A sequence is a list of numbers that follows a rule. In this topic, you will learn how to continue a sequence, describe the rule linking its terms, and find a formula (called the nth term) that lets you work out any term directly, without writing out every term before it.
1. What is a Sequence? #
For example, in the sequence 2, 4, 6, 8, 10, the first term is 2, the second term is 4, and so on.
2. Continuing a Sequence: the Term-to-Term Rule #
To continue a sequence, first work out the term-to-term rule, then apply it again to find the next terms.
| Sequence | Term-to-term rule | Next two terms |
|---|---|---|
| 3, 7, 11, 15, … | Add 4 | 19, 23 |
| 1, 3, 6, 10, 15, … | Add one more each time (add 2, then 3, then 4, then 5, …) | 21, 28 |
| 2, 6, 18, 54, … | Multiply by 3 | 162, 486 |
| 1, 1, 2, 3, 5, 8, … | Add the two terms before it | 13, 21 |
Sequences can also be made of shapes or patterns instead of just numbers. You continue these in the same way: work out what changes from one pattern to the next, then draw or count the next pattern. Look at the L-shaped dot pattern below — the blue dots are kept from the pattern before, and the orange dots are the new ones added each time.
Number of dots: 3, 5, 7, 9, … — term-to-term rule: add 2 each time
3. Relationships Between Sequences #
Sometimes one sequence is connected to another sequence by a simple rule. If you know one sequence, you can use the relationship to work out the other one.
| Sequence A | 3 | 6 | 9 | 12 |
|---|---|---|---|---|
| Sequence B | 5 | 8 | 11 | 14 |
Each term in Sequence B is 2 more than the matching term in Sequence A. So once you know Sequence A, you can find Sequence B by adding 2 to every term.
4. The nth Term of a Sequence #
The nth term is a formula that uses the position number, $n$, to calculate any term in a sequence directly. For example, if $n = 5$, the formula gives you the 5th term straight away, without listing all the terms before it.
4.1 Linear Sequences #
Formula for the nth term of a linear sequence:
$$n\text{th term} = dn + c$$- $d$ = common difference (how much the terms increase or decrease by each time)
- $c$ = a constant number that shifts the sequence up or down
Key idea: $dn$ on its own is just the $d$ times table. For example, $4n$ gives the multiples of 4, and $10n$ gives the multiples of 10.
If you add or subtract the same number to every term of a times table, you still add 4 (or 5, or 10, …) to get from one term to the next — only the starting point changes. This is why every linear sequence with common difference $d$ can be written as $dn + c$: it is just the $d$ times table, shifted up or down by $c$.
Steps to find the nth term of a linear sequence:
- Find the common difference $d$ by subtracting one term from the next. This tells you the $dn$ part (if $d$ is negative, you will use a negative times table).
- Write out the $d$ times table above your sequence, lined up term by term.
- Work out the single number you must add or subtract to turn the times table into your sequence — this number is $c$.
- Write the nth term as $dn + c$.
Find the nth term of 2, 5, 8, 11, 14, …
Step 1: Common difference $d = 5 – 2 = 3$, so the nth term is based on $3n$.
Steps 2–3: write the 3 times table above the sequence, and see how to get from one to the other.
Check: when $n = 1$: $3(1) – 1 = 2$ ✓ when $n = 4$: $3(4) – 1 = 11$ ✓
Find the nth term of 12, 9, 6, 3, …
Step 1: $d = 9 – 12 = -3$. Because $d$ is negative, we write out the negative 3 times table: $-3, -6, -9, -12$.
Steps 2–3: work out how to get from the negative times table to the sequence.
Going from $-3$ to $12$ can be tricky to picture — it helps to use a number line and go via 0:
Adding 3 then 12 is the same as adding 15 in one jump, so $c = 15$ and the nth term is $-3n + 15$.
Check: when $n = 1$: $-3(1) + 15 = 12$ ✓ when $n = 4$: $-3(4) + 15 = 3$ ✓
Using the nth term $3n – 1$ from above:
Find the 20th term: substitute $n = 20$: $3(20) – 1 = 59$
Is 47 a term in this sequence? Set $3n – 1 = 47$, so $3n = 48$, so $n = 16$. Since $n = 16$ is a whole number, yes — 47 is the 16th term.
Is 100 a term in this sequence? Set $3n – 1 = 100$, so $3n = 101$, so $n = 33.67$. Since $n$ is not a whole number, no — 100 is not a term in this sequence.
4.2 Simple Quadratic Sequences #
Steps to find the nth term:
- Find the first differences between the terms.
- Find the second differences (the differences between the first differences). These should be constant.
- Divide the second difference by 2 to get the coefficient of $n^2$, call this $a$.
- Subtract $an^2$ from every term in the original sequence. This gives a new, simpler sequence.
- Find the rule for this new sequence (it will be a constant number, or a linear nth term).
- Combine: nth term $= an^2 + $ (rule found in step 5).
Find the nth term of 6, 19, 38, 63, 94, …
Steps 1–2 — first and second differences:
Row 2 (orange) is the first difference: $19-6=13$, $38-19=19$, $63-38=25$, $94-63=31$ — not constant, so it isn’t linear.
Row 3 (highlighted) is the second difference: $19-13=6$, $25-19=6$, $31-25=6$ — always 6, so it’s a quadratic sequence.
Step 3: $a = 6 \div 2 = 3$, so the sequence starts with $3n^2$
Step 4 — subtract $3n^2$ from every term: write out $3n^2$ (square numbers $\times 3$) underneath the sequence, and subtract.
Step 5: the new sequence 3, 7, 11, 15, 19 is linear (it goes up by 4 each time) — so find its nth term using the section 4.1 method:
Step 6 — combine the quadratic part and the linear part: nth term $= 3n^2 + (4n – 1) = 3n^2 + 4n – 1$
Check: when $n = 5$: $3(5)^2 + 4(5) – 1 = 75 + 20 – 1 = 94$ ✓
Using the nth term $3n^2 + 4n – 1$ from above, find the 10th term:
substitute $n = 10$: $3(10)^2 + 4(10) – 1 = 300 + 40 – 1 = 339$
4.3 Simple Cubic Sequences #
Steps to find the nth term:
- Find the first, second, and third differences between the terms. The third difference should be constant.
- Divide the third difference by 6 to get the coefficient of $n^3$, call this $a$.
- Write out $an^3$ (cube numbers $\times a$) underneath the original sequence, and subtract. This gives a new, simpler sequence.
- For a simple cubic sequence, this new sequence will just be a constant number.
- Combine: nth term $= an^3 + $ (constant from step 4).
Find the nth term of 3, 17, 55, 129, 251, …
Step 1 — first, second, and third differences:
First difference: $17-3=14$, $55-17=38$, $129-55=74$, $251-129=122$ — not constant.
Second difference: $38-14=24$, $74-38=36$, $122-74=48$ — still not constant.
Third difference (blue): $36-24=12$, $48-36=12$ — always 12, so it’s a cubic sequence.
Step 2: $a = 12 \div 6 = 2$, so the sequence starts with $2n^3$
Step 3 — subtract $2n^3$ from every term: write out $2n^3$ (cube numbers $\times 2$) underneath the sequence, and subtract.
Step 4: the new sequence is constant: 1, 1, 1, 1, 1
Step 5 — combine: nth term $= 2n^3 + 1$
Check: when $n = 4$: $2(4)^3 + 1 = 128 + 1 = 129$ ✓
Using the nth term $2n^3 + 1$ from above, find the 6th term:
substitute $n = 6$: $2(6)^3 + 1 = 2(216) + 1 = 433$
- A sequence is an ordered list of terms that follow a rule.
- The term-to-term rule tells you how to get from one term to the next.
- The nth term is a formula that lets you find any term directly using its position number, $n$.
- Linear sequence nth term: $dn + c$, where $d$ is the common difference.
- Simple quadratic sequence nth term: contains $n^2$; find it using the second-difference method ($a$ = second difference $\div$ 2).
- Simple cubic sequence nth term: contains $n^3$; find it using the third-difference method ($a$ = third difference $\div$ 6).
