This note covers how to describe, continue and write rules for number sequences: term-to-term rules, linear and quadratic nth terms, and simple geometric sequences. Sequences appear in pattern questions, in algebra and in real contexts such as savings plans and stacking objects, so a reliable method for finding the nth term is worth learning well.
Key ideas
A sequence is a list of numbers in a fixed order. Each number is a term. We write the first term as \(u_1\), the second as \(u_2\), and the nth term as \(u_n\).
There are two ways to describe a sequence.
- A term-to-term rule tells you how to get the next term from the one before, for example "add 4" or "multiply by 3 then subtract 1".
- A position-to-term rule (the nth term) gives any term straight from its position \(n\). For example \(u_n = 3n + 1\) gives \(u_{20} = 61\) without listing the first 19 terms.
Linear (arithmetic) sequences. The terms go up or down by the same amount each time. This amount is the common difference \(d\). The nth term is \[ u_n = dn + (a - d) \] where \(a\) is the first term. In practice: the number in front of \(n\) is the common difference, then you work out what must be added or subtracted to match the first term.
Quadratic sequences (Higher). The first differences are not constant, but the second differences (the differences of the differences) are. For \(u_n = an^2 + bn + c\), the second difference is always \(2a\). Method:
- Find the first differences, then the second differences.
- Halve the second difference to get \(a\).
- Subtract \(an^2\) from each term. What is left is a linear sequence, so find its nth term \(bn + c\).
- Add the two parts together.
Geometric sequences. Each term is the previous term multiplied by a constant ratio \(r\). If the first term is \(a\), then \(u_n = a r^{\,n-1}\). The ratio can be a fraction (the terms shrink) or negative (the signs alternate). Sequences such as \(2^n\) and \(3^n\) are geometric.
Other sequences to recognise.
- Square numbers: \(n^2\) gives 1, 4, 9, 16, ...
- Cube numbers: \(n^3\) gives 1, 8, 27, 64, ...
- Triangular numbers: 1, 3, 6, 10, 15, ... with nth term \(\tfrac{1}{2}n(n+1)\).
- Fibonacci-type sequences: each term is the sum of the previous two.
Testing whether a number is in a sequence. Set the nth term equal to the number and solve for \(n\). The number is a term only if \(n\) is a positive whole number.
Describing a sequence. A full description of a linear sequence gives the common difference. For a geometric sequence, give the ratio. Writing "it goes up in 5s" is enough for a description of the pattern, but "the nth term is \(5n+2\)" is needed when the question asks for the rule.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Term-to-term rule | Gets each term from the one before. Good for generating the next few terms. |
| nth term \(u_n\) | Gives the term in position \(n\) directly. |
| \(u_n = dn + (a-d)\) | Linear sequence with first term \(a\) and common difference \(d\). |
| Second difference \(=2a\) | For a quadratic sequence \(an^2+bn+c\) (Higher). Halve it to find \(a\). |
| \(u_n = a r^{\,n-1}\) | Geometric sequence with first term \(a\) and common ratio \(r\). |
| \(\tfrac12 n(n+1)\) | nth triangular number. |
| \(u_{n+1} = f(u_n)\) | Notation for a term-to-term rule, used with a starting value \(u_1\). |
Worked examples
Example 1 A gym charges a joining fee and then a fixed monthly fee. The total paid after 1, 2, 3, 4 months is £37, £52, £67, £82. (a) Find the nth term. (b) Find the total after 24 months. (c) After how many months does the total first exceed £500?
(a) The common difference is \(52 - 37 = 15\), so the nth term starts with \(15n\).
When \(n=1\): \(15 \times 1 = 15\). The first term is 37, and \(37 - 15 = 22\).
So \(u_n = 15n + 22\).
(b) \(u_{24} = 15 \times 24 + 22 = 360 + 22 = 382\).
(c) Solve \(15n + 22 \gt 500\). Then \(15n \gt 478\), so \(n \gt 31.87\). The smallest whole number is 32. Check: \(u_{32} = 480 + 22 = 502\) and \(u_{31} = 465 + 22 = 487\).
Answer: (a) \(15n+22\) (b) £382 (c) 32 months (£502)
Example 2 (Higher) A quadratic sequence begins 4, 11, 22, 37, 56, ... (a) Find the nth term. (b) Show that 95 is not a term of the sequence.
(a) First differences: 7, 11, 15, 19. Second differences: 4, 4, 4.
The second difference is 4, so \(2a = 4\) and \(a = 2\). The \(n^2\) part is \(2n^2\).
Subtract \(2n^2\) (which is 2, 8, 18, 32, 50) from each term: \(4-2=2\), \(11-8=3\), \(22-18=4\), \(37-32=5\), \(56-50=6\).
The remainder 2, 3, 4, 5, 6 has nth term \(n + 1\).
So \(u_n = 2n^2 + n + 1\). Check with \(n=5\): \(50 + 5 + 1 = 56\). Correct.
(b) Set \(2n^2 + n + 1 = 95\). Then \(2n^2 + n - 94 = 0\).
The discriminant is \(1^2 - 4 \times 2 \times (-94) = 753\), which is not a perfect square, so \(n\) is not a whole number. Alternatively, test nearby values: \(u_6 = 72 + 6 + 1 = 79\) and \(u_7 = 98 + 7 + 1 = 106\). The terms jump from 79 to 106 and miss 95.
Answer: (a) \(2n^2+n+1\) (b) 95 lies between \(u_6=79\) and \(u_7=106\), so it is not a term.
Example 3 (a) A sequence is defined by \(u_{n+1} = 2u_n - 3\) with \(u_1 = 5\). Find \(u_2\), \(u_3\) and \(u_4\). (b) A geometric sequence has first term 3 and common ratio 2. Find the nth term and the 9th term.
(a) \(u_2 = 2 \times 5 - 3 = 7\).
\(u_3 = 2 \times 7 - 3 = 11\).
\(u_4 = 2 \times 11 - 3 = 19\).
(b) Use \(u_n = a r^{\,n-1}\) with \(a = 3\) and \(r = 2\): \(u_n = 3 \times 2^{\,n-1}\).
\(u_9 = 3 \times 2^{8} = 3 \times 256 = 768\).
Answer: (a) 7, 11, 19 (b) \(3\times 2^{\,n-1}\); \(u_9 = 768\)
Common errors
- Common error: using the common difference as the whole nth term. For 7, 12, 17 the rule is \(5n+2\), not \(5n\). Always test \(n=1\).
- Common error: in a quadratic sequence, using the second difference as \(a\) instead of half of it. If the second difference is 6, then \(a = 3\).
- Common error: forgetting to subtract \(an^2\) before finding the linear part, so the remaining numbers are not a straight-line pattern.
- Common error: writing a geometric nth term as \(a r^n\) when the first term is \(a\). The power should be \(n-1\).
- Common error: saying a number is in a sequence because \(n\) works out as a decimal. \(n\) must be a positive whole number.
Exam tip: Once you have found an nth term, substitute \(n=1\), \(n=2\) and \(n=3\) to check that you reproduce the first three given terms. It takes ten seconds and catches most slips.
Check yourself
- Find the nth term of the sequence 11, 7, 3, -1, ... and the 30th term.
Answer
The common difference is \(-4\), so start with \(-4n\). When \(n=1\), \(-4\), and \(11-(-4) = 15\). So \(u_n = 15 - 4n\). Then \(u_{30} = 15 - 120 = -105\). - Is 250 a term of the sequence with nth term \(7n - 3\)?
Answer
Solve \(7n - 3 = 250\). Then \(7n = 253\) and \(n = 36.14...\), which is not a whole number. So 250 is not a term. - (Higher) Find the nth term of the quadratic sequence 3, 10, 21, 36, ...
Answer
First differences: 7, 11, 15. Second differences: 4, so \(a = 2\). Subtract \(2n^2\) (2, 8, 18, 32): remainders 1, 2, 3, 4, which is \(n\). So \(u_n = 2n^2 + n\). Check \(n=4\): \(32 + 4 = 36\). - A geometric sequence has \(u_1 = 80\) and common ratio \(\tfrac12\). Find \(u_6\).
Answer
\(u_6 = 80 \times (\tfrac12)^5 = 80 \div 32 = 2.5\).