Geometric Sequences (AA HL)
In a geometric sequence, each term is found by multiplying the previous one by a fixed common ratio, so the values grow (or shrink) multiplicatively rather than by a constant step. This page sets out the two core formulas - for a single term and for the sum of several terms - alongside worked examples and the errors that cost the most marks. It's part of the broader Sequences & Series topic.
20 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference SL1.3. Both formulas are in the formula booklet, so the skill being tested is recognising which one a question actually needs, and correctly identifying \(u_1\) and \(r\) from the wording.
nth term
\[u_n = u_1 r^{\,n-1}\]
Where \(u_1\) is the first term and \(r\) is the common ratio.
✓ In the formula bookletSum of n terms
\[S_n = \dfrac{u_1(r^n - 1)}{r - 1}, \quad r \neq 1\]
Works for any \(r \neq 1\), including \(r > 1\) or \(r < -1\).
✓ In the formula bookletNeed the full syllabus wording and formula-booklet reference table, or a refresher on your GDC's sequence tools? See Sequences & Series.
Worked examples
Consider \(\displaystyle\sum_{k=0}^{n-1} 2\left(\tfrac{3}{2}\right)^{k}.\)
(a) Find its exact value.
(b) Hence evaluate it when \(n = 5.\)
Worked solution
(a) The sum is geometric with \(a=2\) and \(r=\tfrac32.\) A1
\(\displaystyle\sum_{k=0}^{n-1}2\left(\tfrac32\right)^{k}=2\cdot\dfrac{\left(\tfrac32\right)^{n}-1}{\tfrac32-1}.\) M1
\(=4\left(\left(\tfrac32\right)^{n}-1\right).\) A1
(b) \(\left(\tfrac32\right)^5=\tfrac{243}{32}\), so the sum is \(4\left(\tfrac{243}{32}-1\right).\) M1
\(=4\cdot\tfrac{211}{32}=\tfrac{211}{8}=26.375.\) A1
Three numbers \(4,\ p,\ 9\) form a geometric sequence with \(p > 0\).
Find \(p\) and the common ratio.
Worked solution
For a GP, the middle term squared equals the product of its neighbours: \(p^2 = 4 \times 9.\) M1
\(p^2 = 36.\) A1
\(p = 6\) (taking \(p>0\)). A1
\(r = \dfrac{6}{4} = \tfrac{3}{2}.\) A1
The first, second and fifth terms of an arithmetic sequence are, in that order, three consecutive terms of a geometric sequence. The arithmetic sequence has first term \(a\) and common difference \(d\neq0.\)
Find the common ratio of the geometric sequence.
Worked solution
The terms are \(a,\ a+d,\ a+4d.\) Geometric \(\Rightarrow (a+d)^2 = a(a+4d).\) M1
\(a^2 + 2ad + d^2 = a^2 + 4ad.\) A1
From \(d^2=2ad\), divide by \(d\) (\(d\neq0\)): \(d=2a.\) M1
\(d = 2a.\) A1
\(r = \dfrac{a+d}{a} = \dfrac{a + 2a}{a}.\) M1
\(r = 3.\) A1
Common mistakes
- Using \(S_\infty\) when \(|r| \geq 1\). The sum to infinity formula only applies to a convergent geometric series. If \(|r| \geq 1\) the series has no finite sum - say so instead of forcing the formula.
- Confusing arithmetic and geometric. Check whether consecutive terms share a common difference (arithmetic) or a common ratio (geometric) before picking a formula - mixing them up gives a completely wrong answer.
- Losing the negative root. When a geometric-mean condition gives \(r^2\) or \(p^2\), there are usually two roots. Discard the invalid one only after checking any stated restriction (such as \(p>0\)), rather than assuming the positive root by default.
Ready to practise properly?
20 geometric-sequence questions, marked instantly like the real exam.
Quick answers
What is the formula for the nth term of a geometric sequence?
\(u_n = u_1 r^{\,n-1}\), where \(u_1\) is the first term and \(r\) is the common ratio.
What is the formula for the sum of a geometric series?
\(S_n = \dfrac{u_1(r^n - 1)}{r - 1}\), valid for any ratio \(r \neq 1\), including negative or large values of \(r\).