How to calculate arithmetic progression formula. Arithmetic progression: what is it? Progression difference: definition

Or arithmetic is a type of ordered numerical sequence, the properties of which are studied in a school algebra course. This article discusses in detail the question of how to find the sum of an arithmetic progression.

What kind of progression is this?

Before moving on to the question (how to find the sum of an arithmetic progression), it is worth understanding what we are talking about.

Any sequence of real numbers that is obtained by adding (subtracting) some value from each previous number is called an algebraic (arithmetic) progression. This definition, when translated into mathematical language, takes the form:

Here i is the serial number of the element of the row a i. Thus, knowing just one starting number, you can easily restore the entire series. The parameter d in the formula is called the progression difference.

It can be easily shown that for the series of numbers under consideration the following equality holds:

a n = a 1 + d * (n - 1).

That is, to find the value of the nth element in order, you should add the difference d to the first element a 1 n-1 times.

What is the sum of an arithmetic progression: formula

Before giving the formula for the indicated amount, it is worth considering a simple special case. Given a progression of natural numbers from 1 to 10, you need to find their sum. Since there are few terms in the progression (10), it is possible to solve the problem head-on, that is, sum all the elements in order.

S 10 = 1+2+3+4+5+6+7+8+9+10 = 55.

It is worth considering one interesting thing: since each term differs from the next one by the same value d = 1, then pairwise summation of the first with the tenth, the second with the ninth, and so on will give the same result. Really:

11 = 1+10 = 2+9 = 3+8 = 4+7 = 5+6.

As you can see, there are only 5 of these sums, that is, exactly two times less than the number of elements of the series. Then multiplying the number of sums (5) by the result of each sum (11), you will arrive at the result obtained in the first example.

If we generalize these arguments, we can write the following expression:

S n = n * (a 1 + a n) / 2.

This expression shows that it is not at all necessary to sum all the elements in a row; it is enough to know the value of the first a 1 and the last a n, as well as the total number of terms n.

It is believed that Gauss first thought of this equality when he was looking for a solution to a problem given by his school teacher: sum the first 100 integers.

Sum of elements from m to n: formula

The formula given in the previous paragraph answers the question of how to find the sum of an arithmetic progression (the first elements), but often in problems it is necessary to sum a series of numbers in the middle of the progression. How to do it?

The easiest way to answer this question is by considering the following example: let it be necessary to find the sum of terms from the m-th to the n-th. To solve the problem, you should present the given segment from m to n of the progression in the form of a new number series. In this representation, the mth term a m will be the first, and a n will be numbered n-(m-1). In this case, applying the standard formula for the sum, the following expression will be obtained:

S m n = (n - m + 1) * (a m + a n) / 2.

Example of using formulas

Knowing how to find the sum of an arithmetic progression, it is worth considering a simple example of using the above formulas.

Below is a numerical sequence, you should find the sum of its terms, starting from the 5th and ending with the 12th:

The given numbers indicate that the difference d is equal to 3. Using the expression for the nth element, you can find the values ​​of the 5th and 12th terms of the progression. It turns out:

a 5 = a 1 + d * 4 = -4 + 3 * 4 = 8;

a 12 = a 1 + d * 11 = -4 + 3 * 11 = 29.

Knowing the values ​​of the numbers at the ends of the algebraic progression under consideration, as well as knowing what numbers in the series they occupy, you can use the formula for the sum obtained in the previous paragraph. It will turn out:

S 5 12 = (12 - 5 + 1) * (8 + 29) / 2 = 148.

It is worth noting that this value could be obtained differently: first find the sum of the first 12 elements using the standard formula, then calculate the sum of the first 4 elements using the same formula, then subtract the second from the first sum.

So, let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many of them as you like (in our case, there are them). No matter how many numbers we write, we can always say which one is first, which one is second, and so on until the last, that is, we can number them. This is an example of a number sequence:

Number sequence
For example, for our sequence:

The assigned number is specific to only one number in the sequence. In other words, there are no three second numbers in the sequence. The second number (like the th number) is always the same.
The number with number is called the th term of the sequence.

We usually call the entire sequence by some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member: .

In our case:

Let's say we have a number sequence in which the difference between adjacent numbers is the same and equal.
For example:

etc.
This number sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius back in the 6th century and was understood in a broader sense as an infinite numerical sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which was studied by the ancient Greeks.

This is a number sequence, each member of which is equal to the previous one added to the same number. This number is called the difference of an arithmetic progression and is designated.

Try to determine which number sequences are an arithmetic progression and which are not:

a)
b)
c)
d)

Got it? Let's compare our answers:
Is arithmetic progression - b, c.
Is not arithmetic progression - a, d.

Let's return to the given progression () and try to find the value of its th term. Exists two way to find it.

1. Method

We can add the progression number to the previous value until we reach the th term of the progression. It’s good that we don’t have much to summarize - only three values:

So, the th term of the described arithmetic progression is equal to.

2. Method

What if we needed to find the value of the th term of the progression? The summation would take us more than one hour, and it is not a fact that we would not make mistakes when adding numbers.
Of course, mathematicians have come up with a way in which it is not necessary to add the difference of an arithmetic progression to the previous value. Take a closer look at the drawn picture... Surely you have already noticed a certain pattern, namely:

For example, let’s see what the value of the th term of this arithmetic progression consists of:


In other words:

Try to find the value of a member of a given arithmetic progression yourself in this way.

Did you calculate? Compare your notes with the answer:

Please note that you got exactly the same number as in the previous method, when we sequentially added the terms of the arithmetic progression to the previous value.
Let’s try to “depersonalize” this formula - let’s put it in general form and get:

Arithmetic progressions can be increasing or decreasing.

Increasing- progressions in which each subsequent value of the terms is greater than the previous one.
For example:

Descending- progressions in which each subsequent value of the terms is less than the previous one.
For example:

The derived formula is used in the calculation of terms in both increasing and decreasing terms of an arithmetic progression.
Let's check this in practice.
We are given an arithmetic progression consisting of the following numbers: Let's check what the th number of this arithmetic progression will be if we use our formula to calculate it:

Since then:

Thus, we are convinced that the formula operates in both decreasing and increasing arithmetic progression.
Try to find the th and th terms of this arithmetic progression yourself.

Let's compare the results:

Arithmetic progression property

Let's complicate the problem - we will derive the property of arithmetic progression.
Let's say we are given the following condition:
- arithmetic progression, find the value.
Easy, you say and start counting according to the formula you already know:

Let, ah, then:

Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but what if we are given numbers in the condition? Agree, there is a possibility of making a mistake in the calculations.
Now think about whether it is possible to solve this problem in one step using any formula? Of course yes, and that’s what we’ll try to bring out now.

Let us denote the required term of the arithmetic progression as, the formula for finding it is known to us - this is the same formula we derived at the beginning:
, Then:

• the previous term of the progression is:
• the next term of the progression is:

Let's sum up the previous and subsequent terms of the progression:

It turns out that the sum of the previous and subsequent terms of the progression is the double value of the progression term located between them. In other words, to find the value of a progression term with known previous and successive values, you need to add them and divide by.

That's right, we got the same number. Let's secure the material. Calculate the value for the progression yourself, it’s not at all difficult.

Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, was easily deduced by one of the greatest mathematicians of all time, the “king of mathematicians” - Karl Gauss...

When Carl Gauss was 9 years old, a teacher, busy checking the work of students in other classes, assigned the following task in class: “Calculate the sum of all natural numbers from to (according to other sources to) inclusive.” Imagine the teacher’s surprise when one of his students (this was Karl Gauss) a minute later gave the correct answer to the task, while most of the daredevil’s classmates, after long calculations, received the wrong result...

Young Carl Gauss noticed a certain pattern that you can easily notice too.
Let's say we have an arithmetic progression consisting of -th terms: We need to find the sum of these terms of the arithmetic progression. Of course, we can manually sum all the values, but what if the task requires finding the sum of its terms, as Gauss was looking for?

Let us depict the progression given to us. Take a closer look at the highlighted numbers and try to perform various mathematical operations with them.


Have you tried it? What did you notice? Right! Their sums are equal

Now tell me, how many such pairs are there in total in the progression given to us? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two terms of an arithmetic progression is equal, and similar pairs are equal, we obtain that the total sum is equal to:
.
Thus, the formula for the sum of the first terms of any arithmetic progression will be:

In some problems we do not know the th term, but we know the difference of the progression. Try to substitute the formula of the th term into the sum formula.
What did you get?

Well done! Now let's return to the problem that was asked to Carl Gauss: calculate for yourself what the sum of numbers starting from the th is equal to and the sum of the numbers starting from the th.

How much did you get?
Gauss found that the sum of the terms is equal, and the sum of the terms. Is that what you decided?

In fact, the formula for the sum of the terms of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people made full use of the properties of the arithmetic progression.
For example, imagine Ancient Egypt and the largest construction project of that time - the construction of a pyramid... The picture shows one side of it.

Where is the progression here, you say? Look carefully and find a pattern in the number of sand blocks in each row of the pyramid wall.


Why not an arithmetic progression? Calculate how many blocks are needed to build one wall if block bricks are placed at the base. I hope you won’t count while moving your finger across the monitor, you remember the last formula and everything we said about arithmetic progression?

In this case, the progression looks like this: .
Arithmetic progression difference.
The number of terms of an arithmetic progression.
Let's substitute our data into the last formulas (calculate the number of blocks in 2 ways).

Method 1.

Method 2.

And now you can calculate on the monitor: compare the obtained values ​​with the number of blocks that are in our pyramid. Got it? Well done, you have mastered the sum of the nth terms of an arithmetic progression.
Of course, you can’t build a pyramid from blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:

Training

Tasks:

1. Masha is getting in shape for summer. Every day she increases the number of squats by. How many times will Masha do squats in a week if she did squats at the first training session?
2. What is the sum of all odd numbers contained in.
3. When storing logs, loggers stack them in such a way that each top layer contains one log less than the previous one. How many logs are in one masonry, if the foundation of the masonry is logs?

Answers:

1. Let us define the parameters of the arithmetic progression. In this case
   (weeks = days).

   Answer: In two weeks, Masha should do squats once a day.

2. First odd number, last number.
   Arithmetic progression difference.
   The number of odd numbers in is half, however, let’s check this fact using the formula for finding the th term of an arithmetic progression:

   Numbers do contain odd numbers.
   Let's substitute the available data into the formula:

   Answer: The sum of all odd numbers contained in is equal.

3. Let's remember the problem about pyramids. For our case, a , since each top layer is reduced by one log, then in total there are a bunch of layers, that is.
   Let's substitute the data into the formula:

   Answer: There are logs in the masonry.

Let's sum it up

1. - a number sequence in which the difference between adjacent numbers is the same and equal. It can be increasing or decreasing.
2. Finding formula The th term of an arithmetic progression is written by the formula - , where is the number of numbers in the progression.
3. Property of members of an arithmetic progression- - where is the number of numbers in progression.
4. The sum of the terms of an arithmetic progression can be found in two ways:
   
   , where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

Let's sit down and start writing some numbers. For example:

You can write any numbers, and there can be as many of them as you like. But we can always say which one is first, which one is second, and so on, that is, we can number them. This is an example of a number sequence.

Number sequence is a set of numbers, each of which can be assigned a unique number.

In other words, each number can be associated with a certain natural number, and a unique one. And we will not assign this number to any other number from this set.

The number with number is called the th member of the sequence.

We usually call the entire sequence by some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member: .

It is very convenient if the th term of the sequence can be specified by some formula. For example, the formula

sets the sequence:

And the formula is the following sequence:

For example, an arithmetic progression is a sequence (the first term here is equal, and the difference is). Or (, difference).

nth term formula

We call a formula recurrent in which, in order to find out the th term, you need to know the previous or several previous ones:

To find, for example, the th term of the progression using this formula, we will have to calculate the previous nine. For example, let it. Then:

Well, is it clear now what the formula is?

In each line we add to, multiplied by some number. Which one? Very simple: this is the number of the current member minus:

Much more convenient now, right? We check:

Decide for yourself:

In an arithmetic progression, find the formula for the nth term and find the hundredth term.

Solution:

The first term is equal. What is the difference? Here's what:

(This is why it is called difference because it is equal to the difference of successive terms of the progression).

So, the formula:

Then the hundredth term is equal to:

What is the sum of all natural numbers from to?

According to legend, the great mathematician Carl Gauss, as a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last numbers is equal, the sum of the second and penultimate is the same, the sum of the third and 3rd from the end is the same, and so on. How many such pairs are there in total? That's right, exactly half the number of all numbers, that is. So,

The general formula for the sum of the first terms of any arithmetic progression will be:

Example:
Find the sum of all two-digit multiples.

Solution:

The first such number is this. Each subsequent number is obtained by adding to the previous number. Thus, the numbers we are interested in form an arithmetic progression with the first term and the difference.

Formula of the th term for this progression:

How many terms are there in the progression if they all have to be two-digit?

Very easy: .

The last term of the progression will be equal. Then the sum:

Answer: .

Now decide for yourself:

1. Every day the athlete runs more meters than the previous day. How many total kilometers will he run in a week if he ran km m on the first day?
2. A cyclist travels more kilometers every day than the previous day. On the first day he traveled km. How many days does he need to travel to cover a kilometer? How many kilometers will he travel during the last day of his journey?
3. The price of a refrigerator in a store decreases by the same amount every year. Determine how much the price of a refrigerator decreased each year if, put up for sale for rubles, six years later it was sold for rubles.

Answers:

1. The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
   .
   Answer:
2. Here it is given: , must be found.
   Obviously, you need to use the same sum formula as in the previous problem:
   .
   Substitute the values:

   The root obviously doesn't fit, so the answer is.
   Let's calculate the path traveled over the last day using the formula of the th term:
   (km).
   Answer:

3. Given: . Find: .
   It couldn't be simpler:
   (rub).
   Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

This is a number sequence in which the difference between adjacent numbers is the same and equal.

Arithmetic progression can be increasing () and decreasing ().

For example:

Formula for finding the nth term of an arithmetic progression

is written by the formula, where is the number of numbers in progression.

Property of members of an arithmetic progression

It allows you to easily find a term of a progression if its neighboring terms are known - where is the number of numbers in the progression.

Sum of terms of an arithmetic progression

There are two ways to find the amount:

Where is the number of values.

Where is the number of values.

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In mathematics, any collection of numbers that follow each other, organized in some way, is called a sequence. Of all existing sequences of numbers, two interesting cases are distinguished: algebraic and geometric progressions.

What is an arithmetic progression?

It should be said right away that algebraic progression is often called arithmetic, since its properties are studied by the branch of mathematics - arithmetic.

This progression is a sequence of numbers in which each next member differs from the previous one by a certain constant number. It is called the difference of an algebraic progression. For definiteness, we denote it by the Latin letter d.

An example of such a sequence could be the following: 3, 5, 7, 9, 11 ..., here you can see that the number 5 is greater than the number 3 by 2, 7 is greater than 5 by 2, and so on. Thus, in the example presented, d = 5-3 = 7-5 = 9-7 = 11-9 = 2.

What are the types of arithmetic progressions?

The nature of these ordered sequences of numbers is largely determined by the sign of the number d. The following types of algebraic progressions are distinguished:

• increasing when d is positive (d>0);
• constant when d = 0;
• decreasing when d is negative (d<0).

The example given in the previous paragraph shows an increasing progression. An example of a decreasing sequence is the following sequence of numbers: 10, 5, 0, -5, -10, -15 ... A constant progression, as follows from its definition, is a collection of identical numbers.

nth term of progression

Due to the fact that each subsequent number in the progression under consideration differs by a constant d from the previous one, its nth term can be easily determined. To do this, you need to know not only d, but also a 1 - the first term of the progression. Using a recursive approach, one can obtain an algebraic progression formula for finding the nth term. It looks like: a n = a 1 + (n-1)*d. This formula is quite simple and can be understood intuitively.

It is also not difficult to use. For example, in the progression given above (d=2, a 1 =3), we define its 35th term. According to the formula, it will be equal to: a 35 = 3 + (35-1)*2 = 71.

Formula for amount

When given an arithmetic progression, the sum of its first n terms is a frequently encountered problem, along with determining the value of the nth term. The formula for the sum of an algebraic progression is written in the following form: ∑ n 1 = n*(a 1 +a n)/2, here the symbol ∑ n 1 indicates that the 1st to nth terms are summed.

The above expression can be obtained by resorting to the properties of the same recursion, but there is an easier way to prove its validity. Let's write down the first 2 and last 2 terms of this sum, expressing them in numbers a 1, a n and d, and we get: a 1, a 1 +d,...,a n -d, a n. Now note that if we add the first term to the last, it will be exactly equal to the sum of the second and penultimate terms, that is, a 1 +a n. In a similar way, it can be shown that the same sum can be obtained by adding the third and penultimate terms, and so on. In the case of a pair of numbers in the sequence, we obtain n/2 sums, each of which is equal to a 1 +a n. That is, we obtain the above formula for the algebraic progression for the sum: ∑ n 1 = n*(a 1 +a n)/2.

For an unpaired number of terms n, a similar formula is obtained if you follow the described reasoning. Just remember to add the remaining term, which is in the center of the progression.

Let's show how to use the above formula using the example of a simple progression that was introduced above (3, 5, 7, 9, 11 ...). For example, it is necessary to determine the sum of its first 15 terms. First, let's define a 15. Using the formula for the nth term (see the previous paragraph), we get: a 15 = a 1 + (n-1)*d = 3 + (15-1)*2 = 31. Now we can apply the formula for the sum of an algebraic progression: ∑ 15 1 = 15*(3+31)/2 = 255.

It is interesting to cite an interesting historical fact. The formula for the sum of an arithmetic progression was first obtained by Carl Gauss (the famous German mathematician of the 18th century). When he was only 10 years old, his teacher asked him to find the sum of numbers from 1 to 100. They say that little Gauss solved this problem in a few seconds, noticing that by summing the numbers from the beginning and end of the sequence in pairs, you can always get 101, and since there are 50 such sums, he quickly gave the answer: 50*101 = 5050.

Example of problem solution

To complete the topic of algebraic progression, we will give an example of solving another interesting problem, thereby strengthening the understanding of the topic under consideration. Let a certain progression be given for which the difference d = -3 is known, as well as its 35th term a 35 = -114. It is necessary to find the 7th term of the progression a 7 .

As can be seen from the conditions of the problem, the value of a 1 is unknown, therefore it will not be possible to use the formula for the nth term directly. The recursion method is also inconvenient, which is difficult to implement manually, and there is a high probability of making a mistake. Let's proceed as follows: write out the formulas for a 7 and a 35, we have: a 7 = a 1 + 6*d and a 35 = a 1 + 34*d. Subtract the second from the first expression, we get: a 7 - a 35 = a 1 + 6*d - a 1 - 34*d. It follows: a 7 = a 35 - 28*d. It remains to substitute the known data from the problem statement and write down the answer: a 7 = -114 - 28*(-3) = -30.

Geometric progression

To reveal the topic of the article more fully, we provide a brief description of another type of progression - geometric. In mathematics, this name is understood as a sequence of numbers in which each subsequent term differs from the previous one by a certain factor. Let's denote this factor by the letter r. It is called the denominator of the type of progression under consideration. An example of this number sequence would be: 1, 5, 25, 125, ...

As can be seen from the above definition, algebraic and geometric progressions are similar in idea. The difference between them is that the first one changes more slowly than the second one.

Geometric progression can also be increasing, constant or decreasing. Its type depends on the value of the denominator r: if r>1, then there is an increasing progression, if r<1 - убывающая, наконец, если r = 1 - постоянная, которая в этом случае может также называться постоянной арифметической прогрессией.

Geometric progression formulas

As in the case of algebraic, the formulas of a geometric progression are reduced to determining its nth term and the sum of n terms. Below are these expressions:

• a n = a 1 *r (n-1) - this formula follows from the definition of geometric progression.
• ∑ n 1 = a 1 *(r n -1)/(r-1). It is important to note that if r = 1, then the above formula gives uncertainty, so it cannot be used. In this case, the sum of n terms will be equal to the simple product a 1 *n.

For example, let’s find the sum of only 10 terms of the sequence 1, 5, 25, 125, ... Knowing that a 1 = 1 and r = 5, we get: ∑ 10 1 = 1*(5 10 -1)/4 = 2441406. The resulting value is a clear example of how quickly the geometric progression grows.

Perhaps the first mention of this progression in history is the legend with the chessboard, when a friend of one Sultan, having taught him to play chess, asked for grain for his service. Moreover, the amount of grain should have been as follows: one grain must be placed on the first square of the chessboard, twice as much on the second as on the first, on the third twice as much as on the second, and so on. The Sultan willingly agreed to fulfill this request, but he did not know that he would have to empty all the bins of his country in order to keep his word.