Geometric Sequences and Series Examples
Example 1:
Find the common ratio and the first four terms of the geometric sequence
 2 n 3 

.
 4 
The easiest part of this problem is finding the first four terms of the
sequence. You do this by plugging in the numbers 1, 2, 3, and 4 into the
formula.
 2 n 3 
The first term is gotten by plugging 1 into 

 4 
213 2 2
1
1
1
a1 

 2


4
4
2  4 4  4 16
 2 n 3 
The second term is gotten by plugging 2 into 

 4 
2 23 2 1
1
1
1
a2 

 1


4
4
2 4 24 8
 2 n 3 
The third term is gotten by plugging 3 into 

 4 
2 3 3 2 0 1
a3 


4
4 4
 2 n 3 
The fourth term is gotten by plugging 4 into 

 4 
2 43 21 2 1
a4 

 
4
4 4 2
This gives us the beginning of the sequence:
1 1 1 1
, , , ,…
16 8 4 2
Now we can fairly easily answer the question “What is the common ratio?”.
The common ratio is the number that we multiply each term by to get the
next term. Another way to think about this is using the formula r 
an1
. To
an
be more specific, you can find r by dividing the any term by the term that
occurred right before it (for example: by dividing the second term by the
first term.
1
 
1 1 1 16 16
8
In our problem, r        
2
8
 1  8 16 8 1
 
 16 
Thus our common ratio is 2.
Example 2:
Find the nth term of the geometric sequence whose initial term is 3 and
common ratio is 4.
The answer to this problem just requires using the basic formula for the nth
term of a geometric sequence. The basic formula for the nth term of a
geometric sequence is a n  a1 r ( n 1) where a1 is the first (initial) term and r is
the common ratio.
For this problem we will plug 3 in for a1 and 4 in for r . This gives us a
formula for the nth term of a n  3(4) ( n 1) . This formula cannot be further
simplified because of the exponent on the 4.
Example 3:
1 1
1
1
1
, 2 , 3 , 4 , 5 , . . .,
5 5
5
5
5
Find the fifth partial sum S 5
For the sequence
Find the sixth partial sum S 6
For this problem, we need to use the formula for the finite sum of a
geometric series. The formula for the finite sum of a geometric series is
1 r n 
 where a1 is the first (initial) term, r is the common ratio, and
S n  a1 
 1 r 
n is number of terms being added.
For our particular problem a1 
1
. We now need to determine r .
5
 1 
 2
1 1 1 5 1
5
r 
 
  . All that is left now is to determine n . For the
 1  25 5 25 1 5
 
5
fifth partial sum, n  5 . We now plug into the equation and get
  1 5 
  3124  
1 

1   


1


 1   5    1  3125   1   3125   781
S 5   
 

 
4
  5   4   3125
 5  1   1    5 


  
   

5


5 
 5 

For the sixth partial sum, n  6 . We now plug into the equation and get
  1 6 
  15624  
1 

1   


1


 1   5    1  15625   1   15625   3906
S 6   
 

 
4
  5   4   15625
 5  1   1    5 


  
   

5


5 
 5 

Example 4:
n
5
Find the indicated sum  3  .
n 1  3 
18
Since the top number on the sigma sign is 18, this is another finite sum. This
is also a geometric sequence. We will use the same formula that we did
1 r n 
 . We need to know how many terms we are adding (n),
before, S n  a1 
 1 r 
the first term (a1), and the common ratio (r). We will do the same thing that
we did in example 1 to find the first term and the common ratio.
1
5
5
The first term is a1  3   3   5 . The second term is
3
3
2
25 25
5
a 2  3   3 

. We could continue this through 18, but we do not
9
3
3
need to. All we need is the first two terms to determine the common ratio.
As before, to determine the common ratio, we divide the second term by the
 25 
 
5
3
first term. This will give us r     . The only thing that is left is for
5
3
us to determine n. n is the total number of terms being added together.
Since we would plug in the integers 1 through 18, n must be 18.
We will plug all this into the formula for the finite sum of a geometric
  5 18   518 
1    1

  3    318 
series. This will give us S18  5
5
 73840.50315 . Although
5   2 
 1
    

  3 
3


exact answers are usually nicer, number like 518 are a little big to deal with.