Sequences & Series (AI SL)
A sequence is an ordered list of numbers following a pattern, and a series is what you get when you add its terms together. This topic covers the two patterns the AI syllabus cares about most: arithmetic sequences, which add a constant amount each time, and geometric sequences, which multiply by a constant ratio each time. Both come with formulas for finding any term and for summing a run of terms, and both show up constantly in real contexts - seating plans, salaries, bouncing balls, and (once you add interest) savings and loans.
What the syllabus says
This topic maps onto two points in the official IB Applications & Interpretation syllabus.
| Code | Syllabus content |
|---|---|
| SL1.2 | Arithmetic sequences and series. Use of the formulae for the \(n\)th term and the sum of the first \(n\) terms. Use of sigma notation. Applications, including simple interest over a number of years, and analysis, interpretation and prediction from a real-world context. |
| SL1.3 | Geometric sequences and series. Use of the formulae for the \(n\)th term and the sum of the first \(n\) terms. Use of sigma notation. Applications, including the spread of disease, salary increase/decrease, and population growth. |
Both sit in the Number & Algebra unit, and SL1.4 (Financial Maths) builds directly on the geometric series formulas from SL1.3.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is an arithmetic sequence?
An arithmetic sequence is a list of numbers where each term is found by adding a fixed amount, the common difference \(d\), to the previous term. The terms increase (or decrease) by the same amount every step.
e.g. \(u_1=15,\ d=2\): the sequence is \(15,17,19,21,\dots\)
What is a geometric sequence?
A geometric sequence is a list of numbers where each term is found by multiplying the previous term by a fixed amount, the common ratio \(r\). This produces exponential growth (\(r>1\)) or decay (\(0 e.g. \(u_1=3,\ r=0.8\): the sequence is \(3, 2.4, 1.92, 1.536,\dots\)
What is the common difference?
The common difference \(d\) is the fixed amount added between consecutive terms of an arithmetic sequence, found by subtracting any term from the one after it. It can be positive, negative, or zero.
e.g. for \(22,25,28,31,\dots\), \(d = 25-22 = 3\).
What is the common ratio?
The common ratio \(r\) is the fixed multiplier between consecutive terms of a geometric sequence, found by dividing any term by the one before it. If \(|r|<1\) the sequence decays towards zero.
e.g. for \(200, 210, 220.5,\dots\), \(r = 210 \div 200 = 1.05\).
What is sigma notation?
Sigma notation (\(\Sigma\)) is shorthand for "add up every term the formula produces" between a starting and ending index, without writing every term out by hand.
e.g. \(\displaystyle\sum_{k=1}^{4}(3k-1) = 2+5+8+11 = 26\).
Key formulas
Four formulas cover this whole topic - two for arithmetic, two for geometric. The tables below summarise all of them at a glance - the explanations underneath go into more depth on each one.
Formula reference
All four formulas below are given in the IB formula booklet, so the skill being tested is choosing and substituting into the right one, not memorising them.
| Formula | Used for | Booklet? |
|---|---|---|
| \(u_n = u_1 + (n-1)d\) | \(n\)th term of an arithmetic sequence | ✓ Yes |
| \(S_n = \dfrac{n}{2}(2u_1 + (n-1)d)\) | Sum of an arithmetic series | ✓ Yes |
| \(u_n = u_1 r^{n-1}\) | \(n\)th term of a geometric sequence | ✓ Yes |
| \(S_n = \dfrac{u_1(r^n-1)}{r-1},\ r\ne1\) | Sum of a geometric series | ✓ Yes |
Arithmetic vs geometric
Both describe patterned growth, but one is additive and the other multiplicative - which changes how fast they grow and how you sum them.
| Feature | Arithmetic | Geometric |
|---|---|---|
| Step between terms | Add \(d\) | Multiply by \(r\) |
| \(n\)th term | \(u_n = u_1 + (n-1)d\) | \(u_n = u_1 r^{n-1}\) |
| Sum of \(n\) terms | \(S_n = \tfrac{n}{2}(2u_1+(n-1)d)\) | \(S_n = \dfrac{u_1(r^n-1)}{r-1}\) |
| Growth shape | Straight line | Exponential curve |
Arithmetic sequences and series
Look for a constant amount being added each term - a fixed step size is the signal to use these formulas.
nth term
\[u_n = u_1 + (n-1)d\]
Gives the value of any single term directly, without listing every term before it.
✓ In the formula bookletSum of n terms
\[S_n = \dfrac{n}{2}(2u_1 + (n-1)d)\]
Adds the first \(n\) terms without summing them one by one.
✓ In the formula bookletReal-world modelling
Seating rows, salaries with a fixed annual rise, and loan repayments that increase by a fixed amount are all arithmetic.
Identify \(u_1\) and \(d\) from the context before reaching for a formula.
Not in the formula booklet - modelling skillGeometric sequences and series
Look for a constant amount being multiplied each term - percentage growth or decay is the clearest signal.
nth term
\[u_n = u_1 r^{n-1}\]
A percentage increase of \(x\%\) gives \(r = 1 + \tfrac{x}{100}\); a decrease gives \(r = 1 - \tfrac{x}{100}\).
✓ In the formula bookletSum of n terms
\[S_n = \dfrac{u_1(r^n-1)}{r-1}\]
Valid for \(r\ne1\); works whether the series is growing or shrinking.
✓ In the formula bookletReal-world modelling
Population growth, radioactive-style decay, and compound interest are all geometric - anything that changes by a fixed percentage each step.
Identify \(u_1\) and \(r\) from the context before reaching for a formula.
Not in the formula booklet - modelling skillWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
A cinema has 15 seats in row 1 and 2 more seats in each subsequent row. There are 20 rows.
(a) Find the number of seats in the last row.
(b) Find the total number of seats.
(c) Which row has 41 seats?
Worked solution
(a) Arithmetic with \(u_1 = 15,\ d = 2,\ n = 20.\) M1
\(u_{20} = 15 + (20-1)(2) = 15 + 38 = 53\) seats. A1
(b) \(S_{20} = \tfrac{20}{2}(u_1 + u_{20})\)
\(10(15 + 53) = 10(68)\) A1 \(= 680\) seats. A1
(c) Step 1 - Solve \(u_n = 41\): \(15 + 2(n-1) = 41 \Rightarrow 2(n-1) = 26 \Rightarrow n-1 = 13 \Rightarrow n\) M1 \(= 14.\) Row 14. A1
A ball dropped from 3 m rebounds to 80% of its previous height each bounce.
(a) Find the height after the 4th bounce.
(b) Find the total distance fallen in the first 4 drops.
Worked solution
(a) Step 1 - Rebound heights are geometric with ratio \(0.8\): height after the \(n\)th bounce is \(3(0.8)^n.\) \(3(0.8)^4 = 3(0.4096)\) M1 \(\approx 1.23\text{ m}.\) A1
(b) Step 1 - The drops themselves form a geometric series with \(u_1 = 3,\ r = 0.8,\ n = 4.\) M1
Step 2 - \(S_4 = \dfrac{3(0.8^4 - 1)}{0.8 - 1}\) M1 \(= 8.856\text{ m}.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Using \(n\) instead of \(n-1\) in the exponent or multiplier. Both the \(n\)th-term formulas use \((n-1)\), not \(n\) - the first term \(u_1\) corresponds to zero steps, not one.
- Expanding sigma notation wrong. \(\displaystyle\sum_{k=1}^{n}\) means substitute \(k=1,2,\dots,n\) into the expression and add every result - not evaluate the expression once at \(k=n\).
- Mixing up \(d\) and \(r\). A constant difference means arithmetic (\(d\)); a constant ratio means geometric (\(r\)). Confusing the two applies the wrong pair of formulas entirely.
- Rounding a percentage into the wrong ratio. A 5% increase gives \(r = 1.05\), not \(r=0.05\) or \(r=1.5\) - convert the percentage to a decimal first, then add it to (or subtract it from) 1.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
For very large or very small numbers - avoids typing long strings of zeros and prevents rounding errors, useful when a geometric sequence with \(r>1\) produces a huge term after many steps.
- Scientific notation means \(a\times10^n\), e.g. \(3.2\times10^8\) or \(4.5\times10^{-3}\).
- Use 2nd → , (EE) to enter the ×10 part: type 3.2 2nd , 8 to enter \(3.2\times10^8\). Do NOT type ×10^ separately.TI-84
- Use the EE key (or type ×10^ from the keyboard template) to enter scientific notation. Or just type 3.2×10^8 using the ^ key.Nspire
- Use the ×10ˣ key (EXP key) - type 3.2 then EXP then 8. Do NOT type ×10^ manually.Casio
- To display answers in scientific notation: on TI-84 press MODE and choose SCI; on Casio set the display mode in SET UP.
Tip: A common mistake is typing ×10^ instead of using the EE/EXP key - this gives ×10×... (multiplication, then a power) rather than proper scientific notation.
Faster and safer than algebra for finding which term reaches a target value, especially once the unknown \(n\) sits in a geometric exponent and needs logarithms.
- Graph \(f(x)\) first so you can see how many solutions exist and roughly where they are.
- Rearrange so everything is on one side: \(f(x) = 0\) - or graph both sides as separate functions and find intersections.
- MATH → Solver: enter the expression, type a starting guess close to one root, press ALPHA + ENTER. Move the guess to near a different root and repeat for each solution.TI-84
- Type nSolve(f(x)=0, x, guess) - include a guess or interval e.g. nSolve(f(x)=0, x, 2) or nSolve(f(x)=0, x, {1,5}) to target a specific root.Nspire
- Run-Matrix → SolveN(f(x), x) returns all real roots at once; or use the Equation app for a visual approach.Casio
- Always verify each solution by substituting back into the original equation.
Tip: The solver finds ONE root near your starting guess - change the guess to find others. The graph shows you how many to expect.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Sequences & series questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between an arithmetic and a geometric sequence?
An arithmetic sequence adds a fixed common difference \(d\) between consecutive terms, so it grows in a straight line. A geometric sequence multiplies by a fixed common ratio \(r\), so it grows (or decays) exponentially. Spotting a constant difference versus a constant ratio tells you which formulas to use.
How do I find which term reaches a target value?
Set the nth-term formula equal to the target value and solve for \(n\). For an arithmetic sequence this is straightforward algebra; for a geometric sequence you'll usually need logarithms, since \(n\) sits in the exponent.
Can I use my GDC for this topic?
Yes - every question on this topic is calculator-permitted. Your GDC can generate and sum sequences directly with sum(seq(...)), and can solve numerically for an unknown term or unknown n, which is especially useful once logarithms are involved. See the GDC guide for model-specific instructions.
Sub-topics
Sequences & Series broken down into its individual skills, each with its own focused page.
Related topics
More Number & Algebra topics from the same AI SL syllabus unit, in case you want to keep going.