Chapter 11 Sec 3
Geometric Sequences
Algebra 2 Chapter 11 Sections 3 – 5
Geometric Sequence
• A geometric sequence is a sequence in
which each term after the first is found by
multiplying the previous term by a constant
r called the common ratio.
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Algebra 2 Chapter 11 Sections 3 – 5
Example 1
Find the eighth term of a geometric sequence for
which a1 = – 3 and r = – 2.
an = a1 · r n – 1
a8 = (–3) · (–2) 8 – 1
a8 = (–3) · (–128)
a8 = 384
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Algebra 2 Chapter 11 Sections 3 – 5
Example 2
Write an equation for the n term of a geometric
a 12
sequence 3, 12, 48, 192… a1  3 and r  2   4
a1 3
an = a1 · rn – 1
an = (3) · (4) n – 1
So the equations is an = 3(4) n – 1
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Algebra 2 Chapter 11 Sections 3 – 5
Example 3
Find the tenth term of a geometric sequence for which
a4 = 108 and r = 3.
an = a1 · r
a4 = a1 ·
n–1
(3) 4 – 1
108 = a1 ·
(3) 3
108 = 27a1
an = a1 · r n – 1
a10 = 4 · (3) 10 – 1
a10 = 4 · (3) 9
a10 = 78,732
4 = a1
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Algebra 2 Chapter 11 Sections 3 – 5
Geometric Means
As we saw with arithmetic means, you are given two terms
of a geometric sequence and are asked to find the terms
between, these terms between are called geometric means.
Find the three geometric means between 3.12 and 49.92.
24.96
6.24
12.48 –24.96
– 6.24 _____,
3.12, _____,
_____, 49.92
a1
a2
a3
a4
a5
an = a1 · r n – 1
a5 = 3.12 · r 5 – 1
49.92 = 3.12 r 4
16 = r 4
±2 = r So…
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Chapter 11 Sec 4
Geometric Series
Algebra 2 Chapter 11 Sections 3 – 5
Geometric Series
Geometric Sequence
Geometric Series.
1, 2, 4, 8, 16
1 + 2 + 4 + 8 + 16
4, –12, 36
4 + (–12) + 36
Sn represents the sum of the first n terms of a series. For
example, S4 is the sum of the first four terms.
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Algebra 2 Chapter 11 Sections 3 – 5
Example 1
6
Evaluate
5 2
n 1

a1 1  r
Sn 
1 r
a1  5  2
n6
r2
n 1
n
11
a6  5  2
5
6 1


5 1 2
Sn 
1 2
6

5 63
Sn 
 315
1
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Algebra 2 Chapter 11 Sections 3 – 5
Only have the first and last terms?
You can use the formula for finding the nth term in
(an = a1 · r n – 1 ) conjunction with the sum formula

a1  a1r n 
 S n 
. when you don’t know n.
1 r 

n


a

a
r
1
1
n
–
1
 S n 
.
an · r = a1 · r
·r
1 r 

an · r = a1 · r n
a1  an r 

 Sn 
.
1 r 

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Algebra 2 Chapter 11 Sections 3 – 5
Example 3
Find a1 in a geometric series for which S8 = 39,360
and r = 3.

a1 1  r
Sn 
1 r

n

a1 1  3
39,360 
1 3
8

39,360  3280a1
12  a1
 6560a1
39,360 
2
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Chapter 11 Sec 5
Infinite Geometric
Series
Algebra 2 Chapter 11 Sections 3 – 5
Infinite Geometric Series
Any geometric series with an infinite number of terms.
1 1 1 1
Consider the infinite geometric series     ...
2 4 8 16
You have already learned to find the sum Sn of the first n terms,
this is called partial sum for an infinite series.
Notice that as n increases, the
partial sum levels off and
approaches a limit of one. This
leveling-off behavior is
characteristic of infinite
geometric series for which
| r | < 1.
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Algebra 2 Chapter 11 Sections 3 – 5
Sum of an Infinite Series
Lets use the formula for the sum of a finite series to find a
formula for an infinite series.
a1  a1r n
a1 a1r n
Sn 
Sn 

1 r
1 r 1 r
If –1 < r < 1 , the value if rn will approach 0 as n increases.
Therefore the partial sum of the infinite series will approach
a1 a1 0
a1
Sn 

or
1 r 1 r
1 r
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Algebra 2 Chapter 11 Sections 3 – 5
Example 1
Find the sum of each infinite geometric series, if it exists.
1 3 9
a.    ... First find the3 value of r to determine if the sum
exists.
2 8 32
3
3
8
r   , Since  1, the sum exists.
1
3
1 4
a1  and a2  so
4
2
8
2
1
a1
Sn 
 2
1 r 1 3
4
1
 2 2
1
4
b.1  2  4  8  ...
2
r
 2, Since  2  1, the sum
1
a1  1 and a2  2 so
does not exist.
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Algebra 2 Chapter 11 Sections 3 – 5
Example 2: Sigma Time…
Evaluate

 1
24  

 5
n 1
a1
Sn 

1 r

a1  24
n 1
24
 1
1   
 5
 1
24  

 5
n 1
1
r
5
24

 20
6
5
n 1
 20
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Algebra 2 Chapter 11 Sections 3 – 5
Repeeeeating Decimal
typo intentional
Write 0.39 as a fraction.
S = 0.39
S = 0 .393939393939… then
100S = 39.393939393939… Subtract 100S – S
– S = 0 .393939393939…
99S = 39
39 13
S

99 33
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Algebra 2 Chapter 11 Sections 3 – 5
Daily Assignment
•
Chapter 11 Sections 3 – 5
•
Study Guide (SG)
•
Pg 145 – 150 Odd
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