GEOMETRIC
SEQUENCES
These are sequences where the ratio of
successive terms of a sequence is always
the same number. This number is called
the common ratio.
1, 2, 4, 8, 16 . . .
r=2
Notice in this sequence that if we find the ratio of any
term to the term before it (divide them) we always get 2.
2 is then called the common ratio and is denoted with the
letter r.
To get to the next term in the sequence we would
multiply by 2 so a recursive formula for this sequence is:
an  2an 1
2 2 2 2
r=2
a=1
1, 2, 4, 8, 16 . . .
Each time you want another term in the sequence you’d
multiply by r. This would mean the second term was the
first term times r. The third term is the first term
multiplied by r multiplied by r (r squared). The fourth term
is the first term multiplied by r multiplied by r multiplied by
r (r cubed). So you can see to get the nth term we’d take
the first term and multiply r (n - 1) times.
an  ar
n 1
Try this to get the 5th term.
a5  12
51
 16
Let’s look at a formula for a geometric sequence and see
what it tells us.
you can see what the
n 1
common ratio will be
This factor gets us
in the formula
started in the right
 23 
place. With n = 1 we’d
get -2 for the first term
Subbing in the set of positive integers we get:
-2, -6, -18, -54 …
What is the
common
ratio?
3n-1 would generate the powers of 3. With
the - 2 in front, the first term would be
-2(30) =- 2. What would you do if you
wanted the sequence -4, -12, -36, -108, . . .?
r=3
 43 
n 1
Find the nth term of the geometric sequence when a = -2 and r =4
If we use 4n-1 we will generate a sequence whose
common ratio is 4, but this sequence starts at 1 (put 1
in for n to get first term to see this). We want ours to
start at -2. We then need the “compensating factor”.
We need to multiply by -2.
 24 
n 1
Check it out by putting in the first few positive
integers and verifying that it generates our
sequence.
Sure enough---it starts at -2
and has a common ratio of 4
-2, -8, -32, -128, . . .
Find the 8th term of 0.4, 0.04. 0.004, . . .
an  ar
To find the common ratio,
take any term and divide it
by the term in front
n 1
0.04
r
 0.1
0.4
an  0.40.1
n 1
a8  0.40.1
81
 0.00000004
If we want to add n terms in a geometric sequence, we
use the formula below:
number of terms
first term
1 r
Sn  a
, r  0, 1
1 r
sum of n terms
15
Find the sum:
 43
n 1
n
common ratio
n 1
1 3
S15  4
1 3
15
4 + 12 + 36 + 108 + . . . + 4(3)14 = 28,697,812
1
an   
2
Let’s look at the sum of the geometric sequence
1 1 1 1
   
2 4 8 16
1
2
n
Let’s look at this on
the number line

1
4

1 1

8 16
0
1
Each time we add another term we’d be going half the
distance left. As n   the sum  1.
means infinity
If the common ratio was not a fraction between -1 and 1,
then the sequence would keep getting larger and larger
and would   as n  . If the common ratio is a
fraction between -1 and 1, the sum as n   is as follows:

a
k 1
ar 

1 r
k 1
first term
common ratio
Let’s try this for the previous sequence:
1
1
S  2  2  1
1
1
1
2
2
1 1 1 1
   
2 4 8 16
1
x
2
1
x
2
1
x
2
Let’s try one more:
8a

1
8 

k 1  3 
8
  12
S 
1  r1 2
3
3
k 1
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au