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Sequence And Series
- 1. DONE BY,
AFSAL M NAHAS
XI B
KV PATTOM
Slide 11-1
Copyright © 2007 Pearson Education, Inc.
- 2. S E QUE NCE
A ND S E RIE S
Slide 11-2
Copyright © 2007 Pearson Education, Inc.
- 8. 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
Copyright © 2007 Pearson Education, Inc.
- 9. 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
Copyright © 2007 Pearson Education, Inc.
- 10. 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
Copyright © 2007 Pearson Education, Inc.
- 11. 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
Copyright © 2007 Pearson Education, Inc.
- 12. Given an arithmetic sequence with a15 = 38 and d = −3, find a1.
a1 → First term
x
38 an → nth term
n → number of terms
15
NA Sn → sum of n terms
d → common difference
-3
an = a1 + ( n − 1) d
38 = x + ( 15 − 1) ( −3 )
X = 80
Slide 11-12
Copyright © 2007 Pearson Education, Inc.
- 13. Find S63 of − 19, − 13, −7,...
-19 a1 → First term
an → nth term
353 ??
n → number of terms
63
Sn → sum of n terms
x
d → common difference
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
Copyright © 2007 Pearson Education, Inc.
- 14. Try this one: Find a16 if a1 = 1.5 and d = 0.5
1.5 a1 → First term
an → nth term
x
n → number of terms
16
NA Sn → sum of n terms
d → common difference
0.5
an = a1 + ( n − 1) d
a16 = 1.5 + ( 16 − 1) 0.5
a16 = 9
Slide 11-14
Copyright © 2007 Pearson Education, Inc.
- 15. Find n if an = 633, a1 = 9, and d = 24
a1 → First term
9
633 an → nth term
n → number of terms
x
NA Sn → sum of n terms
d → common difference
24
an = a1 + ( n − 1) d
633 = 9 + ( x − 1) 24
633 = 9 + 24x − 24
X = 27
Slide 11-15
Copyright © 2007 Pearson Education, Inc.
- 16. Find d if a1 = −6 and a29 = 20
a1 → First term
-6
20 an → nth term
n → number of terms
29
NA Sn → sum of n terms
d → common difference
x
an = a1 + ( n − 1) d
20 = −6 + ( 29 − 1) x
26 = 28x
13
x=
14
Slide 11-16
Copyright © 2007 Pearson Education, Inc.
- 17. Find two arithmetic means between –4 and 5
-4, ____, ____, 5
a1 → First term
-4
an → nth term
5
n → number of terms
4
Sn → sum of n terms
NA
d → common difference
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
Copyright © 2007 Pearson Education, Inc.
- 18. Find three arithmetic means between 1 and 4
1, ____, ____, ____, 4
a1 → First term
1
an → nth term
4
n → number of terms
5
Sn → sum of n terms
NA
d → common difference
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
Copyright © 2007 Pearson Education, Inc.
- 19. 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 )
n → number of terms
x
440 =
2
440 Sn → sum of n terms
880 = x ( 7 + 3x )
d → common difference
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
Copyright © 2007 Pearson Education, Inc.
- 21. 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.
- 22. 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
- 23. 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
Copyright © byPearson Education, Inc.
Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 24. 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
Copyright © byPearson Education, Inc.
Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 25. 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
Copyright © byPearson Education, Inc.
Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 26. 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
Copyright © byPearson Education, Inc.
Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 27. 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
Copyright © byPearson Education, Inc.
Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 28. 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
Copyright © byPearson Education, Inc.
Copyright © 2007 Houghton Mifflin Company, Inc. All rights reserved.
- 29. 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
Copyright © 2007 Pearson Education, Inc.
- 30. 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
Copyright © 2007 Pearson Education, Inc.
- 31. 11.3 Geometric Sequences and Series
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
Copyright © 2007 Pearson Education, Inc.
- 32. 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
Copyright © 2007 Pearson Education, Inc.
- 33. 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
Copyright © 2007 Pearson Education, Inc.
- 34. 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
Copyright © 2007 Pearson Education, Inc.
- 35. 11.3 Geometric Sequences and Series
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
Copyright © 2007 Pearson Education, Inc.
- 36. 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
Copyright © 2007 Pearson Education, Inc.
- 37. Vocabulary of Sequences (Universal)
a1 → First term
an → nth term
n → number of terms
Sn → sum of n terms
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
Copyright © 2007 Pearson Education, Inc.
- 38. 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
Copyright © 2007 Pearson Education, Inc.
- 39. 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
Copyright © 2007 Pearson Education, Inc.
- 40. 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
n → number of terms 4
−27 = x 3
Sn → sum of n terms NA
−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
Copyright © 2007 Pearson Education, Inc.
- 41. 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
Copyright © 2007 Pearson Education, Inc.
- 42. Find a9 of 2, 2, 2 2,...
a1 → First term 2
x
an → nth term
n → number of terms 9
Sn → sum of n terms NA
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
Copyright © 2007 Pearson Education, Inc.
- 43. If a5 = 32 2 and r = − 2, find a 2
____, ____, ____,____,32 2
a1 → First term x
an → nth term 32 2
n → number of terms 5
Sn → sum of n terms NA
−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
Copyright © 2007 Pearson Education, Inc.
- 44. *** Insert one geometric mean between ¼ and 4***
*** denotes trick question
1
,____,4
4
a1 → First term 1/4
an → nth term 4
n → number of terms 3
Sn → sum of n terms NA
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
Copyright © 2007 Pearson Education, Inc.
- 45. 111
Find S7 of + + + ...
248
a1 → First term 1/2
an → nth term NA
n → number of terms 7
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
Copyright © 2007 Pearson Education, Inc.
- 52. 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
Copyright © 2007 Pearson Education, Inc.
- 53. 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
Copyright © 2007 Pearson Education, Inc.
- 54. 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
Copyright © 2007 Pearson Education, Inc.
- 55. 2111
+++ + ...
Find the sum, if possible:
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
Copyright © 2007 Pearson Education, Inc.
- 56. 248
+ + + ...
Find the sum, if possible:
777
48
r = 7 = 7 = 2 → −1 ≤ r ≤ 1 → No
24
77
NO SUM
Slide 11-56
Copyright © 2007 Pearson Education, Inc.
- 57. 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
Copyright © 2007 Pearson Education, Inc.
- 58. 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
Copyright © 2007 Pearson Education, Inc.
- 59. 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
Copyright © 2007 Pearson Education, Inc.
- 60. 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
Copyright © 2007 Pearson Education, Inc.
- 61. 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
Copyright © 2007 Pearson Education, Inc.
- 62. 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
Copyright © 2007 Pearson Education, Inc.
- 63. 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
Copyright © 2007 Pearson Education, Inc.