Arithmetic Sequences and Series

Arithmetic Sequence
Examples Find the general term for the following sequences
both recursively and explicitly:

2,6,10,14,18,22, ...

5,3,1,1,3,...

1,4,7,10,13,16,...

1,10,21,32,43,54,...

3,0,3,6,9,12,...
Solution
All of these have one thing in common. To get to the next term we
add a fixed number.
Add four to obtain the next term. Thus
a_{n+1} = a_{n}
+ 4, a_{1} = 2
To find an explicit expression, we use the following reasoning. To
get the first term, we start with 2 and add no 4's. To get to the
second term we start with 2 and add one 4. To get to the third
term, we start with 2 and add two 4's. To get the the fourth term
we start with 2 and add three 4's. To get to the n^{th}
term, we start with 2 and add n  1 four. Hence
a_{n} = 2 + 4(n  1)
Add two to obtain the next term. Thus
a_{n+1} = a_{n}
+ 2, a_{1} = 5
To find an explicit expression, we use the same reasoning as in part
"A". To get the first term, we start with 5 and add no
2's. To get to the second term we start with 5 and add one
2. To get to the third term, we start with 5 and add two
2's. To get the the fourth term we start with 5 and add three
2's. To get to the n^{th} term, we start with 5 and add n
 1 twos. Hence
a_{n} = 5 + 2(n  1)
As in the preceding exercises,
a_{n+1} = a_{n}
+ 3, a_{1} = 1
and
a_{n} = 1 + 3(n  1)
We have
a_{n+1} = a_{n}
+ 11, a_{1} = 1
and
a_{n} = 1 + 11(n 
1)
Finally,
a_{n+1} = a_{n}
 3, a_{1} = 3
and
a_{n} = 3  3(n
 1)
Definition
A sequence with general term
a_{n+1} = a_{n} + d
is called an arithmetic sequence. 
This definition defines an arithmetic sequence recursively. The next
theorem shows how to find an explicit form for an arithmetic sequence.
Theorem
An arithmetic sequence with
a_{n+1} = a_{n}
+ d
has explicit form
a_{n} = a_{1} + (n  1)d 
Proof: (by induction)
For n = 1, we have
a_{1} = a_{1} + (1  1)d
(true)
Assume that the theorem is true for n = k  1, hence
a_{k1} = a_{1} + (k  1  1)d = a_{1} + (k  2)d
Then
a_{k} = a_{k1} + d = a_{1} + (k  2)d + d
= a_{1} + kd  2d + d = a_{1} + kd  d = a_{1} +
(k  1)d
Hence by mathematical induction, the theorem is true.
Example:
Suppose that a_{1} = 4 and d = 2
then the sequence
is
4,6,8,...,(4 + (n  1)d),...
Example
Suppose that the 13th term of an arithmetic sequence
is 46 and the fourth term is 100. Find the expression for the general
term.
Solution We have
46 = a_{1}
+ d(13  1) = a_{1} + 12d
and
100 = a_{1}
+ d(4  1) = a_{1} + 3d
Subtracting the two equations gives
54 = 9d
or
d = 6
Putting this back into the first equation gives
46 = a_{1}
+ 12(6)
or
a_{1}
= 118 We can conclude that
a_{n}
= 118  6(n  1)

The Arithmetic Series
The following theorem provides us with an easy way to calculate the arithmetic
series.
Theorem
If
a_{n}
= a_{1} + (n  1)d
is an arithmetic sequence then the sum of the sequence is
S_{n} =
S _{i=1}^{n } a_{n}
= n/2 (a_{1} + a_{n}) 
Proof:
S_{n} = S _{i=1}^{n }
a_{n} = a_{1} + a_{2} + ... + a_{n1} + a_{n
}
= a_{1} + (a_{1}
+ d) + ... +(a_{1} + (n  2)d) + (a_{1} + (n  1)d)
The sum can also be written by
working backwards from the last term, that is to get to the previous term,
subtract d.
S_{n} = S
_{i=1}^{n } a_{n} = a_{n} + a_{n1}
+ ... + a_{2} + a_{1
}
= (a_{1} + (n  1)d) + (a_{1}
+ (n  2)d) + ... + (a_{1} + d)+ a_{1}
Since these are the same we can add them together to get 2S.
S_{n} = a_{1 }
+ (a_{1} +
d) + ... + (a_{1} + (n 
2)d) + (a_{1} + (n  1)d)
S_{n} = (a_{1} +
(n  1)d) + (a_{1} + (n  2)d) + ... + (a_{1} +
d) + a_{1
}
2S_{n} =
[a_{1} + (a_{1} + (n  1)d)] + [a_{1} + (a_{1}
+ (n  1)d)] + ...
+[a_{1} + (a_{1} + (n  1)d)] + [a_{1} + (a_{1}
+ (n  1)d)]
= [a_{1} + a_{n}]
+ [a_{1} + a_{n}] +... + [a_{1} + a_{n}] +
[a_{1} + a_{n}]
= n[a_{1} + a_{n}]
Hence
S_{n} = n/2 [a_{1}
+ a_{n}]
Example:
Find
3 + 7 + 11 + 15 + ... + 35
Solution:
We have
a_{1} = 3, a_{n} =
35, d = 4
To find n we note that
35 = 3 + (n  1)4
so that
32 = (n  1)4
Dividing gives
8 = n  1
Hence
n = 9
Now we are ready to use the formula
S_{n} = 9/2 (3 + 35) = 171
Exercise:
Suppose that the sum of the first 18 terms of an
arithmetic sequence is 45 and
d = 9
find the first term.
Application
Suppose that you play black jack at Harrah's on
June 1 and lose $1,000. Tomorrow you bet and lose $15 less. Each day
you lose $15 less that your previous loss. What
will your total losses be for the 30 days of June?
Solution
This is an arithmetic series with
a_{1} = 1000
and
d = 15
We can calculate
a_{30} = 1000  15(30  1) =
565
Now we use the formula
S_{30} = 30/2 (1000 + 565) =
23,475
You will lose a total of $23,475 during June.
Back to the College Algebra Part II (Math 103B) Site
Back to the LTCC Math Department Page
email Questions and
Suggestions
