Download GEOMETRIC SEQUENCES and SERIES

C2 CHAPTER 7
GEOMETRIC SEQUENCES and SERIES
GEOMETRIC SEQUENCES
These are sequences where you keep multiplying by the same
number to get the next term. This number is called the
common ratio, r
e.g. 3, 6, 12, 24, 48, 96, 192, …
r=2
Ex 7A
The general form of a geometric sequence is:
2
3
4
th
n–1
n
a, ar, ar , ar , ar , ..... ,ar
n–1
term = ar
They are often referred to as GEOMETRIC PROGRESSIONS
(GP’s)
Example 1
The 6th term of a GP = 1215 and the 9th term = 32805
Find the first term and the common ratio.
5
t6 = ar = 1215
8
t9 = ar = 32805

t9
t6
8
=
ar
5
ar
=
32805
1215
3
 r = 27  r = 3
5
a  3 = 1215  a =
JMcC
1215
3
5
= 5
1
C2 CHAPTER 7
Example 2
The numbers 2, x and (x + 4) form the first three terms of a
GP with all positive terms. Calculate the possible values of x.
u
u
r = 2 = 3
u1
u2

x
=
2
x + 4
2
 x = 2(x + 4)
x
2
 x – 2x – 8 = 0
(x – 4)(x + 2) = 0
 x = 4 (x  – 2 since all terms must be positive)
Ex 7B
GROWTH & DECAY PROBLEMS
Example
The population of a colony of bees is growing at 5% per year.
Initially there are 3000 bees. Find:
a)
the number of bees 4 years later.
b)
how many years before the population exceeds 5000
a) This is a GP with a = 3000, r = 1.05 and n = 5
4
4
u5 = ar = 3000  1·05 = 3646·51875  3647
b) After n years  un + 1 > 5000
n
n
3000  1·05 > 5000  1·05 >
n log 1·05 > log
n >
log 1·666
log 1·05
5
3
 log 1·05
n
> log
5
3
5
3
= 10·47  after 11 years
Ex 7C
JMcC
2
C2 CHAPTER 7
THE SUM OF A GEOMETRIC SERIES
2
3
n–2
+ ar
2
3
n–2
+ ar
Sn = a + ar + ar + ar + ....... + ar
rSn =
ar + ar + ar + ....... + ar
n–1
n–1
n
+ ar
Subtracting gives:
n
Sn – rSn = a – ar
n
(1 – r)Sn = a(1 – r )
n
Sn =
n
a(1 – r )
or
1 – r
Sn =
Use if |r| < 1
a(r – 1)
r – 1
Use if |r| > 1
Example 1
Find the sum of the series 5 + 15 + 45 + 135 + …(12 terms)
a = 5, r = 3, n = 12
S12 =
5(3
12
– 1)
3 – 1
= 1 328 600
Example 2
8
Find

6  ( – 2)
r
r=1
2
8
= 6[( – 2) + ( – 2) + ... + ( – 2) ]
GP with a = r = – 2, n = 8
8
 ( – 2)(( – 2) – 1) 
 – 510 
= 6 
=
6


 = 1020
( – 2) – 1


 – 3 
JMcC
3
C2 CHAPTER 7
Example 3
Find the least value of n such that the sum 2 + 6 + 18 + 54 + …
to n terms would exceed 5 million
a = 2, r = 3
Sn =
2(3
n
– 1)
3 – 1
n
= 3 – 1
n
3 – 1 > 5 000 000
n
3 > 5 000 001
n log 3 > log 5 000 001
n >
log 5 000 001
= 14·040
log 3
least n = 15
[S14 = 4 782 968,
S15 = 14 348 906
Ex 7D
SUM TO INFINITY OF A GEOMETRIC PROGRESSION
n
Consider
n
r
Sn =
a(1 – r )
1 – r
n
 0 so 1 – r
as n   with r < 1
 1
S =
a
1 – r
The series converges to a finite sum (it is a convergent series)
[Remember |r|<1]
JMcC
4
C2 CHAPTER 7
The simplest example of this is:
1 +
1
2
+
1
4
1
+
8
+
1
16
+ ........
The total will never exceed 2
S =
1
1 – 0·5
= 2
e.g. S20 = 1·999998093,
S30 = 1·999999998
Example
Find the sum to infinity of the series:
1 –
1
3
+
1
9
–
1
27
a = 1, r = –
+ ...
1
3


1  
4 3
S = 1   1 –  –
=
1
=

 
3  
3 4


Ex 7E
Mixed Ex 7F
JMcC
5