## Monday, March 5, 2007

### 9.3 Geometric Sequences and Series

9.3 Geometric Sequences and Series

Geometric Sequences - consecutive terms have a common ratio.

A. Definition of Geometric Sequences:

A sequence is a geometric sequence if the ratios of consecutive terms are the same.

(a2/a1) = (a3/a2) = (a4/a3) = ... = r, r cannot equal 0

The number r is the common ratio of the sequence.

Example 1: given 1, -2, 4, -8, ... , find the common ratio

(-2/1) = (4/ -2) = (-8/ 4) = -2

Example 2: write the first five terms of the geometric sequence.
a1 = 10 , r = 2

(a2/a1) = r = 2
(a2/10) = 2
a2 = 20

(a3/a2) = 2
(a3/20) = 2
a3 = 40

(a4/a3) = 2
(a4/40) = 2
a4 = 80

(a5/a4) = 2
(a5/80) = 2
a5 = 160

B. The nth Term of a Geometric Sequence:
the nth term of a geometric sequence has the form

an = a1 rn-1

where r is the common ratio of consecutive terms of the sequence. So, every geometric sequence can be written in the following form.
a1, a2, a3, a4, a5... , an, ... which is the same as

a1, a1r, a1r2, a1 r3, a1 r4 , ..., a1 rn-1, ...

Therefore
a1 = a1
a2 = a1r
a3 = a1r2
a4 = a1r3
a5 = a1r4
an, = a1 rn-1

Example 3: write the first five terms of the geometric sequence. Determine the common ratio and write the nth term of the sequence as a function of n.

a1 = 81, ak+1 = (1/3) ak
a2 = a1+1 = (1/3) a1 = (1/3) (81) = 27
a3 = a2+1 = (1/3) a2 = (1/3) (27) = 9
a4 = a3+1 = (1/3) a3 = (1/3) (9) = 3
a5 = a4+1 = (1/3) a4 = (1/3) (3) = 1

Therefore the common ratio is (27/81) = (9/27) = (3/9) = (1/3 )
an = a1 rn-1 = 81 (1/3)n-1 = (1/3)-4 (1/3)n-1 = (1/3)-4 +n-1 = (1/3)n-5

Example 4: Find the 7th term of the following geometric sequence: 3, 36, 432

r = (36/3) = 432/36) = 12
a1 = 3
a7 = a1r7-1 = 3 (12)6 = 3 (2985984) = 8957952

C. The Sum of a Finite Geometric Sequence

The sum of the geometric sequence
a1, a1r, a1r2, a1 r3, a1 r4 , ..., a1 rn-1
with the common ratio r is not equal to 1 is

Sn = a1 ((1 - rn)/(1 - r))

Example 5: Finding the sum of a finite geometric sequence

(-2)1-1 + (-2)2-1 + (-2)3-1 + (-2)4-1 + (-2)5-1 + (-2)6-1 + (-2)7-1 + (-2)8-1 + (-2)9-1 =
(-2)0 + (-2)1 + (-2)2 + (-2)3 + (-2)4 + (-2)5 + (-2)6 + (-2)7 + (-2)8 =
1 + -2 + 4 + -8 + 16 + -32 + 64 + -128 + 256 = 171
or
Sn = a1 ((1 - rn)/(1 - r))
S9 = a1 ((1 - r9)/(1 - r)) = 1(1 - (-2)9)/(1 –-2) = (1 - -512) / (3) = (513) / (3) = 171

C. The Sum of a Infinite Geometric Sequence

then the infinite geometric series

a1, a1r, a1r2, a1 r3, a1 r4 , ..., a1 rn-1 , ...

has the sum

S = a1 / (1 - r)

Example 6: Find the sum of the infinite geometric series

recall: an = a1 rn-1 so therefore a1= 1 and r = 2

Sn = (1- rn) / (1 - r)
since rn = 2 = ∞
Sn = (1- ∞ ) / (1-2) = -∞ / (- 1) = ∞

Example 7: Find the sum of the infinite geometric series

S = a1 / (1 - r)
Since the first term n=0 then 4(.2)0 = 4 therefore

= 4 / (1 - .2) = 4 / .8 = 5

9.3 Homework #54; page 645; # 11 - 23 odd, 31 - 45 odd, 55 - 83 odd
Quiz on Sections 9.1 - 9.3 next class