### 7.3 - San Juan College

```SECTION 7.3
GEOMETRIC SEQUENCES
GEOMETRIC
SEQUENCES
(a) 3, 6, 12, 24, 48, 96
(b) 12, 4, 4/3, 4/9, 4/27, 4/81
(c) .2, .6, 1.8, 5.4, 16.2, 48.6
Geometric Sequences have a
“common ratio”.
(a) r = 2
(b) r = 1/3
(c) r = 3
GEOMETRIC SEQUENCE
RECURSION FORMULA
a n = ra n - 1
This formula relates each term in the
sequence to the previous term in the
sequence.
a n = 2a n - 1
b n = 1/3b n - 1
c n = 3c n - 1
EXAMPLE:
Given that a 1 = 5 and the recursion
formula a n = 1.5a n - 1, determine
the the value of a 5 .
a 2 = 1.5(5) = 7.5
a 3 = 1.5(7.5) = 11.25
a 4 = 1.5(11.25) = 16.875
a 5 = 1.5(16.875) = 25.3125
Again, recursion formulas have a
Explicit Formulas are much better
for finding nth terms.
GEOMETRIC SEQUENCE
EXPLICIT FORMULA
a 2 = ra 1
a 3 = ra 2 = r(ra 1 ) = r2 a 1
a 4 = ra 3 = r(r2 a 1 ) = r3 a 1
In general, an = rn - 1a1
PREVIOUS EXAMPLE:
Given that a 1 = 5 and r = 1.5,
determine the the value of a 5 .
a 5 = 1.5 4 (5) = 25.3125
EXAMPLE:
Given that {an} = 64, 48, 36 . . .
determine the value of a8
First, determine r
r = 48/64 = .75
a8 =
.757
(64)
a8 =
2187
256
EXAMPLE:
If a person invests \$500 today at 6%
interest compounded monthly, how
much will the investment be worth at
the end of 10 years (that is, at the end
of 120 months)?
The 6% is an annual rate.
The corresponding monthly rate is
.06/12 = .005
EXAMPLE:
a1 = 500(1.005) Amt at end of mth 1
a2 = 500(1.005)2 Amt at end of mth 2


a120 = 500(1.005)120 Amt at end of mth
120
EXAMPLE:
a120 = 500(1.005)120 Amt at end of mth
120
\$909.70
GEOMETRIC SEQUENCE
SUM FORMULA
Let a1, a2, a3 be a geometric sequence
Then Sn = a1+ a2 + a3 + . . . + an is the
sum of the first n terms of that
sequence.
Sn can also be written as
S n = a 1 + a 1 r + a 1 r2 + . . . + a 1 rn - 1
GEOMETRIC SEQUENCE
SUM FORMULA
S n = a 1 + a 1 r + a 1 r2 + . . . + a 1 rn - 1
rSn =
a 1 r + a 1 r2 + . . . + a 1 rn - 1 + a 1 rn
Sn - rSn = a1 + 0 + 0 + . . . +
0 + - a 1 rn
Sn (1 - r) = a1 (1 - rn)
n
Sn =
a1 (1 - r )
1- r
or
a1 ( r
n
- 1)
r - 1
EXAMPLE:
Determine the sum of the first 20
terms of the geometric sequence
36, 12, 4, 4/3, . . .
a1 = 36
r = 1/3
EXAMPLE:
 1 

36   - 1
 3 



=
1
-1
3
20
S20
53.9999999
EXAMPLE:
If you were offered 1¢ today, 2¢
tomorrow, 4¢ the third day and so
on for 20 days or a lump sum of
\$10,000, which would you choose?
S20 = .01
2
20
- 1
2 - 1
= \$10,485.75
This formula is for the sum of the
first n terms of a geometric
sequence.
Can we find the sum of an entire
sequence?
For example: 1 + 3 + 9 + 27 + . . .
SUMS OF ENTIRE
GEOMETRIC SEQUENCES
But we can for a sequence such as
1
2
+
n

1
 1 
1 -   
 2 
2
1 -
1
2
1
4
=
+
1
8
+
1
16
 1
1 -  
 2
+ . . .
n
1
as n

GEOMETRIC SEQUENCE
SUM FORMULA
Any geometric sequence with
r< 1
As n,
S =
n
Sn =
a1
1 - r
a1(1 - r )
1 - r

r< 1
a1
1 - r
EXAMPLE:
Evaluate the sum of the geometric
series:
16 + 12 + 9 + 27/4 + . . .
r = 3/4
S =
16
1 -
64
3
4
EXAMPLE:
A ball is dropped from a height of
16 feet. At each bounce it rises to
a height of three-fourths the
previous height. How far will it
have traveled (up and down) by
the time it comes to rest?
EXAMPLE:
Down series: 16 + 12 + 9 + . . .
Up series:
SD =
12 + 9 + 27/4 . . .
16
1 -
3
4
SU =
12
1 -
3
4
64 + 48 = 112 ft.

k
 1
  3  =
k =1
 1
+  
 3
3
1
2
 1
+  
 3
3
 1
+  
 3
4
+ . . .
Geometric Series
Recall:
a1
1 - r
1
1
1
3
=
=
1
2
2
1 3
3
3
EXAMPLE
8
3 =
k
3 + 3 2 + 33 + . . . + 3 8
Geometric
Sequence
k = 1
n
Recall:
Sn =
S8 =
a1 (1 - r )
1 - r
3(1 - 6561)
1 - 3
= 9840
CONCLUSION OF SECTION 7.3
```