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

(a_5/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

a_n = a₁ 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.
a₁, a₂, a₃, a₄, a₅... , a_n, ... which is the same as

a₁, a₁r, a₁r², a₁ r³, a₁ r⁴ , ..., a₁ r^n-1, ...

Therefore
a₁ = a₁
a₂ = a₁r
a₃ = a₁r²
a₄ = a₁r³
a₅ = a₁r⁴
a_n, = a₁ 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.

a₁ = 81, a_k+1 = (1/3) a_k
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 )
a_n = a₁ 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
a₁ = 3
a₇ = a₁r^7-1 = 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 r^n-1
with the common ratio r is not equal to 1 is

S_n = 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
S_n = 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 r^n-1 , ...

has the sum

S = a₁ / (1 - r)

Example 6: Find the sum of the infinite geometric series


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

S_n = (1- rⁿ) / (1 - r)
since rⁿ = 2^∞ = ∞
S_n = (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