Geometric Sequences (AA SL)
A geometric sequence is multiplied by the same fixed value, the common ratio, from one term to the next, so it grows or shrinks rather than climbing steadily like an arithmetic sequence. This page covers the term formula, the sum of \(n\) terms, and the special case of a sum to infinity, with worked examples and the mistakes that lose the most marks. It's part of the broader Sequences & Series topic.
21 questions on this sub-topic.
The three formulas
Covered under IB syllabus reference SL1.3. All three formulas are in the formula booklet, so the skill is spotting which one a question actually wants - and checking whether a sum to infinity is even valid.
nth term
\(u_n = u_1 r^{n-1}\)
Use when you need one specific term - "find the 5th term", "find the first term that exceeds 1000".
Sum of n terms
\(S_n = \dfrac{u_1(r^n-1)}{r-1},\ r\neq1\)
Use when you need a total over a finite number of terms - "find the sum of the first 12 terms".
Sum to infinity
\(S_\infty = \dfrac{u_1}{1-r},\ |r|<1\)
Only valid when \(|r|<1\) - state this before using the formula, since it's often worth its own mark.
Need the full syllabus wording and formula-booklet reference table? See Sequences & Series.
Worked examples
A geometric sequence has \(u_1=2\) and \(r=3\).
(a) Find \(u_5\).
(b) Find a formula for \(u_n\).
Worked solution
(a) Find \(u_5\). \(u_5 = u_1 r^4 = 2(3)^4\) M1
\(= 162.\) A1
(b) Formula for \(u_n\). \(u_n = 2\cdot 3^{\,n-1}.\) A1
A geometric series has \(a = 8\) and \(r = \tfrac{1}{2}\).
Find \(S_\infty\).
Worked solution
A sum to infinity exists only when \(|r|<1\). Here \(|r| = \tfrac12 < 1\), so the series converges. R1
\(S_\infty = \dfrac{a}{1-r}\): M1
\(S_\infty = \frac{8}{1-\tfrac12} = \frac{8}{\tfrac12} = 16.\) A1
A geometric sequence has \(u_2 = 6\) and \(u_5 = 48\).
(a) Find \(r.\)
(b) Find \(u_1.\)
Worked solution
(a) Divide the terms to remove \(u_1\): \(\dfrac{u_5}{u_2} = \dfrac{u_1 r^4}{u_1 r} = r^3 = \dfrac{48}{6} = 8.\) M1
So \(r = 2.\) A1
(b) Back-substitute for \(u_1\): \(u_2 = u_1 r \Rightarrow 6 = 2u_1 \Rightarrow u_1\) M1
\(= 3.\) A1
The first three terms of a geometric sequence are \(16,\ 20,\ 25\).
(a) Find the common ratio.
(b) Find the value of the 8th term, giving your answer to 3 significant figures.
(c) Find the smallest value of \(n\) for which the sum of the first \(n\) terms exceeds \(500\).
Worked solution
(a) \(r=\dfrac{20}{16}\) M1
\(r=1.25\) (check: \(\tfrac{25}{20}=1.25\)). A1
(b) \(u_8=16(1.25)^{7}\) M1
\(=76.3\) (3 s.f.). A1
(c) \(S_n=\dfrac{16(1.25^n-1)}{1.25-1}=64(1.25^n-1).\) M1
\(64(1.25^n-1)>500\Rightarrow 1.25^n>8.8125\Rightarrow n>\dfrac{\ln 8.8125}{\ln 1.25}=9.75.\) M1
\(S_9\approx412.8<500
Common mistakes
- Quoting \(S_\infty\) without checking \(|r|<1\). If \(|r|\geq1\) the series diverges and no sum to infinity exists - the convergence check is often worth its own mark, not just a formality.
- Mixing up arithmetic and geometric formulas. \(u_n=u_1+(n-1)d\) adds; \(u_n=u_1r^{n-1}\) multiplies - using the wrong one produces a completely different sequence.
- Sign errors when \(r\) is negative. With a negative ratio the terms alternate sign, so \(r^{n-1}\) needs care - especially when squaring or comparing terms to find \(r\) from two given values.
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.
When does a geometric series have a sum to infinity?
Only when \(|r| < 1\), so the terms shrink towards zero. Then \(S_\infty = \dfrac{u_1}{1-r}\).