Geometric Sequences
Geometric Sequence EXAMPLE: 4, 12, 36, 108, 324, …
ratio = 12/4 = 3, or 36/12 = 3
To find subsequent terms, you multiply successive 3s: one 3 for the second term, two 3s for the
third term, three 3s for the fourth term, so 3n-1 for the nth term. Since we are multiplying by the
same value, the equation has the initial amount multiplied by successive ratios.
Recursive Formula
Explicit Formula
Use when you want to find the next few terms
Use when you want to find any term
n 1
n
1
an  an1  r , given a1
a ar
NEXT = PREVIOUS TERM  r, given the FIRST TERM
ANY TERM = FIRST TERM  RATIO^( TERM # - 1)
common ratio, r - the same amount is multiplied from one term to the next
To find the common ratio, divide consecutive terms: r = a2/ a1 or a3/ a2, etc.
The explicit formula of a geometric sequence is an exponential function with base r.
f ( x)  a1r x 1
an = any term  a1 = first term, a2 = second term, etc.
an-1 = previous term
r = common ratio  what you multiply each term by to get the next
term
n = term number
Examples of Geometric Sequences:
Finite: 11, 22, 44, 88, 176
[r = 2]
Infinite: 4, .8, .16, .032, .0064, …
[r = .2]
Terms of Geometric Sequences
Recursive Formula:
an  an1  r , given a1
Explicit Formula:
an  a1r n 1
Find the first 4 terms of the sequence of numbers defined by the given recursive formula.
1.
an  5an1; a1  8 *(- 5)previous term
a1  8
a2  5  8  40
a3  5  40  200
a4  5  200  1000
8, -40, 200, -1000
2.
an  1 an1; a1  7 *(½)previous term
2
a1   7
a2  1  ______  _______
2
a3  1  ______  _______
2
a4  1  ______  _______
2
________, ________, ________, ________
Find the first 4 terms of the sequence of numbers defined by the given explicit rule.
3. an  6(2)
n 1
a1  f (1)  6(2) 0  6
a2  f (2)  6(2)1  12
a3  f (3)  6(2) 2  24
a4  f (4)  6(2)3  48
6, 12, 24, 48
4. an  20  5 
n 1
a1  f (1) 
a2  f (2) 
a3  f (3) 
a4  f (4) 
________, ________, ________, ________
For each arithmetic sequence, write the recursive and explicit formula. Find the 20 th term.
5. -3, -12, -48, -192, -768, …
6. 1700, 170, 17, 1.7, 0.17, …
**You are given a1 = -3. Find r = -12/-3 = 4
**Given a1 = 1700. Find r: 170/1700 =.1
Recursive: an  4an 1; a1  3
Explicit:
an  3(4) n 1
Recursive: an  ____________; a1  ____
Explicit: an = _____________________
Geometric Series and Sigma Notation
Geometric Sequence
4, 8, 16, 32, 64, 128
-.34, -3.4, -340, -3400, …
60, 30, 15, 7.5, 3.75, …
Geometric Series
Type of Sequence
4+8+16+32+64+128
Finite
-.34+-3.4+-340+-3400+ …
Infinite
60+30+15+7.5+3.75+…
Infinite
Is there a sum?
Yes
No
Yes
When the ratio of an infinite geometric series is between 0 and 1 or 0 and -1, the sum of the terms
is getting closer and closer to a sum.
Sigma Notation (Summation Notation)
The Greek letter  is a mathematical symbol for finding the sum of the first n terms of a sequence
ak.
n
a
k 1
k
 a1  a2  a3  a4  ...  an
Examples: Find each sum. You will need to find each individual term and add them.
5
2(3)
1. 
k 1
k 1
= 2  6  18  54  162  242
Find each term: f(1) = 2, f(2) = 6, f(3) = 18, f(4) = 54, f(5) = 162
4
72
2. 
k 1
k 1
 7  14  28  56
 105
Find each term: f(1) = -7, f(2) = -14, f(3) = -28, f(4) = -56
7
k 1
40(3)

3.
k 3
Find the sum.
 360  1080  3240  9720  29160
 43,560
Find the sum.
Sum of Geometric Series
Sum of a Finite Geometric Series
Use when the series ends
Sum of an Infinite Geometric Series
Use when the ratio is 0<|r|<1
a1 (1  r n )
Sn 
1 r
S 
a1
1 r
You cannot find the sum of an infinite geometric series if the ratio, r, is above 1 or below -1.
Examples: Find the sum of each series. You will need the first term a1, the last term an, and n.
10
5(3)
4. 
a1  f (1)  5(3)0  5
r  3, n  10
k 1
5(1  310 )
Sn 
 147, 620
(1  3)
k 1
a1  f (1)  8(0.2) 0  8

5. First 7 terms of
 8(0.2)
i 1
i 1
r  0.2, n  7
8(1  0.27 )
Sn 
 9.999872
1  0.2
*this is a partial sum of an infinite series
r  3
6. Find the sum of -6, -3, -1.5, -.75, …
6
a1  6
S 
*this is the sum of ALL terms of the infinite series
or 1.5
6
 12
1  .5
3
 .5
Applications
Fish Population: The caretaker of a local fish
pond does not allow fishing unless there is an
abundance of fish available. He estimates when
the fish count passes 1,000, he will open the
pond to the public. The season-opening fish
population was estimated at 60 fish, but weekly
checks showed the fish count was doubling each
week. How many weeks will it take before the
pond will open?
Courtesy of Freedigitalphotos.net
One Solution: This is an example of a geometric sequence in which each week the population is
multiplied by 2, which means r = 2. We need to find when the sum of the fish reaches 1,000. We
can write the formula in explicit form: an  60(2) n 1 . We want to find when an = 1000.
1000  60(2) n 1
50  (2) n 1
3
log 2 50  n  1
3
log 50
3 1  n
log 2
n  4.06  1  5.06
 
 
Alternate Method:
So, the pond should open in a little over 5 weeks.
Week
Fish Pop
1
60
2
120
3
240
4
480
5
960
6
1920
Alternate Method: y1  60(2)n 1 and y2  1000 , graph and find the x-coordinate of the intersection.
iPod Purchase: An NCVPS teacher has $349 saved towards a new $399 IPod. He invests his savings
from a second job in a bank account that pays 1.5% interest each week. In how many weeks will he
be able to buy the IPod?
One Solution: The initial amount is $349 and it is increasing by (1+.015) each week. This situation
can be modeled by the formula an  349(1.015) n 1 . The goal is $399. Solve 399  349(1.015) n1 Use
any method detailed above. It will take 10 weeks to reach $399.