Math 114 Geometric Sequences and Series
The follow examples of numbers are called geometric sequences. You multiple each number in
the sequence by the same number to get the next number in the sequence. Each number in the
sequence is called a term. Some examples of arithmetic sequences:
Example 1: 1,3,9,...
Example 2: 7,
7 7
, ,...
2 4
Example 3: 1, 2, 4,..., 100
A geometric sequence of numbers has a common ratio. It is the number we multiply each term
in the sequence by to get the next term in the sequence. We call the common ratio r .
In Example 1, r  3 . In Example 2: r 
1
. In Example 3, r  2 .
2
a 
An easy way to calculate r is to divide the second term by the first term  2  .
 a1 
a 3
For Example 1, 2   3 .
a1 1
We can represent the first term in the sequence by a1 , the 2nd term by a2 and the 3rd term by a3 .
7
7
7
7
The nth term is an . In Example 2, a1  7 a2 
and so on.
a3 
a4 
a5 
2
4
8
16
We can write any sequence of numbers as a1 , a2 , a3 ,..., an . This is called a finite sequence
because it has a particular number of terms. Example: 2, 8, 32, 128, 512, 2048. There are 6
terms so n = 6. A sequence that is not finite is infinite. We cannot count the terms in an infinite
sequence. Example: 2, 8, 32, 128, 512, 2048 , 8192,… (Dots mean the terms go on and on.)
To get any term in an geometric sequence, use the formula
an  a1  r
n 1
Example: Find the 20th term of the sequence with first 3 terms 3, -6, 12.


a
6
n  20, a1  3, r  2  r  2 
 2  .
Solution:
a1
2


a20  a1  r 201
a20  3 2   3(2)19  1572864
___________________________________________________________________________
201
Geometric Series: You add the terms in a geometric sequence to get a geometric series.
Example (finite sequence): 2 + 8 + 32 + 128 + 512 + 2048.
Example (infinite sequence): 2 + 8 + 32 + 128 + 512 + 2048 + 8192 + …
To get the sum of the first n terms of a geometric series, use the formula
a1 (1  r n )
Sn 
1 r
Example: Find the sum of the first 20 terms of the sequence with first 3 terms 3, -6, 12.
n  20, a1  3, r  2 and a20  1572864 from what we calculated
Solution:
above.
a1 (1  r 20 )
S20 
1 r
3(1  (2) 20 ) 3(1  1048576)


 1048575
1  (2)
3
INFINITE Geometric Series: Under a certain condition, you can get the sum of an infinite
geometric series. The condition is that r , the common ratio, must be between -1 and 1.
If 1  r  1 (| r | 1) , then the sum of an infinite geometric series is S 
a1
.
1 r
We say the infinite geometric series converges to S.
Example: Find the sum of the infinite geometric series 1 
1 1 1
   ...
2 4 8
 1 
1
a2  2  1
 . r is between – 1 and 1 ( 1 
 1 ) so we can
CHECK r .
=
1
2
2
a1
calculate the sum, S.
S
a1
2
1
1
2


 . The sum, S, is .
3
1  r   1    3  3
1

  2   2 
    