Arithmetic
Sequences
1
USING AND WRITING SEQUENCES
The numbers in sequences are called terms.
You can think of a sequence as a
function whose domain is a set of
consecutive integers. If a domain is
not specified, it is understood that
the domain starts with 1.
2
USING AND WRITING SEQUENCES
DOMAIN:
n
an
RANGE:
1
3
2
6
3
9
4
12
5
15
The domain gives
the relative position
of each term.
The range gives the
terms of the sequence.
This is a finite sequence having the rule
an = 3n,
where an represents the nth term of the sequence.
3
Writing Terms of Sequences
Write the first six terms of the sequence an = 2n + 3.
SOLUTION
a 1 = 2(1) + 3 = 5
1st term
a 2 = 2(2) + 3 = 7
2nd term
a 3 = 2(3) + 3 = 9
3rd term
a 4 = 2(4) + 3 = 11
4th term
a 5 = 2(5) + 3 = 13
5th term
a 6 = 2(6) + 3 = 15
6th term
4
Writing Terms of Sequences
Write the first six terms of the sequence f (n) = (–2) n – 1 .
SOLUTION
f (1) = (–2) 1 – 1 = 1
1st term
f (2) = (–2) 2 – 1 = –2
2nd term
f (3) = (–2) 3 – 1 = 4
3rd term
f (4) = (–2) 4 – 1 = – 8
4th term
f (5) = (–2) 5 – 1 = 16
5th term
f (6) = (–2) 6 – 1 = – 32
6th term
5
An introduction…………
1, 4, 7, 10, 13
d=3
9, 1,  7,  15 d = -8
6.2, 6.6, 7, 7.4 d = .4
,   3,   6 d = 3
ARITHMETIC
2, 4, 8, 16, 32
r =2
1

9,  3, 1,  1/ 3 r = 3
1
1, 1/ 4, 1/16, 1/ 64 r = 4
, 2.5, 6.25 r = 2.5
GEOMETRIC
MULTIPLY
ADD
(by the same #)
(by the same #)
To get the next term To get the next term
6
Vocabulary of Sequences (Universal)
a1  First term
an  nth term
an-1 previous term
an+1
next term
n  number of terms
d  common difference
Finite VS. Infinite
7
Arithmetic Sequence: sequence whose
consecutive terms have a common difference.
Example: 3, 5, 7, 9, 11, 13, ...
The terms have a common difference of 2.
(known as d)
To find the common difference you use
an+1 – an
Example: Is the sequence arithmetic? If so, find d.
–45, –30, –15, 0, 15, 30
d = 15
8
Find the next 4 terms of –9, -2, 5, …
2  9  5  2  7
7 is referred to as d
Next four terms……
12, 19, 26, 33
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, -31k
10
4, 10, 16, 22
a1  4
 4  0(6)
a2  4  6
 4  1(6)
a3  4  6  6
 4  2(6)
a4  4  6  6  6  4  3(6)
The nth term of an arithmetic sequence is given by:
an  a1  (n  1)d
The nth term
in the sequence
First
term
The term #
The common
difference
Find the 10th term:
a10  4  (10  1)6  4  54  58
11
Find the 14th term of
the sequence:
4, 7, 10, 13,……
an  a1  (n  1)d
a14  4  (14  1)3
 4  (13)3
 43
12
In the arithmetic sequence
4,7,10,13,…, which term has a
value of 301?
an  a1  (n  1)d
301  4  (n  1)3
301  4  3n  3
301  1  3n
100

n
300  3n
13
Given an arithmetic sequence with
a15  38 and d  3, find a1.
an  a1  n  1 d
38  x  15  1 3 
X = 80
14
Find d if a1  6 and a29  20
20  6   29  1 x
26  28x
13
x
14
15
Example: If the common difference is 4
and the fifth term is 15, what is the 10th
term of an arithmetic sequence?
an = a1 + (n – 1)d
d = 4, a5 = 15, n = 5,
15 = a1 + (5 – 1)4
15 = a1 +16
a1 = –1
a1=?
a10 = –1 + (10 – 1)4
= -1 + 36
a10 = 35
16
Explicit vs. Recursive Formulas
Explicit Formula – used to find the nth
term of the arithmetic sequence in which
the common difference and 1st term are
known.
Ex: 4, 6, 8, 10…
Use a1 and d in sequence formula:
an = 4 + (n – 1)2
an = 2n + 2
17
Find the explicit formula for the
following arithmetic sequence:
3, 8, 13, 18…
an = a1 + (n – 1)d
a1 = 3 d = 5 n = ?
an = 3 + (n – 1)5
an = 3 + 5n – 5
an = -2 + 5n OR an = 5n – 2
18
Explicit vs. Recursive Formulas
Recursive Formula – (includes a1) used
to find the next term of the sequence by
adding the common difference to the
previous term.
an = an-1 + d
a1 = ___
an+1 = an + d
a1 = ___
an = an-1 + 2
Ex: 4, 6, 8, 10…
a1 = 4
19
Find the recursive formula for the
following arithmetic sequence:
3, 8, 13, 18…
an = an-1 + d
a1 = 3 d = 5
an = an-1 + 5
a1 = 3
20
Using Recursive & Explicit Formulas
an = an-1 + 6
a1 = 4
a2 = 4 + 6 = 10
a3 = 10 + 6 = 16
a4 = 16 + 6 = 22
a5 = 22 + 6 = 28
1. Create the 1st 5 terms:
4, 10, 16, 22, 28
2. Find the explicit formula:
an = a1 + (n – 1)d
an = 4 + (n – 1)6
an = 4 + 6n – 6
an = 6n – 2
21
Using Recursive & Explicit Formulas
an = 7 – 2n
1. Create the
a1 = 7 – 2(1) = 5
a2 = 7 – 2(2) = 3
a3 = 7 – 2(3) = 1
a4 = 7 – 2(4) = –1
a5 = 7 – 2(5) = –3
st
1
5 terms:
5, 3, 1, –1, –3
2. Find the
recursive formula:
an = an-1 – 2
a1 = 5
22
An arithmetic mean of two numbers, a and
b, is simply their average. Use the formula
and information given to find the common
difference to create the sequence.
Insert 3 arithmetic
means between 8 & 16.
8, 10 , 12 , 14 ,16
Let 8 be the 1st term
Let 16 be the 5th term
Let 5 be N
d is missing
an  a1  (n  1)d
16  8  (5  1)d
d 2
23
Find two arithmetic means
between –4 and 5 -4, ____, ____, 5
an  a1  n  1 d
5  4   4  1 x 
x3
The two arithmetic means are –1 and 2,
24
since –4, -1, 2, 5 forms an arithmetic sequence
Find 3 arithmetic means between 1 & 4
1, ____, ____, ____, 4
an  a1  n  1 d
4  1   5  1 x 
3
x
4
The 3 arithmetic means are
since
7 10 13
1, , ,
4 4 4
7 10 13
,
,
4 4 4
,4 forms a sequence
25
Geometric
Sequences
26
Vocabulary of Sequences (Universal)
a1  First term
an  nth term
an-1 previous term
an+1
next term
n  number of terms
d  common difference
Finite VS. Infinite
27
Find the next 3 terms of 2, 3, 9/2, __, __, __
3 – 2 vs. 9/2 – 3… not arithmetic
a
• Use n 1 to determine common ratio
an
3 9/2
3

 1.5  geometric  r 
2
3
2
28
How is the
formula derived?
1st term: 2
The nth term of a
geometric sequence is given by:
a ar
n1
3
2nd term : 2   3
n
1
2
3 3 9
3rd term : 2   
2 2 2
4th term:
5th term:
6th term:
 3 3 3
3 3 3 3  3 3 3 3 3

2



2




 2 2  2
 2 2 2  2 2  2  2  2  2   2


29
1
2
If a1  , r  , find a9 .
2
3
9 1
n1
 1  2 
x

 2  3 
n
1
  
7
8
2
2
128

x

8
8
3
23
6561
a ar
30
2
Find a2  a 4 if a1  3 and r 
3
4
8


2 -3, ____,
3
9

2
____,
____
Since r  ...
3
 8  10
a 2  a 4  2  


9
 9 
31
Find a9 of 2, 2, 2 2,...
an  a1r
a=
2
r=
2
1
n=9
n1
 2
2  2
9 1
x 2
x
8
x  16 2
32
If a5  32 2 and r   2, find a 2
____, ____, ____,____,32 2
n1
an  a1r
32 2  x   2 

5 1
32 2  x  2
32 2  4x
8
2 x

4
 
a2  8 2  2

8 2  2

1
 8 4
 16
21
33
Explicit vs. Recursive Formulas
Explicit Formula – used to find the nth term of
the geometric sequence in which the common
ratio and 1st term are known.
Ex: 4, 12, 36, 108…
Use a1 and r in sequence formula:
n-1
Ex: an = a1*r
an = 4 * 3
n-1
34
Find the explicit formula for the
following geometric sequence:
3, 6, 12, 24…
an = a1*r
n-1
a1 = 3 r =2
an = 3 *2
n-1
35
Explicit vs. Recursive Formulas
Recursive Formula (includes a1) – used to find
the next term of the sequence by multiplying
the common ratio to the previous term.
an = an-1 (r)
a1 = ___
a1 (r) = a2
a2 (r) = a3
a3 (r) = a4
an+1 = r(an)
a1 = ___
Ex: –1, 4, –16, 64 …
an = an-1 (–4)
a1 = –1
36
Find the recursive formula for the
following geometric sequence:
3, 6, 12, 24…
an = an-1 * r
a1 = 3 r = 2
an = an-1 * 2
a1 = 3
37
Using Recursive & Explicit Formulas
an = an-1 (3)
a1 = –1
a2 = –1(3) = –3
a3 = –3(3) = –9
a4 = –9(3) = –27
a5 = –27(3) = –81
1. Create the 1st 5 terms:
–1, –3, –9, –27, – 81
2. Find the explicit formula:
n-1
an = a1 (r)
n-1
an = –1(3)
an = –3
n-1
38
Using Recursive & Explicit Formulas
an = 2(4)
n–1
1. Create the
a1 = 2(4)1-1 = 2
a2 = 2(4)2-1 = 8
3-1
a3 = 2(4) = 32
4-1
a4 = 2(4) = 128
a5 = 2(4)5-1 = 512
st
1
5 terms:
2, 8, 32, 128, 512
2. Find the
recursive formula:
an = 4an-1
a1 = 2
39
A geometric mean(s) of numbers are the
terms between any 2 nonsuccessive terms of
a geometric sequence. Use the terms given
to find the common ratio and find the
missing terms called the geometric means.
Ex: Find two
geometric means
between –2 and 54
an  a1r
n1
54   2  x 
6 –18
-2, ____,
____, 54
The 2 geometric means
are 6 and -18
27  x
3  x
3
40
41
*** Insert one geometric mean between ¼ and 4***
*** denotes trick question
1
,____,4
4
an  a1r
n1
1 31
1 2
4  r  4  r  16  r 2  4
4
4
r
1
, 1, 4
4
1
,  1, 4
4
41
Series
42
Vocabulary of Sequences (Universal)
a1  First term
an  nth term
an-1 previous term
an+1
next term
n  number of terms
d  common difference
Finite VS. Infinite
43
USING SERIES
When the terms of a sequence are added,
the resulting expression is a series. A
series can be finite or infinite.
FINITE SEQUENCE
INFINITE SEQUENCE
3, 6, 9, 12, 15
3, 6, 9, 12, 15, . . .
FINITE SERIES
INFINITE SERIES
3 + 6 + 9 + 12 + 15
...
3 + 6 + 9 + 12 + 15 + . . .
You can use summation notation to write a series. For example, for the
finite series shown above, you can write
5
3 + 6 + 9 + 12 + 15 = ∑ 3i
i=1
44
UPPER BOUND
TERM NUMBER
B
a
 n
SIGMA
(SUM OF TERMS)
n A
# of Terms:
B–A+1
NTH TERM
SEQUENCE
(EXPLICIT FORMULA)
LOWER BOUND
TERM NUMBER
45
  j  2  18
4
j1
 1  2   2  2    3  2   4  2 
7
  2a 
a4
 44
  2  4     2 5    2  6    2  7 
46
An arithmetic series is
a series associated
with an arithmetic
sequence.
It can be
infinite or finite.
47
1, 4, 7, 10, 13, ….
3, 7, 11, …, 51
Infinite Arithmetic
(constantly getting
larger or smaller)
Finite Arithmetic
No Sum
n
Sn   a1  an 
2
1, 2, 4, …, 64
1, 2, 4, 8, …
1 1 1
3,1, , , ...
3 9 27
48
Find the sum of the 1st
100 natural numbers. 1 + 2 + 3 + 4 + … + 100
a1  1
an  100
n  100
n
S n   a1  an 
2
S100
100

(1  100)
2
 5050
49
Find the sum of the 1st
14 terms of the series: 2 + 5 + 8 + 11 + 14 + 17 +…
n
S n  a1  an 
2
To find a14 , you need
d 3
a1  2 n  14 a14  2  (14  1)d
a14 = 2 + (14 - 1)(3) = 41
14
S14  2  a14 
2
S14 =
14
2  41  301
2
50
13 Find the sum of the series
 (4n  5)  9 13 17  ....
n 1
n
Need 13th term:
S n   a1  an 
2
4(13) + 5 = 57
a1  9 d  4 n  13
13
13
S13  9  57 
S13  9  a13 
2
2
13

(66)  429
2
51
Finding the Sum from Summation Notation
n
S n   a1  an 
2
4
  j  2 n = 4 a1 = 3 a4 = 6
j1
3, 4, 5, 6
4
S 4  3  6  2(9)  18
2
7
  2a 
a4
8, 10, 12, 14
n = (7 – 4) + 1 a4 = 8
a7 = 14
4
S 4  8  14  2(22)  44
2
52
19
  4b  3 
b 4
23
a4 =19
19, 23, 27, 31…79
a19 = 79
n = (19 - 4) + 1 = 16
16
S16  (19  79)  8(98)  784
2
  2x  1
15, 17, 19, …47
x 7
a7 =15
a23 = 47
n = (23-7) + 1 = 17
17
S17  (15  47)  8.5(62)  527
2
53
An geometric series is a
series associated with a
geometric sequence.
They can be
infinite or finite.
Finite and infinite have
different formulas
depending on the
value of r.
54
1, 4, 7, 10, 13, ….
Infinite Arithmetic
(constantly getting
larger or smaller)
3, 7, 11, …, 51
Finite Arithmetic
No Sum
n
Sn   a1  an 
2
a1 (1  r )
Sn 
1 r
n
1, 2, 4, …, 64
1, 2, 4, 8, …
Finite Geometric
Infinite Geometric
r < -1 OR r > 1
(constantly getting
larger or smaller)
“diverges”
Infinite Geometric
1 1 1
3, 1, , , ...
-1 < r < 1
3 9 27
“converges”
No Sum
a1
S
1 r
55
1 1 1
Find S7 of    ...
2 4 8
Finite : Find sum of 1st 7 terms
1 1
1
1
8
4
n
r  
a1 (1  r ) a1 
Sn 
1 1 2
2 r ?
1 r
n7
2 4
1
1 7
(1  ( ) ) 127
2
2
S7 

1
128
1
2
56
Sums of Infinite Series
Made Finite
(referred to as
partial sums)
Infinite Series
Finding the Sum
of Infinite Sequences
“Converges” vs. “Diverges”
57
58
Find the sum, if possible:
Geometric
~need to find r~
1 1
1
2
4
r  
1 1 2
2
S
1
1
1
2
1 1 1
1     ...
2 4 8
Is -1 < r < 1?
Yes
(Infinite Series - converges)
a1
S

1
1 r

2
1
2
59
Find the sum, if possible:
2 2  8  16 2  ...
8
16 2
r

2 2
8
2 2
Is -1 < r < 1?
No
(Infinite series - Diverges)
NO SUM
60
Find the sum, if possible:
4
2
2
9
3
r


2
1
3
3
S
2 4 8
1  
 ...
3 9 27
Is -1 < r < 1?
Yes
(Infinite Series – Converges)
1
1

3
2
3
a1
S
1 r
61
Find the sum, if possible:
2 4 8
   ...
7 7 7
4 8
7
7
r  2
Is -1 < r < 1?
No
2 4
(Infinite Series–Diverges)
7 7
NO SUM
62
Find the sum, if possible:
5
10  5   ...
2
5
5
1
2
r
 
10 5 2
10
S
1
1
2
-1 < r < 1
Yes
(Infinite Series–Converges)
 20
a1
S
1 r
63
Finding the Sum from Sigma Notation
n

3
3   ...
3
3

6    6   6   6 

5
5
5
b 0  5 
23
3
r
5
so “converges”
0
1
2
a1  6  15
S
3
1 r 1
5
  2x  1 27  1  28  1  29  1  ...  223  1
x 7
15, 17, 19, ...47
n
23  7  1
Sn   a1  an  
15  47  527
2
2
64
Rewrite using sigma notation: 3 + 6 + 9 + 12
1st term
4th term
Arithmetic, d= 3
an  a1  n  1 d
an  3  n  1 3
an  3n
Explicit
formula
4
 3n
n=1
65
Rewrite using sigma notation: 16 + 8 + 4 + 2 + 1
1st term
5th term
Geometric, r = ½
an  a1r
n1
 1
an  16  
2
Explicit
formula
n1
5

1
16 
2
n=1
66
n 1
Rewrite the following using sigma notation:
3 9 27


 ... Numerator is geometric, r = 3
5 10 15
Denominator is arithmetic d= 5
NUMERATOR: 3  9  27  ...  an  3  3 
n1
DENOMINATOR:
5  10  15  ...  an  5  n  1 5  an  5n
n1

SIGMA NOTATION:

n1
3 3 
5n
67