The big idea: Geometric = multiply (or divide) by the same number every step. Always × or ÷, never + or −.
That fixed multiplier is the common ratio, r.
Example: 2, 6, 18, 54, … multiplies by 3 each time, so r = 3.
- Sequence
- A list of numbers in a set order. Each number is a term.
- Common ratio (r)
- The constant multiplier between one term and the next.
- First term (u₁)
- The term you start from, in position 1.
- Term (uₙ)
- The term in position n.
How to tell it is geometric
- Divide each term by the one before it.
- If every ratio is the same → geometric.
- 3, 6, 12, 24 has ratios 2, 2, 2 → geometric (r = 2).
- 3, 6, 9, 12 has a constant gap of 3 → that one ADDS, so it is arithmetic.
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Find r from two terms: Sometimes the exam gives you two terms and asks for r: divide the later value by the earlier, then take the (steps)-th root.
See it on numbers
u₂ = 6 and u₅ = 48.
Find the common ratio.
Step by step
- Value-ratio — divide the later value by the earlier.
- Steps — subtract the positions.
- Take the (steps)-th root: each step multiplies by the same r, so undo the 3 steps with a cube root.
Final answer
r = 2 (the sequence is 6, 12, 24, 48).
Press a chip to watch it build: the value-ratio 8 is shared across 3 equal ×r steps, so r is the cube root, 2. Try the steeper example, or the even-steps case where r comes out ±.
Interactive diagram
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IB-style question — common ratio from two terms
The 3rd term of a geometric sequence is 18 and the 6th term is 486.
Find the common ratio.
Step by step
- Value-ratio — divide the later value by the earlier.
- Steps — subtract the positions.
- Take the (steps)-th root.
Final answer
r = 3.
IB-style question — when r has two answers
u₁ = 3 and u₃ = 12.
Find the common ratio.
Step by step
- Value-ratio — divide the later value by the earlier.
- Steps — subtract the positions. This time it's an EVEN number.
- Take the 2-step root — that's a square root, and a square root has TWO answers (+ and −).
Final answer
r = 2 gives 3, 6, 12, … and r = −2 gives 3, −6, 12, … — both have u₁ = 3 and u₃ = 12! If the question says the terms are positive, pick r = 2.
Square root → two answers (±): A square root has two answers, + and −.
You get this whenever the two terms are an even number of steps apart (u₁ → u₃, u₂ → u₄, …).
An odd gap — like the cube-root examples above — gives just one r.
When you land on ±, choose using any clue in the question, e.g. "all terms positive".
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The nth-term formula — your main tool: To jump straight to any term, start at u₁ and multiply by r one fewer time than the position.
- the first term
- the common ratio (multiplier)
- the position (which term you want)
IB-style question — find any term
A geometric sequence has u₁ = 2 and r = 3.
Find the 5th term.
Step by step
- Write the formula.
- Substitute u₁ = 2, r = 3, n = 5. You multiply by r only (5 − 1) = 4 times.
- Work it out.
Final answer
u₅ = 162.
IB-style question — build the general term from two terms
The 3rd term is 18 and the 6th term is 486.
Find an expression for the nth term uₙ.
Step by step
- Find r first — value-ratio, then root (from the last section).
- Step back to the first term: u₃ = u₁r², so divide by r².
- Put u₁ = 2 and r = 3 into the formula.
Final answer
uₙ = 2 × 3ⁿ⁻¹.
Why the power is n − 1, not n: You start at u₁, so reaching the 5th term takes only 4 multiplications, not 5 — the first term is already there.
That is why the formula multiplies by r exactly (n − 1) times.
The whole trick: which term is it?: Whatever you're given first is u₁ — term 1, 0 jumps. Every ×r is one jump, so uₙ = u₁ rⁿ⁻¹. The only thing to work out is which term they want:
• "the 5th bounce" → 6th term → 5 jumps (the drop is term 1) • "after 3 years" → 4th term → 3 jumps (year 0 is term 1)
Press a chip and watch the ×r arrows build one at a time. Count the jumps from the start to the term you want — the number of arrows is the power. It works the same whether the values go down (ball, car) or up (savings).
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Explore the labelled diagram, charts and maps for this topic in full study mode.
In the exam: draw the arrows: Sketch the start box, then one ×r arrow per bounce / year, and count to the term you want.
The number of arrows is the power — there is no arrow on the start, because it is already term 1.
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Three terms — set middle² = product: A common exam question gives three terms in a row containing an unknown and says they are geometric — set the middle squared equal to the product of its neighbours and solve.
IB-style question — show the condition
u₁, u₂ and u₃ are three consecutive terms of a geometric sequence.
Show that u₂² = u₁u₃.
Step by step
- Geometric means consecutive ratios are equal.
- Cross-multiply.
- So the middle term squared equals the product of its neighbours.
Final answer
u₂² = u₁u₃ — the middle term is the geometric mean of the two outer ones.
IB-style question — find the unknown
The terms x, x + 6 and x + 18 are consecutive terms of a geometric sequence.
Find x.
Step by step
- Middle squared = product of the neighbours.
- Expand both sides.
- The x² cancels — solve what is left.
Final answer
x = 6 (the sequence is 6, 12, 24, with r = 2).
Common mistakes
- Adding instead of multiplying (that is arithmetic).
- Dividing the value-ratio by the steps instead of rooting it.
- Using middle = average of the neighbours (that is arithmetic).
Do this instead
- Geometric multiplies by r each step.
- r = the (steps)-th root of the value-ratio.
- Three terms geometric ⇒ middle² = product (u₂² = u₁u₃).
When the condition becomes a quadratic: A harder exam twist: the middle² = product condition turns into a quadratic, so the unknown has two values — solve it, then use any stated condition to choose between them.
IB-style question — find both values
The terms k, k + 12 and 4k are consecutive terms of a geometric sequence.
Find the possible values of k and the common ratio for each.
Step by step
- Middle² = product of the neighbours.
- Expand and bring everything to one side.
- Divide by 3, then factorise.
- Two solutions — find r for each (r = middle ÷ first).
Final answer
k = 12 (r = 2) or k = −4 (r = −2) — both give a valid geometric sequence.
Use a condition to choose: If the question adds a condition, use it to pick one value. Here "all terms positive" keeps k = 12 (k = −4 gives −4, 8, −16, which alternate in sign).
A condition like "the series has a sum to infinity" instead keeps the value with |r| < 1 — see Sum to infinity.