An arithmetic sequence changes by the same amount, the common difference d, at every step. Its nth term is uₙ = a + (n − 1)d, where a is the first term. At A-Level the conditions are rarely given directly: you form simultaneous equations from information about terms, use the symmetric form for three terms, and solve inequalities to find how many terms satisfy a condition.
Arithmetic sequences
- nth term: uₙ = a + (n − 1)d
- uₙ = dn + (a − d), which is linear in n
- Three consecutive terms: m − d, m, m + d
- Consecutive terms x, y, z are in AP when y − x = z − y
The nth term formula
Each term is the previous term plus d, so after n − 1 steps from the first term, uₙ = a + (n − 1)d. Expanding gives uₙ = dn + (a − d): the coefficient of n is the common difference. You can read a and d from a given formula, or build the formula from two known terms. To find which term equals a given value, set uₙ equal to it and solve for n, which must be a positive whole number.
Conditions that become simultaneous equations
Conditions are often indirect, such as “the sixth term is twice the second term” or “the sum of the third and fifth terms is 30”. Write each term in terms of a and d, form one equation from each condition, and solve them simultaneously. Turning every term into a + (k − 1)d first keeps the algebra straightforward.
Three numbers in arithmetic progression
Three numbers in arithmetic progression can be written as m − d, m and m + d. Their sum is 3m, which gives the middle number immediately. A product or a sum of squares then gives a quadratic in d. If the three terms are given as expressions in an unknown, use the fact that the middle term minus the first equals the last minus the middle.
Inequalities and the first term beyond a value
To find the first term greater than (or less than) a value, solve the inequality uₙ > value for n. Because n must be a whole number, round to the correct integer and check the term on each side of the boundary. For a decreasing sequence, the first negative term is the first n for which uₙ < 0.
Worded problems
A quantity that changes by a fixed amount each period, such as a salary rising by a set sum each year, forms an arithmetic sequence. Read carefully which period each value refers to: “year 3” means u₃. Two pieces of information give two equations for a and d; then evaluate a term or solve an inequality to finish.
Worked examples
Example 1. For uₙ = 5n − 2, state the first term and common difference, and find which term equals 63.
- u₁ = 5 − 2 = 3, and the coefficient of n gives d = 5.
- 5n − 2 = 63, so 5n = 65.
Answer: a = 3, d = 5; the 13th term.
Marks: B1 for a and d, M1 for 5n = 65, A1 for n = 13.
Example 2. Given that u₆ = 2u₂ and u₄ = 18, find a and d.
- a + 5d = 2(a + d), so a = 3d.
- u₄ = a + 3d = 6d = 18, so d = 3 and a = 9.
Answer: a = 9, d = 3.
Marks: M1 for a = 3d, M1 for 6d = 18, A1 for both.
Example 3. Three numbers in arithmetic progression have sum 21 and product 231. Find them.
- Write them as m − d, m, m + d. Sum: 3m = 21, so m = 7.
- Product: 7(49 − d²) = 231, so 49 − d² = 33 and d² = 16.
- d = ±4 gives the same three numbers.
Answer: 3, 7 and 11.
Marks: M1 for m = 7, M1 for d² = 16, A1 for the numbers.
Example 4. Find the first negative term of 87, 83, 79, …
- uₙ = 87 + (n − 1)(−4) = 91 − 4n.
- 91 − 4n < 0 gives n > 22.75, so n = 23.
- u₂₃ = 91 − 92 = −1.
Answer: The 23rd term, −1.
Marks: M1 for the inequality, M1 for n = 23, A1 for −1.
Example 5. A salary rises by a fixed amount each year. In year 3 it is £29 000 and in year 8 it is £41 500. Find the starting salary and the annual rise.
- u₈ − u₃ = 5d = 12 500, so d = 2500.
- u₃ = a + 2d = 29 000, so a = 24 000.
Answer: Starting salary £24 000, annual rise £2500.
Marks: M1 for 5d = 12 500, M1 for a, A1 for both.
Common mistakes
- Using nd instead of (n − 1)d. The first term has no difference added, so the nth term is a + (n − 1)d.
- Giving a non-integer number of terms. n counts terms, so round to a whole number and check the boundary term.
- Mixing up which year is which. “Year 3” means u₃, not u₂ or the value after 3 changes.
- Forgetting d can be negative. A decreasing sequence has a negative common difference.
- Not using the symmetric form. Writing three terms as m − d, m, m + d makes the sum condition immediate.
How arithmetic sequences are examined
Arithmetic sequences open the sequences and series chapter of Pure 2 (WMA12) in Edexcel International A-Level; in Edexcel A-Level (9MA0) they are taught in Year 2. Questions ask you to:
- find a and d from two given terms
- find which term equals a value, or the first term beyond a value
- find an unknown when three expressions are consecutive terms
- model a real situation, such as a salary or a loan repayment
Questions in these styles, with mark schemes, are in the Pure 2 sequences and series exam questions. The sequences and series worksheet and the A-Level sequences and series revision engine give more practice.
Next, arithmetic series add the terms together. For the GCSE introduction, see sequences.
Frequently asked questions
How do you find the nth term of an arithmetic sequence?
uₙ = a + (n − 1)d, where a is the first term and d is the common difference.
How do you find the common difference?
Subtract any term from the next one. If you know two terms that are k places apart, their difference divided by k is d.
How do you show three expressions are in arithmetic progression?
Show that the second minus the first equals the third minus the second.