# What is arithmetic progression?

## What is arithmetic progression?

In today’s article, I am going to provide you with a equipo of solved arithmetic progression exercises, with the aim of learning what an arithmetic progression is and preparing you to take the next step with geometric progressions.

Remember that Knowing what a progression is cánido help you better understand what an annuity is. Sure, annuities are mostly based on geometric progressions, but I think sometimes it’s better to go from fácil to more complicated.

Therefore, first of all, it is better that you understand what a arithmetic progression and then you perro move on to geometric progressions.

I hope you find it useful.

## What is a progression?

A progression cánido be seen as an uninterrupted series or sequence of numbers.

Some aspects that you have to consider are the following:

1. Each number belonging to the sequence is called a term.
2. In every progression there must be a first term.
3. Each progression has a criterion, which helps us to determine each of the following terms of the sequence.
4. The terms that make up the progression cánido be obtained by difference (arithmetic progression) or by quotient (geometric progression).

## Examples of progressions

Some examples of progressions are:

• 1, 2, 3, 4, 5, 6, 7, 8, 9,…
• 1, 3, 5, 7, 9, 11, 13, 15,…
• 2, 4, 6, 8, 10, 12, 14, 16,…
• 0, 1, 1, 2, 3, 5, 8, 13, 21, 34,…

## Types of progressions

We perro find arithmetic progressions and geometric progressions.

## What is an arithmetic progression?

Arithmetic progressions are progressions whose terms integrate by difference. I know that it is possible that it is better that I explain it to you with an example.

Do you remember that in elementary school they made us find the next number in a sequence?

What number or term follows in the following sequence? 2, 4, 6, 8, __

Sure, it is a very fácil sequence, but how would you go about finding the answer?

One way to do it is by finding the constant difference that exists between the terms of the sequence, that is, by subtraction.

Following the previous example, the following subtractions cánido be done:

• 4 – 2 = 2
• 6 – 4 = 2
• 8 – 6 = 2

Therefore, by subtraction we have been able to find the constant difference of the arithmetic progression. In this case, the constant difference is equal to 2.

So, to find the next term in the sequence, all you have to do is add 2 to the last term of the sequence you have, which is 8. Therefore, the next term of the sequence is 10.

It should be noted that it has been a very fácil sequence and it has not been necessary to have a elabora, but What would happen if I ask you to give me the term 100 of that sequence. In the example above I have asked you for the fifth term, but what is the 100th term or the 1000th term?

Well, for that we cánido use a elabora that I am going to espectáculo you later. For now, I want you to stick with the fact that in arithmetic progressions you are going to use difference (subtraction) to get the different terms of the progression.

Note: remember that we are talking about a constant difference.

For example, in the previous case, it does not matter which terms you choose from the sequence, since the difference will always be 2.

## literals

Before continuing, it is important that we agree on the literals that are going to be used and that you know their meaning.

By the way, I’m going to continue using the example: 2, 4, 6, 8, __

• a = first term of the sequence.

Remember that I said at the beginning that you should always have a first term.

Continuing with the previous example, the first term is the value 2.

• d = constant difference. Simply put, it is the value found by subtracting its terms.

In the previous example it was the value 2 (4 – 2).

• n = number of terms that the progression contains. Continuing with the example, the sequence has 4 terms and we are asked for the fifth term.
• I = x. It is the term of the sequence that we are looking for or the last term of the sequence.

Note: Remember that the progression cánido be finite or infinite.

## Así expression of the arithmetic progression

So that you cánido see the way in which the different terms of an arithmetic progression are found, then, I am going to give you the following table.

Continuing with the same example (2, 4, 6, 8, __), then, it is as follows:

## Arithmetic progression elabora

As you cánido see, in order to find the nth term of a sequence, all you need is the following elabora:

• a = first term of the sequence.
• d = constant difference.
• n = number of terms that the progression contains.
• I = term that we are looking for or the last term of the sequence.

Therefore, if you want to find the 100th term of the sequence: 2, 4, 6, 8, __

2 + (100 – 1)2 = 200

## Clearances of the arithmetic progression elabora

Just as they perro ask us to provide the last number of the sequence (I), they perro also ask us to provide the first term or the constant difference.

Therefore, next I am going to provide you with the different clearances of the arithmetic progression.

### How to find the first term?

If you want to find the first term of a progression, then the clearance is as follows:

### How to find the constant difference?

The clearance of the constant difference is as follows:

### How to find the number of terms of an arithmetic progression?

The elabora (clearance) that will help you find the number of terms that an arithmetic progression has is the following:

## What if the constant difference is positive?

If the constant difference of the sequence is positive, then it is an increasing progression.

### example of increasing progression

Suppose you have the sequence: 4, 8, 12, 16, 20,

If we obtain the constant difference, we will obtain as a result that the constant difference is equal to 4 TRUE? In fact, just by looking at the sequence we cánido see that the terms are increasing by 4 by four.

Of course, here they have given us the succession directly, but there are times when they will only give us the data.

For example:

In this way, we already know that the progression has the number 4 as its first term; it is a growing sequence; increases from 4 to four and has 5 terms.

## What if the constant difference is negative?

In this case, the opposite happens to the previous case.

So that if the constant difference is negative, then the progression is decreasing.

### Example of decreasing progression

Suppose you are given the following data:

Therefore, the arithmetic progression has as its first term the number 40; has a constant difference of –5 (so it is decreasing) and has 5 terms.

Thus, the sequence is as follows:

40, 35, 30, 25, 20

By the way, to know any value of the sequence, you cánido use the expression that I gave you before.

Let’s suppose that you are asked to find the next term of the previous sequence (sixth term).

Note: remember that we are going to use I = a + (n – 1) d

40 + ( 6 – 1) -5 = 15

As you perro see, the next term of the sequence is 15.

In fact, since I’m using fácil sequences, it’s very easy to predict the next term.

That way you perro verify that it works.

## What is the sum of an arithmetic progression?

As its name already indicates, it is the sum of all the terms that a certain progression (sequence) has. In fact, there is a very interesting theorem that tells us the following:

“In every limited arithmetic progression, the sum of the means equidistant from the extremes is equal to the sum of the extremes”

To exemplify the theorem, I will take the following sequence as an example: 5, 10, 15, 20, 25, 30

Now, let’s start with the simplest part of the previous theorem… Who are the ends of a progression? Well, they are the first and last terms of the sequence, right? After all, we are talking about extremes.

As we have already seen in the elabora, the foreground of a sequence is equal to the letter a. He Last term is equal to the letter I. You already know them because they are in the elabora.

Note: remember that I used those letters, which I have seen in books, but in theory you cánido take whatever letters you want.

As long as you know what each letter means.

However… What are the means equidistant from the extremes? Well, in this case, we are talking about two terms that are the same distance from the extremes.

Following the example sequence, the second term (10) of the sequence and the fifth term (25) of the sequence comply with the above.

After all, the second term is next to the first term, and the fifth term is next to the last term.

Likewise, the third and fourth terms are also means equidistant from the extremes.

### Does the theorem hold?

Now we are going to verify that the theorem is true.

• The sum of the extremes is equal to 35. a = 5; I = 30.

Therefore, 5 + 30 equals 35.

• Now, I am going to use the second term and the fifth term of the sequence as means equidistant from the extremes.

The second term is equal to 10; the fifth term is equal to 25.

Therefore, 10 + 25 = 35. As you perro see, it is also true.

You cánido try to make the sum of the other equidistant means so that you cánido see that it is fulfilled.

## Arithmetic Progression Sum Elabora

Since we have seen that the previous theorem is true, now I am going to provide you with two formulas that will help you find the sum of an arithmetic progression.

### If you know the last term of the sequence

If you know the last term of the progression, then you perro use the following elabora to find the sum of the progression:

As an example I will use the following sequence: 5, 10, 15, 20, 25, 30

As you perro see, we already have all the data we need, that is:

Therefore: S = (5 + 30) 6 / 2

The answer is equal to 105. In fact, if you think it’s not true, then you cánido do the sum: 5 + 10 + 15 + 20 + 25 + 30 = 105.

That way, if they put you to add all the terms of a sequence that has 100 or 1000 terms, it won’t take you all day to do the sum.

### If you do not know the last term of the sequence

If you do not know the last term of the sequence and do not want to obtain it, then you perro use the following elabora:

I am going to use the same sequence as in the previous example, but I am going to imagine that I do not know the last term (30).

Therefore, the sequence looks like this: 5, 10, 15, 20, 25, __

So the data we have is:

Therefore: S = (6/2)(2*5 + (6 -1) 5) = 105.

As you perro see, it gives us the same result.

Although, you cánido choose the elabora you want to use.

After all, you cánido get the data you need.

## Solved exercises of arithmetic progression

Well, at this point you already know enough to be able to start solving some exercises.

Therefore, next, I am going to provide you with some solved exercises of arithmetic progression.

I recommend that you try to solve them first and then check your result.

### Exercise 1

The data for the first year are as follows:

• a = 3.
• d = 3.
• n=10.
• I = ? What is the last term of the sequence?

To find the last term of the progression, what you have to do is the following:

I = a + (n – 1) d

I = 3 + (10 – 1) 3

Therefore, the last term of the progression is equal to 30. In fact, it is the table of 3, but so that you perro see that it is true, I am going to put the ten terms of the sequence: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

### Exercise 2

For exercise two, the data is as follows:

• to = ?
• d = 7.
• n=5.
• I = 42.

To find the first term of an arithmetic progression, then, you have to do the following:

a = I – (n – 1) d

a = 42 – (5 – 1) 7

Therefore, the first term is equal to 14.

In such a way that the sequence is as follows: 14, 21, 28, 35, 42

### Exercise 3

For exercise 3, I will use the following data:

• a = 50.
• I = 10.
• n=6.
• d = ?

So that you cánido find the constant difference in a progression, what you have to do is the following:

d = (I – a) / (N – 1)

D = (10-50) / (6-1)

Therefore, the answer of the third exercise of arithmetic progression is equal to – 8.

### Exercise 4

The last exercise has the following data:

• a = 79.
• d = -9.
• I = 25.
• n = ?

In order for you to find the number of terms in the sequence, you have to do the following:

n = ( (Ia) /(d) ) + 1

n = ((25 – 79) / -9) + 1

Therefore, the number of terms in the sequence is equal to 7.

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