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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 a_n , 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 a_n = 4a_n-1

When we apply the formula we just got for a_n , 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 46^th term of the infinite sequence with a_n = 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 46^th term, it's easier to perform the calculation instead of adding 5 each time until we reach the 46^th term.

To find the 46^th 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

a_1., 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 a_n = 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 29^th 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 29^th term in the sequence, we use the formula to find the nth term in a sequence. Substitute 29 for n in the formula a_n = 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 95^th 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 a_n = a₁ + (n – 1)d and substitute 95 for n and 6 for d and solve for a_n 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 30^th term and we need to find the 15^th term. How do we approach such a problem?

First, we have to determine the common difference by substituting the value of the 30^th term for an, the value of the first term for a₁ and 30 for n into the formula a_n = a₁ + (n – 1)d.

Example: The first term of an arithmetic sequence is 6 and the 45^th 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 a_n = a₁ + (n – 1)d. Since the 45^th term is 490, we substitute 490 for a_n . 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