249
Section 12.2 Arithmetic Sequences
Objective #1: Determining if a sequence is arithmetic.
If you start with a number a and add a fixed constant d over and over
again, you would get what is called an arithmetic sequence.
Definition
Let a and d be real numbers. An Arithmetic Sequence may be defined
recursively as
1)
a1 = a and an = an – 1 + d.
The number d is called the common difference.
Determine if the following sequences are arithmetic:
Ex. 1a
9, 17, 25, 33, ….
Ex. 1b
1, 2, 3, 5, 8, …
Solution:
a)
Since the difference of each successive terms is 8 and the first
term is 9, the sequence is arithmetic. We can define it as follows:
a1 = 9
an = a n – 1 + 8
b)
Since the difference between the first two terms is 1, but the
difference between 5th and 4th terms is 3, the sequence is not
arithmetic.
Ex. 2a
{sn} = {5 – 2n}
Ex. 2b
{sn} = {ln(n2)}
Solution:
a)
s1 = 5 – 2(1) = 3
The difference between the nth term and the (n – 1)th term is:
sn – sn – 1 = 5 – 2n – [5 – 2(n – 1)] = 5 – 2n – 5 + 2n – 2 = – 2
Thus, the common difference is – 2, so the sequence is
arithmetic. We can define it as follows:
s1 = 3
sn = s n – 1 – 2
b)
Since, ln(n2) = 2ln(n), then sn= {2ln(n)}
s1 = 2ln(1) = 0
The difference between the nth term and the (n – 1)th term is:
sn – sn – 1 = 2ln(n) – 2ln(n – 1) = 2(ln(n) – ln(n – 1)) = 2ln(
which is not constant, so the sequence is not arithmetic.
€
n
)
n−1
250
Objective #2
Deriving a formula for arithmetic sequences.
To derive the formula for arithmetic sequence, let’s consider the first few
terms of the sequence.
a1 = a
a2 = a 1 + d
a3 = a2 + d = (a1 + d) + d = a1 + 2d = a1 + (3 – 1)d
a4 = a3 + d = (a1 + 2d) + d = a1 + 3d = a1 + (4 – 1)d
a5 = a4 + d = (a1 + 3d) + d = a1 + 4d = a1 + (5 – 1)d
⋮
an = an – 1 + d = …
⋮
= a1 + (n – 1)d
Theorem
Let a1 and d be real numbers. The nth term of an arithmetic sequence is
given by the formula: 2) an = a1 + (n – 1)d.
Find the indicated term in each arithmetic sequence:
Ex. 3
75th term of 9, 17, 25, 33, ….
Solution:
The first term is a1 = 9 and d = 8. By formula #2, the nth term is
an = 9 + (n – 1)(8). Thus, the 75th term is
a75 = 9 + (75 – 1)(8) = 9 + 592 = 601
Find the following:
Ex. 4
The 8th term is – 5 and the 25th term is 37.5.
a)
Find the first term and the common difference.
b)
Give a recursive formula for the sequence.
c)
What is the nth term of the sequence?
Solution:
a)
From formula #2, an = a1 + (n – 1)d. Since a8 = – 5 and
a25 = 37.5, then
1)
a8 = a1 + 7d = – 5
2)
a25 = a1 + 24d = 37.5
Subtracting the second equation from the first yields:
– 17d = – 42.5
(divide by – 17)
d = 2.5
Now, replace d = 2.5 in equation #1 to solve for a1:
a1 + 7(2.5) = – 5 ⇒
a1 + 17.5 = – 5 ⇒
a1 = – 22.5
Thus, the first term is – 22.5 and the common difference is 2.5.
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b)
Plugging in a1 = – 22.5 and d = 2.5 into formula #1 yields:
a1 = – 22.5
an = an – 1 + 2.5
c)
Plugging in a1 = – 22.5 and d = 2.5 into formula #2 yields:
an = – 22.5 + (n – 1)(2.5) = – 22.5 + 2.5n – 2.5 = 2.5n – 25.
Thus, an = 2.5n – 25
Objective #3
Find the sum of an arithmetic sequence.
Theorem: Sum of the First n Terms of an Arithmetic Sequence
Let {an} be an arithmetic sequence with the first term a1 and common
difference d. The sum Sn of the first n terms of {an} is given by:
3)
Sn =
4)
Sn =
n
2
n
2
[2a1 + (n – 1)d]
or
(a1 + an)
If you know the first term and the common difference, use formula #3. If
€ the first and last term of the sequence, use formula #4.
you know
Proof:
By€definition,
Sn = a1 + a2 + a3 + … + an (use formula #2 for each term)
= a1 + (a1 + d) + (a1 + 2d) + … + (a1 + (n – 1)d)
(regroup terms)
= a1 + a1 + a1 + … + a1 + d + 2d + … + (n – 1)d
(n terms of a1)
= na1 + d + 2d + … + (n – 1)d
(factor out d)
= na1 + d(1 + 2 + … + (n – 1))
(rewrite as a summation)
n−1
= na1 + d
( ∑k )
k=1
(n−1)(n−1+1)
= na1 + d
2
d(n−1)n
= na1 +
2
€
2na1
d(n−1)n
=
+
2
€ 2
n
= [2a1 + d(n – 1)]
€ 2
n
= [2a1 + (n – 1)d]
€2
n
= [a1 + a1 + (n – 1)d]
2
n
= (a1 + an)
2
(
€
€
€
€
€
(apply formula #6 from section 12.1)
)
(simplify)
(rewrite na1 =
(factor out
n
2
2na1
2
)
)
(reorder d(n – 1) to get formula #3)
€
(rewrite 2a1 as a1 + a1)
€
(replace a1 + (n – 1)d by an)
(formula #4)
252
Find the sum of the first n terms of the following arithmetic sequence:
Ex. 5
{an} = {9 – 2n}
Solution:
The first term is a1 = 9 – 2(1) = 7 and the last term is an = 9 – 2n.
Using formula #4, we get:
Sn =
Ex. 6
€
n
2
(a1 + an) =
n
2
(7 + 9 – 2n) =
n
(16
2
– 2n) = 8n – n2
74 + 79 + 84 + 89 + … + 589
Solution:
€ is a = 74, the
€ last term a = 589 and the common
The first term
1
n
difference d = 5. Plugging into formula #2 and solving for n, we get:
an = a1 + (n – 1)d
589 = 74 + (n – 1)(5)
(subtract 74 from both sides)
515 = 5(n – 1)
(divide both sides by 5)
103 = n – 1
(add 1 to both sides)
n = 104
Using formula #4, we get:
S104 =
n
2
(a1 + an) =
104
2
(74 + 589) = 52(663) = 34,476
Solve the following:
Ex. 7 A theater has 39 seats in the first row, 41 seats in the second row, 43
€ seats in the€third row and so forth. If there are 33 rows, how many
people can the theater seat?
Solution:
The first term is a1 = 39, the common difference is d = 2 and n = 33.
Using formula #3, we get:
S33 =
n
2
[2a1 + (n – 1)d] =
33
2
[2(39) + (104 – 1)2]
= 16.5[78 + 206] = 4686
The theater can hold 4686 people.
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