Download n – 1

Arithmetic Sequences
How do I define an arithmetic
sequence and how do I use the
formula to find different terms of
the sequence?
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.
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.
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
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
Arithmetic Sequences and Series
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.
The common difference is the number d.
To find the common difference you use an+1 – an
Example: Is the sequence arithmetic?
–45, –30, –15, 0, 15, 30
Yes, the common difference is 15
How do you find any term in this sequence?
To find any term in an arithmetic sequence, use the
formula
an = a1 + (n – 1)d
where d is the common difference.
The first term of an arithmetic sequence
is a1 . We add d to get the next term.
There is a pattern, therefore there is a
formula we can use to give use any term
that we need without listing the whole
sequence .
The nth term of an arithmetic sequence is given by:
an  a1  (n  1)d
The last # in the
sequence/or the #
you are looking for
First
term
The position
the term is in
The common
difference
Find the 14th term of the
arithmetic sequence
4, 7, 10, 13,……
an  a1  (n  1)d
a14  4  (14  1)3
 4  (13)3
 4  39
 43
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
300  3n
100  n
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
sum of n terms of arithmetic sequence  Sn 
n
 a1  an 
2
Given an arithmetic sequence with a15  38 and d  3, find a1.
x
a1  First term
38 an  nth term
15
n  number of terms
NA Sn  sum of n terms
-3
d  common difference
an  a1  n  1 d
38  x  15  1 3 
X = 80
Find d if a1  6 and a29  20
-6
a1  First term
20 an  nth term
29
n  number of terms
NA Sn  sum of n terms
x
d  common difference
an  a1  n  1 d
20  6   29  1 x
26  28x
13
x
14
Find n if an  633, a1  9, and d  24
9
a1  First term
633 an  nth term
x
n  number of terms
NA Sn  sum of n terms
24
d  common difference
an  a1  n  1 d
633  9   x  1 24
633  9  24x  24
X = 27
Try this one: Find a16 if a1  1.5 and d  0.5
1.5 a1  First term
x
16
an  nth term
n  number of terms
NA Sn  sum of n terms
0.5
d  common difference
an  a1  n  1 d
a16  1.5  16  1 0.5
a16  9
Example: Find a formula for the nth term of the
arithmetic sequence in which the common difference
is 5 and the first term is 3.
an = a1 + (n – 1)d
a1 = 3 d = 5
an = 3 + (n – 1)5
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
We need to determine what the first term is...
d=4
and a5 = 15
a5 = a1 + (5 – 1)4 = 15
a1 = –1
a10 = –1 + (10 – 1)4
a10 = 35
An arithmetic mean of two numbers, a and
b, is simply their average. Using the
arithmetic mean we can also form a
sequence.
Insert three arithmetic means
between 8 and 16.
Let 8 be the 1st term
Let 16 be the 5th term
an  a1  (n  1)d
16  8  (5  1)d
d 2
8, 10 , 12 , 14 ,16
Find two arithmetic means between –4 and 5
-4, ____, ____, 5
-4
a1  First term
5
an  nth term
n  number of terms
4
NA
x
Sn  sum of n terms
d  common difference
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
Find three arithmetic means between 1 and 4
1, ____, ____, ____, 4
1
a1  First term
4
an  nth term
5
NA
x
n  number of terms
Sn  sum of n terms
d  common difference
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
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
UPPER BOUND
(NUMBER)
B
SIGMA
(SUM OF TERMS)
a
n
n A
NTH TERM
(SEQUENCE)
LOWER BOUND
(NUMBER)
An arithmetic series is a series
associated with an arithmetic
sequence.
The sum of the first n terms:
n
Sn   a1  an 
2
n
Sn  (2a1  (n  1)d )
2
Find the sum of the first 100
natural numbers.
1 + 2 + 3 + 4 + … + 100
a1  1
an  100
n  100
n
Sn   a1  an 
2
S100
100

(1  100)
2
 5050
Find the sum of the first 14
terms of the arithmetic series
2 + 5 + 8 + 11 + 14 + 17 +…
a1  2
d  3 n  14
n
Sn  (2a1  (n  1)d )
2
14
S14  (2(2)  (14  1)3)
2
S14  7(3  13(3))
 7(43)
 301
4
  j  2   1  2   2  2  3  2   4  2  18
j1
7
  2a    2  4   2 5    2  6    2 7  44
a4

4
n 0
 
 
 
 
 
4
3
2

0.5

2

0.5

2

0.5

2
0.5  2  0.5  2  0.5  2
n
 33.5
0
1

Find the sum of the series
13
 (4n  5)  9  13  17  ....
n 1
a1  9
d  4 n  13
n
Sn  (2a1  (n  1)d )
2
13
S13  (2(9)  (13  1)4)
2
13

(66)  429
2
  4b  3    4  4  3   4 5  3   4  6  3  ...   4 19  3
19
b 4
Sn 
n
19  4  1
a

a

 1 n
19  79   784
2
2
  2x  1   2 7  1   2 8  1   2 9  1  ...   2  23  1
23
x 7
n
23  7  1
Sn   a1  an  
15  47   527
2
2