Find the nth term of a sequence Find the index of a given term of a sequence Given a geometric series, be able to calculate the nth partial sum Identify a geometric series as convergent or divergent.

1. 11.3 Geometric Sequences &
Series
Chapter 11 Further Topics in Algebra

2. Concepts and Objectives
⚫ Geometric Sequences and Series
⚫ Find the nth term of a sequence
⚫ Find the index of a given term of a sequence
⚫ Given a geometric series, be able to calculate Sn, the nth
partial sum and vice versa
⚫ Identify whether a geometric series converges and its
limit

3. Geometric Sequences
⚫ A geometric sequence is a sequence in which each term
equals a constant multiplied by the preceding term.
⚫ The constant for a geometric sequence is called the
common ratio, r, because the ratio between any two
adjacent terms equals this constant.
⚫ Like arithmetic sequences, formulas for calculating tn for
geometric sequences can be found by linking the term
number to the term value.

4. Geometric Sequences (cont.)
⚫ Consider the geometric sequence:
3, 6, 12, 24, 48, …

5. Geometric Sequences (cont.)
⚫ Consider the geometric sequence:
3, 6, 12, 24, 48, …
This sequence has t1 = 3 and common ratio r = 2. Thus:
1 3t =
2 3 2t =
2
3 3 2 2 3 2t = =
3
4 3 2 2 2 3 2t = =
1
3 2n
nt −
=

6. Geometric Sequences (cont.)
⚫ The nth term of a geometric sequence equals the first
term multiplied by (n – 1) common ratios. That is,
⚫ A geometric sequence is actually just an example of an
exponential function. The only difference is that the
domain of a geometric sequence is  rather than all real
numbers.
1
1
n
nt t r −
=

7. Examples
1. Calculate t100 for the geometric sequence with first term
t1 = 35 and common ratio r = 1.05.

8. Examples
1. Calculate t100 for the geometric sequence with first term
t1 = 35 and common ratio r = 1.05.
( )( )100 1
100 35 1.05t −
=
( )( )99
35 1.05 4383.375262= =

9. Examples
2. A geometric sequence has t1 = 17 and r = 2. If
tn = 34816, find n.

10. Examples
2. A geometric sequence has t1 = 17 and r = 2. If
tn = 34816, find n.
To solve for n, we will take the log of each side:
( )( )1
34816 17 2n−
=
1
2048 2n−
=
1
log2048 log2n−
=
( )log2048 1 log2n= −
log2048
1
log2
n= −
11 1n= −
12n =

11. Geometric Series
⚫ If we wanted the sum of the first 100 terms of our first
geometric sequence, we could write it as follows:
⚫ Now, suppose we multiplied both sides of this equation
by –2, or the opposite of our common ratio, and added
the two equations together:
= + + + + + +2 3 98 99
100 3 3 2 3 2 3 2 ... 3 2 3 2S
= + + + + + +
− = − − − − − − −
2 3 98 99
100
2 3 98 99 100
100
3 3 2 3 2 3 2 ... 3 2 3 2
2 3 2 3 2 3 2 ... 3 2 3 2 3 2
S
S

12. Geometric Series (cont.)
Solving for S, gives us
which certainly suggests a formula.
= + + + + + +
− = − − − − − − −
2 3 98 99
100
2 3 98 99 100
100
3 3 2 3 2 3 2 ... 3 2 3 2
2 3 2 3 2 3 2 ... 3 2 3 2 3 2
S
S
− = + + + + + + − 100
100 1002 3 0 0 0 ... 0 0 3 2S S
− = − 100
100 1002 3 3 2S S
( ) ( )− = −
 −
=  
− 
100
100
100
100
1 2 3 1 2
1 2
3
1 2
S
S

13. Geometric Series (cont.)
⚫ The nth partial sum of a geometric series is given by the
formula
⚫ For some reason, I’m always tempted to try to factor the
fraction further. It doesn’t factor.
 −
=  
− 
1
1
1
n
n
r
S t
r

14. Geometric Series
⚫ Example: Find S34 for the geometric series with t1 = 7
and r = 1.03.

15. Geometric Series
⚫ Example: Find S34 for the geometric series with t1 = 7
and r = 1.03.
Using the formula, we have:
 −
=  
− 
34
34
1 1.03
7
1 1.03
S
 404.111

16. Geometric Series
⚫ Example: 50238.14 is the approximate value of a partial
sum in the geometric series with t1 = 150 and r = 1.04.
Which term is it?

17. Geometric Series
⚫ Example: 50238.14 is the approximate value of a partial
sum in the geometric series with t1 = 150 and r = 1.04.
Which term is it?
 −
=  
− 
1
1
1
n
n
r
S t
r
 −
=  
− 
1 1.04
50238.14 150
1 1.04
n
( )( )−
= −
50238.14 0.04
1 1.04
150
n
=1.04 14.39683...n
1–1.04 = –0.04

18. Geometric Series
(cont.) Taking the log of each side gets n out of the
exponent.
So n = 68. (n always has to be a positive integer.)
=log1.04 log14.39683...n
=log1.04 log14.39683...n
= =
log14.39683...
68.000001...
log1.04
n

19. Convergent Geometric Series
⚫ It should be obvious that the partial sums of a geometric
sequence such as the last example will continue to
increase as n increases.
⚫ Now, let’s look at a different sequence:
The first six partial sums would look like:
1 1 1 1
2, 1, , , , , ...
2 4 8 16
=1 2S =2 3S =3
7
2
S =4
15
4
S =5
31
8
S =6
63
16
S

20. Convergent Geometric Series
⚫ If we were to graph this sequence of
partial sums, we can see that it
approaches the line y = 4.
⚫ Using some algebra, we can transform
the series:
−
  
−  
   = = −    − 
 
2
1
1
12
2 4
1 21
2
n
n
nS

21. Convergent Geometric Series
⚫ With this rewritten formula, we can see that as n
increases, (½)n-2 gets closer and closer to 0. (Check out
the value of ½ raised to larger and larger powers.)
⚫ Therefore, we say that the limit of Sn as n increases
without bound (or approaches infinity) is 4 or
⚫ In order for the common ratio term to go to 0 as n
increases, the denominator of the partial sums formula
must be a proper fraction. That is, . This is called a
convergent geometric series. A series that does not
converge diverges.
→
=lim 4n
n
S
1r

22. Convergent Geometric Series
⚫ The formula for the sum of a convergent geometric
series is
Example: In our previous sequence, t1 = 2 and r = ½:
= 
−
1
, where 1
1
t
S r
r
= = =
−
2 2
4
1 1
1
2 2
S

23. Classwork
⚫ College Algebra
⚫ Page 1023: 8-20 (even); page 1014: 34-40, 44-48 (even);
page 1005: 68-76 (even)