B. Find the nth Term of a Geometric Sequence
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continue applying additional factors of r for each successive term.
a1 ϭ a1r0
6 ϭ 3 # 21
a2 ϭ a1r1
12 ϭ 3 # 22
a3 ϭ a1r2
24 ϭ 3 # 23
a4 ϭ a1r3
48 ϭ 3 # 24
a5 ϭ a1r4
current term
initial
term
S
3 ϭ 3 # 20
S
784
exponent on common ratio
From this pattern, we note the exponent on r is always 1 less than the subscript of
the current term: 5 Ϫ 1 ϭ 4, which leads us to the formula for the nth term of a geometric sequence.
The n th Term of a Geometric Sequence
The nth term of a geometric sequence is given by
an ϭ a1rnϪ1
where r is the common ratio.
EXAMPLE 3
ᮣ
Finding a Specific Term in a Sequence
Identify the common ratio r, and use it to write the expression for the nth term.
Then find the 10th term of the sequence: 3, Ϫ6, 12, Ϫ24, p .
Solution
ᮣ
By inspection we note that a1 ϭ 3 and r ϭ Ϫ2. This gives
an ϭ a1rnϪ1
ϭ 31Ϫ22 nϪ1
n th term formula
substitute 3 for a1 and Ϫ2 for r
To find the 10th term we substitute n ϭ 10:
a10 ϭ 31Ϫ22 10Ϫ1
ϭ 31Ϫ22 9 ϭ Ϫ1536
substitute 10 for n
simplify
Now try Exercises 33 through 46
EXAMPLE 4
ᮣ
ᮣ
Determining the Number of Terms in a Geometric Sequence
1
.
Find the number of terms in the geometric sequence 4, 2, 1, p , 64
Solution
ᮣ
Observing that a1 ϭ 4 and r ϭ 12, we have
an ϭ a1 rnϪ1
1 nϪ1
ϭ 4a b
2
n th term formula
substitute 4 for a1 and
1
for r
2
Although we don’t know the number of terms in the sequence, we do know the last
1
1
. Substituting an ϭ 64
or nth term is 64
gives
1
1 nϪ1
ϭ 4a b
64
2
1
1 nϪ1
ϭa b
256
2
substitute
1
for an
64
1
divide by 4 amultiply by b
4
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Section 9.3 Geometric Sequences
785
From our work in Chapter 5, we attempt to write both sides as exponentials with a
like base, or apply logarithms. Since 256 ϭ 28, we equate bases.
1 8
1 nϪ1
a b ϭa b
2
2
S8 ϭ n Ϫ 1
9ϭn
write
1
1 8
as a b
256
2
like bases imply exponents must be equal
solve for n
This shows there are nine terms in the sequence.
Now try Exercises 47 through 58
ᮣ
Note that in both Examples 3 and 4, the nth term had the form of an exponential
1 nϪ1
equation 1y ϭ a # bn 2 after simplifying: an ϭ 31Ϫ22 nϪ1 and an ϭ 4a b . This is in
2
fact a characteristic of geometric sequences, with the common ratio r corresponding to
the base b. This means the graph of a geometric sequence will always be a set of discrete points that lie on an exponential graph (see Worthy of Note). After entering
1 nϪ1
(from Example 4), Figure 9.35 shows the table of values for this
u1n2 ϭ 4a b
2
sequence, with the graph in Figure 9.36.
As before, we can see the graph of the sequence more distinctly by entering the
natural numbers 1 through 8 in L1, then defining L2 as u(L1). The resulting graph is
shown in Figure 9.37.
WORTHY OF NOTE
The sequence from Example 3 is an
alternating sequence, and the
exponential characteristics of its
graph can be seen by taking a look
at the graph of an ϭ |31Ϫ22 nϪ1 | .
Figure 9.35
Figure 9.36
Figure 9.37
5
5
10
0
Ϫ1.5
10
0
Ϫ1.5
One additional advantage of using a list is that we can easily verify whether or not
a common ratio r exists. We do this by duplicating L2 in L3 (define L3 ϭ L2), then
deleting the first entry of L3 and the last entry of L2 (to keep the same number of terms
akϩ1
in each list). See Figure 9.38. This enables us to find the ratio
for the entire list
ak
by defining L4 as the ratio L3/L2 (which automatically computes the ratio of successive terms). See Figures 9.39 and 9.40.
Figure 9.38
Figure 9.39
Figure 9.40
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CHAPTER 9 Additional Topics in Algebra
EXAMPLE 5
ᮣ
Graphing Geometric Sequences
Enter the natural numbers 1 through 6 in L1, and the terms of the sequence
0.25, 1.25, 6.25, 31.25, 156.25, 781.25 in L2. Then determine if the sequence is
geometric by
a. Graphing the related points to see if they appear to form along an exponential
graph.
b. Finding the successive ratios between terms.
c. If the sequence is geometric, find the nth term and graph the sequence.
Solution
ᮣ
a. The plotted points are shown in Figure 9.41 and appear to lie along an
exponential graph.
Figure 9.41
Figure 9.42
1000
8
0
Ϫ250
b. Using the approach described prior to
Example 5, we find there is a common
ratio of r ϭ 5 (Figure 9.42).
c. With a1 ϭ 0.25 and r ϭ 5, the nth term
must be
an ϭ a1rnϪ1
ϭ 0.25152 nϪ1
Figure 9.43
1000
n th term formula
8
0
substitute for a1 and r
The nth term for this sequence is
an ϭ 0.25152 nϪ1. The graph is shown in
Figure 9.43.
Ϫ250
Now try Exercises 59 through 62
If the term a1 is unknown but a term ak is given, the nth term can be written
an ϭ akrnϪk,
(the subscript on the term ak and the exponent on r sum to n).
EXAMPLE 6
ᮣ
Finding the First Term of a Geometric Sequence
Given a geometric sequence where a4 ϭ 0.075 and a7 ϭ 0.009375, find the
common ratio r and the value of a1.
Solution
ᮣ
Since a1 is not known, we express a7 as the product of a known term and the
appropriate number of common ratios: a7 ϭ a4r3 17 ϭ 4 ϩ 3, as required).
a7 ϭ a4 # r3
0.009375 ϭ 0.075r3
0.125 ϭ r3
r ϭ 0.5
a1 is unknown
substitute 0.009375 for a7 and 0.075 for a4
divide by 0.075
solve for r
ᮣ
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Section 9.3 Geometric Sequences
Having found r, we can now solve for a1
a7 ϭ a1r6
0.009375 ϭ a1 10.52 6
0.009375 ϭ a1 10.0156252
a1 ϭ 0.6
n th term formula
substitute 0.009375 for a7 and 0.5 for r
simplify
solve for a1
The first term is a1 ϭ 0.6 and the common ratio is r ϭ 0.5.
B. You’ve just seen how
we can find the nth term of a
geometric sequence
Now try Exercises 63 through 68
ᮣ
C. Find the n th Partial Sum of a Geometric Sequence
As with arithmetic series, applications of geometric series often involve computing sums
of consecutive terms. We can adapt the method for finding the sum of an arithmetic
sequence to develop a formula for adding the first n terms of a geometric sequence.
For the nth term an ϭ a1rnϪ1, we have Sn ϭ a1 ϩ a1r ϩ a1r2 ϩ a1r3 ϩ p ϩ
nϪ1
a1r . If we multiply Sn by Ϫr then add the original series, the “interior terms” sum
to zero.
Ϫ rSn ϭ Ϫa1r ϩ 1Ϫa1r2 2 ϩ 1Ϫa1r 3 2 ϩ p ϩ 1Ϫa1rnϪ1 2 ϩ 1Ϫa1rn 2
------S --S --S
S a rnϪ1
ϩ Sn ϭ a1 ϩ a1r ϩS a1r 2 ϩ p ϩ a1rnϪ2 ϩ
1
Sn Ϫ rSn ϭ
a1 ϩ
0
ϩ
0
ϩ 0 ϩ
0
ϩ
0
ϩ 1Ϫa1rn 2
We then have Sn Ϫ rSn ϭ a1 Ϫ a1rn, and can now solve for Sn:
Sn 11 Ϫ r2 ϭ a1 Ϫ a1rn
factor out Sn
a1 Ϫ a1r
solve for Sn (divide by 1 Ϫ r )
1Ϫr
The result is a formula for the nth partial sum of a geometric sequence.
n
Sn ϭ
The n th Partial Sum of a Geometric Sequence
Given a geometric sequence with first term a1 and common ratio r, the nth partial
sum (the sum of the first n terms) is
a1 11 Ϫ rn 2
a1 Ϫ a1rn
Sn ϭ
ϭ
,r 1
1Ϫr
1Ϫr
In words: The sum of the first n terms of a geometric sequence is the difference of
the first and 1n ϩ 12st term, divided by 1 minus the common ratio.
EXAMPLE 7
Solution
ᮣ
Computing a Partial Sum
ᮣ
3i (the first nine powers
Use the preceding summation formula to find the sum:
iϭ1
of 3). Verify the result on a graphing calculator.
The initial terms of this series are 3 ϩ 9 ϩ 27 ϩ p , and we note a1 ϭ 3, r ϭ 3,
and n ϭ 9. We could find the first nine terms and add, but using the partial sum
formula is much faster and gives
Sn ϭ
S9 ϭ
a1 11 Ϫ rn 2
1Ϫr
311 Ϫ 39 2
1Ϫ3
9
͚
sum formula
substitute 3 for a1, 9 for n, and 3 for r
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ϭ
31Ϫ19,6822
Ϫ2
ϭ 29,523
simplify
result
To verify, we enter u1n2 ϭ 3 on the Y= screen,
and find the sum of the first nine terms of the
sequence on the home screen as before. See figure.
n
C. You’ve just seen how
we can find the nth partial sum
of a geometric sequence
Now try Exercises 69 through 92
ᮣ
D. The Sum of an Infinite Geometric Series
To this point we’ve considered only partial sums of a geometric series. While it is
impossible to add an infinite number of these terms, some of these “infinite sums” appear
to have what is called a limiting value. The sum appears to get ever closer to this value
but never exceeds it—much like the asymptotic behavior of some graphs. We will define
the sum of this infinite geometric series to be this limiting value, if it exists. Consider
the illustration in Figure 9.44, where a standard sheet of typing paper is cut in half. One
of the halves is again cut in half and the process is continued indefinitely, as shown.
1 1 p
, 32, with a1 ϭ 12 and r ϭ 12.
Notice the “halves” create an infinite sequence 21, 14, 18, 16
1
1
1
1
1
The corresponding infinite series is 2 ϩ 4 ϩ 8 ϩ 16 ϩ 32 ϩ p ϩ 21n ϩ p .
Figure 9.44
1
2
1
8
1
2
1
4
1
4
1
8
1
16
1
16
1
32
1
32
1
64
1
64
and so on
Figure 9.45
1
64
1
16
1
32
1
4
1
8
1
2
WORTHY OF NOTE
The formula for the sum of an
infinite geometric series can
also be derived by noting that
Sq ϭ a1 ϩ a1r ϩ a1r2 ϩ
a1r3 ϩ p can be rewritten as
Sq ϭ a1 ϩ r1a1 ϩ a1r ϩ
a1r2 ϩ a1r3 ϩ p 2 ϭ a1 ϩ rSq.
Sq Ϫ rSq ϭ a1
Sq 11 Ϫ r2 ϭ a1
a1
.
Sq ϭ
1Ϫr
If we arrange one of the halves from each stage as shown in Figure 9.45, we would be
rebuilding the original sheet of paper. As we add more and more of these halves together,
we get closer and closer to the size of the original sheet. We gain an intuitive sense that this
series must add to 1, because the pieces of the original sheet of paper must add to 1 whole
sheet. To explore this idea further, consider what happens to 1 12 2 n as n becomes large.
1 4
n ϭ 4: a b ϭ 0.0625
2
1 8
n ϭ 8: a b Ϸ 0.004
2
1 12
n ϭ 12: a b Ϸ 0.0002
2
Further exploration with a calculator seems to support the idea that as
n S q, 1 12 2 n S 0, although a definitive proof is left for a future course. In fact, it can
be shown that for any ͿrͿ 6 1, rn becomes very close to zero as n becomes large.
a1 Ϫ a1rn
a1
a1rn
n
In symbols: as n S q, r S 0. For Sn ϭ
ϭ
Ϫ
, note that if
1Ϫr
1Ϫr
1Ϫr
ͿrͿ 6 1 and “we sum an infinite number of terms,” the second term becomes zero, leaving
a1
.
only the first term. In other words, the limiting value (represented by Sq) is Sq ϭ
1Ϫr
Infinite Geometric Series
Given a geometric sequence with first term a1 and 0 r 0 6 1, the sum of the related
infinite series is given by
a1
Sq ϭ
;r 1
1Ϫr
If ͿrͿ 7 1, no finite sum exists.