13.1, 13.3 Arithmetic and Geometric Sequences and Series 13.5 Sums of Infinite Series Objective 1.To identify an arithmetic or geometric sequence and find.
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13.1, 13.3 Arithmetic and Geometric Sequences and Series13.5 Sums of Infinite Series
Objective
1. To identify an arithmetic or geometric sequence and find a formula for its n-th term;
2. To find the sum of the first n terms for arithmetic and geometric series.
1. Arithmetic Sequence and Series
1, 4, 7,10,13
9,1, 7, 15
6.2, 6.6, 7, 7.4
, 3, 6
Arithmetic Sequences
ADDTo get next term
2, 4, 8,16, 32
9, 3,1, 1/ 3
1,1/ 4,1/16,1/ 64
, 2.5 , 6.25
Geometric Sequences
MULTIPLYTo get next term
Arithmetic Series
Sum of Terms
35
12
27.2
3 9
Geometric Series
Sum of Terms
62
20 / 3
85 / 64
9.75
a1 a2 a3 a4 a5 a6anan - 1
+ d + d + d + d + d
r
+ d
r r r r r
The notation of a sequence can be simply denoted as
{an}, meaning a1, a2, …, an, …
For any three consecutive terms in an arithmetic sequence,
… , ak – 1 , ak , ak + 1 , …
ak is called the arithmetic/geometric mean of ak – 1 and ak + 1 if the sequence is arithmetic/geometric.
If three consecutive terms,
… , ak – 1 , ak , ak + 1 , …
in an arithmetic sequence, then
ak = ak – 1 + dak = ak + 1 – d
ak + ak = ak – 1 + ak + 1
2ak = ak – 1 + ak + 1
Or, ak is the arithmetic average of ak – 1 and ak + 1.
k 1 k 1
k
a aa
2
If three consecutive terms,
… , ak – 1 , ak , ak + 1 , …
in a geometric sequence, then
Or, ak is the geometric average of ak – 1 and ak + 1.
k k k 1 k 1a a a a
k k 1
k 1 k
a ar
a a
2k k 1 k 1a a a
Geometric mean
k k k 1 k 1a a a a
2k k 1 k 1a a a
Arithmetic meanak + ak = ak – 1 + ak + 1
2ak = ak – 1 + ak + 1
Find the next four terms of –9, –2, 5, …
Arithmetic Sequence
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
7)2(5)9(2
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
Vocabulary of Sequences (Universal)
1a First term
na nth term
nS sum of n terms
n number of terms
d common difference
n 1
n 1 n
nth term of arithmetic sequence
sum of n terms of arithmetic sequen
a a n 1 d
nS a a
2ce
d
na n Sn 2
1Or 1
Given an arithmetic sequence with 15 1a 38 and d 3, find a .
1a First term
na nth term
nS sum of n terms
n number of terms
d common difference
x
15
38
NA
-3
n 1a a n 1 d
138 a 1 15 3
a1 = 80
63Find S of 19, 13, 7,...
1a First term
na nth term
nS sum of n terms
n number of terms
d common difference
-19
63
x
x
6
n 1a a n 1 d
63a 19 6 13 6 353
353
n 1 n
nS a a
2
63
633 3S
219 5
63 1 1S 052
15021]167[63]63119[6362
1631963Or 63
S
16 1Find a if a 1.5 and d 0.5 Try this one:
1a First term
na nth term
nS sum of n terms
n number of terms
d common difference
1.5
16
x
NA
0.5
n 1a a n 1 d
16 1.5 0.a 16 51
16a 9
n 1Find n if a 633, a 9, and d 24
1a First term
na nth term
nS sum of n terms
n number of terms
d common difference
9
x
633
NA
24
n 1a a n 1 d
633 9 n 1 24
633 9 24n 24
n = 27
1 29Find d if a 6 and a 20
1a First term
na nth term
nS sum of n terms
n number of terms
d common difference
-6
29
20
NA
x
n 1a a n 1 d
20 6 29 1 d
26 28d
13
d14
Find two arithmetic means between –4 and 5
-4, ____, ____, 5
1a First term
na nth term
nS sum of n terms
n number of terms
d common difference
-4
4
5
NA
x
n 1a a n 1 d
5 4 4 1 d
The two arithmetic means are –1 and 2, since –4, -1, 2, 5
forms an arithmetic sequence
d 3
Find three arithmetic means between 1 and 4
1, ____, ____, ____, 4
1a First term
na nth term
nS sum of n terms
n number of terms
d common difference
1
5
4
NA
x
n 1a a n 1 d
4 1 5 1 d
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
3
d4
Find n for the series in which 1 na 5, d 3, S 440
1a First term
na nth term
nS sum of n terms
n number of terms
d common difference
5
x
y
440
3
n 1a a n 1 d
n 1 n
nS a a
2
5 n 1 3
n
440 5 3n 22
n 7 3n440
2
880 n 7 3n
20 3n 7n 880
n = 16
( 16)(3 55) 0n n 55
16, or 3
n n 3n 2
Find the number of multiples of 4 between 35 and 225.
The difference of any two consecutive multiples of 4 is 4. Those multiples of 4 form an arithmetic sequence. Since 35 < 36 (the first multiple of 4 we are looking for), we can write the nth term of an arithmetic sequence as
36 ( 1) 4na n
35 225na
35 36 ( 1) 4 225n
1 4( 1) 189n
1 4 4 189n
3 4 193n
3 19348.25
4 4n
48n
1, 4, 7,10,13
9,1, 7, 15
6.2, 6.6, 7, 7.4
, 3, 6
Arithmetic Sequences
ADDTo get next term
2, 4, 8,16, 32
9, 3,1, 1/ 3
1,1/ 4,1/16,1/ 64
, 2.5 , 6.25
Geometric Sequences
MULTIPLYTo get next term
Arithmetic Series
Sum of Terms
35
12
27.2
3 9
Geometric Series
Sum of Terms
62
20 / 3
85 / 64
9.75
2. Geometric Sequences and Series
a1 a2 a3 a4 a5 a6anan - 1
+ d + d + d + d + d
r
+ d
r r r r r
Vocabulary of Sequences (Universal)
1a First term
na nth term
nS sum of n terms
n number of terms
r common ratio
n 1n 1
n
n 1
nth term of geometric sequence
sum of the first n terms of geometric seq
a a r
1 rue ce S a
1n
r
1 n
n
or
sum of the first n terms of geometric sequenca ra
S1
er
Find the next three terms of 2, 3, 9/2, ___, ___, ___
3 – 2 vs. 9/2 – 3… not arithmetic3 9 / 2 3
1.5 geometric r2 3 2
3 3 3 3 3 3
2 2 2
92, 3, , , ,
2
9 9 9
2 2 2 2 2 2
92, 3, , ,
27 81 243
4 8,
2 16
1 9
1 2If a , r , find a .
2 3
1a First term
na nth term
nS sum of n terms
n number of terms
r common ratio
1/2
x
9
NA
2/3
n 1n 1a a r
9 1
9
1 2a
2 3
8
9 8
2a
2 3
7
8
2
3
128
6561
Find two geometric means between –2 and 54
-2, ____, ____, 54
1a First term
na nth term
nS sum of n terms
n number of terms
r common ratio
-2
54
4
NA
x
n 1n 1a a r
4 154 2 r
327 r
r 3
The two geometric means are 6 and -18, since –2, 6, -18, 54
forms an geometric sequence
2 4 1
2Find a a if a 3 and r
3
-3, ____, ____, ____
2Since r ...
3
4 83, 2, ,
3 9
2 4
8 10a a 2
9 9
9Find a of 2, 2, 2 2,...
1a First term
na nth term
nS sum of n terms
n number of terms
r common ratio
x
9
NA
2
2 2 2r 2
22
n 1n 1a a r
9
9
1
a 2 2
8
2 2
16 2
5 2 4If a 32 2 and r 2, find a to a .
____, , ____,________ ,32 2
1a First term
na nth term
nS sum of n terms
n number of terms
r common ratio
x
5
NA
32 2
2n 1
n 1a a r
5
1
1
32 2 a 2
4
132 2 a 2
132 2 4a
1a 8 2
2a 8 2( 2) 16
23a 8 2( 2) 16 2
34a 8 2( 2) 32
*** Insert one geometric mean between ¼ and 4***
*** denotes trick question
1,____,4
4
1a First term
na nth term
nS sum of n terms
n number of terms
r common ratio
1/4
3
NA
4
xn 1
n 1a a r
3 114
4r 2r
14
4 216 r 4 r
1,1, 4
4
1, 1, 4
4
7
1 1 1Find S of ...
2 4 8
1a First term
na nth term
nS sum of n terms
n number of terms
r common ratio
1/2
7
x
NA
11184r
1 1 22 4
n
n 1
1 rS a
1 r
7
7
11
1s
2 1
212
71
1 22
1
12
7
11
2
63
64
1, 4, 7, 10, 13, …. Infinite Arithmetic No Sum
3, 7, 11, …, 51 Finite Arithmetic
n 1 n
1
nS a a
2n 1
n a d2
1, 2, 4, …, 64 Finite Geometric
n1 n
n 1
a ra1 rS a
1 r 1 r
1, 2, 4, 8, … Infinite Geometricr 1 or r -1
No Sum
1 1 13,1, , , ...
3 9 27Infinite Geometric
-1 < r < 11a
S1 r
13.5 Infinite Geometric Series
|r| 1
|r| < 1
Find the sum, if possible: 1 1 1
1 ...2 4 8
1 112 4r
11 22
1 r 1 Yes
1a 1S 2
11 r 12
Find the sum, if possible: 2 2 8 16 2 ...
8 16 2r 2 2
82 2 r 1 No
NO SUM
Find the sum, if possible: 2 1 1 1
...3 3 6 12
1 113 6r
2 1 23 3
1 r 1 Yes
1
2a 43S
11 r 312
Find the sum, if possible: 2 4 8
...7 7 7
4 87 7r 22 47 7
r 1 No
NO SUM
Find the sum, if possible: 5
10 5 ...2
55 12r
10 5 2 1 r 1 Yes
1a 10S 20
11 r 12
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
32
32/5
40
32
32/5
40S 45
504
10
1554
0S 2 5 500 50
504
15
450
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
75
225/4
100
75
225/4
10S 80
100
4 43
1
0
10
3
An old grandfather clock is broken. When the pendulum is swung it follows a swing pattern of 25 cm, 20 cm, 16 cm, and so on until it comes to rest. What is the total distance the pendulum swings before coming to rest?
25
20
16
25
20
16
S
254
15
2 250
Find what values of x does the infinite series converge?
22 4 8x x ...
7 49 343
This infinite geometric series with r = 2x/7. By the theorem of convergence for the infinite geometric series, the series converges when | r | < 1. Or,
21
7
x
7
2x
7 7
2 2x or
Find what values of x does the infinite series converge?
23 91 (x 1) (x 1) ...
2 4
This infinite geometric series with r = 3(x – 1)/2. By the theorem of convergence for the infinite geometric series, the series converges when | r | < 1. Or,
3( 1) 1
2x
21
3x
2 21
3 3x or
1 5
3 3x So
Challenge Question
Sum of non-arithmetic and non-geometric series
Find the formula for the sum of the first n terms of the series, or Sn, then find the sum
1 1 1 13 5 7 (2 1)
3 9 27 3n nS n
The integer part of the each term forms a arithmetic series and the fraction part of each term forms a geometric series:
1 1 1 13 5 7 (2 1)
3 9 27 3n nS n
We apply two formula, one for arithmetic series and one for geometric series to get the overall formula for Sn:
1 11 1
1 1 3 33 2 ( 2)
12 3 213
n n
n
nS n n n
Challenge Question
Textbook P. 503 #30 (Sum of non-arithmetic and non-geometric series.)
Find the formula for the sum of the first n terms of the series, or Sn, then find the sum of the infinite series by using limit.
1 1 1 1
1 3 3 5 5 7 (2 1)(2 1)nSn n
Each term can be decomposed to half of the difference of two fractions – numerators are all 1 and the denominators are the two integers, respectively:
1 1 1 1 1 1 1 1 1
2 1 3 3 5 5 7 2 1 2 1nSn n
Starting from the second group, each first term within each parentheses can be cancel out by the second terms in the preceding group, respectively.
The only terms are not canceled out are 1 and –1/(2n + 1). So,
1 1 1 1 1 1 1 1 1
2 1 3 3 5 5 7 2 1 2 1nSn n
1 1 1 21
2 2 1 2 2 1 2 1n
n nS
n n n
Therefore the limit of the sum is
lim lim2 1n
n n
nS
n
lim
2 1n
nn
nn
1
lim1
2n
n
1
2
Given an arithmetic sequence {an}, how can you make a geometric sequence?
Challenge Question
We make a geometric sequence as . Since{ }nae
1k ka a d
1k ka a de e 1k ka a de e e
1
k
k
ad
a
ee
e
It shows that the new built sequence is a geometric sequence of common ratio r = ed .
Given a geometric sequence {an}, how can you make an arithmetic sequence?
Challenge Question
We make an arithmetic sequence as . Since{ln| |}na
1 1k k ka r a r a
It shows that the new built sequence is an arithmetic sequence of common difference d = ln|r| .
1ln lnk ka r a
1ln ln lnk ka r a
1ln ln lnk ka a r
B
nn A
a
UPPER BOUND(NUMBER)
LOWER BOUND(NUMBER)
SIGMA(SUM OF TERMS) NTH TERM
(SEQUENCE)
Index must be the same variable
Sigma Notation
The relationship between Sn , Sn – 1 and an
1 2 11
n
n n n kk
S a a a a a
1
1 1 2 11
n
n n kk
S a a a a
1n n nS S a
1n n na S S
1
1 1
n n
n k kk k
a a a
Rewrite using sigma notation: 3 + 6 + 9 + 12
Arithmetic, d= 3
n 1a a n 1 d
na 3 n 1 3
na 3n4
1n
3n
Rewrite using sigma notation: 16 + 8 + 4 + 2 + 1
Geometric, r = ½ n 1
n 1a a r n 1
n
1a 16
2
n 1
n
5
1
116
2
Rewrite using sigma notation: 19 + 18 + 16 + 12 + 4
Not Arithmetic, Not Geometric
n 1na 20 2
n 1
n
5
1
20 2
19 + 18 + 16 + 12 + 4 -1 -2 -4 -8
Rewrite the following using sigma notation:3 9 27
...5 10 15
Numerator is geometric, r = 3Denominator is arithmetic d= 5
NUMERATOR: n 1
n3 9 27 ... a 3 3
DENOMINATOR: n n5 10 15 ... a 5 n 1 5 a 5n
SIGMA NOTATION: 1
1
n
n 5n
3 3
j
4
1
j 2
21 2 2 3 2 24 18
7
4a
2a 42 2 5 2 6 72 44
n
n 0
4
0.5 2
00.5 2 10.5 2 20.5 2 30.5 2 40.5 2
33.5
CAUTION
4
j 1
j (1 2 3 4)2 2
CAUTION
4 4 4
7
a a
7
a
7
( a) a a (4 5 6 7) 2 22 2 2 22 44
Some Properties of Sigma Notation
( )B B B
k k k kk A k A k A
a b a b
( 1)B
k A
c B A c
10
3k
m m m m
B B
k kk A k A
ca c a
1.
2.
3.
Ex.
15 15 152 2 2 2 2
5 5 5
( ) (5 6 15) (5 6 15 )k k k
k k k k
Ex.
8 times
(10 3 1) 8 (10 3)m m m
19 193 3 3 3 3
6 6
3 3 3(6 7 19 )k k
k k
Ex.
Some Useful Series
1
n
k
c nc
1
( 1)
2
n
k
n nk
1.
3. 4. 2
1
( 1)(2 1)
6
n
k
n n nk
23
1
( 1)
2
n
k
n nk
5.
1
1
1
nnk
k
aa a
a
2.
0
n
b
36
5
0
36
5
13
65
23
65
...
1aS
1 r
6
153
15
2
j
3
7
2j 1 2 1 2 8 1 2 9 1 ...7 2 123
n 1 n
2n 1S a a 15
2
3
2
747
527
2
j
3
7
2j 1
23 23
j j j7 7 7 1 1
23 23 6
j j
232j 1 2 j 7 1 1 2 j 2 j 17
1 123 23 6 62
( ) ( )2 2 17 24
26 7
23 17 527
1
b
9
4
4b 3
4 3 4 5 3 4 6 3 ...4 4 319
n 1 n
1n 1S a a 19
2
9
2
479
784
1
b
9
4
4b 3
19 19
b b b4 4 4 1 1
19 19 3
b b
194b 3 4 b 4 1 3 4 b 4 b 48
1 119 19 3 33
( ) ( )4 4 48 20
26 4
28 48 784
Assignment
13.1 P. 477 #4 – 36 (x4), 17 – 41(odd), 44 – 4534, 38, 47, 49, 51.
13.3 P. 489 #2 – 12 (even), 11, 15, 18 – 21, 23, 27, 30
13.5 P. 502 #2 – 18 (even), 21 – 35 (odd)
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