Detailed theory with examples (2019). The sum of the members of an arithmetic progression. Arithmetic progression. average level

So let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many as you like (in our case, them). No matter how many numbers we write, we can always say which one is the first, which is the second, and so on to 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 one.
The number with the number is called the th member of the sequence.

We usually call the entire sequence 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 numerical 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 endless number sequence. The name "arithmetic" was carried over from the theory of continuous proportions, which the ancient Greeks were engaged in.

This is a numerical sequence, each term of which is equal to the previous one, added to the same number. This number is called the difference of the arithmetic progression and is denoted by.

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

a)
b)
c)
d)

Understood? Let's compare our answers:
Is an 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 member. Exists two the way to find it.

1. Method

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

So, the th member 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 be mistaken when adding numbers.
Of course, mathematicians have come up with a way in which you do not need to add the difference of the arithmetic progression to the previous value. Look closely at the drawing you have drawn ... Surely you have already noticed a certain pattern, namely:

For example, let's see how the value of the th member of this arithmetic progression is added:


In other words:

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

Calculated? Compare your notes to the answer:

Pay attention that you got exactly the same number as in the previous method, when we successively added the members of the arithmetic progression to the previous value.
Let's try to "depersonalize" this formula - we will bring it into a general form and get:

Arithmetic progression equation.

Arithmetic progressions are ascending and sometimes decreasing.

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

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

The derived formula is used in calculating the 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 turn out if we use our formula to calculate it:

Since, then:

Thus, we made sure that the formula works in both decreasing and increasing arithmetic progression.
Try to find the th and th terms of this arithmetic progression on your own.

Let's compare the results obtained:

Arithmetic progression property

Let's complicate the task - we will derive the property of the 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, a, 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 if we are given numbers in the condition? Admit it, there is a chance of making a mistake in the calculations.
Now think about whether it is possible to solve this problem in one action using any formula? Of course, yes, and it is her that we will try to withdraw now.

Let's denote the required term of the arithmetic progression as, we know the formula for finding it - this is the same formula we derived at the beginning:
, then:

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

Let's summarize the previous and subsequent members of the progression:

It turns out that the sum of the previous and subsequent members of the progression is the doubled value of the member of the progression located between them. In other words, to find the value of a member of the progression with known previous and consecutive values, it is necessary to add them up and divide by.

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

Well done! You know almost everything about progression! There is only one formula left to learn, which, according to legend, was easily deduced for himself by one of the greatest mathematicians of all time, the "king of mathematicians" - Karl Gauss ...

When Karl Gauss was 9 years old, a teacher, busy checking the work of students in other grades, set the following task in the lesson: "Calculate the sum of all natural numbers from up to (according to other sources up to) inclusive." Imagine the teacher's surprise when one of his students (it was Karl Gauss) gave the correct answer to the problem in a minute, while most of the daredevil's classmates, after long calculations, received the wrong result ...

Young Karl Gauss noticed a certain pattern that you can easily notice.
Let's say we have an arithmetic progression consisting of -th members: We need to find the sum of the given members of the arithmetic progression. Of course, we can manually sum all the values, but what if in the task it is necessary to find the sum of its members, as Gauss was looking for?

Let's draw a given progression. Look closely at the highlighted numbers and try to perform various mathematical operations with them.


Have you tried it? What have you noticed? Right! Their sums are equal

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

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

Well done! Now let's return to the problem that was given to Karl Gauss: calculate yourself what is the sum of the numbers starting from the -th, and the sum of the numbers starting from the -th.

How much did you get it?
Gauss found that the sum of the members is equal, and the sum of the members. Is that how you decided?

In fact, the formula for the sum of the members of an arithmetic progression was proved by the ancient Greek scientist Diophantus in the 3rd century, and throughout this time, witty people were using the properties of an arithmetic progression to the utmost.
For example, imagine Ancient Egypt and the largest construction site of that time - the construction of the pyramid ... The figure shows one side of it.

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


Isn't it an arithmetic progression? Calculate how many blocks are needed to build one wall if block bricks are placed in the base. I hope you won't count by running your finger across the monitor, do you remember the last formula and everything we said about the arithmetic progression?

In this case, the progression looks like this:.
Difference of arithmetic progression.
The number of members of the arithmetic progression.
Let's substitute our data into the last formulas (we will count 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. Did it come together? Well done, you have mastered the sum of the terms of the 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:

Workout

Tasks:

1. Masha is getting in shape by summer. Every day she increases the number of squats by. How many times will Masha squat in weeks, if at the first workout she did squats.
2. What is the sum of all the odd numbers contained in.
3. When storing logs, lumberjacks 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 logs serve as the basis of the masonry.

Answers:

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

   Answer: After two weeks, Masha should squat once a day.

2. First odd number, last number.
   Difference of arithmetic progression.
   The number of odd numbers in is half, however, we will check this fact using the formula for finding the -th term of an arithmetic progression:

   The numbers do contain odd numbers.
   Substitute the available data into the formula:

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

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

   Answer: There are logs in the masonry.

Let's summarize

1. - a numerical sequence in which the difference between adjacent numbers is the same and equal. It can be increasing and decreasing.
2. Finding formula The th member of the 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 the progression.
4. The sum of the members 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 as you like. But you can always say which one is the first, which is the 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 the only one. And we will not assign this number to any other number from this set.

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

We usually call the entire sequence 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 given by some formula. For example, the formula

specifies 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). Or (, difference).

Nth term formula

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

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

Well, what is the formula now?

In each line we add to, multiplied by some number. For what? 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? And here's what:

(it is because it is called the difference, which is equal to the difference of the consecutive members of the progression).

So the formula is:

Then the hundredth term is:

What is the sum of all natural numbers from to?

According to legend, the great mathematician Karl Gauss, being 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 the last but one is the same, the sum of the third and third from the end is the same, and so on. How many such pairs will there be? 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 would be:

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

Solution:

The first such number is. Each next is obtained by adding to the previous number. Thus, the numbers of interest to us form an arithmetic progression with the first term and the difference.

The th term formula for this progression is:

How many members are in the progression if they all have to be double digits?

Very easy: .

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

Answer: .

Now decide for yourself:

1. Every day, the athlete runs more m than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
2. A cyclist drives more kilometers every day than the previous one. On the first day, he drove km. How many days does he need to travel to cover the km? How many kilometers will he travel in the last day of the journey?
3. The price of a refrigerator in a store decreases by the same amount every year. Determine how much the price of the refrigerator has decreased every 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 members of this progression:
   .
   Answer:
2. It is given here:, it is necessary to find.
   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 distance traveled for the last day using the th term formula:
   (km).
   Answer:

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

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN

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

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

For example:

The formula for finding the n-th term of an arithmetic progression

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

Property of members of an arithmetic progression

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

The sum of the members 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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When studying algebra in a comprehensive school (grade 9), one of the important topics is the study of number sequences, which include progressions - geometric and arithmetic. In this article, we will consider the arithmetic progression and examples with solutions.

What is an arithmetic progression?

To understand this, it is necessary to give a definition of the considered progression, as well as to give the basic formulas that will be further used in solving problems.

It is known that in some algebraic progression the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.

Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1. We substitute in it the known data from the condition, that is, the numbers a 1 and a 7, we have: 18 = 6 + 6 * d. From this expression, you can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, the answer to the first part of the problem.

To restore a sequence up to 7 terms, you should use the definition of an algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d, and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2 = 8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16, a 7 = 18.

Example # 3: making a progression

Let us complicate the condition of the problem even more. Now it is necessary to answer the question of how to find the arithmetic progression. You can give the following example: given two numbers, for example, - 4 and 5. It is necessary to compose an algebraic progression so that three more terms fit between these.

Before starting to solve this problem, it is necessary to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 = -4 and a 5 = 5. Having established this, we proceed to the problem, which is similar to the previous one. Again, for the n-th term, we use the formula, we get: a 5 = a 1 + 4 * d. From where: d = (a 5 - a 1) / 4 = (5 - (-4)) / 4 = 2.25. Here we received not an integer value of the difference, but it is a rational number, so the formulas for the algebraic progression remain the same.

Now add the found difference to a 1 and restore the missing members of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 = 2.75 + 2.25 = 5, which coincided with the condition of the problem.

Example # 4: the first term of the progression

Let's continue to give examples of arithmetic progression with a solution. In all the previous problems, the first number of the algebraic progression was known. Now consider a problem of a different type: let there be given two numbers, where a 15 = 50 and a 43 = 37. It is necessary to find the number from which this sequence begins.

The formulas used so far assume knowledge of a 1 and d. Nothing is known about these numbers in the problem statement. Nevertheless, we write out expressions for each member about which there is information: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. Received two equations, in which 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.

This system is easiest to solve if you express a 1 in each equation, and then compare the resulting expressions. The first equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 = a 43 - 42 * d = 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d = 37 - 42 * d, whence the difference d = (37 - 50) / (42 - 14) = - 0.464 (only 3 decimal places are given).

Knowing d, you can use any of the 2 above expressions for a 1. For example, the first one: a 1 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.

If you have doubts about the result, you can check it, for example, determine the 43 term of the progression, which is specified in the condition. We get: a 43 = a 1 + 42 * d = 56.496 + 42 * (- 0.464) = 37.008. A small error is due to the fact that the calculations used rounding to thousandths.

Example # 5: amount

Now let's look at some examples with solutions for the sum of an arithmetic progression.

Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How do you calculate the sum of these 100 numbers?

Thanks to the development of computer technology, it is possible to solve this problem, that is, to add up all the numbers sequentially, which the computer will do as soon as a person presses the Enter key. However, the problem can be solved in the mind, if we pay attention that the presented series of numbers is an algebraic progression, and its difference is 1. Applying the formula for the sum, we get: S n = n * (a 1 + an) / 2 = 100 * (1 + 100) / 2 = 5050.

It is curious to note that this problem is called "Gaussian", since at the beginning of the 18th century the famous German, while still only 10 years old, was able to solve it in his head in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add in pairs the numbers on the edges of the sequence, you always get one result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since of these amounts will be exactly 50 (100/2), then to get the correct answer, it is enough to multiply 50 by 101.

Example # 6: sum of members from n to m

Another typical example of the sum of an arithmetic progression is the following: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its members from 8 to 14 will equal.

The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then adding them sequentially. Since there are few terms, this method is not laborious enough. Nevertheless, it is proposed to solve this problem by the second method, which is more universal.

The idea is to obtain a formula for the sum of the algebraic progression between the terms m and n, where n> m are integers. Let us write out two expressions for the sum for both cases:

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

Since n> m, it is obvious that the 2 sum includes the first. The last conclusion means that if we take the difference between these sums, and add to it the term a m (in the case of taking the difference, it is subtracted from the sum S n), then we get the necessary answer to the problem. We have: S mn = S n - S m + am = n * (a 1 + an) / 2 - m * (a 1 + am) / 2 + am = a 1 * (n - m) / 2 + an * n / 2 + am * (1- m / 2). In this expression it is necessary to substitute the formulas for a n and a m. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d * (3 * m - m 2 - 2) / 2.

The resulting formula is somewhat cumbersome; nevertheless, the sum of S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.

As can be seen from the solutions presented, all problems are based on knowledge of the expression for the nth term and the formula for the sum of the set of the first terms. Before proceeding with the solution of any of these problems, it is recommended to carefully read the condition, clearly understand what needs to be found, and only then proceed to the solution.

Another tip is to strive for simplicity, that is, if you can answer a question without using complex mathematical calculations, then you need to do just that, since in this case the probability of making a mistake is less. For example, in an example of an arithmetic progression with solution # 6, one could stop at the formula S mn = n * (a 1 + an) / 2 - m * (a 1 + am) / 2 + am, and break the general problem into separate subtasks (in this case, first find the members an and am).

If there are doubts about the result obtained, it is recommended to check it, as was done in some of the above examples. We figured out how to find the arithmetic progression. If you figure it out, it's not that difficult.

Infinite arithmetic progression a 1 , a 2 , ..., a n, ... consists of distinct natural numbers.

a) Is there such a progression in which among the numbers a 1 , a 2 , ..., a 7 are exactly three numbers divisible by 36?

b) Is there such a progression in which among the numbers a 1 , a 2 , ..., a 30 are exactly 9 numbers divisible by 36?

c) For what is the greatest natural n it could be that among the numbers a 1 , a 2 , ..., a 2n more multiples of 36 than numbers a 2n + 1 , a 2n + 2 , ..., a 5n ?

Solution.

a) A suitable example is a progression with the first term 18 and the difference 18. Among the first seven members (18, 36, 54, 72, 90, 108, 126), exactly three are divisible by 36.

b) Denote by d difference of arithmetic progression a 1 , a 2 , ..., a n, .... It follows from the condition that d- natural number. Let be m and n- integers, m > n, Gcd ( d, 36) denotes the greatest common divisor of numbers d and 36. We have

Therefore, the difference a m − a n is divisible by 36 if and only if the difference m − n is divisible by So, if among the members of the arithmetic progression a 1 , a 2 , ..., a n, ... are multiples of 36, then these are terms with numbers of the form where q- number of the first term, multiple of a p runs through all non-negative integers. Therefore, among any k a 1 , a 2 , ..., a n, ... exactly one will be divisible by 36. If then among the numbers a 1 , a 2 , ..., a 30 will be at least 10 multiples of 36. If then among the numbers a 1 , a 2 , ..., a 30 there will be no more than 8 numbers divisible by 36. This means that there is no such progression in which among the numbers a 1 , a 2 , ..., a 30 is exactly 9 numbers divisible by 36.

c) Denote by [ x] integer part of a number x- the largest integer not exceeding x... As proved in point b), among any k consecutive members of the progression a 1 , a 2 , ..., a n, ... exactly one will be divisible by 36, where d- the difference of the arithmetic progression.

Hence, among the numbers a 1 , a 2 , ..., a 2n no more than numbers will be multiples of 36. Similarly, among the numbers a 2n + 1 , a 2n + 2 , ..., a 5n multiples of 36 will be at least numbers. The inequality is satisfied if and only if Let this equality be satisfied. Then the difference between the numbers and is less than 1. We get that and Means, and Since the number k does not exceed 36, whence it follows that Consider a progression with the first term 27 and the difference 1. Then among the numbers a 1 , a 2 , ..., a 46 exactly two are divisible by 36 ( a 10 = 36 and a 46 = 72). Among the numbers a 47 , a 48 , ..., a 115 is exactly one divisible by 36 ( a 82 = 108). This example shows that n can be equal to 23.

Answer: a) Yes, for example, progression 18, 36, 54, 72, 90, 108, 126, ...; b) no; c) 23.