Sequence And Series

DONE BY,

AFSAL M NAHAS
XI B
KV PATTOM

Slide 11-1
Copyright © 2007 Pearson Education, Inc.

S E QUE NCE
A ND S E RIE S

Slide 11-2

3

4

5

6

7

An introduction…………

1, 4, 7, 10, 13 35 62
2, 4, 8, 16, 32
9, 1, − 7, − 15 −12 9, − 3, 1, − 1/ 3 20 / 3
6.2, 6.6, 7, 7.4 27.2 85 / 64
1, 1/ 4, 1/16, 1/ 64
π, π + 3, π + 6 3π + 9 π, 2.5π, 6.25π 9.75π
Arithmetic Sequences Geometric Sequences
ADD MULTIPLY
To get next term To get next term

Arithmetic Series Geometric Series
Sum of Terms Sum of Terms

Slide 11-8

Find the next four terms of –9, -2, 5, …
Arithmetic Sequence
−2 − −9 = 5 − −2 = 7
7 is referred to as the common difference (d)
Common Difference (d) – what we ADD to get next term

Next four terms……12, 19, 26, 33

Slide 11-9

Find the next four terms of 0, 7, 14, …

Arithmetic Sequence, d = 7
21, 28, 35, 42

Find the next four terms of x, 2x, 3x, …
Arithmetic Sequence, d = x
4x, 5x, 6x, 7x

Find the next four terms of 5k, -k, -7k, …

Arithmetic Sequence, d = -6k
-13k, -19k, -25k, -32k

Slide 11-10

Vocabulary of Sequences (Universal)
a1 → First term
an → nth term
n → number of terms
Sn → sum of n terms
d → common difference

nth term of arithmetic sequence → an = a1 + ( n − 1) d
n
sum of n terms of arithmetic sequence → Sn = ( a1 + an )
2

Slide 11-11

Given an arithmetic sequence with a15 = 38 and d = −3, find a1.
a1 → First term
x
38 an → nth term
15
NA Sn → sum of n terms
-3
an = a1 + ( n − 1) d
38 = x + ( 15 − 1) ( −3 )

X = 80

Slide 11-12

Find S63 of − 19, − 13, −7,...

-19 a1 → First term
an → nth term
353 ??
63
x
6

n
an = a1 + ( n − 1) d ( a1 + an )
Sn =
2
?? = −19 + ( 63 − 1) ( 6 ) 63
( −19 + 353 )
=
S63
?? = 353 2
S63 = 10521

Slide 11-13

Try this one: Find a16 if a1 = 1.5 and d = 0.5
1.5 a1 → First term
an → nth term
x
16
0.5

an = a1 + ( n − 1) d
a16 = 1.5 + ( 16 − 1) 0.5
a16 = 9

Slide 11-14

Find n if an = 633, a1 = 9, and d = 24

a1 → First term
9
633 an → nth term
x
24
an = a1 + ( n − 1) d

633 = 9 + ( x − 1) 24

633 = 9 + 24x − 24
X = 27

Slide 11-15

Find d if a1 = −6 and a29 = 20

a1 → First term
-6
20 an → nth term
29
x
an = a1 + ( n − 1) d

20 = −6 + ( 29 − 1) x

26 = 28x
13
x=
14

Slide 11-16

Find two arithmetic means between –4 and 5

-4, ____, ____, 5
a1 → First term
-4
an → nth term
5
4
NA
x
an = a1 + ( n − 1) d
5 = −4 + ( 4 − 1) ( x )
x=3
The two arithmetic means are –1 and 2, since –4, -1, 2, 5
forms an arithmetic sequence

Slide 11-17

Find three arithmetic means between 1 and 4

1, ____, ____, ____, 4
a1 → First term
1
an → nth term
4
5
NA
x
an = a1 + ( n − 1) d
4 = 1 + ( 5 − 1) ( x )
3
x=
4
The three arithmetic means are 7/4, 10/4, and 13/4
since 1, 7/4, 10/4, 13/4, 4 forms an arithmetic sequence
Slide 11-18

Find n for the series in which a1 = 5, d = 3, Sn = 440
a1 → First term
5
x
440 = ( 5 + 5 + ( x − 1) 3 )
an → nth term
y 2
x ( 7 + 3x )
x
440 =
2
440 Sn → sum of n terms
880 = x ( 7 + 3x )
3
0 = 3x 2 + 7x − 880
an = a1 + ( n − 1) d
Graph on positive window
y = 5 + ( x − 1) 3
X = 16
n
Sn = ( a1 + an )
2
x
440 = ( 5 + y )
2

Slide 11-19

20

An infinite sequence is a function whose domain
is the set of positive integers.
a1, a2, a3, a4, . . . , an, . . .

terms

The first three terms of the sequence an = 2n2 are
a1 = 2(1)2 = 2
finite sequence
a2 = 2(2)2 = 8
a3 = 2(3)2 = 18.

Slide 11-21 21
Copyright © byPearson Education, Inc.
Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.

A sequence is geometric if the ratios of consecutive
terms are the same.
2, 8, 32, 128, 512, . . .
8 =4 geometric sequence
2
32 = 4
8
The common ratio, r, is 4.
128 = 4
32
512 = 4
128

Slide 11-22
Copyright © 2007Houghton MifflinInc.
Copyright © by Pearson Education, Company, Inc. All rights reserved. 22

The nth term of a geometric sequence has the form
an = a1rn - 1
where r is the common ratio of consecutive terms of the
sequence.
r = 75 = 5
15
a1 = 15
15, 75, 375, 1875, . . .
a2 = a3 = a4 =
15(5) 15(52) 15(53)

The nth term is 15(5n-1).

Slide 11-23 23

Example: Find the 9th term of the geometric sequence
7, 21, 63, . . .

r = 21 = 3
a1 = 7
7

an = a1rn – 1 = 7(3)n – 1

a9 = 7(3)9 – 1 = 7(3)8
= 7(6561) = 45,927

The 9th term is 45,927.

Slide 11-24 24

The sum of the first n terms of a sequence is
represented by summation notation.
upper limit of summation
n

∑a = a + a + a3 + a4 + L + an
i 1 2
i =1

lower limit of summation
index of
summation

5

∑ 4n = 41 + 42 + 43 + 44 + 45
n=1
= 4 + 16 + 64 + 256 + 1024
= 1364
Slide 11-25 25

The sum of a finite geometric sequence is given by

)
(
n
1− rn .
Sn = ∑ a1r i −1
= a1
1− r
i =1

5 + 10 + 20 + 40 + 80 + 160 + 320 + 640 = ?

n=8
r = 10 = 2
a1 = 5 5

)( )(
( )( )
1 − r n = 5 1 − 28 = 5 1 − 256 = 5 −255 = 1275
Sn = a1
1− 2 −1
1− r 1− 2

Slide 11-26 26

The sum of the terms of an infinite geometric
sequence is called a geometric series.

If |r| < 1, then the infinite geometric series

a1 + a1r + a1r2 + a1r3 + . . . + a1rn-1 + . . .
∞
a1
has the sum S = ∑ a1r = i
.
1− r
i =0

If r ≥ 1 , then the series does not have a sum.

Slide 11-27 27

Example: Find the sum of 3 − 1 + 1 − 1 +L
39
r = −1
3

 a1  = 3
S =
()
1− r  1− − 1
 
3

= 3 = 3 = 3⋅ 3 = 9
1+ 1 4 44
33

The sum of the series is 9 .
4

Slide 11-28 28

11.3 Geometric Sequences and Series

A geometric sequence is a sequence in which each
term after the first is obtained by multiplying the
preceding term by a constant nonzero real number.

 1, 2, 4, 8, 16 … is an example of a geometric
sequence with first term 1 and each subsequent
term is 2 times the term preceding it.
 The multiplier from each term to the next is
called the com onratio and is usually denoted
m
by r.

Slide 11-29

11.3 Finding the Common Ratio

 In a geometric sequence, the common ratio can be
found by dividing any term by the term preceding it.

The geometric sequence 2, 8, 32, 128, …
has common ratio r = 4 since

8 32 128
= = = ... = 4
28 32

Slide 11-30


nth Term of a Geometric Sequence
In the geometric sequence with first term a1 and
common ratio r, the nth term an, is

an = a1r n −1

Slide 11-31

11.3 Using the Formula for the nth Term

ple Find a 5 and a n for the geometric
Exam
sequence 4, –12, 36, –108 , …

Solution Here a 1= 4 and r = 36/–12 = – 3. Using
n −1
n=5 in the formula an = a1r
5 −1
a5 = 4 ⋅ (−3) = 4 ⋅ (−3) = 324
4

In general
n −1 n −1
an = a1r = 4 ⋅ (−3)
Slide 11-32

11.3 Modeling a Population of Fruit Flies

Exam ple A population of fruit flies grows in such a
way that each generation is 1.5 times the previous
generation. There were 100 insects in the first
generation. How many are in the fourth generation.

Solution The populations form a geometric sequence
with a1= 100 and r = 1.5 . Using n=4 in the formula
for an gives
a4 = a1r 3 = 100(1.5)3 = 337.5

or about 338 insects in the fourth generation.

Slide 11-33

11.3 Geometric Series

 A g om trics rie is the sum of the terms of a
ee es
geometric sequence .

In the fruit fly population model with a 1 = 100 and
r = 1.5, the total population after four generations
is a geometric series:

a1 + a2 + a3 + a4
= 100 + 100(1.5) + 100(1.5) 2 + 100(1.5)3
≈ 813

Slide 11-34


Sum of the First n Terms of an Geometric
Sequence
If a geometric sequence has first term a1 and common
ratio r, then the sum of the first n terms is given by

a1 (1 − r )
n
Sn = r ≠1 .
where
1− r

Slide 11-35

11.3 Finding the Sum of the First n Terms

6

∑ 2 ⋅ 3i
Example Find
i=1

SolutionThis is the sum of the first six terms of a
geometric series with a1 = 2 ⋅ 3 = 6 and r = 3.
1

From the formula for Sn ,

6(1 − 3 ) 6(1 − 729) 6(−728)
6
S6 = = = = 2184 .
1− 3 −2 −2

Slide 11-36

Vocabulary of Sequences (Universal)
a1 → First term
an → nth term
r → common ratio

nth term of geometric sequence → an = a1r n−1
( )
a1 r n − 1 
sum of n terms of geometric sequence → Sn =  
r −1

Slide 11-37

Find the next three terms of 2, 3, 9/2, ___, ___, ___
3 – 2 vs. 9/2 – 3… not arithmetic
3 9/2 3
= = 1.5 → geometric → r =
2 3 2

99 39 3 39 3 3 3
2, 3, , × , × × , × × ×
22 22 2 22 2 2 2

9 27 81 243
2, 3, , ,,
2 4 8 16

Slide 11-38

1 2
If a1 = , r = , find a9 .
2 3
a1 → First term 1/2
x
an → nth term
n → number of terms 9
Sn → sum of n terms NA
r → common ratio 2/3

an = a1r n−1
9 −1
 1  2 
x =   
 2  3 
28 27 128
x= 8= 8=
2×3 3 6561

Slide 11-39

Find two geometric means between –2 and 54

-2, ____, ____, 54
a1 → First term -2
an = a1r n−1
54
an → nth term
54 = ( −2 ) ( x )
4 −1
−27 = x 3
−3 = x
r → common ratio x

The two geometric means are 6 and -18, since –2, 6, -18, 54
forms an geometric sequence

Slide 11-40

2
Find a 2 − a 4 if a1 = −3 and r =
3

-3, ____, ____, ____

2
Since r = ...
3

−4 −8
−3, − 2, ,
39
 −8  −10
a 2 − a 4 = −2 −  = 9
9

Slide 11-41

Find a9 of 2, 2, 2 2,...

a1 → First term 2
x
an → nth term
r → common ratio 2 22
r= = =2
2
2
an = a1r n−1

( 2)
9 −1
x= 2

2 ( 2)
8
x=

x = 16 2
Slide 11-42

If a5 = 32 2 and r = − 2, find a 2
____, ____, ____,____,32 2

a1 → First term x
an → nth term 32 2
−2
r → common ratio
an = a1r n−1

()
5 −1
32 2 = x − 2

2 = x( − 2)
4
32
32 2 = 4x
8 2=x
Slide 11-43

*** Insert one geometric mean between ¼ and 4***
*** denotes trick question
1
,____,4
4
an → nth term 4
r → common ratio x
an = a1r n−1 1
, 1, 4
4
1 3−1 12
4 = r → 4 = r → 16 = r 2 → ±4 = r
1
4 4 , − 1, 4
4
Slide 11-44

111
Find S7 of + + + ...
248
an → nth term NA
11
Sn → sum of n terms x
1
r= 4 = 8 =
r → common ratio
112
( )
a1 r n − 1  24
Sn =  
r −1
 1   1 7   1   1 7 
    − 1      − 1 
   
2  2  2  2 
   63
   =
x= =
1 1 64
−1 −
2 2
Slide 11-45

46

47

48

49

50

51

1, 4, 7, 10, 13, …. No Sum
Infinite Arithmetic
n
Sn = ( a1 + an )
Finite Arithmetic
3, 7, 11, …, 51
2
( )
a1 r n − 1
Sn =
Finite Geometric
1, 2, 4, …, 64
r −1

1, 2, 4, 8, … Infinite Geometric No Sum
r>1
r < -1

11 1 a1
Infinite Geometric
3,1, , , ... S=
3 9 27 -1 < r < 1 1− r

Slide 11-52

111
Find the sum, if possible: 1 + + + + ...
248
11
2 = 4 = 1 → −1 ≤ r ≤ 1 → Yes
r=
112
2

a1 1
S= = =2
1
1− r
1−
2

Slide 11-53

2 2 + 8 + 16 2 + ...
Find the sum, if possible:

8
16 2
= 2 2 → −1 ≤ r ≤ 1 → No
r= =
8
22

NO SUM

Slide 11-54

2111
+++ + ...
3 3 6 12
11
3 = 6 = 1 → −1 ≤ r ≤ 1 → Yes
r=
212
33
2
a1 4
3
S= = =
13
1− r
1−
2

Slide 11-55

248
+ + + ...
777
48
r = 7 = 7 = 2 → −1 ≤ r ≤ 1 → No
24
77

NO SUM

Slide 11-56

5
Find the sum, if possible: 10 + 5 + + ...
2
5
5 2 = 1 → −1 ≤ r ≤ 1 → Yes
r= =
10 5 2

a1 10
S= = = 20
1
1− r
1−
2

Slide 11-57

The Bouncing Ball Problem – Version A

A ball is dropped from a height of 50 feet. It rebounds 4/5 of
it’s height, and continues this pattern until it stops. How far
does the ball travel?
50

40 40

32 32

32/5 32/5

50 40
S= + = 450
4 4
1− 1−
5 5

Slide 11-58

The Bouncing Ball Problem – Version B

A ball is thrown 100 feet into the air. It rebounds 3/4 of
it’s height, and continues this pattern until it stops. How far
does the ball travel?

100 100

75 75

225/4 225/4

100 100
S= + = 800
3 3
1− 1−
4 4

Slide 11-59

11.3 Infinite Geometric Series

 If a 1, a 2, a 3, … is a geometric sequence and the
sequence of sums S1, S2, S3, …is a convergent
sequence, converging to a number S∞. Then S∞ is
said to be the sum of the infinite geometric series

a1 + a2 + a3 + ... = S∞

Slide 11-60

11.3 An Infinite Geometric Series

Given the infinite geometric sequence
111 1
2, 1, , , , ,...
2 4 8 16
the sequence of sums is S1 = 2, S2 = 3, S3 = 3.5, …

The calculator screen shows
more sums, approaching a value
of 4. So
11
2 + 1 + + + ... = 4
24

Slide 11-61

11.3 Infinite Geometric Series

Sum of the Terms of an Infinite Geometric
Sequence
The sum of the terms of an infinite geometric sequence
with first term a1 and common ratio r, where –1 < r < 1
is given by
a1
S∞ = .
1− r

Slide 11-62

11.3 Finding Sums of the Terms of Infinite
Geometric Sequences
i
∞
 3
ple Find ∑  
Exam
i =1  5 

3
3
and r =
SolutionHere a1 = so
5
5
3
i
∞
 3 a1 5 =3
∑ 5  = 1− r = 3 2 .
i =1  
1−
5

Slide 11-63

64

Sequence And Series

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