# 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*, which denotes the_{n}*nth*term of the sequence. Similarly*a*_{1}denotes the first term in the sequence,*a*_{2 }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*a*_{1}= 2 and*a*_{2}is 4 times*a*_{1}. It is also 6 more than*a*_{1}but that is not the pattern throughout the sequence. If it was the pattern,*a*_{3}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*4_{n}=*a*_{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.*a*

_{1 }= 2

*a*

_{2}= 4(

*a*

_{1}) = 4(2) = 8

*a*

_{3}= 4(

*a*

_{2}) = 4(8) = 32

*a*

_{4}= 4(

*a*

_{3}) = 4(32) = 128

*a*

_{5}= 4(

*a*

_{4}) = 4(128) = 512

**Example:**Find the first 6 terms and the 46

^{th}term of the infinite sequence with

*a*= 5

_{n}*n*– 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*a*_{1},*a*_{2},*a*_{3},*a*_{4},*a*_{5 }and*a*_{6}.
Therefore,

*a*

_{1}= 5(1) – 11 = 5 – 11 = -6

*a*

_{2}= 5(2) – 11 = 10 – 11 = -1

*a*

_{3}= 5(3) – 11 = 15 – 11 = 4

*a*

_{4}= 5(4) – 11 = 20 – 11 = 9

*a*

_{5}= 5(5) – 11 = 25 – 11 = 14

*a*

_{6}= 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,

*a*_{46}= 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.},

*a*

_{1}+

*d*,

*a*

_{1}+ 2

*d*,

*a*

_{1}+ 3

*d*,

*a*

_{1}+ 4

*d*, …...

*a*

_{1}+ (

*n*– 1)

*d*, where

*a*

_{1}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}*a*_{1}+ (*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,

*a*

_{1}= 8,

*a*

_{2}= 8 – 2 = 6,

*a*

_{3 }= 6 – 2 = 4,

*a*

_{4}= 4 – 2 = 2,

*a*

_{5}= 2 – 2 = 0,

*a*

_{6 }= 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}_{ }=*a*_{1}+ (*n*– 1)*d*to get*a*

_{29 }= 8 + (29 – 1)(-2)

*a*

_{29}= 8 + (28)(-2)

*a*

_{29}= 8 – 56

*a*

_{29}

*= -*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

*a*_{1}= 4 and*a*_{2}*=*10, the common difference is 10 – 4 = 6.
Now
we use the formula

*a*_{n}_{ }=*a*_{1}+ (*n*– 1)*d*and substitute 95 for*n*and 6 for*d*and solve for*a*_{n }to get*a*

_{95}= 4 + (95 – 1)(6)

*a*

_{95}= 4 + 94(6)

*a*

_{95}= 4 + 564

*a*

_{95}

^{ }= 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*a*_{1 }and 30 for*n*into the formula*a*_{n}_{ }=*a*_{1}+ (*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}_{ }=*a*_{1}+ (*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*a*_{1}and 45 for*n*to get
490
= 6 + (45 – 1)

*d*
490
= 6 + 44

*d*
484
= 44

*d*
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

*a*

_{1}= 6

*a*

_{2}= 6 + 11 = 17

*a*

_{3}= 17 + 11 = 28

*a*

_{4}= 28 + 11 = 39

*a*

_{5}= 39 + 11 = 50

___________________________________________________________________________________________________

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

*a*_{1 }= 8 but we don't know*a*_{2}or*a*_{3}. We know that*a*_{4}= 23. There is a jump from*a*_{1}to*a*_{2}, from*a*_{2 }to*a*_{3}and from*a*_{3}to*a*_{4}. 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,*a*_{2}and*a*_{3}are then*a*

_{2}=

*a*

_{1}+ 5 = 8 + 5 = 13,

*a*

_{3}=

*a*

_{2 }+ 5 = 13 + 5 = 18

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