Arithmetic Sequences and Series

1. Arithmetic Sequence

Examples

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

1. 2,6,10,14,18,22, ...

2.  -5,-3,-1,1,3,...

3. 1,4,7,10,13,16,...

4. -1,10,21,32,43,54,...

5. 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.

1. Add four to obtain the next term. Thus

an+1 = an + 4,     a1 = 2

an = 2 + 4(n - 1)

2.  Add two to obtain the next term. Thus

an+1 = an + 2,     a1 = -5

an = -5 + 2(n - 1)

3. As in the preceding exercises,

an+1 = an + 3,     a1 = 1

and

an = 1 + 3(n - 1)

4. We have

an+1 = an + 11,     a1 = -1

and

an = -1 + 11(n - 1)

5. Finally,

an+1 = an - 3,     a1 = 3

and

an = 3 -  3(n - 1)

 Definition   A sequence with general term           an+1 = an + 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            an+1 = an + d   has explicit form             an = a1 + (n - 1)d

Proof:  (by induction)

For n = 1, we have

a1 = a1 + (1 - 1)d  (true)

Assume that the theorem is true for n = k - 1, hence

ak-1 = a1 + (k - 1 - 1)d = a1 + (k - 2)d

Then

ak = ak-1 + d = a1 + (k - 2)d + d

= a1 + kd - 2d + d = a1 + kd - d = a1 + (k - 1)d

Hence by mathematical induction, the theorem is true.

Example:

Suppose that a1 = 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 = a1 + d(13 - 1) = a1 + 12d

and

100 = a1 + d(4 - 1) = a1 + 3d

Subtracting the two equations gives

-54 = 9d

or

d = -6

Putting this back into the first equation gives

46 = a1 + 12(-6)

or

a1 = 118

We can conclude that

an = 118 - 6(n - 1)

2. The Arithmetic Series

The following theorem provides us with an easy way to calculate the arithmetic series.

 Theorem If           an =  a1 + (n - 1)d is an arithmetic sequence then the sum of the sequence is          Sn =  S i=1n  an =  n/2 (a1 + an)

Proof:

Sn = S i=1 an = a1 + a2 + ... + an-1 + an

= a1 + (a1 + d)  + ... +(a1 + (n - 2)d) + (a1 + (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.

Sn = S i=1 an = an + an-1 + ... + a2 + a1

= (a1 + (n - 1)d) + (a1 + (n - 2)d) + ... + (a1 + d)+ a1

Since these are the same we can add them together to get 2S.

Sn = a1                    +    (a1 + d)        + ... + (a1 + (n - 2)d) + (a1 + (n - 1)d)
Sn = (a1 + (n - 1)d) + (a1 + (n - 2)d) + ... + (a1 + d)           +  a1

2Sn = [a1 + (a1 + (n - 1)d)] + [a1 + (a1 + (n - 1)d)] + ...
+[a1 + (a1 + (n - 1)d)] + [a1 + (a1 + (n - 1)d)]

=  [a1 + an] + [a1 + an] +... + [a1 + an] + [a1 + an

= n[a1 + an]

Hence

Sn =  n/2 [a1 + an]

Example:

Find

3 + 7 + 11 + 15  + ... + 35

Solution:

We have

a1 = 3, an = 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

Sn = 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

a1 = 1000

and

d = -15

We can calculate

a30 = 1000 - 15(30 - 1) = 565

Now we use the formula

S30 = 30/2 (1000 + 565) = 23,475

You will lose a total of \$23,475 during June.