## Thursday, October 11, 2012

There are many types of sequences seen in mathematics.  A sequence refers to a list of numbers written in a specific order. Sequences can either be finite, which contains a set number of terms, or infinite, where the sequence continues on forever. A finite sequence is a function whose domain is the set of all natural numbers for a specified number n. An infinite sequence is a function whose domain is the set of all natural numbers. An easier way to denote terms of a sequence is to use an , which denotes the nth term of the sequence. Similarly a1 denotes the first term in the sequence, a2 denotes the second term of the sequence and so on.

The type of sequences focused on in this post is the arithmetic sequence.  Suppose the sequence is 2, 6, 10, 14, 18, 22, ….... Notice that each term is 4 more than the previous term. A sequence such as this is known as an arithmetic sequence because each term is found by adding the same number to the previous term. An arithmetic sequence is a sequence in the form a1., a1 + d, a1 + 2d, a1 + 3d, a1 + 4d, …... a1 + (n – 1)d, where a1 is the first term of the sequence and d is the common difference between the terms.
The nth term of an arithmetic sequence is given by the formula "an" = a1 + (n – 1)d.
Consider the following arithmetic sequences below. Notice the common difference by subtracting the first term from the second, the second term from the third and so on.
Finite arithmetic sequence: 1, 8, 15, 22, 29
The common difference d is 7 since 8 - 1 = 7, 15 - 8 = 7, 22 - 15 = 7 and 29 – 22 = 7
Infinite arithmetic sequence: 6, 3, 0, -3, 6, -9, ….....
The common difference d is -3 since 3 - 6 = -3, -3 - 0 = -3, -6 - (-3) = -3, -9 - (-6) = -3 and so on.

In an arithmetic sequence, the first term is 8 and the common difference is -2. What are the first 6 terms of the sequence and what is the 29th term?  To solve this problem we start with 8 and add the common difference to each term thereafter until we get all 6 terms.
Therefore,
a1 = 8, a2 = 8 – 2 = 6, a3 = 6 – 2 = 4, a4 = 4 – 2 = 2, a5 = 2 – 2 = 0, a6 = 0 – 2 = -2
The first 6 terms in the series are 8, 6, 4, 2, 0, -2.
To find the 29th term in the sequence, we use the formula to find the nth term in a sequence. Substitute 29 for n in the
formula "an" = a1 + (n – 1)d to get
a29 = 8 + (29 – 1)(-2)
a29 = 8 + (28)(-2)
a29 = 8 – 56
a29 = -48
Sometimes we don't know the first several terms of a sequence. Maybe we know the first term and the 50th term and we need to find the 15th term. How do we approach such a problem?  First, we have to determine the common difference by substituting the value of the 50th term for an, the value of the first term for a1 and 30 for n into the formula "an" = a1 + (n – 1)d

Example: The first term of an arithmetic sequence is 6 and the 45th term is 490. What are the first 5 terms of the sequence?
To solve this problem we need to find the common difference. We use the formula an = a1 + (n – 1)d. Since the 45th term is 490, we substitute 490 for an . Since the first term is 6, we substitute 6 for a1 and 45 to get

490 = 6 + (45 – 1)d
490 = 6 + 44d
484 = 44d
11 = d

Since the common difference is 11, the first 5 terms of the sequence are 6, 17, 28, 39, 50 and are determined as follows
a1 = 6 , a2 = 6 + 11 = 17, a3 = 17 + 11 = 28, a4 = 28 + 11 = 39, a5 = 39 + 11 = 50.
Suppose we wish to find the sum of the first 10 terms of the sequence 1, 3, 5, 7, 9, 11, … We can easily solve this by adding the first 10 terms together. But what if we want the sum of the first 100 terms? Adding them would become tedious, so there is a formula we can use to solve such problems quickly.
The sum of the first n terms of an arithmetic sequence is given by
Sn = n(a1 + "an")/2
Sn = 2 , where a1 is the first term, an is the nth term and n is the number of terms.

Example: Find the sum of the first 40 terms of the sequence 7, 12, 17, 22, 27, …..
To solve this problem we need to find the common difference d. Once we find d, we can substitute the values for a1, n and d into the formula an = a1 + (n – 1)d so we can find a40. Once we find a40 we can use the formula for Sn to find the sum.
Therefore,
a40 = 7 + (40 – 1)(5)
a40 = 7 + 39(5)
a40 = 7 + 195
a40 = 202
Now we can use the formula for Sn to get
Sn = 40(7 + 202)/2
Sn = 20(209)
Sn = 4,180