## Monday, May 14, 2012

Great time to get the books before final exam, especially now, on the lulu homepage, there will be a name of a savings coupon that they can add your cart to save. This promotion ends Friday May, 18 at 11:59 PM. you must go to the homepage www.lulu.com, but my books are at http://www.lulu.com/spotlight/KKauffman1969

Here's part of the second book, chapter on sequences and series.

# Chapter 7

## Sequences and Geometric Applications

When we think of a sequence, we think of an ordered list. For example, a history teacher might discuss a sequence of battles during World War II. A baseball historian might discuss the sequence of events leading up to the first organized professional baseball league. In terms of 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.

Some examples of finite and infinite sequences are as follows:

Finite: 2, 5, 8, 11, 14, 17, 20, 23, 26

1, -1, -3, -5, -7, -9, -11, -13, -15

Infinite: 1, 2, 3, 4, 5, 6, 7, …....

4, 8, 12, 16, 20, 24, 28, …...

Consider the following infinite sequence:

2, 8, 32, 128, 512, …..

We can define the nth term of the sequence by finding the relationship between the terms. Notice a1 = 2 and a2 is 4 times a1. It is also 6 more than a1 but that is not the pattern throughout the sequence. If it was the pattern, a3 would be 8 + 6 = 14, but the third term is 32, which is 4 times 8. The formula holds true for the next term as well since 32 times 4 equals 128. Therefore the formula for the nth term in the sequence is an = 4an-1

When we apply the formula we just got for an , we obtain the first 5 numbers in the series and therefore we know our formula is correct.

a1 = 2

a2 = 4(a1) = 4(2) = 8

a3 = 4(a2) = 4(8) = 32

a4 = 4(a3) = 4(32) = 128

a5 = 4(a4) = 4(128) = 512

Example: Find the first 6 terms and the 46th term of the infinite sequence with an = 5n – 11.

To find the first 6 terms of the sequence, we need to substitute 1, 2, 3, 4, 5 and 6 for n in the formula. Recall that the first 6 terms of the sequence are denoted as a1, a2, a3, a4, a5 and a6.

Therefore,

a1 = 5(1) – 11 = 5 – 11 = -6

a2 = 5(2) – 11 = 10 – 11 = -1

a3 = 5(3) – 11 = 15 – 11 = 4

a4 = 5(4) – 11 = 20 – 11 = 9

a5 = 5(5) – 11 = 25 – 11 = 14

a6 = 5(6) – 11 = 30 – 11 = 19

Therefore, the first 6 terms of the sequence are -6, -1, 4, 9, 14, 19

Note the pattern is each term is 5 greater than the previous term. Once that pattern has been established, we can just add 5 to the previous term and don't have to perform each calculation. In the next part where we have to find the 46th term, it's easier to perform the calculation instead of adding 5 each time until we reach the 46th term.

To find the 46th term of the sequence, we need to substitute 46 for n in the formula.

Therefore, a46 = 5(46) – 11 = 230 -11 = 219.

### Arithmetic Sequence

In another example, suppose the sequence is 2, 8, 14, 20, 26, 32, 38, ….... Notice that each term is 6 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: 3, 10, 17, 24, 31, 38

The common difference d is 7 since 10 – 3 = 7, 17 – 10 = 7, 24 – 17 = 7, 31 – 24 = 7 and 38 – 31 = 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.

Example: 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

Example: The first 4 terms of an arithmetic sequence are 4, 10, 16 and 22. What is the 95th term?

To solve this problem we have to find the common difference between each term. Since a1 = 4 and a2 = 10, the common difference is 10 – 4 = 6.

Now we use the formula an = a1 + (n – 1)d and substitute 95 for n and 6 for d and solve for an to get

a95 = 4 + (95 – 1)(6)

a95 = 4 + 94(6)

a95 = 4 + 564

a95 = 568

Sometimes we don't know the first several terms of a sequence. Maybe we know the first term and the 30th 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 30th 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 for n 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

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Suppose we know two terms of an arithmetic sequence and we wish to insert numbers between them. The numbers are known as arithmetic means. If a single number is inserted between the two numbers, it's known as the arithmetic mean between the two numbers.

For example, the numbers 4 and 7 are the arithmetic means between 1 and 10 in the sequence 1, 4, 7, 10. The common difference between the terms is 3. The number 6 is the arithmetic mean between 3 and 9 in the sequence 3, 6, 9. The common difference between the terms is 3.

How do we determine the arithmetic means? Consider the following examples.

Example: Find two arithmetic means between 8 and 23.

To solve this problem, we have to determine how many terms there will be and then find the common difference between the terms.

We know a1 = 8 but we don't know a2 or a3. We know that a4 = 23. There is a jump from a1 to a2, from a2 to a3 and from a3 to a4. Therefore there are 3 equal distances going from 8 to 23. The total distance from 8 to 23 is 15 since 23 – 8 = 15. Now we take 15 divided by 3 to get 5. Therefore the common difference is 5. The arithmetic means, a2 and a3 are then

a2 = a1 + 5 = 8 + 5 = 13, a3 = a2 + 5 = 13 + 5 = 18