The big idea: Arithmetic = add (or subtract) the same number every step. Always + or −, never × or ÷.
That fixed step is the common difference, d.
Example: 5, 8, 11, 14, … adds 3 each time, so d = 3.
- Sequence
- A list of numbers in a set order. Each number is a term.
- Term (uₙ)
- One number in the sequence. uₙ is the term in position n.
- First term (u₁)
- The term you start from, in position 1.
- Common difference (d)
- The constant gap between one term and the next.
A decreasing sequence: In 20, 17, 14, 11, … each term is 3 less than the one before, so d = −3.
A negative common difference means the sequence decreases.
How to tell it is arithmetic
- Look at the gaps between consecutive terms.
- If every gap is the same → arithmetic.
- 2, 6, 10, 14 has gaps 4, 4, 4 → arithmetic (d = 4).
- 2, 6, 18, 54 has gaps 4, 12, 36 → not arithmetic (that one multiplies).
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Find d from the gap: The step d is the same every time — so subtract the two values, then divide by the number of steps between them.
See it: 5, ?, 11
An arithmetic sequence is 5, ?, 11 — the 1st and 3rd terms.
Find the common difference d.
Step by step
- Same d is added each step: 5 → ? → 11.
- Subtract the two values.
- Count the steps between them (3rd − 1st).
- Share that gap across the steps.
Final answer
d = 3, so the sequence is 5, 8, 11.
Press a chip to watch it build: two equal steps share the gap of 6, so each step (the common difference) is 3. Try the further-apart example too.
Interactive diagram
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Step it up — terms further apart
The 3rd term is 17 and the 7th term is 41.
Find the common difference.
Step by step
- Subtract the two values.
- Count the steps between them (7th − 3rd).
- Share that gap across the steps.
Final answer
d = 6.
Count steps, not terms: The 3rd and 7th terms are 4 steps apart (7 − 3), not 7.
Always divide by the number of steps, not the term numbers.
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The nth-term formula — your main tool: One formula jumps straight to any term — without writing them all out.
- the first term
- the common difference
- the position (which term you want)
IB-style question — find any term
An arithmetic sequence has u₁ = 5 and d = 3.
Find the 10th term.
Step by step
- Write the formula.
- Substitute u₁ = 5, d = 3, n = 10. You add d only (10 − 1) = 9 times.
- Work it out.
Final answer
u₁₀ = 32.
IB-style question — build the general term from two terms
The 2nd term is 6 and the 5th term is 18.
Find an expression for the nth term uₙ.
Step by step
- Find d first — gap ÷ steps (from the last section).
- Step back to the first term: from u₂, subtract one d.
- Put u₁ = 2 and d = 4 into the formula and tidy up.
Final answer
uₙ = 4n − 2.
Why (n − 1)? Count the +d jumps from the start: The number on d is just how many +d jumps you've made from the start. You begin at u₁ (0 jumps), so the 10th term is only 9 jumps along — not 10. That is the whole reason for the (n − 1).
So always spot u₁ first: whatever you're given first is term 1, with 0 jumps. Count from there.
Counting TERMS → use n − 1
- Term 1 is the start — 0 jumps.
- The nth term is n − 1 jumps along.
- uₙ = u₁ + (n − 1)d
Counting YEARS / STEPS → use n
- "Now" (year 0) is the start — 0 jumps.
- Each year adds d once → n jumps.
- after n years = u₁ + nd
Read it straight off: Sometimes the exam hands you the rule itself (like uₙ = 20 − 4n) instead of u₁ and d — so you read them straight off it, no working.
IB-style question — read u₁ and d from an explicit rule
The nth term of an arithmetic sequence is uₙ = 20 − 4n.
Write down the first term and the common difference.
Step by step
- Common difference: just read off the number in front of n. No working needed.
- First term: put n = 1 into the rule.
- Not convinced d is just the coefficient? Find a second term and subtract — same answer.
Final answer
u₁ = 16 and d = −4.
Why d is the coefficient of n: Expand the formula: uₙ = u₁ + (n − 1)d = dn + (u₁ − d).
The number multiplying n is always d — that is why you can read it straight off.
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Set the term equal and solve: Sometimes you are given a term's value and asked which term it is — that is, its position n in the sequence.
Picture the sequence as a numbered list. Set uₙ equal to the value and solve for n (the position).
| Position n | 1 | 2 | 3 | 4 | 5 | … | n = ? |
|---|---|---|---|---|---|---|---|
| Term uₙ | 4 | 9 | 14 | 19 | 24 | … | 99 |
IB-style question — which term equals a value
An arithmetic sequence has u₁ = 4 and d = 5.
Which term is equal to 99?
Step by step
- Write down the formula.
- Put the values in (u₁ = 4, d = 5) and set it equal to 99.
- Expand and simplify.
- Solve for n.
Final answer
The 20th term equals 99.
n must be a whole number: A value is only a term if solving gives a positive whole number.
If n comes out as a decimal (like 7.5), the value is not in the sequence.
Equal gaps: A classic exam twist: three terms in a row contain an unknown.
Their gaps must be equal, so set u₂ − u₁ = u₃ − u₂ and solve.
IB-style question — find the unknown that makes it arithmetic
The first three terms of an arithmetic sequence are:
u₁ = k + 2
u₂ = 2k + 3
u₃ = 5k − 2
Find the value of k.
Step by step
- Equal differences: u₂ − u₁ = u₃ − u₂.
- Simplify each side.
- Solve for k.
Final answer
k = 3 (the sequence is 5, 9, 13, with d = 4).
Common mistakes
- Setting the terms equal instead of the differences.
- Adding d n times instead of (n − 1) times.
- Taking d as the constant in uₙ = a + bn instead of the coefficient b.
Do this instead
- Set u₂ − u₁ = u₃ − u₂ (the gaps are equal).
- Use uₙ = u₁ + (n − 1)d — the first term is already u₁.
- The number multiplying n is always the common difference.