Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 1 Number and algebra · 1.2 Arithmetic sequences and series
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Recognise an arithmetic sequence and find its common difference | SL, HL | "Write down the common difference" (1 mark) |
| Use uₙ = u₁ + (n − 1)d to find a term, the position of a term, or u₁ and d from two terms | SL, HL | "u₄ = 23 and u₁₀ = 53. Find u₁ and d" (4 marks) |
| Use both forms of Sₙ to find the sum of the first n terms | SL, HL | "Find the sum of the first 30 terms" (2 to 3 marks) |
| Find n when a term or a sum is given, including the least n for which a sum exceeds a value | SL, HL | Paper 2: "Find the least number of terms for which Sₙ > 1000" (3 marks) |
| Read and write sigma notation for arithmetic series, and evaluate it | SL, HL | "Find the value of ∑ (3r + 4) from r = 1 to 25" (3 marks) |
| Use a GDC or spreadsheet to generate a sequence and its sums, stating u₁ and d | SL, HL | Paper 2: any of the above, with u₁ and d written down |
| Apply arithmetic sequences to simple interest and other contexts | SL, HL | Paper 2 section A or B, 5 to 8 marks in parts |
| Approximate a common difference from real data that is only nearly arithmetic, predict, and comment | SL, HL | Paper 2: a table of data, "estimate…", "predict…", "comment on the reliability" |
Before you start
You need to substitute into a formula accurately, including negative numbers and brackets, and to solve a linear equation, a pair of simultaneous linear equations, and a quadratic equation or inequality. The graph of a straight line and its gradient, from your earlier study, is the picture behind this whole subtopic.
1The idea in one paragraph
A sequence is an ordered list of numbers, called terms. In an arithmetic sequence you get from each term to the next by adding the same number every time, the common difference, d. Because every step is the same, you do not need to walk along the list to reach the 100th term: you start at the first term and add d ninety-nine times, which is one multiplication. Adding up terms is just as quick, because the list, written forwards and backwards, pairs off into equal totals. Two short formulas, both in the formula booklet, do the whole job; the marks are in choosing the right one and substituting carefully.
2Arithmetic sequences and the common difference
Terms are written u₁, u₂, u₃, …, where the small number is the term's position, and uₙ is the term in position n. The sequence 7, 11, 15, 19, … has u₁ = 7, u₂ = 11 and so on.
The common difference is any term minus the one before it:
d = u₂ − u₁ = u₃ − u₂ = … and the sequence is arithmetic only if these are all the same.
d can be anything. It is negative when the sequence decreases (20, 17, 14, … has d = −3), and it can be a fraction or a decimal (2, 2.5, 3, … has d = 0.5). Always subtract in the order later minus earlier, or the sign of d comes out wrong.
To test whether a sequence is arithmetic, work out every difference. 5, 9, 13, 18 is not arithmetic: the differences are 4, 4, 5.
Figure 1 plots 7, 11, 15, 19, … with the term number across and the value up. Each step along is one term, each step up is d, so the points lie on a straight line whose gradient is d. That is the picture to keep: an arithmetic sequence is a straight line sampled at n = 1, 2, 3, …
3The nth term
To reach u₆ from u₁ you take five steps of d, not six, because six terms have only five gaps between them. Figure 2 counts them.
In general the nth term is n − 1 steps from the first:
uₙ = u₁ + (n − 1)d (in the formula booklet)
This single formula answers four kinds of question, depending on which letter is missing.
Find a term. For 7, 11, 15, …, find u₂₀.
Find a general formula for uₙ. Substitute u₁ and d and simplify.
The coefficient of n is d, just as the gradient of the line in Figure 1 is d. You can check the formula on a term you know: n = 1 gives 7. ✓
Find the position of a term. Which term of 7, 11, 15, … is 203?
Is 150 a term? 4n + 3 = 150 gives n = 36.75. A position must be a positive whole number, so 150 is not a term of this sequence. Saying why earns the reasoning mark.
Find u₁ and d from two terms. An arithmetic sequence has u₅ = 17 and u₁₂ = 45. Write each as an equation and subtract.
A quicker way to see the first line of the subtraction: from u₅ to u₁₂ is seven steps, so 7d = 45 − 17. Count the gaps between the positions, 12 − 5 = 7, not the terms.
4Arithmetic series and the sum formula
A series is what you get when you add the terms of a sequence. Sₙ means the sum of the first n terms:
Sₙ = u₁ + u₂ + … + uₙ
Here is why there is a quick way to find it. Write the series for 7 + 11 + 15 + 19 + 23 + 27 twice, once forwards and once backwards, and stack each pair. Figure 3 does exactly that.
Every column has the same total, 7 + 27 = 34, because as one row goes up by d the other goes down by d. Six columns of 34 make 204, which is two copies of the series, so S₆ = 102. In general, n columns each equal to u₁ + uₙ give 2Sₙ = n(u₁ + uₙ), and replacing uₙ by u₁ + (n − 1)d gives the second form.
Sₙ = n/2 (u₁ + uₙ) = n/2 (2u₁ + (n − 1)d) (both in the formula booklet)
Choose the form by what you know. If you know the last term, n/2 (u₁ + uₙ) is shorter. If you know d but not the last term, use n/2 (2u₁ + (n − 1)d).
Sum of a given number of terms. Find the sum of the first 40 terms of 7, 11, 15, …
Sum up to a given last term. Find 2 + 5 + 8 + … + 146. You need n first, from the nth-term formula.
The step people skip is finding n. It is never "146 ÷ 3" or "146 − 2": let the formula count.
Least n for a sum to pass a value. How many terms of 3 + 7 + 11 + … are needed for the sum to exceed 1000?
Check both sides of the boundary: S₂₂ = 22 × 45 = 990, which is not enough, and S₂₃ = 23 × 47 = 1081, which is. The answer is 23 terms. On Paper 2 you can solve the quadratic with the GDC, or look down a table of Sₙ; either way, write the two values that bracket 1000, because they are the evidence.
One more link between the two ideas: the nth term is the difference between consecutive sums, uₙ = Sₙ − Sₙ₋₁. If a question gives Sₙ as a formula and asks for a term, that is the route.
5Sigma notation
Sigma notation is a compact way to write a sum. The symbol ∑ is the Greek capital S, for "sum". Figure 4 labels the parts of one sum.
The letter r is a counter. It starts at the lower limit, goes up one at a time to the upper limit, and each value is put into the rule; the results are added. So
∑ (5r − 2), from r = 1 to 30, = 3 + 8 + 13 + … + 148.
A rule of the form (constant × r + constant) always gives an arithmetic series, with d equal to the coefficient of r. To evaluate one, find the first term, the last term and the number of terms, then use n/2 (u₁ + uₙ).
Count the terms carefully when the lower limit is not 1. The number of terms is upper − lower + 1.
Writing a series in sigma notation runs the other way. For 4 + 9 + 14 + … + 99: d = 5 and u₁ = 4, so uᵣ = 4 + (r − 1) × 5 = 5r − 1. The last term 99 needs 5r − 1 = 99, so r = 20. The series is ∑ (5r − 1) from r = 1 to 20, and it adds up to 20/2 × (4 + 99) = 1030.
6Using technology
On Paper 2, a GDC can list a sequence and add it up. On most models there is a seq command in the list menu that builds the terms from a rule, and a sum command that adds a list, so sum(seq(5X − 2, X, 1, 30)) gives 2265. A table of values of uₙ or Sₙ, generated from the formula, is often the quickest way to find a least n. Spreadsheets do the same thing with one formula filled down a column.
The guide is explicit about one thing: if you use technology, you are still expected to identify the first term and the common difference. Write down u₁, d and n, and the formula you used, before you write the GDC's answer. A bare number from a calculator can earn the accuracy mark and lose the method marks around it.
On Paper 1 there is no calculator, so the numbers will be chosen to be manageable. Expect factorising or simple arithmetic, not decimals.
7Applications: simple interest and other steady changes
Simple interest is interest paid only on the original amount, so it adds the same sum every year. That makes the balance an arithmetic sequence.
€2,500 is deposited at 3.2% simple interest per year. The interest each year is 3.2% of €2,500, which is €80, every year. Figure 5 plots the balance.
Notice the indexing. If you let u₁ be the balance at the end of year 1, then u₁ = 2580 and d = 80, and uₙ = 2580 + 80(n − 1), which is the same as 2500 + 80n. If you let u₁ be the amount deposited, then the balance after n years is uₙ₊₁. Both are correct; mixing them is how an answer ends up one year out. Decide what u₁ means, write it down, and stick to it. (Interest that is added to the balance and then earns interest itself is compound interest, which is geometric and is the subject of 1.4.)
Other contexts. Anything that changes by a fixed amount per step is arithmetic: seats in rows that each have 3 more than the row in front, a training plan that adds 2 km a week, a stack of cups each 0.8 cm taller than the last.
A theatre has 24 seats in the front row and each row has 3 more seats than the row in front. There are 30 rows.
Read the question for which it wants: the last row (a term) or the whole theatre (a sum).
8When real data is only nearly arithmetic
Real quantities rarely go up by exactly the same amount each time. The guide expects you to spot data that is close to arithmetic, estimate a common difference, use it to predict, and say how far to trust the prediction.
Figure 6 shows invented data: the number of members of a climbing club at the end of each month.
| Month, n | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Members | 212 | 229 | 245 | 263 | 278 | 296 |
| Difference | 17 | 16 | 18 | 15 | 18 |
The differences are not equal but they are close, between 15 and 18, so an arithmetic model is reasonable. Estimate d by the mean of the differences. The mean of the differences is always the total change divided by the number of steps, because the middle terms cancel when you add the differences up:
Round a count of people to a whole number, and say it is an estimate. Then comment on reliability, because that is usually a separate mark:
- The model fits the six months of data well, so a prediction for month 7 or 8 is reasonable.
- Month 12 is six months beyond the data. This is extrapolation: it assumes growth continues at the same rate, and nothing in the data guarantees that. A club can fill up, lose members in the summer, or run a recruitment drive.
- A prediction within the range of the data is more reliable than one outside it.
9Where marks are lost
Writing u₁ + nd. The nth term is u₁ + (n − 1)d. Six terms have five gaps.
Confusing uₙ and Sₙ. u₁₀ is the tenth term on its own; S₁₀ is the first ten terms added up. Underline which one the question asks for before you choose a formula.
Guessing n instead of calculating it. For 2 + 5 + … + 146, solve 2 + (n − 1)3 = 146 to get n = 49. Dividing 146 by 3 does not count terms.
Miscounting terms in sigma notation. From r = 5 to r = 20 there are 16 terms, not 15. Upper − lower + 1.
Getting the sign of d wrong. d is later term minus earlier term. For 20, 17, 14, …, d = −3, and forgetting the minus makes every later answer wrong.
Accepting a non-integer n. If solving for a position gives n = 36.75, the value is not a term. For a least-n question, round up to the next whole number and show the two sums either side.
Being a year out in a money question. Say whether u₁ is the amount deposited or the balance after one year, and count from there.
Giving only the calculator's answer. On Paper 2, write u₁, d and n down. The guide says you are expected to identify them.
10Work it right
- Write down u₁ and d (and check d is really constant).
- Decide whether the question wants a term (uₙ) or a sum (Sₙ). Underline it.
- If you need n, find it from uₙ = u₁ + (n − 1)d; never guess it.
- For a sum, choose n/2 (u₁ + uₙ) if you know the last term, n/2 (2u₁ + (n − 1)d) if you do not.
- For sigma notation, find the first term, last term and number of terms (upper − lower + 1).
- For "least n", solve the inequality, round up, and show the sums just below and just above the target.
- In context, say what u₁ means, round sensibly (people and seats are whole numbers, money to the cent), and answer in words.
- For data that is only nearly arithmetic, estimate d as (last − first) ÷ number of steps, predict, and comment on extrapolation.
11Try it
Marks in brackets. Questions 1 to 3 are Paper 1 style: no calculator. Questions 4 and 5 are Paper 2 style.
Q1. An arithmetic sequence has u₁ = 11 and u₂ = 8.
(a) Write down the common difference. 1 mark
(b) Find u₃₀. 2 marks
(c) Find the sum of the first 30 terms. 2 marks
Q2. In an arithmetic sequence, u₄ = 23 and u₁₀ = 53.
(a) Find the common difference and the first term. 4 marks
(b) Find S₂₀. 2 marks
Q3. Find the value of ∑ (3r + 4), from r = 1 to 25. 3 marks
Q4. Sofia saves €40 in the first week of a savings plan. Each week after that she saves €6 more than the week before.
(a) Find the amount she saves in week 26. 2 marks
(b) Find the total she has saved by the end of week 26. 2 marks
(c) Find the first week at the end of which her total savings are more than €5,000. 3 marks
Q5. The population of a village (invented data) at the start of each year is shown.
| Year | 2019 | 2020 | 2021 | 2022 | 2023 |
|---|---|---|---|---|---|
| Population | 1480 | 1523 | 1561 | 1607 | 1644 |
(a) Explain why an arithmetic sequence is a reasonable model for these data, and estimate its common difference. 2 marks
(b) Use your model, with u₁ = 1480 for 2019, to predict the population at the start of 2030. 2 marks
(c) Comment on the reliability of your prediction. 1 mark
12In one breath
An arithmetic sequence adds the same common difference d at every step, so its graph against n is a straight line with gradient d. The nth term is uₙ = u₁ + (n − 1)d, with n − 1 because n terms have n − 1 gaps, and it finds a term, a position, or u₁ and d from two terms by subtracting. Writing the series forwards and backwards pairs it into n equal totals, which gives Sₙ = n/2 (u₁ + uₙ) = n/2 (2u₁ + (n − 1)d); use the first when you know the last term. Find n from the nth-term formula, never by guessing, and for a least-n question round up and show the sums either side. Sigma notation adds a rule from a lower to an upper limit, with upper − lower + 1 terms. Simple interest is arithmetic because the interest is the same every year. When data only nearly fits, estimate d as the total change over the number of steps, predict, and warn that extrapolation is less reliable the further it runs.
Answers
Q1. (a) d = 8 − 11 = −3. (b) u₃₀ = 11 + 29 × (−3) = 11 − 87 = −76. (c) S₃₀ = 30/2 × (11 + (−76)) = 15 × (−65) = −975. (a) A1. (b) M1 for substituting n = 30 and d = −3 into the nth-term formula, A1 for −76. (c) M1 for a correct sum formula with their values, A1 for −975. Follow through from (b) in (c). Using d = 3 loses the A1 in (a) but the method marks in (b) and (c) can still be earned.
Q2. (a)
(b) S₂₀ = 20/2 × (2 × 8 + 19 × 5) = 10 × (16 + 95) = 1110. (a) M1 for one correct equation in u₁ and d, M1 for a second and an attempt to eliminate, A1 for d = 5, A1 for u₁ = 8. (b) M1 for substitution into a sum formula, A1 for 1110. Follow through from (a).
Q3. The first term is 3(1) + 4 = 7, the last is 3(25) + 4 = 79, and there are 25 terms. Sum = 25/2 × (7 + 79) = 25/2 × 86 = 1075. A1 for the first and last terms (or u₁ = 7 and d = 3), M1 for a sum formula with n = 25, A1 for 1075. Adding only the first few terms scores zero.
Q4. u₁ = 40 and d = 6. (a) u₂₆ = 40 + 25 × 6 = €190. (b) S₂₆ = 26/2 × (40 + 190) = 13 × 230 = €2,990. (c)
Her total first exceeds €5,000 at the end of week 36. (a) M1 A1. (b) M1 A1 (follow through from (a)). (c) M1 for setting Sₙ > 5000 (or = 5000) with a correct Sₙ, A1 for 35.1 or for the two sums 4970 and 5220, A1 for week 36. An answer of 35 scores M1 A1 A0.
Q5. (a) The yearly increases are 43, 38, 46 and 37, which are roughly equal, so the population is rising by about the same amount each year. d ≈ (1644 − 1480) ÷ 4 = 164 ÷ 4 = 41. (b) 2030 is n = 12, so u₁₂ = 1480 + 11 × 41 = 1931 people. (c) 2030 is seven years beyond the last data point, so the prediction is an extrapolation and assumes the village keeps growing by about 41 people a year; it should be treated with caution. (a) R1 for the differences being approximately constant, A1 for d ≈ 41 (any sensible estimate from the differences, 40 to 42, is accepted). (b) M1 for u₁₂ with their d, A1 for 1931 (follow through). (c) R1 for identifying extrapolation or a changing growth rate. "It is reliable because the model fits" with no mention of the time gap scores zero.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 1.2 Arithmetic sequences and series. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.