College Algebra & Trigonometry
4th EDITION
and
Precalculus
10TH EDITION
7.2 - 1
11.2
Arithmetic Sequences and Series
Arithmetic Sequences
Arithmetic Series
7.2 - 2
Arithmetic Sequences
A sequence in which each term after the first
is obtained by adding a fixed number to the
previous term is an arithmetic sequence (or
arithmetic progression). The fixed number
that is added is the common difference. The
sequence
5,9,13,17,21,
is an arithmetic sequence since each term
after the first is obtained by adding 4 to the
previous term.
7.2 - 3
Arithmetic Sequences
That is,
954
13  9  4
17  13  4
21  17  4,
and so on. The common difference is 4.
7.2 - 4
Arithmetic Sequences
If the common difference of an arithmetic
sequence is d, then by the definition of an
arithmetic sequence,
d  an 1  an ,
Common difference d
for every positive integer n in the domain of
the sequence.
7.2 - 5
Example 1
FINDING THE COMMON
DIFFERENCE
Find the common difference, d, for the
arithmetic sequence
9, 7, 5, 3, 1,
Solution
We find d by choosing any two adjacent terms
and subtracting the first from the second.
Choosing – 7 and – 5 gives
Be careful when
d  5  ( 7)  2.
subtracting a
negative number.
Choosing – 9 and – 7 would give the same
result.
d  7  ( 9)  2.
7.2 - 6
Example 2
FINDING TERMS GIVEN a1 AND d
Find the first five terms for each arithmetic
sequence.
a. The first term is 7, and the common
difference is – 3.
Solution
a1  7
Start with a1 = 7.
a2  7  ( 3)  4
Add d = – 3.
a3  4  ( 3)  1
Add – 3.
a4  1  ( 3)  2
Add – 3.
a5  2  ( 3)  5
Add – 3.
7.2 - 7
Example 2
FINDING TERMS GIVEN a1 AND d
Find the first five terms for each arithmetic
sequence.
b. a1 = – 12, d = 5
Solution
a1  12
a2  12  5  7
Start with a1.
Add d = 5.
a3  7  5  2
a4  2  5  3
a5  3  5  8
7.2 - 8
Look At It This Way
If a1 is the first term of an arithmetic sequence
and d is the common difference, then the terms
of the sequence are given by
a1  a1
a2  a1  d
a3  a2  d  a1  d  d  a1  2d
a4  a3  d  a1  2d  d  a1  3d
a5  a1  4d
a6  a1  5d,
and, by this pattern,
an  a1  (n  1)d.
7.2 - 9
nth Term of an Arithmetic
Sequence
In an arithmetic sequence with first
term a1 and common difference d,
the nth term, is given by
an  a1  (n  1)d.
7.2 - 10
Example 3
FINDING TERMS OF AN
ARITHMETIC SEQUENCE
Find a13 and an for the arithmetic sequence
– 3, 1, 5, 9, …
Solution
Here a1 = – 3
and
d = 1 – (– 3) = 4.
To find a13
substitute 13 for
n in the formula
for the nth term.
a1  a1  (n  1)d
Work inside
parentheses
first.
a13  a1  (13  1)d
n = 13
a13  3  (12)4
Let a1 = – 3,
d = 4.
a13  3  48
Simplify.
a13  45
7.2 - 11
Example 3
FINDING TERMS OF AN
ARITHMETIC SEQUENCE
Find a13 and an for the arithmetic sequence
– 3, 1, 5, 9, …
Solution
Find an by
substituting
values for a1
and d in the
formula for an.
an  3  (n  1) 4
Let a1 = – 3,
d = 4.
an  3  4n  4
Distributive
property
an  4n  7
Simplify.
7.2 - 12
Example 4
FINDING TERMS OF AN
ARTHEMTIC SEQUENCE
Find a18 and an for the arithmetic sequence having
a2 = 9 and a3 = 15.
Solution
Find d first; d = a3 – a2 = 15 – 9 = 6.
Since
a2  a1  d ,
9  a1  6
Let a2 = 9, d = 6.
a1  3.
7.2 - 13
Example 4
FINDING TERMS OF
ARTHEMTIC SEQUENCE
Find a18 and an for the arithmetic sequence having
a2 = 9 and a3 = 15.
Solution
d = a3 – a2 = 15 – 9 = 6.
Then,
a18  3  (18  1)6
a18  105,
Formula for an;
a1 = 3, n = 18,
d = 6.
7.2 - 14
Example 4
FINDING TERMS OF
ARTHEMTIC SEQUENCE
Find a18 and an for the arithmetic sequence having
a2 = 9 and a3 = 15.
Solution
d = a3 – a2 = 15 – 9 = 6.
and
an  3  (n  1)6
an  3  6n  6
Distributive
property.
an  6n  3.
7.2 - 15
Example 5
FINDING THE FIRST TERM OF AN
ARITHMETIC SEQUENCE
Suppose that an arithmetic sequence has a8 = – 16
and a16 = – 40. Find a1
Solution
Since a16 = a8 + 8d, it follows that
8d  a16  a8  40  ( 16)  24,
And so d = – 3. To find a1, use the equation
a8 = a1 + 7d.
16  a1  7d
Let a8 = – 16.
16  a1  7( 3)
Let d = – 3.
a1  5
7.2 - 16
FINDING THE FIRST TERM OF AN
ARITHMETIC SEQUENCE
The graph of any sequence is a scatter diagram.
To determine the characteristics of the graph of
an arithmetic sequence, start by rewriting the
formula for the nth term.
an  a1  (n  1)d
Formula for the nth term.
 a1  nd  d
Distributive property.
 dn  (a1  d )
Commutative & associative
properties.
 dn  c
Let c = a1 – d.
7.2 - 17
FINDING THE FIRST TERM OF AN
ARITHMETIC SEQUENCE
The points in the graph of an arithmetic sequence
are determined by (n) = dn + c, where n is a
natural number. Thus, the points in the graph of 
must lie on the line
y  dx  c.
Slope
y-intercept
7.2 - 18
FINDING THE FIRST TERM OF AN
ARITHMETIC SEQUENCE
For example, the
sequence shown here is
an arithmetic sequence
because the points that
comprise its graph are
collinear (lie on a line).
The slope determined
by these points is 2, so
the common difference
d equals 2.
7.2 - 19
FINDING THE FIRST TERM OF AN
ARITHMETIC SEQUENCE
On the other hand,
the sequence bn
shown here is not an
arithmetic sequence
because the points
are not collinear.
7.2 - 20
Example 6
FINDING THE nth TERM FROM A
GRAPH
Find a formula for the nth term of the sequence an
shown here. What are the domain and range of
this sequence?
Solution
The points in this graph
lie on a line, so the
sequence is arithmetic.
The equation of the
dashed line shown here
is y = – .5x + 4, so the
nth term of this sequence
is determined by
an  .5n  4.
7.2 - 21
Example 6
FINDING THE nth TERM FROM A
GRAPH
Find a formula for the nth term of the sequence an
shown here. What are the domain and range of
this sequence?
Solution
The sequence is
comprised of the points
(1,3.5),(2,3),(3,2.5),
(4,2),(5,1.5),(6,1).
7.2 - 22
Example 6
FINDING THE nth TERM FROM A
GRAPH
Find a formula for the nth term of the sequence an
shown here. What are the domain and range of
this sequence?
Solution
Thus, the domain of the
sequence is given by
{1, 2, 3, 4, 5, 6}, and
the range is given by
{3.5, 3, 2.5, 2, 1.5, 1}.
7.2 - 23
Arithmetic Series
The sum of the terms of an arithmetic
sequence is an arithmetic series. To
illustrate, suppose that a person borrows
$3000 and agrees to pay $100 per month
plus interest of 1% per month on the unpaid
balance until the loan is paid off. The first
month, $100 is paid to reduce the loan, plus
interest of (.01)3000 = 30 dollars. The second
month, another $100 is paid toward the loan,
and (.01)2900 dollars is paid for interest
7.2 - 24
Arithmetic Series
Since the loan is reduced by $100 each
month, interest payments decrease by
(.01)100 = 1 dollar each month, forming the
arithmetic sequence
30, 29, 28, …, 3, 2, 1.
7.2 - 25
Arithmetic Series
The total amount of interest paid is given by
the sum of the terms of this sequence.
Now we develop a formula to find this sum
without adding all 30 numbers
directly. Since the sequence is arithmetic, we
can write the sum of the first n terms as
Sn  a1  a1  d   a1  2d  
 a1  (n  1)d .
7.2 - 26
Arithmetic Series
We used the formula for the general term in
the last expression. Now we write the same
sum in reverse order, beginning with an and
subtracting d.
Sn  an  an  d   an  2d  
 an  (n  1)d 
7.2 - 27
Arithmetic Series
Adding respective sides of these two equations
term by term, we obtain
Sn  Sn  (a1  an )  (a1  an )
or
 (a1  an ),
2Sn  n(a1  an ),
since there are n terms of a1 + an on the right.
Now solve for Sn to get
n
Sn  (a1  an ).
2
7.2 - 28
Arithmetic Series
Using the formula an = a1 + (n – 1)d, we can
also write this result for Sn as
or
n
Sn  [a1  a1  (n  1)d ],
2
n
Sn  [2a1  (n  1)d ],
2
which is an alternative formula for the sum of
the first n terms of an arithmetic sequence.
7.2 - 29
Sum of the First n Terms of
an Arithmetic Sequence
If an arithmetic sequence has first term and
common difference d, then the sum of the
first n terms is given by
n
n
Sn  (a1  an ) or Sn  [2a1  (n  1)d ].
2
2
7.2 - 30
The first formula is used when the first and last
terms are known; otherwise the second formula
is used.
For example, in the sequence of interest payments
discussed earlier, n = 30, a1 = 30, and an = 1.
Choosing the first formula,
n
Sn  (a1  an ),
2
30
gives S30 
(30  1)  15(31)  465,
2
so a total of $465 interest will be paid over the 30
months.
7.2 - 31
Example 7
USING THE SUM FORMULAS
a. Evaluate S12 for the arithmetic sequence – 9,
– 5, – 1, 3, 7, ….
Solution
We want the sum of the first 12 terms. Using
a1 = – 9, n = 12, and d = 4 in the second formula,
n
Sn  [2a1  (n  1)d ],
2
12
gives S12 
[2( 9)  11(4)]  156.
2
7.2 - 32
Example 7
USING THE SUM FORMULAS
b. Use a formula for Sn to evaluate the sum of
the first 60 positive integers.
Solution
The first 60 positive integers form the arithmetic
sequence 1, 2, 3, 4, …, 60. Thus, n = 60, a1 = 1,
and a60 = 60, so we use the first formula in the
preceding box to find the sum.
n
Sn  (a1  an )
2
60
S60 
(1  60)  1830
2
7.2 - 33
Example 8
USING THE SUM FORMULA
The sum of the first 17 terms of an arithmetic
sequence is 187. If a17 = – 13, find a1 and d.
Solution
17
S17 
(a1  a17 )
2
17
187 
(a1  13)
2
22  a1  13
a1  35
Use the first formula for Sn,
with n = 17.
Let S17 = 187, a17 = – 13.
2
.
Multiply by
17
Add 13; rewrite.
7.2 - 34
Example 8
USING THE SUM FORMULA
The sum of the first 17 terms of an arithmetic
sequence is 187. If a17 = – 13, find a1 and d.
Solution
Since a17 = a1 + (17 – 1)d,
13  35  16d
Let a17 = – 13, a1 = 35.
 48  16d
Subtract 35.
d  3.
Divide by 16; rewrite.
7.2 - 35
n
Any sum of the form
 (d i  c ), where d and c
i 1
are real numbers, represents the sum of the
terms of an arithmetic sequence having first
term a1 = d(1) + c = d + c and common
difference d. These sums can be evaluated
using the formulas in this section.
7.2 - 36
Example 9
USING SUMMATION NOTATION
Evaluate each sum.
10
a.
 (4i  8)
i 1
Solution
This sum contains the first 10 terms of the
arithmetic sequence having
a1  4 1  8  12,
and a10  4 10  8  48.
First term.
Last term.
10
10
Thus,  (4i  8)  S10 
(12  48)  5(60)  300.
2
i 1
7.2 - 37
Example 9
USING SUMMATION NOTATION
Evaluate each sum.
9
b.  (4  3k )
k 3
Solution The first few terms are
[4  3(3)]  [4  3(4)]  [4  3(5)] 
 5  ( 8)  ( 11) 
Thus, a1 = – 5 and d = – 3. If the sequence started
with k = 1, there would be nine terms. Since it starts at
3, two of those terms are missing, so there are seven
terms and n = 7. Use the second formula for Sn.
9
7
(4  3k )  [2( 5)  6( 3)]  98

2
k 3
7.2 - 38