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.

(a

_{2}/a

_{1}) = (a

_{3}/a

_{2}) = (a

_{4}/a

_{3}) = ... = 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.

a

_{1}= 10 , r = 2

(a

_{2}/a

_{1}) = r = 2

(a

_{2}/10) = 2

a

_{2}= 20

(a

_{3}/a

_{2}) = 2

(a

_{3}/20) = 2

a

_{3}= 40

(a

_{4}/a

_{3}) = 2

(a

_{4}/40) = 2

a

_{4}= 80

(a

_{5/a}4) = 2

(a

_{5}/80) = 2

a

_{5}= 160

B. The nth Term of a Geometric Sequence:

the nth term of a geometric sequence has the form

a

_{n}= a

_{1}r

^{n-1}

where r is the common ratio of consecutive terms of the sequence. So, every geometric sequence can be written in the following form.

a

_{1}, a

_{2}, a

_{3}, a

_{4}, a

_{5}... , a

_{n}, ... which is the same as

a

_{1}, a

_{1}r, a

_{1}r

^{2}, a

_{1}r

^{3}, a

_{1}r

^{4}, ..., a

_{1}r

^{n-1}, ...

Therefore

a

_{1}= a

_{1}

a

_{2}= a

_{1}r

a

_{3}= a

_{1}r

^{2}

a

_{4}= a

_{1}r

^{3}

a

_{5}= a

_{1}r

^{4}

a

_{n}, = a

_{1}r

^{n-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.

a

_{1}= 81, a

_{k+1}= (1/3) a

_{k}

a

_{2}= a

_{1+1}= (1/3) a

_{1}= (1/3) (81) = 27

a

_{3}= a

_{2+1}= (1/3) a

_{2}= (1/3) (27) = 9

a

_{4}= a

_{3+1}= (1/3) a

_{3}= (1/3) (9) = 3

a

_{5}= a

_{4+1}= (1/3) a

_{4}= (1/3) (3) = 1

Therefore the common ratio is (27/81) = (9/27) = (3/9) = (1/3 )

a

_{n}= a

_{1}r

^{n-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

a

_{1}= 3

a

_{7}= a

_{1}r

^{7-1}= 3 (12)

^{6}= 3 (2985984) = 8957952

C. The Sum of a Finite Geometric Sequence

The sum of the geometric sequence

a

_{1}, a

_{1}r, a

_{1}r

^{2}, a

_{1}r

^{3}, a

_{1}r

^{4}, ..., a

_{1}r

^{n-1}

with the common ratio r is not equal to 1 is

S

_{n}= a

_{1}((1 - r

^{n})/(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

S

_{n}= a

_{1}((1 - r

^{n})/(1 - r))

S

_{9}= a

_{1}((1 - r

^{9})/(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

a

_{1}, a

_{1}r, a

_{1}r

^{2}, a

_{1}r

^{3}, a

_{1}r

^{4}, ..., a

_{1}r

^{n-1}, ...

has the sum

S = a

_{1}/ (1 - r)

Example 6: Find the sum of the infinite geometric series

recall: an = a

_{1}r

^{n-1}so therefore a

_{1}= 1 and r = 2

S

_{n}= (1- r

^{n}) / (1 - r)

since r

^{n}= 2

^{∞}= ∞

S

_{n}= (1- ∞ ) / (1-2) = -∞ / (- 1) = ∞

Example 7: Find the sum of the infinite geometric series

S

_{∞}= a

_{1}/ (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