How to Find the nth Term of a Linear Sequence: IGCSE Maths
The nth term of a linear sequence is the formula that jumps straight to any term of the pattern, and it is one of the most-tested algebra skills in Edexcel IGCSE Maths. A linear sequence adds the same amount each time: that fixed gap is the common difference, written d. This page shows you how to find the nth-term rule, use the arithmetic-sequence formula with a and d, and test whether a number belongs to a sequence. Work through the worked examples, then try the five rooms of free, auto-marked practice questions below.
What is a linear sequence?
A sequence is just an ordered list of numbers that follows a rule. A linear sequence is one where the gap between each pair of consecutive terms is the same. That gap is called the common difference.
Examples of linear sequences:
- \(6, 10, 14, 18, 22, \ldots\) (common difference \(+4\))
- \(20, 17, 14, 11, 8, \ldots\) (common difference \(-3\))
- \(1, 8, 15, 22, 29, \ldots\) (common difference \(+7\))
The goal of this topic is to write down a formula, called the nth term, that gives you any term you want without listing them all. Once you have the formula, the 100th term, the 500th term, or any other position drops out from a single substitution.
Two ways to describe a sequence
- A term-to-term rule tells you how to get the next term from the one before: "start at \(5\) and add \(4\) each time". Easy to follow, but slow if you want the 100th term.
- A position-to-term rule (the nth term) jumps straight to any position: \(4n + 1\) gives the 100th term in one step.
Sequences can also step by a decimal (for example \(2.5,\ 3,\ 3.5,\ 4, \ldots\) has common difference \(0.5\)), and some term-to-term rules multiply or divide instead of adding: those are not linear, but you continue them the same way, one term at a time.
Substituting into a given nth-term formula
If you are given the nth-term formula, you can find any term you want by substituting the value of \(n\) into the formula.
For example, if the nth term is \(6n - 2\):
- The first term (\(n=1\)) is \(6(1) - 2 = 4\)
- The second term (\(n=2\)) is \(6(2) - 2 = 10\)
- The 50th term (\(n=50\)) is \(6(50) - 2 = 298\)
Always replace \(n\) with the position number, not anything else. The position is always a positive whole number: \(1, 2, 3, \ldots\)
Finding the nth term of a linear sequence
The nth term of a linear sequence has the form:
Here \(a\) is the common difference and \(b\) is whatever number makes the first term come out right.
Three-step method
- Find the common difference by subtracting consecutive terms. This is your value of \(a\).
- Test what \(an\) gives at \(n=1\). The first term of \(an\) alone is just \(a\) itself.
- Find \(b\) by working out what you need to add (or subtract) to get the actual first term.
Worked example: the sequence \(9, 14, 19, 24, \ldots\)
- Common difference is \(+5\), so \(a = 5\).
- The first term of \(5n\) alone is \(5(1) = 5\).
- The actual first term is \(9\), so we need to add \(4\). Therefore \(b = 4\).
The nth term is \(5n + 4\). Quick check: substitute \(n=2\) to get \(14\). Correct.
Arithmetic sequences
A linear sequence is also called an arithmetic sequence. The same rule, written using the first term and the common difference, gives this equivalent formula:
where \(a\) is the first term and \(d\) is the common difference.
Notation: \(a\) and \(d\)
Two letters are used to describe any arithmetic sequence concisely:
| Letter | Meaning | Example: \(9, 14, 19, 24, \ldots\) |
|---|---|---|
| \(a\) | First term | \(a = 9\) |
| \(d\) | Common difference | \(d = 5\) |
From the first term you can build any term by repeatedly adding \(d\):
Notice that the 2nd term uses \(d\) once, the 3rd uses \(d\) twice, and so on. The coefficient of \(d\) is always one less than the position number, which is exactly why the formula is \(a + (n-1)d\).
Which form should I use?
Both \(an + b\) and \(a + (n-1)d\) give the same answers; they are just rearrangements of each other. Use whichever is easier for the question at hand:
- The \(an + b\) form is quicker to use once you have already spotted the common difference and the first term.
- The \(a + (n-1)d\) form is more useful when you are given two specific terms and need to set up simultaneous equations to find \(a\) and \(d\).
Worked examples
💡 Example 1: continue a sequence
Find the next two terms of the sequence \(4, 10, 16, 22, \ldots\)
Common difference: \(10 - 4 = 6\)
Check next: \(16 - 10 = 6\), \(22 - 16 = 6\). Constant.
Next term: \(22 + 6 = 28\)
Term after: \(28 + 6 = 34\)
What's happening?
Subtract pairs of consecutive terms to confirm the gap is the same all the way along. Then keep adding it.
💡 Example 2: use a given formula
The nth term of a sequence is \(8n - 5\). Find the 20th term.
Substitute \(n = 20\):
\[ \begin{aligned} 8(20) - 5 &= 160 - 5 \\ &= 155 \end{aligned} \]What's happening?
Wherever you see \(n\) in the formula, replace it with \(20\). Multiply first, then subtract: BIDMAS still applies.
💡 Example 3: find the nth term
Find the nth term of the sequence \(9, 13, 17, 21, 25, \ldots\)
Common difference: \(13 - 9 = 4\), so \(a = 4\).
The first term of \(4n\) alone is \(4(1) = 4\).
The actual first term is \(9\), so add \(5\). \(b = 5\).
nth term \(= 4n + 5\)
What's happening?
The coefficient of \(n\) is always the common difference. Then adjust by a constant to match the first term.
💡 Example 4: find a, d from two terms
The 5th term of an arithmetic sequence is \(23\) and the 9th term is \(43\). Find the first term and the common difference.
Use the formula \(a + (n-1)d\):
\[ \begin{aligned} a + 4d &= 23 \\ a + 8d &= 43 \end{aligned} \]Subtract: \(4d = 20\), so \(d = 5\).
Substitute back: \(a + 4(5) = 23\), so \(a = 3\).
What's happening?
Two unknowns, two equations: classic simultaneous equations setup. Subtract the equations to eliminate \(a\), solve for \(d\), then substitute back.
💡 Example 5: is a number in the sequence?
Is the number \(146\) a term of the sequence \(6, 13, 20, 27, \ldots\)? If so, which term?
Find the nth term first.
Common difference \(= 7\), so \(a = 7\).
The first term of \(7n\) is \(7\); actual first term is \(6\), so subtract \(1\). \(b = -1\).
nth term \(= 7n - 1\).
Set equal to \(146\):
\[ \begin{aligned} 7n - 1 &= 146 \\ 7n &= 147 \\ n &= 21 \end{aligned} \]\(146\) is the 21st term.
What's happening?
If \(n\) comes out as a positive whole number, the value is in the sequence. If \(n\) is a fraction or negative, the value is not in the sequence at all.
💡 Example 6: a sequence of fractions
Find the nth term of the sequence \(\dfrac{1}{4},\ \dfrac{2}{5},\ \dfrac{3}{6},\ \dfrac{4}{7},\ \ldots\)
Treat the tops and bottoms as two separate sequences.
Numerators: \(1,\ 2,\ 3,\ 4,\ \ldots\) so the numerator is \(n\).
Denominators: \(4,\ 5,\ 6,\ 7,\ \ldots\) so the denominator is \(n + 3\).
nth term \(= \dfrac{n}{n+3}\)
Check \(n = 2\): \(\dfrac{2}{2+3} = \dfrac{2}{5}\). Correct.
What's happening?
Each of the top and the bottom is its own linear sequence, so find an nth term for each and stack them. Do not simplify the terms while you are spotting the pattern: \(\dfrac{3}{6}\) is easier to read as position 3 than as \(\dfrac{1}{2}\).
🔑 Key points
- A linear sequence has a constant common difference between consecutive terms.
- The nth term of a linear sequence has the form \(an + b\), where \(a\) is the common difference.
- The arithmetic-sequence form \(a + (n-1)d\) gives the same answers; useful when given two specific terms.
- To check whether a value is a term, set the formula equal to the value and solve for \(n\). If \(n\) is a positive whole number, yes.
- The common difference can be a decimal: \(2.5,\ 3,\ 3.5, \ldots\) has \(d = 0.5\) and works exactly the same way.
- For a sequence of fractions, find an nth term for the numerators and another for the denominators, then stack them.
- Always test your formula on at least one known term before relying on it.
⚠️ Common pitfalls
- Forgetting that the common difference can be negative. The sequence \(20, 17, 14, \ldots\) has \(a = -3\).
- Writing the nth term as \(an\) only and forgetting the constant. The first term is rarely just \(a\).
- Confusing the position (\(n\)) with the value of the term. \(n\) is always a positive whole number; the value can be anything.
- Using the second term as the first term. Always start counting at \(n=1\).
- Assuming every sequence is linear. If the differences are not constant, this method does not apply.