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

What is this progression?

Before proceeding to consider the question (how to find the sum of an arithmetic progression), it is worth understanding what will be discussed.

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, translated into the language of mathematics, takes the form:

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

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

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

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

What is the sum of an arithmetic progression: formula

Before giving a 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 members in the progression (10), it is possible to solve the problem head-on, that is, to 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 in the series. Then multiplying the number of sums (5) by the result of each sum (11), you will come to the result obtained in the first example.

If we generalize this reasoning, then 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 up 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 set by his school teacher: sum the first 100 integers.

Sum of elements from m to n: formula

The formula given in the previous paragraph gives an answer to the question of how to find the sum of an arithmetic progression (first elements), but often in problems it is necessary to sum up 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 m-th to n-th. To solve the problem, a given segment from m to n of the progression should be presented in the form of a new numerical series. In this representation, the mth term a m will be the first, and a n will be n- (m-1). In this case, applying the standard formula for the sum, you get the following expression:

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 formulas given.

Below is a numerical sequence, you should find the sum of its members, starting with 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 considered algebraic progression, and also knowing which numbers in the row 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.

When studying algebra in a general education school (grade 9), one of the important topics is the study of numerical 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, we have answered 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 make 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 two numbers be given, 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.

The easiest way to solve this system is to 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", because 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 their sequential summation. 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 get 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 given, 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 is required 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 it was done in some of the examples given. We figured out how to find the arithmetic progression. If you figure it out, it's not that difficult.

Before we start deciding arithmetic progression problems, consider what a number sequence is, since an arithmetic progression is a special case of a number sequence.

A numerical sequence is a numerical set, each element of which has its own ordinal number... The elements of this set are called the members of the sequence. The ordinal number of the sequence element is indicated by the index:

The first element of the sequence;

Fifth element of the sequence;

- "nth" element of the sequence, i.e. the item "in the queue" n.

There is a relationship between the value of a sequence element and its ordinal number. Therefore, we can think of a sequence as a function whose argument is the ordinal number of an element of the sequence. In other words, we can say that a sequence is a function of a natural argument:

The sequence can be set in three ways:

1 . The sequence can be set using a table. In this case, we simply set the value of each member of the sequence.

For example, Someone decided to take up personal time management, and to begin with, calculate how much time he spends on VKontakte during the week. Writing down the time in the table, he will receive a sequence consisting of seven elements:

The first line of the table contains the number of the day of the week, the second - the time in minutes. We see that, that is, on Monday, Someone spent 125 minutes on VKontakte, that is, on Thursday - 248 minutes, and, that is, on Friday, only 15.

2 . The sequence can be specified using the nth term formula.

In this case, the dependence of the value of the sequence element on its number is expressed directly in the form of a formula.

For example, if, then

To find the value of an element of a sequence with a given number, we substitute the element number into the formula of the nth term.

We do the same if we need to find the value of a function if the value of the argument is known. We substitute the value of the argument instead into the equation of the function:

If, for example, , then

Once again, I note that in a sequence, unlike an arbitrary numeric function, only a natural number can be an argument.

3 ... A sequence can be specified using a formula that expresses the dependence of the value of the sequence member numbered on the value of the previous members. In this case, it is not enough for us to know only the number of the sequence member in order to find its value. We need to specify the first member or the first few members of the sequence.

For example, consider the sequence ,

We can find the values ​​of the members of the sequence in sequence starting with the third:

That is, each time, to find the value of the n-th member of the sequence, we go back to the previous two. This way of sequencing is called recurrent, from the Latin word recurro- come back.

Now we can define an arithmetic progression. Arithmetic progression is a simple special case of a number sequence.

Arithmetic progression a numerical sequence is called, each member of which, starting from the second, is equal to the previous one, added with the same number.

The number is called difference of arithmetic progression... The difference in arithmetic progression can be positive, negative, or zero.

If title = "(! LANG: d> 0">, то каждый член арифметической прогрессии больше предыдущего, и прогрессия является !} increasing.

For example, 2; 5; eight; eleven;...

If, then each member of the arithmetic progression is less than the previous one, and the progression is diminishing.

For example, 2; -1; -4; -7; ...

If, then all members of the progression are equal to the same number, and the progression is stationary.

For example, 2; 2; 2; 2; ...

The main property of the arithmetic progression:

Let's look at the picture.

We see that

, and at the same time

Adding these two equalities, we get:

.

Divide both sides of the equality by 2:

So, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of two neighboring ones:

Moreover, since

, and at the same time

, then

, and therefore

Each member of the arithmetic progression starting with title = "(! LANG: k> l">, равен среднему арифметическому двух равноотстоящих. !}

Formula of the th member.

We see that for the members of the arithmetic progression, the following relations are fulfilled:

and finally

We got the formula of the nth term.

IMPORTANT! Any member of the arithmetic progression can be expressed in terms of and. Knowing the first term and the difference of the arithmetic progression, you can find any of its members.

The sum of n members of an arithmetic progression.

In an arbitrary arithmetic progression, the sums of the members equidistant from the extreme are equal to each other:

Consider an arithmetic progression with n terms. Let the sum of n members of this progression be.

Let us arrange the members of the progression first in ascending order of numbers, and then in descending order:

Let's add in pairs:

The sum in each parenthesis is equal, the number of pairs is n.

We get:

So, the sum of n terms of an arithmetic progression can be found by the formulas:

Consider solving problems for arithmetic progression.

1 . The sequence is given by the nth term formula: . Prove that this sequence is an arithmetic progression.

Let us prove that the difference between two adjacent members of the sequence is equal to the same number.

We found that the difference between two adjacent members of the sequence does not depend on their number and is constant. Therefore, by definition, this sequence is an arithmetic progression.

2 . You are given an arithmetic progression -31; -27; ...

a) Find 31 members of the progression.

b) Determine if the number 41 is included in this progression.

a) We see that;

Let's write the formula of the nth term for our progression.

In general

In our case , therefore

We get:

b) Suppose 41 is a member of a sequence. Let's find his number. To do this, we solve the equation:

We got a natural value of n, therefore, yes, the number 41 is a member of the progression. If the found value of n were not a natural number, then we would answer that the number 41 is NOT a member of the progression.

3 ... a) Between the numbers 2 and 8, insert 4 numbers so that they, together with the given numbers, make an arithmetic progression.

b) Find the sum of the members of the resulting progression.

a) Insert four numbers between the numbers 2 and 8:

We got an arithmetic progression with 6 members.

Let's find the difference of this progression. To do this, we will use the formula for the nth term:

Now it's easy to find the values ​​of numbers:

3,2; 4,4; 5,6; 6,8

b)

Answer: a) yes; b) 30

4. The truck transports a batch of crushed stone weighing 240 tons, daily increasing the transportation rate by the same number of tons. It is known that during the first day 2 tons of crushed stone were transported. Determine how many tons of rubble were transported on the twelfth day if all the work was completed in 15 days.

According to the condition of the problem, the amount of rubble transported by the truck increases by the same number every day. Therefore, we are dealing with an arithmetic progression.

Let us formulate this problem in terms of an arithmetic progression.

During the first day, 2 tons of crushed stone were transported: a_1 = 2.

All work was completed in 15 days:.

The truck transports a batch of crushed stone weighing 240 tons:

We need to find.

First, find the difference in the progression. Let's use the formula for the sum of n terms of the progression.

In our case:

Yes, yes: the arithmetic progression is not a toy for you :)

Well, friends, if you are reading this text, then the internal cap-evidence tells me that you do not yet know what an arithmetic progression is, but you really (no, like this: SOOOOO!) Want to know. Therefore, I will not torment you with long introductions and will immediately get down to business.

Let's start with a couple of examples. Consider several sets of numbers:

• 1; 2; 3; 4; ...
• 15; 20; 25; 30; ...
• $ \ sqrt (2); \ 2 \ sqrt (2); \ 3 \ sqrt (2); ... $

What do all these sets have in common? At first glance, nothing. But actually there is something. Namely: each next element differs from the previous one by the same number.

Judge for yourself. The first set is simply consecutive numbers, each next one more than the previous one. In the second case, the difference between the adjacent numbers is already equal to five, but this difference is still constant. In the third case, roots in general. However, $ 2 \ sqrt (2) = \ sqrt (2) + \ sqrt (2) $, and $ 3 \ sqrt (2) = 2 \ sqrt (2) + \ sqrt (2) $, i.e. and in this case, each next element simply increases by $ \ sqrt (2) $ (and don't be afraid that this number is irrational).

So: all such sequences are called arithmetic progressions. Let's give a strict definition:

Definition. A sequence of numbers in which each next differs from the previous by exactly the same amount is called an arithmetic progression. The very amount by which the numbers differ is called the difference of the progression and is most often denoted by the letter $ d $.

Designation: $ \ left (((a) _ (n)) \ right) $ - the progression itself, $ d $ - its difference.

And just a couple of important remarks. First, only orderly sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. You cannot rearrange or swap numbers.

Secondly, the sequence itself can be either finite or infinite. For example, the set (1; 2; 3) is obviously a finite arithmetic progression. But if you write something in the spirit (1; 2; 3; 4; ...) - this is already an endless progression. The ellipsis after the four, as it were, hints that there are still quite a few numbers going on. Infinitely many, for example. :)

I would also like to note that progressions are increasing and decreasing. We have already seen the increasing ones - the same set (1; 2; 3; 4; ...). And here are examples of decreasing progressions:

• 49; 41; 33; 25; 17; ...
• 17,5; 12; 6,5; 1; −4,5; −10; ...
• $ \ sqrt (5); \ \ sqrt (5) -1; \ \ sqrt (5) -2; \ \ sqrt (5) -3; ... $

Okay, okay: this last example might seem overly complicated. But the rest, I think you understand. Therefore, we will introduce new definitions:

Definition. An arithmetic progression is called:

1. increasing if each next element is larger than the previous one;
2. decreasing if, on the contrary, each subsequent element is less than the previous one.

In addition, there are so-called "stationary" sequences - they consist of the same repeating number. For example, (3; 3; 3; ...).

There remains only one question: how to distinguish an increasing progression from a decreasing one? Fortunately, it all depends on the sign of the number $ d $, i.e. difference progression:

1. If $ d \ gt 0 $, then the progression is increasing;
2. If $ d \ lt 0 $, then the progression is obviously decreasing;
3. Finally, there is the case $ d = 0 $ - in this case the whole progression is reduced to a stationary sequence of identical numbers: (1; 1; 1; 1; ...), etc.

Let's try to calculate the difference $ d $ for the three decreasing progressions given above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract the number on the left from the number on the right. It will look like this:

• 41−49=−8;
• 12−17,5=−5,5;
• $ \ sqrt (5) -1- \ sqrt (5) = - 1 $.

As you can see, in all three cases, the difference really turned out to be negative. And now that we have more or less figured out the definitions, it's time to figure out how progressions are described and what their properties are.

Progression members and recurrent formula

Since the elements of our sequences cannot be swapped, they can be numbered:

\ [\ left (((a) _ (n)) \ right) = \ left \ (((a) _ (1)), \ ((a) _ (2)), ((a) _ (3 )), ... \ right \) \]

The individual elements of this set are called members of the progression. They are indicated by a number: the first term, the second term, etc.

In addition, as we already know, the neighboring members of the progression are related by the formula:

\ [((a) _ (n)) - ((a) _ (n-1)) = d \ Rightarrow ((a) _ (n)) = ((a) _ (n-1)) + d \]

In short, to find the $ n $ th term in the progression, you need to know the $ n-1 $ th term and the $ d $ difference. Such a formula is called recurrent, since with its help you can find any number, only knowing the previous one (and in fact - all the previous ones). This is very inconvenient, so there is a more tricky formula that reduces any calculations to the first term and the difference:

\ [((a) _ (n)) = ((a) _ (1)) + \ left (n-1 \ right) d \]

Surely you have already met this formula. They love to give it in all sorts of reference books and reshebniks. And in any sensible textbook on mathematics, she goes one of the first.

Nevertheless, I suggest we practice a little.

Problem number 1. Write down the first three terms of the arithmetic progression $ \ left (((a) _ (n)) \ right) $ if $ ((a) _ (1)) = 8, d = -5 $.

Solution. So, we know the first term $ ((a) _ (1)) = 8 $ and the difference of the progression $ d = -5 $. Let's use the formula just given and substitute $ n = 1 $, $ n = 2 $ and $ n = 3 $:

\ [\ begin (align) & ((a) _ (n)) = ((a) _ (1)) + \ left (n-1 \ right) d; \\ & ((a) _ (1)) = ((a) _ (1)) + \ left (1-1 \ right) d = ((a) _ (1)) = 8; \\ & ((a) _ (2)) = ((a) _ (1)) + \ left (2-1 \ right) d = ((a) _ (1)) + d = 8-5 = 3; \\ & ((a) _ (3)) = ((a) _ (1)) + \ left (3-1 \ right) d = ((a) _ (1)) + 2d = 8-10 = -2. \\ \ end (align) \]

Answer: (8; 3; −2)

That's all! Please note: our progression is decreasing.

Of course, $ n = 1 $ could not have been substituted - the first term is already known to us. However, substituting one, we made sure that our formula works even for the first term. In other cases, it all boiled down to trivial arithmetic.

Problem number 2. Write out the first three terms of the arithmetic progression if its seventh term is −40 and the seventeenth term is −50.

Solution. Let's write down the condition of the problem in the usual terms:

\ [((a) _ (7)) = - 40; \ quad ((a) _ (17)) = - 50. \]

\ [\ left \ (\ begin (align) & ((a) _ (7)) = ((a) _ (1)) + 6d \\ & ((a) _ (17)) = ((a) _ (1)) + 16d \\ \ end (align) \ right. \]

\ [\ left \ (\ begin (align) & ((a) _ (1)) + 6d = -40 \\ & ((a) _ (1)) + 16d = -50 \\ \ end (align) \ right. \]

I put the sign of the system because these requirements must be fulfilled simultaneously. And now note that if we subtract the first from the second equation (we have the right to do this, since we have a system), we get this:

\ [\ begin (align) & ((a) _ (1)) + 16d- \ left (((a) _ (1)) + 6d \ right) = - 50- \ left (-40 \ right); \\ & ((a) _ (1)) + 16d - ((a) _ (1)) - 6d = -50 + 40; \\ & 10d = -10; \\ & d = -1. \\ \ end (align) \]

That's how easy we found the difference in the progression! It remains to substitute the found number into any of the equations of the system. For example, in the first:

\ [\ begin (matrix) ((a) _ (1)) + 6d = -40; \ quad d = -1 \\ \ Downarrow \\ ((a) _ (1)) - 6 = -40; \\ ((a) _ (1)) = - 40 + 6 = -34. \\ \ end (matrix) \]

Now, knowing the first term and the difference, it remains to find the second and third terms:

\ [\ begin (align) & ((a) _ (2)) = ((a) _ (1)) + d = -34-1 = -35; \\ & ((a) _ (3)) = ((a) _ (1)) + 2d = -34-2 = -36. \\ \ end (align) \]

Ready! The problem has been solved.

Answer: (-34; -35; -36)

Pay attention to an interesting property of the progression that we discovered: if we take the $ n $ th and $ m $ th terms and subtract them from each other, we get the difference of the progression multiplied by the number $ n-m $:

\ [((a) _ (n)) - ((a) _ (m)) = d \ cdot \ left (n-m \ right) \]

A simple but very useful property that you should definitely know - with its help, you can significantly speed up the solution of many problems in progressions. Here's a prime example:

Problem number 3. The fifth term of the arithmetic progression is 8.4, and its tenth term is 14.4. Find the fifteenth term of this progression.

Solution. Since $ ((a) _ (5)) = 8.4 $, $ ((a) _ (10)) = 14.4 $, and you need to find $ ((a) _ (15)) $, then we note following:

\ [\ begin (align) & ((a) _ (15)) - ((a) _ (10)) = 5d; \\ & ((a) _ (10)) - ((a) _ (5)) = 5d. \\ \ end (align) \]

But by condition $ ((a) _ (10)) - ((a) _ (5)) = 14.4-8.4 = $ 6, therefore $ 5d = $ 6, whence we have:

\ [\ begin (align) & ((a) _ (15)) - 14.4 = 6; \\ & ((a) _ (15)) = 6 + 14.4 = 20.4. \\ \ end (align) \]

Answer: 20.4

That's all! We did not need to compose some systems of equations and calculate the first term and the difference - everything was solved in just a couple of lines.

Now let's consider another type of tasks - to find negative and positive members of the progression. It is no secret that if the progression increases, while the first term is negative, then sooner or later positive terms will appear in it. And on the contrary: the members of the decreasing progression will sooner or later become negative.

At the same time, it is far from always possible to grope this moment "head-on", sequentially going through the elements. Often, problems are designed in such a way that without knowing the formulas, the calculations would take several sheets - we would just fall asleep while we found the answer. Therefore, we will try to solve these problems in a faster way.

Problem number 4. How many negative terms are in the arithmetic progression -38.5; −35.8; ...?

Solution. So, $ ((a) _ (1)) = - 38.5 $, $ ((a) _ (2)) = - 35.8 $, from where we immediately find the difference:

Note that the difference is positive, so the progression increases. The first term is negative, so at some point we really will stumble upon positive numbers. The only question is when it will happen.

Let's try to find out: how long (i.e. up to what natural number $ n $) the negativity of the terms is preserved:

\ [\ begin (align) & ((a) _ (n)) \ lt 0 \ Rightarrow ((a) _ (1)) + \ left (n-1 \ right) d \ lt 0; \\ & -38.5+ \ left (n-1 \ right) \ cdot 2.7 \ lt 0; \ quad \ left | \ cdot 10 \ right. \\ & -385 + 27 \ cdot \ left (n-1 \ right) \ lt 0; \\ & -385 + 27n-27 \ lt 0; \\ & 27n \ lt 412; \\ & n \ lt 15 \ frac (7) (27) \ Rightarrow ((n) _ (\ max)) = 15. \\ \ end (align) \]

The last line requires clarification. So, we know that $ n \ lt 15 \ frac (7) (27) $. On the other hand, we will only be satisfied with integer values ​​of the number (moreover: $ n \ in \ mathbb (N) $), so the largest allowed number is exactly $ n = 15 $, and in no case is 16.

Problem number 5. In arithmetic progression $ (() _ (5)) = - 150, (() _ (6)) = - 147 $. Find the number of the first positive term of this progression.

It would be exactly the same problem as the previous one, but we don't know $ ((a) _ (1)) $. But the neighboring terms are known: $ ((a) _ (5)) $ and $ ((a) _ (6)) $, so we can easily find the difference of the progression:

In addition, we will try to express the fifth term in terms of the first and the difference according to the standard formula:

\ [\ begin (align) & ((a) _ (n)) = ((a) _ (1)) + \ left (n-1 \ right) \ cdot d; \\ & ((a) _ (5)) = ((a) _ (1)) + 4d; \\ & -150 = ((a) _ (1)) + 4 \ cdot 3; \\ & ((a) _ (1)) = - 150-12 = -162. \\ \ end (align) \]

Now we proceed by analogy with the previous task. We find out at what point in our sequence there will be positive numbers:

\ [\ begin (align) & ((a) _ (n)) = - 162+ \ left (n-1 \ right) \ cdot 3 \ gt 0; \\ & -162 + 3n-3 \ gt 0; \\ & 3n \ gt 165; \\ & n \ gt 55 \ Rightarrow ((n) _ (\ min)) = 56. \\ \ end (align) \]

The smallest integer solution to this inequality is 56.

Please note: in the last task, everything was reduced to a strict inequality, so the $ n = 55 $ option will not suit us.

Now that we have learned how to solve simple problems, let's move on to more complex ones. But first, let's study another very useful property of arithmetic progressions, which will save us a lot of time and unequal cells in the future. :)

Arithmetic mean and equal indents

Consider several consecutive members of the increasing arithmetic progression $ \ left (((a) _ (n)) \ right) $. Let's try to mark them on the number line:

Members of an arithmetic progression on a number line

I specifically noted arbitrary terms $ ((a) _ (n-3)), ..., ((a) _ (n + 3)) $, not any $ ((a) _ (1)) , \ ((a) _ (2)), \ ((a) _ (3)) $, etc. Because the rule, which I will now talk about, works the same for any "segments".

And the rule is very simple. Let's remember the recursion formula and write it down for all marked members:

\ [\ begin (align) & ((a) _ (n-2)) = ((a) _ (n-3)) + d; \\ & ((a) _ (n-1)) = ((a) _ (n-2)) + d; \\ & ((a) _ (n)) = ((a) _ (n-1)) + d; \\ & ((a) _ (n + 1)) = ((a) _ (n)) + d; \\ & ((a) _ (n + 2)) = ((a) _ (n + 1)) + d; \\ \ end (align) \]

However, these equalities can be rewritten differently:

\ [\ begin (align) & ((a) _ (n-1)) = ((a) _ (n)) - d; \\ & ((a) _ (n-2)) = ((a) _ (n)) - 2d; \\ & ((a) _ (n-3)) = ((a) _ (n)) - 3d; \\ & ((a) _ (n + 1)) = ((a) _ (n)) + d; \\ & ((a) _ (n + 2)) = ((a) _ (n)) + 2d; \\ & ((a) _ (n + 3)) = ((a) _ (n)) + 3d; \\ \ end (align) \]

Well, so what? And the fact that the terms $ ((a) _ (n-1)) $ and $ ((a) _ (n + 1)) $ lie at the same distance from $ ((a) _ (n)) $. And this distance is equal to $ d $. The same can be said about the members $ ((a) _ (n-2)) $ and $ ((a) _ (n + 2)) $ - they are also removed from $ ((a) _ (n)) $ the same distance equal to $ 2d $. You can continue indefinitely, but the meaning is well illustrated by the picture.

The members of the progression lie at the same distance from the center

What does this mean for us? This means that you can find $ ((a) _ (n)) $ if the neighboring numbers are known:

\ [((a) _ (n)) = \ frac (((a) _ (n-1)) + ((a) _ (n + 1))) (2) \]

We came up with an excellent statement: every member of the arithmetic progression is equal to the arithmetic mean of the neighboring terms! Moreover: we can deviate from our $ ((a) _ (n)) $ left and right not one step, but $ k $ steps - and still the formula will be correct:

\ [((a) _ (n)) = \ frac (((a) _ (n-k)) + ((a) _ (n + k))) (2) \]

Those. we can easily find some $ ((a) _ (150)) $ if we know $ ((a) _ (100)) $ and $ ((a) _ (200)) $, because $ (( a) _ (150)) = \ frac (((a) _ (100)) + ((a) _ (200))) (2) $. At first glance, it may seem that this fact does not give us anything useful. However, in practice, many problems are specially "sharpened" for the use of the arithmetic mean. Take a look:

Problem number 6. Find all values ​​of $ x $ for which the numbers $ -6 ((x) ^ (2)) $, $ x + 1 $ and $ 14 + 4 ((x) ^ (2)) $ are consecutive members of the arithmetic progression (in order).

Solution. Since the indicated numbers are members of the progression, the condition of the arithmetic mean is satisfied for them: the central element $ x + 1 $ can be expressed in terms of adjacent elements:

\ [\ begin (align) & x + 1 = \ frac (-6 ((x) ^ (2)) + 14 + 4 ((x) ^ (2))) (2); \\ & x + 1 = \ frac (14-2 ((x) ^ (2))) (2); \\ & x + 1 = 7 - ((x) ^ (2)); \\ & ((x) ^ (2)) + x-6 = 0. \\ \ end (align) \]

The result is a classic quadratic equation. Its roots: $ x = 2 $ and $ x = -3 $ - these are the answers.

Answer: −3; 2.

Problem number 7. Find the $$ values ​​for which the numbers $ -1; 4-3; (() ^ (2)) + 1 $ make an arithmetic progression (in that order).

Solution. Again, we express the middle term in terms of the arithmetic mean of the neighboring terms:

\ [\ begin (align) & 4x-3 = \ frac (x-1 + ((x) ^ (2)) + 1) (2); \\ & 4x-3 = \ frac (((x) ^ (2)) + x) (2); \ quad \ left | \ cdot 2 \ right .; \\ & 8x-6 = ((x) ^ (2)) + x; \\ & ((x) ^ (2)) - 7x + 6 = 0. \\ \ end (align) \]

Again the quadratic equation. And again there are two roots: $ x = 6 $ and $ x = 1 $.

Answer: 1; 6.

If in the process of solving a problem you get out some brutal numbers, or you are not completely sure of the correctness of the answers found, then there is a wonderful technique that allows you to check: did we solve the problem correctly?

For example, in problem no. 6 we received answers -3 and 2. How to check that these answers are correct? Let's just plug them in and see what happens. Let me remind you that we have three numbers ($ -6 (() ^ (2)) $, $ + 1 $ and $ 14 + 4 (() ^ (2)) $), which must form an arithmetic progression. Substitute $ x = -3 $:

\ [\ begin (align) & x = -3 \ Rightarrow \\ & -6 ((x) ^ (2)) = - 54; \\ & x + 1 = -2; \\ & 14 + 4 ((x) ^ (2)) = 50. \ end (align) \]

Received numbers -54; −2; 50, which differ by 52, is undoubtedly an arithmetic progression. The same thing happens for $ x = 2 $:

\ [\ begin (align) & x = 2 \ Rightarrow \\ & -6 ((x) ^ (2)) = - 24; \\ & x + 1 = 3; \\ & 14 + 4 ((x) ^ (2)) = 30. \ end (align) \]

Again a progression, but with a difference of 27. Thus, the problem is solved correctly. Those interested can check the second problem on their own, but I'll say right away: everything is correct there too.

In general, while solving the last problems, we came across another interesting fact, which also needs to be remembered:

If three numbers are such that the second is the arithmetic mean of the first and the last, then these numbers form an arithmetic progression.

In the future, understanding this statement will allow us to literally "construct" the necessary progressions, based on the condition of the problem. But before we get down to such "construction", we should pay attention to one more fact, which directly follows from what has already been considered.

Grouping and sum of elements

Let's go back to the number axis again. Let us note there several members of the progression, between which, perhaps. there are a lot of other members:

The number line has 6 elements marked

Let's try to express "left tail" in terms of $ ((a) _ (n)) $ and $ d $, and "right tail" in terms of $ ((a) _ (k)) $ and $ d $. It's very simple:

\ [\ begin (align) & ((a) _ (n + 1)) = ((a) _ (n)) + d; \\ & ((a) _ (n + 2)) = ((a) _ (n)) + 2d; \\ & ((a) _ (k-1)) = ((a) _ (k)) - d; \\ & ((a) _ (k-2)) = ((a) _ (k)) - 2d. \\ \ end (align) \]

Now, note that the following sums are equal:

\ [\ begin (align) & ((a) _ (n)) + ((a) _ (k)) = S; \\ & ((a) _ (n + 1)) + ((a) _ (k-1)) = ((a) _ (n)) + d + ((a) _ (k)) - d = S; \\ & ((a) _ (n + 2)) + ((a) _ (k-2)) = ((a) _ (n)) + 2d + ((a) _ (k)) - 2d = S. \ end (align) \]

Simply put, if we consider as a start two elements of the progression, which in total are equal to some $ S $ number, and then we start walking from these elements in opposite directions (towards each other or vice versa to move away), then the sums of the elements that we will stumble upon will also be equal$ S $. This can be most clearly represented graphically:

Equal indentation gives equal amounts

Understanding this fact will allow us to solve problems of a fundamentally higher level of complexity than those that we considered above. For example, such:

Problem number 8. Determine the difference of the arithmetic progression in which the first term is 66, and the product of the second and twelfth terms is the smallest possible.

Solution. Let's write down everything we know:

\ [\ begin (align) & ((a) _ (1)) = 66; \\ & d =? \\ & ((a) _ (2)) \ cdot ((a) _ (12)) = \ min. \ end (align) \]

So, we do not know the difference of the progression $ d $. Actually, the whole solution will be built around the difference, since the product $ ((a) _ (2)) \ cdot ((a) _ (12)) $ can be rewritten as follows:

\ [\ begin (align) & ((a) _ (2)) = ((a) _ (1)) + d = 66 + d; \\ & ((a) _ (12)) = ((a) _ (1)) + 11d = 66 + 11d; \\ & ((a) _ (2)) \ cdot ((a) _ (12)) = \ left (66 + d \ right) \ cdot \ left (66 + 11d \ right) = \\ & = 11 \ cdot \ left (d + 66 \ right) \ cdot \ left (d + 6 \ right). \ end (align) \]

For those in the tank: I took out the common factor of 11 from the second parenthesis. Thus, the sought product is a quadratic function with respect to the variable $ d $. Therefore, consider the function $ f \ left (d \ right) = 11 \ left (d + 66 \ right) \ left (d + 6 \ right) $ - its graph will be a parabola with branches up, since if we expand the brackets, then we get:

\ [\ begin (align) & f \ left (d \ right) = 11 \ left (((d) ^ (2)) + 66d + 6d + 66 \ cdot 6 \ right) = \\ & = 11 (( d) ^ (2)) + 11 \ cdot 72d + 11 \ cdot 66 \ cdot 6 \ end (align) \]

As you can see, the coefficient at the leading term is 11 - this is a positive number, so we really are dealing with a parabola with branches up:

quadratic function graph - parabola

Please note: this parabola takes its minimum value at its vertex with the abscissa $ ((d) _ (0)) $. Of course, we can calculate this abscissa according to the standard scheme (there is also the formula $ ((d) _ (0)) = (- b) / (2a) \; $), but it would be much more reasonable to notice that the desired vertex lies on the axis symmetry of the parabola, so the point $ ((d) _ (0)) $ is equidistant from the roots of the equation $ f \ left (d \ right) = 0 $:

\ [\ begin (align) & f \ left (d \ right) = 0; \\ & 11 \ cdot \ left (d + 66 \ right) \ cdot \ left (d + 6 \ right) = 0; \\ & ((d) _ (1)) = - 66; \ quad ((d) _ (2)) = - 6. \\ \ end (align) \]

That is why I was in no hurry to open the brackets: in the original form, the roots were very, very easy to find. Therefore, the abscissa is equal to the arithmetic mean of the numbers −66 and −6:

\ [((d) _ (0)) = \ frac (-66-6) (2) = - 36 \]

What does the discovered number give us? With it, the required product takes the smallest value (by the way, we haven't counted $ ((y) _ (\ min)) $ - we don't need this). At the same time, this number is the difference between the original progression, i.e. we found the answer. :)

Answer: −36

Problem number 9. Insert three numbers between the numbers $ - \ frac (1) (2) $ and $ - \ frac (1) (6) $ so that they together with the given numbers form an arithmetic progression.

Solution. Basically, we need to make a sequence of five numbers, with the first and last numbers already known. Let's denote the missing numbers by the variables $ x $, $ y $ and $ z $:

\ [\ left (((a) _ (n)) \ right) = \ left \ (- \ frac (1) (2); x; y; z; - \ frac (1) (6) \ right \ ) \]

Note that the number $ y $ is the "middle" of our sequence - it is equidistant from both the numbers $ x $ and $ z $, and from the numbers $ - \ frac (1) (2) $ and $ - \ frac (1) ( 6) $. And if at the moment we cannot get $ y $ from the numbers $ x $ and $ z $, then the situation is different with the ends of the progression. Remembering the arithmetic mean:

Now, knowing $ y $, we will find the remaining numbers. Note that $ x $ lies between the numbers $ - \ frac (1) (2) $ and the $ y = - \ frac (1) (3) $ just found. That's why

Reasoning similarly, we find the remaining number:

Ready! We found all three numbers. Let's write them down in the answer in the order in which they should be inserted between the original numbers.

Answer: $ - \ frac (5) (12); \ - \ frac (1) (3); \ - \ frac (1) (4) $

Problem number 10. Insert several numbers between the numbers 2 and 42, which together with these numbers form an arithmetic progression, if you know that the sum of the first, second and last of the inserted numbers is 56.

Solution. An even more difficult task, which, however, is solved according to the same scheme as the previous ones - through the arithmetic mean. The problem is that we don't know exactly how many numbers to insert. Therefore, for definiteness, let us assume that after inserting everything there will be exactly $ n $ numbers, and the first of them is 2, and the last is 42. In this case, the desired arithmetic progression can be represented as:

\ [\ left (((a) _ (n)) \ right) = \ left \ (2; ((a) _ (2)); ((a) _ (3)); ...; (( a) _ (n-1)); 42 \ right \) \]

\ [((a) _ (2)) + ((a) _ (3)) + ((a) _ (n-1)) = 56 \]

Note, however, that the numbers $ ((a) _ (2)) $ and $ ((a) _ (n-1)) $ are obtained from the numbers 2 and 42 at the edges by one step towards each other, i.e. ... to the center of the sequence. This means that

\ [((a) _ (2)) + ((a) _ (n-1)) = 2 + 42 = 44 \]

But then the expression written above can be rewritten as follows:

\ [\ begin (align) & ((a) _ (2)) + ((a) _ (3)) + ((a) _ (n-1)) = 56; \\ & \ left (((a) _ (2)) + ((a) _ (n-1)) \ right) + ((a) _ (3)) = 56; \\ & 44 + ((a) _ (3)) = 56; \\ & ((a) _ (3)) = 56-44 = 12. \\ \ end (align) \]

Knowing $ ((a) _ (3)) $ and $ ((a) _ (1)) $, we can easily find the difference of the progression:

\ [\ begin (align) & ((a) _ (3)) - ((a) _ (1)) = 12 - 2 = 10; \\ & ((a) _ (3)) - ((a) _ (1)) = \ left (3-1 \ right) \ cdot d = 2d; \\ & 2d = 10 \ Rightarrow d = 5. \\ \ end (align) \]

It remains only to find the rest of the members:

\ [\ begin (align) & ((a) _ (1)) = 2; \\ & ((a) _ (2)) = 2 + 5 = 7; \\ & ((a) _ (3)) = 12; \\ & ((a) _ (4)) = 2 + 3 \ cdot 5 = 17; \\ & ((a) _ (5)) = 2 + 4 \ cdot 5 = 22; \\ & ((a) _ (6)) = 2 + 5 \ cdot 5 = 27; \\ & ((a) _ (7)) = 2 + 6 \ cdot 5 = 32; \\ & ((a) _ (8)) = 2 + 7 \ cdot 5 = 37; \\ & ((a) _ (9)) = 2 + 8 \ cdot 5 = 42; \\ \ end (align) \]

Thus, already at the 9th step we will come to the left end of the sequence - the number 42. In total, it was necessary to insert only 7 numbers: 7; 12; 17; 22; 27; 32; 37.

Answer: 7; 12; 17; 22; 27; 32; 37

Word problems with progressions

In conclusion, I would like to consider a couple of relatively simple tasks. Well, how simple: for most students who study mathematics at school and have not read what is written above, these tasks may seem like a tin. Nevertheless, it is precisely such problems that come across in the OGE and USE in mathematics, so I recommend that you familiarize yourself with them.

Problem number 11. The brigade produced 62 parts in January, and in each next month it produced 14 more parts than in the previous one. How many parts did the team make in November?

Solution. Obviously, the number of parts, scheduled by month, will represent an increasing arithmetic progression. Moreover:

\ [\ begin (align) & ((a) _ (1)) = 62; \ quad d = 14; \\ & ((a) _ (n)) = 62+ \ left (n-1 \ right) \ cdot 14. \\ \ end (align) \]

November is the 11th month of the year, so we need to find $ ((a) _ (11)) $:

\ [((a) _ (11)) = 62 + 10 \ cdot 14 = 202 \]

Consequently, 202 parts will be manufactured in November.

Problem number 12. The bookbinding workshop bound 216 books in January, and each next month it bound 4 more books than the previous one. How many books did the workshop bind in December?

Solution. All the same:

$ \ begin (align) & ((a) _ (1)) = 216; \ quad d = 4; \\ & ((a) _ (n)) = 216+ \ left (n-1 \ right) \ cdot 4. \\ \ end (align) $

December is the last, 12th month of the year, so we are looking for $ ((a) _ (12)) $:

\ [((a) _ (12)) = 216 + 11 \ cdot 4 = 260 \]

This is the answer - 260 books will be bound in December.

Well, if you have read this far, I hasten to congratulate you: you have successfully completed the "Young Fighter Course" in arithmetic progressions. You can safely proceed to the next lesson, where we will study the formula for the sum of a progression, as well as important and very useful consequences from it.

First level

Arithmetic progression. Detailed theory with examples (2019)

Number sequence

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:

Numerical 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 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 was occupied by the ancient Greeks.

This is a numerical sequence, each member 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 sum up - 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 in the progression? 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. Take a closer look at the drawn picture ... 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 find by yourself in this way the value of a member of a given arithmetic progression.

Calculated? Compare your notes to the answer:

Please note 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 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 it out 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 operates 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, is it 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, in order 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 engaged in checking the work of students in other grades asked the following problem 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 the 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 of 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?
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 back in the 3rd century, and throughout this time, witty people were using the properties of an arithmetic progression with might and main.
For example, imagine Ancient Egypt and the most ambitious construction site of that time - the construction of the pyramid ... The picture 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 (let's 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 ascending and decreasing.
2. Finding formula-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.

Numerical 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 specified 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 successive 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 the 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 members 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 we are interested in 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 a 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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