Transcript Lesson 15 - Geometric Series

Geometric Sequences and
Series
Part III
Geometric Sequences and Series
The sequence
1, 2, 4, 8, . . . 2
63
is an example of a Geometric sequence
A sequence is geometric if
each term
r
previous term
where r is a constant called the common ratio
In the above sequence, r = 2
Geometric Sequences and Series
A geometric sequence or geometric progression (G.P.)
is of the form
a, ar , ar 2 , ar 3 , . . .
The nth term of an G.P. is
un  ar
n 1
Geometric Sequences and Series
Exercises
1. Use the formula for the nth term to find the term
indicated of the following geometric sequences
(a)
2, 8, 32, . . .
Ans:
(b) 12,  3,
3
, . . .
4
6th term
2(4) 5  2048
5th term
4
3
 1
Ans: 12   
64
 4
(c) 0.2, 0  02, 0  002, . . .
7th term
Ans:
0  2(0.1) 6  0.0000002
Geometric Sequences and Series
Summing terms of a G.P.
e.g.1 Evaluate
5
3( 2)

n 1
n
Writing out the terms helps us to recognize the G.P.
3(2)  3(2) 2  3(2) 3  3(2) 4  3(2) 5
With a calculator we can see that the sum is 186.
But we need a formula that can be used for any G.P.
The formula will be proved next but you don’t need to
learn the proof.
Geometric Sequences and Series
Summing terms of a G.P.
With 5 terms of the general G.P., we have
TRICK Multiply
by r:
S 5  a  ar  ar 2  ar 3  ar 4
rS 5  ar  ar 2  ar 3  ar 4  ar 5
Subtracting the expressions gives
S 5  rS 5  a  ar  ar 2  ar 3  ar 4
 ar  ar 2  ar 3  ar 4  ar 5
Move the lower row 1 place to the right
Geometric Sequences and Series
Summing terms of a G.P.
With 5 terms of the general G.P., we have
S 5  a  ar  ar  ar  ar
2
Multiply by r:
3
4
rS 5  ar  ar  ar  ar  ar
2
3
4
5
Subtracting the expressions gives
S 5  rS 5  a  ar  ar  ar  ar
2
3
4
5
 ar  ar  ar  ar  ar
2
and subtract
3
4
Geometric Sequences and Series
Summing terms of a G.P.
With 5 terms of the general G.P., we have
S 5  a  ar  ar  ar  ar
2
Multiply by r:
3
4
rS 5  ar  ar  ar  ar  ar
2
3
4
5
Subtracting the expressions gives
S 5  rS 5  a  ar  ar  ar  ar
2
3
4
5
 ar  ar  ar  ar  ar
2
S 5  rS 5  a
3
4
 ar
5
Geometric Sequences and Series
Summing terms of a G.P.
So,
S 5  rS 5  a  ar
5
Take out the common factors
S5 ( 1  r )  a ( 1  r 5 )
and divide by ( 1 – r )

a( 1  r )
S5 
1 r
5
Similarly, for n terms we get
a( 1  r )
Sn 
1 r
n
Geometric Sequences and Series
Summing terms of a G.P.
The formula
a( 1  rn )
Sn 
1 r
gives a negative denominator if r > 1
Instead, we can use
a ( r n 1 )
Sn 
r 1
Geometric Sequences and Series
Summing terms of a G.P.
For our series
3(2)  3(2)  3(2)  3(2)  3(2)
2
a  6, r  2 and n  5
a ( r n 1 )
Using
Sn 
r 1
6( 2 1)
Sn 
21
5
6 ( 31 )

1
 186
3
4
5
Geometric Sequences and Series
Summing terms of a G.P.
EX
Find the sum of the first 20 terms of the
geometric series, 2  6  18  54  . .
leaving your answer in index form
 3 6
Solution:
a  2, r 
 3
12
n
20
a ( 1 r )
Sn 
1 r
 S 20

2 1   3

1   3
.

We’ll simplify this answer without using a calculator
Geometric Sequences and Series
Summing terms of a G.P.
 S 20

2 1   3 20

1   3
1

There are 20 minus signs here
and 1 more outside the bracket!
2 1  3 20

42


1  3 20
2

Geometric Sequences and Series
Summing terms of a G.P.
e.g. 3
In a geometric sequence, the sum of the 3rd
and 4th terms is 4 times the sum of the 1st
and 2nd terms. Given that the common
ratio is not –1, find its possible values.
Solution: As there are so few terms, we don’t need
the formula for a sum
3rd term + 4th term = 4( 1st term + 2nd term )

ar  ar  4(a  ar )
2
3
Divide by a since the 1st term, a, cannot be zero:

r 2  r 3  4(1  r )
3
2

r  r  4r  4  0
Geometric Sequences and Series
Summing terms of a G.P.
We need to solve the cubic equation
3
2
r  r  4r  4  0
Should use the factor theorem: We will do this soon !!
f (1)  1  1  4  4  0  (r  1) is not a factor
f (1)  1  1  4  4  0  (r  1) is a factor
f ( 2)  8  4  8  4  0
 ( r  2) is a factor
f ( 2)  8  4  8  4  0  ( r  2) is a factor
( r  1)( r  2)( r  2) are the factors
Geometric Sequences and Series
Summing terms of a G.P.
The solution to this cubic equation is therefore
3
2
r
r
 4r  4  0
 (r  1)( r  2 )(r  2 )  0
Since we were told
r  1
we get
r  2
Geometric Sequences and Series
SUMMARY

A geometric sequence or geometric progression
(G.P.) is of the form
a, ar , ar 2 , ar 3 , . . .

The nth term of an G.P. is
un  ar

n 1
The sum of n terms is
a (1 r
Sn 
1 r
n
)
or
a ( r 1 )
Sn 
r 1
n
Geometric Sequences and Series
Sum to Infinity
IF |r|<1 then
a ( 1  ( 1) )
S 
1  (r )
n
a
S 
1 r
0
Because (<1)∞ = 0
Geometric Sequences and Series
Exercises
1. Find the sum of the first 15 terms of the
following G.P., giving the answers in index
form
2 + 8 + 32 + . . .
2. Find the sum of the first 15 terms of the G.P.
4 2+1+. . .
giving your answer correct to
3 significant figures.
Geometric Sequences and Series
Exercises
1. Solution:
2 + 8 + 32 + . . .
a  2, r  4, n  15
 S 15
a ( r n 1 )
Sn 
r 1
2 ( 4 15  1 )
2(4 15  1 )

 S 15 
41
3
2. Solution:
4 2+1+. . .
a  4, r  0  5, n  15
 S 15

4 1   0  5 15

1   0  5

a (1 r n )
Sn 
1 r

S15  2  67
( 3 s.f. )