Geometric Sequences (College Board AP® Precalculus): Study Guide

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

Updated on

Geometric sequences

What is a geometric sequence?

  • A geometric sequence is a sequence in which successive terms have a common ratio

    • This means each term is obtained by multiplying the previous term by the same constant

    • This common ratio represents a constant proportional change

    • The common ratio is usually denoted by r

  • E.g. the sequence 3, 6, 12, 24, 48, is geometric with a common ratio of r=2

    • Each term is 2 times the previous term

      • 3×2=6

      • 6×2=12

      • 12×2=24

      • etc.

  • The common ratio can be any nonzero value:

    • r>1: the terms grow away from zero (for positive-valued sequences)

    • 0<r<1: the terms shrink toward zero (for positive-valued sequences)

    • r<0: the terms alternate in sign

How can I find the common ratio of a geometric sequence?

  • To find the common ratio of a geometric sequence

    • Divide any term by the term before it

      • r=gn+1gn

  • If the ratios between consecutive terms are not all equal, then the sequence is not geometric

What is the general term of a geometric sequence?

  • The general term (also called the nth term) of a geometric sequence can be written in two forms

  • Using the initial value

    • gn=g0rn

      • where g0 is the initial value (the term when n=0)

      • and r is the common ratio

  • Using any known term

    • gn=gkr(nk)

      • where gk is the value of the kth term

      • and r is the common ratio

    • This form is useful when you don't know g0 but you do know a different term

  • Both forms express the same idea

    • Start from a known term and multiply by the common ratio the appropriate number of times

How do these formulas work in practice?

  • E.g. a geometric sequence has initial value g0=5 and a common ratio of r=3

    • The general term is gn=5·3n

    • So g0=5, g1=5×31=15, g2=5×32=45, g3=5×33=135,

  • Or e.g. you are told that g2=48 and r=12

    • Using the second form: gn=48·(12)(n2)

    • You can verify that to check the answer

      • g2=48·(12)22=48·(12)0=48·1=48

    • And also find other terms

      • E.g. the initial term is  g0=48·(12)02=48·(12)2=48·4=192

How does a geometric sequence grow compared to an arithmetic sequence?

  • An increasing arithmetic sequence increases equally with each step

    • the same amount is added each time

  • An increasing geometric sequence (with positive values) increases by a larger amount with each successive step

    • because the same ratio is applied to an ever-larger value

  • E.g. consider the geometric sequence 2, 6, 18, 54, 162,  (ratio r=3)

    • The increases between terms are 4, 12, 36, 108, 

      • getting larger each time

  • Compare this to the arithmetic sequence 2, 6, 10, 14, 18,  (difference d=4)

    • The increases between terms are always 4

  • This distinction between additive change (arithmetic) versus multiplicative change (geometric) is a key idea that carries over into the study of linear and exponential functions

What does the graph of a geometric sequence look like?

  • Like all sequences, the graph of a geometric sequence consists of discrete points at whole number values of n

  • For a geometric sequence with r>1 (and positive initial value), the points curve upward with increasing steepness

  • For a geometric sequence with 0<r<1 (and positive initial value), the points curve downward, getting closer and closer to zero

Two graphs comparing geometric sequences: left graph increases with r>1, right graph decreases with 0<r<1, both plotted against n.
Graphs of increasing (r>1) and decreasing (0<r<1) geometric sequences

Worked Example

Graph on a grid with horizontal axis labeled n and vertical axis labeled gₙ. Points are plotted at (1,9), (3,3) and (4,1), and then also at n=5, n=6 and n=7 for values of gₙ that continue to decrease.

Values of the terms of a geometric sequence gn are graphed in the figure. Which of the following is an expression for the nth term of the geometric sequence?

(A)  gn=3(13)(n3)

(B)  gn=9(3)(n2)

(C)  gn=9(13)(n3)

(D)  gn=27(13)n

Answer:

Consider the values of gn for the first three points on the graph

  • I.e. for n=2, n=3 and n=4

9,  3,  1, ...

They are going down by a factor of 13 each time

  • I.e., 9×13=3 and 3×13=1

That means that the common ratio is 13

  • That rules out option (B), which has a common ratio of 3

  • Don't be fooled by the fact that when n=2,  g2=9(3)(22)=9·30=9·1=9

    • That point agrees with the graph

    • But that formula will not give correct values for n=3, 4, ...

Test out the values of the other three options when n=2

option (A):  g2=3(13)(23)=3(13)1=3·3=9

option (C):  g2=9(13)(23)=9(13)1=9·3=27

option (D):  g2=27(13)2=27·19=3

Only option (A) gives the correct value for g2

(A)  gn=3(13)(n3)

Examiner Tips and Tricks

If you remember the general form for a sequence

gn=gkr(nk)

then you should be able to spot right away that option (A) in the above Worked Example gives the correct form for the sequence in the graph with k=3, where r=13 and g3=3.

Worked Example

Values of the terms of a geometric sequence gn are given in the table below.

n

0

1

2

3

4

gn

6

18

54

162

486

(a) Find the common ratio r of the sequence.

(b) Write an expression for the general term gn.

(c) Find the value of g7.

Answer:

(a)

The common ratio is found by dividing consecutive terms:

r=g1g0=186=3

This can be verified by checking with other terms

  • g2g1=5418=3, g3g2=16254=3, etc.

r=3

(b)

Use the formula gn=g0rn

  • with g0=6 and r=3:

gn=6·3n

(c)

Substitute n=7 into the equation from part (b)

g7=6·37=6·2187=13,122

g7=13,122

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Build on this topic

Roger B

Author: Roger B

Expertise: Development Editor

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Mark Curtis

Reviewer: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.