) | \(=\) | \(5n + 2\) |
| \(5 \times 20 + 2\) | \(=\) | \(102\) |
| \(5n + 2\) | \(=\) | \(100\) |
| \(n\) | \(=\) | \(19.6\) |
(i) \(T = 5n + 2\)
(ii) \(102\)
(iii) No: \(n = 19.6\) is not a whole number.
(i) Find a rule for the cost \(C\).
(ii) Find the cost of \(6\) hours.
(iii) A hire cost \(\$205\). How many hours?
Each hour adds \(61 - 43 = 18\). At \(h = 1\), \(18 \times 1 = 18\) but \(C = 43\), so add \(25\) (the deposit).
| \(C\) | \(=\) | \(18h + 25\) |
| \(18 \times 6 + 25\) | \(=\) | \(133\) |
| \(18h\) | \(=\) | \(205 - 25 = 180\) |
| \(h\) | \(=\) | \(10\) |
(i) \(C = 18h + 25\)
(ii) \(\$133\)
(iii) \(10\) hours
(i) \(2, 6, 18, 54, \ldots\)
(ii) \(3, 4, 7, 11, 18, \ldots\)
(i) Each term is \(3\) times the one before: \(162\) and \(486\).
(ii) Each term is the sum of the two before it: \(11 + 18 = 29\), then \(18 + 29 = 47\).
(i) Find a rule for pattern B.
(ii) For which \(n\) are the terms equal, and what is that term?
B goes down by \(3\): at \(n = 1\), \(-3 \times 1 = -3\) but \(T = 34\), so add \(37\).
| \(T\) | \(=\) | \(37 - 3n\) |
| \(3n + 1\) | \(=\) | \(37 - 3n\) |
| \(6n\) | \(=\) | \(36\) |
| \(n\) | \(=\) | \(6\) |
(i) \(T = 37 - 3n\)
(ii) \(n = 6\), where both terms are \(3 \times 6 + 1 = 19\).
Common pitfalls
Frequently asked questions
How do I find the rule for a number pattern?
Find the step, write T = step × n, then add or subtract the number that makes the 1st term correct. Check with another term.
What does the nth term mean?
The term in position n. T₁ is the first term, T₁₀ the tenth.
How do I check whether a number is in a pattern?
Set the rule equal to the number and solve for n. It is a term only if n is a positive whole number.
What if the pattern goes down?
The step is negative, so the rule starts with a negative multiple of n, such as T = −3n + 13.
Do all patterns have a rule like T = 5n + 2?
No. Only patterns with a constant step. Others, such as 2, 6, 18, 54, are made in different ways.