Unified State Exam schedule 2019 official FIPI - adjusted table for all subjects for high school students. The order of holding the Unified State Exam is determined by the main and reserve days. For school graduates who do not successfully pass the test, additional exams are also provided in autumn period. Adjustment of the schedule conducting the Unified State Exam 2019 is underway Federal Institute pedagogical dimensions in accordance with approved standards and methods, as a result of which the final and official final version of the schedule is formed. Latest changes Unified State Exam schedule-2019FIPI are published 2 months before the start of the exam.

If the days for the Unified State Exam coincide, the student must come to take the exam on the reserve day. A reserve date is also used in case of absence due to a valid reason or illness. If violations were identified during the Unified State Exam, then you must submit a complaint directly to the commission at the delivery point. In this case, the results for a group of students may be canceled and the retake scheduled for a reserve day. If there is a repeated violation on the reserve day, the decision to re- passing the Unified State Exam accepted by the regional center, or postponed to September. So far, there have been no precedents for double violations.

Early passing of the Unified State Exam in 2019 is provided for those who:

• Conscripted into the army;
• Enters a foreign university;
• Sent for treatment;
• Goes to sports competitions, Olympics, competitions;

06/05/2019 – social studies.

06/07/2019 – in physics and literature.

06/09/2019 – Russian language.

06/13/2019 – English, German, biology.

06/19/2019 – chemistry and history.

09/05/2019 – Russian language.

09/08/2019 – mathematics.

Reserve days

04/10/2019 – history, English, computer science, geography.

04/12/2019 – physics, biology, literature, social studies, German and other foreign languages.

04/14/2019 – Russian and mathematics.

06/20/2019 – geography and computer science.

06/21/2019 – literature, chemistry, physics. In social studies.

06/22/2019 – in biology, foreign language, history. .

06/23/2019 – retake in English.

06/28/2019 – mathematics, both levels (professional and basic).

06/29/2019 – Russian language.

07/1/2019 – other items.

09/16/2019 – all items.

This schedule is preliminary; changes may be made before the final approved version is released. Changes are being made in connection with adjustments to the rules for conducting the exam, as well as according to the recommendations of the Ministry of Education.

The Unified State Exam 2019 schedule from FIPI shows that it is necessary to begin intensive preparation in January.

Every schoolchild in our country is required to take the unified tests. state exams, which demonstrate the level of knowledge acquired at school and become the basis for further development education – admission to university. Such an important event requires a long preparation, and therefore every student strives to find out the schedule in advance Unified State Examinations 2017.

Features of the Unified State Exam 2017

Until 2017, tests were the main form of knowledge testing. In 2016 uniform test questions recognized as outdated, because even without knowing the correct answer, the student had the opportunity to guess it from the options provided. Starting from 2017, it was decided to return to the survey form for passing exams, that is, most subjects will be taken as it was customary in the “noughties” - in the period before 2009. In addition, the student can expect a whole series of innovations. Let's tell you more about them.

Firstly, to two mandatory exams a third is added - it should become history. True, the name of the third item has not yet been firmly established, but at the beginning academic year this information will already be made public. That is, you will have to take the Russian language, mathematics and, most likely, history - more precisely, it will be known to Unified State Exam date 2017.

Secondly, RAO ( Russian Academy Education) insists on introducing point scale essay grades. To today The essay was evaluated according to only two criteria: pass or fail. This, according to representatives of the Russian Academy of Education, negatively affects the knowledge of students and gives advantages to those students who are simply too lazy to study literature - it is much easier to get a “pass” in an essay than an “A”.

Thirdly, on Unified State Exam results The grades on the certificate will also be affected. The higher the scores for school subjects, the higher final grade for the state exam.

Fourthly, if the points scored do not reach the threshold level, students will be given the opportunity to retake the Unified State Exam twice more. It will also be possible to take a retake if the student for some reason is not satisfied with the points he scored.

So on Unified State Examination for schoolchildren you have to take one of your choice. They can be taken several times until the student finds the result satisfactory.

Dates for the Unified State Examination in 2017

The Unified State Exam schedule in 2017 consists of two parts - early and main exams.

Early period for passing the Unified State Examination

• Geography, Computer Science and ICT

• Russian language / compulsory subject

• history, chemistry

• mathematics / compulsory subject

• Geography, literature

• foreign languages(oral examination)

• foreign languages, biology, physics

• social studies, literature

Starting next week, reserve time begins for all exams included in the Unified State Examination list.

• reserve: geography, chemistry, computer science and ICT, foreign languages ​​(oral), history

• reserve: foreign languages, literature, physics, social studies, biology

• reserve: Russian language, mathematics B, P

• Foreign language, history, social studies (reserve)

• Foreign language (oral), geography, physics, biology (reserve).

However, to exercise the right to early delivery Not every student is in a hurry to take an exam. Therefore, most students will be interested in the second section of the 2017 Unified State Exam schedule - the main period.

• Geography, Computer Science and ICT

• mathematics B

• mathematics P

• social science

• physics, literature

• Russian language

• foreign languages, biology

• foreign languages ​​(oral)

• foreign languages ​​(oral)

• chemistry, history

Reserve days for the Unified State Exam begin on Tuesday.

• reserve: geography, computer science and ICT

• reserve: literature, chemistry, physics, social studies

• reserve: biology, history foreign languages

• reserve: foreign languages

• reserve: mathematics B, mathematics P

• reserve: Russian language

• reserve: for all subjects

Additional period (September)

Retaking the Unified State Examinations in 2017

In addition to the main and reserve days, the Unified State Examination process itself also provides for a third period - a retake. The right to retake is granted to every student - both those who did not reach the minimum threshold and those who simply want to improve their own results and score more points. True, to improve own level will require remarkable confidence in own strength and knowledge.

The Unified State Exam retake usually takes place in September, most often in the first half of the month. However, the schedule for a possible retake will be known only by August 2017.

Extra points

Additional points may be added for exam scores. So, 10 points can be added to:

• for a certificate with only A's;
• for prizes won at Olympiads in school subjects;
• for achievements in sports.

Considering the possible addition of points, it is worth thinking about taking the Unified State Exam in advance: taking part in olympiads and competitions in all subjects, not just specialized ones; increase your level of knowledge, striving for excellent grades; participate in sports life schools.

a) Is there a finite arithmetic progression consisting of five natural numbers, such that the sum of the largest and smallest terms of this progression is equal to 99?

b) A finite arithmetic progression consists of six natural numbers. The sum of the largest and smallest terms of this progression is 9. Find all the numbers that make up this progression.

c) Average arithmetic members the finite arithmetic progression consisting of natural numbers is 6.5. Which greatest number members may be in this progression?

Solution.

a) The sum of the first and fifth terms of this progression is 2 a + 4d and is even number. Since 99 is an odd number, the sum of the largest and smallest terms of a finite arithmetic progression of 5 natural numbers cannot be equal to 99.

b) The sum of the first and sixth terms of this progression is 2 a + 5d= 9. Since d d- a natural number, we get d= 1. Then a= 2. Searched numbers: 2, 3, 4, 5, 6, 7.

c) The arithmetic mean of a progression is equal to half the sum of its extreme terms, so we get. Hence, Natural numbers from 1 to 12 constitute a progression, the arithmetic mean of whose terms is equal to 6.5, and the number of terms is 12. Therefore, the greatest possible quantity numbers are 12.

Answer: a) no; b) 2, 3, 4, 5, 6, 7; c)12.

Source: Unified State Examination - 2014. Main wave.

Dans n

a) Can the sum of all these numbers be equal to 10?

n, if the sum of all given numbers is less than 1000?

n, if the sum of all given numbers is 129.

Solution.

a) Yes, it can. The numbers 1, 2, 3, 4 form an arithmetic progression, and their sum is 10.

b) For the sum of terms of an arithmetic progression, the following inequality is true:

So, where do we find the sum of the arithmetic progression 1, 2, ..., 44 is equal to 990 n is equal to 44.

c) For the sum of terms of an arithmetic progression, the following is true:

Thus, the number is a divisor of the number 258. If then, therefore, Since we find that or Progressions of 3 and 6 terms with a sum of 129 exist: for example, 42, 43, 44 and 19, 20, 21, 22, 23, 24.

Answer: a) yes; b) 44; c) 3, 6.

Source: Unified State Exam in Mathematics 04/23/2013. Early wave. Option 901.

a 1 , a 2 , ..., a 7 Exactly three numbers are divisible by 100?

a 1 , a 2 , ..., a 49 Exactly 11 numbers are divisible by 100?

n a 1 , a 2 , ..., a 2n more multiples of 100 than among numbers a 2n + 1 , a 2n + 2 , ..., a 5n ?

Solution.

a) A suitable example is a progression with a first term of 50 and a difference of 50. Among its first seven terms (50, 100, 150, 200, 250, 300, 350), exactly three are divisible by 100.

b) Let us denote by d a 1 , a 2 , ..., a n d- natural number. Let m And n- natural numbers, m > n, gcd( d, 100) denotes the largest common divisor numbers d and 100. We have

Therefore, the difference a m − a n is divisible by 100 if and only if the difference m − n a 1 , a 2 , ..., a n, ... are multiples of 100, then these are terms with numbers of the form where q p k a 1 , a 2 , ..., a n, ... exactly one will be divisible by 100. If then among the numbers a 1 , a 2 , ..., a 49 will be at least 12 numbers that are multiples of 100. If then among the numbers a 1 , a 2 , ..., a 49 will be no more than 10 numbers divisible by 100. This means that there is no progression in which among the numbers a 1 , a 2 , ..., a 49 Exactly 11 numbers are divisible by 100.

c) Denote by [ x] integer part of the number x x k successive terms of the progression a 1 , a 2 , ..., a n, ... exactly one will be divisible by 100, where d

So, among the numbers a 1 , a 2 , ..., a 2n No more numbers will be multiples of 100. Likewise, among the numbers a 2n + 1 , a 2n + 2 , ..., a 5n no less than numbers will be multiples of 100. An 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 100, it follows that Consider a progression with the first term 69 and the difference 1. Then among the numbers a 1 , a 2 , ..., a 132 exactly two is divisible by 100 ( a 32 = 100 and a 132 = 200). Among the numbers a 133 , a 134 , ..., a 330 is exactly one divisible by 100 ( a 232 = 300). This example shows that n may be 66.

Answer: a) Yes, for example, progression 50, 100, 150, 200, 250, 300, 350, ...; b) no; c) 66.

a) Is there such a progression in which among the numbers a 1 , a 2 , ..., a 7 Exactly three numbers are 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 which greatest natural n it could turn out that among the numbers a 1 , a 2 , ..., a 2n more multiples of 36 than among 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 its first seven terms (18, 36, 54, 72, 90, 108, 126) exactly three are divisible by 36.

b) Let us denote by d arithmetic progression difference a 1 , a 2 , ..., a n, .... From the condition it follows that d- natural number. Let m And n- natural numbers, 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 terms of an 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, a multiple of a p runs through all non-negative integers. Therefore, among any k successive terms of the progression 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 numbers that are multiples of 36. If then among the numbers a 1 , a 2 , ..., a 30 will be no more than 8 numbers divisible by 36. This means that there is no progression in which among the numbers a 1 , a 2 , ..., a 30 Exactly 9 numbers are divisible by 36.

c) Denote by [ x] integer part of the number x- the largest integer not exceeding x. According to what was proven in point b) among any k successive terms of the progression a 1 , a 2 , ..., a n, ...exactly one will be divisible by 36, where d- difference of arithmetic progression.

So, among the numbers a 1 , a 2 , ..., a 2n No more numbers will be multiples of 36. Likewise, among the numbers a 2n + 1 , a 2n + 2 , ..., a 5n multiples of 36 will be no less than numbers. An 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, 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 is 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 may be 23.

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

· Task prototype ·

a) Can S equal 8?

b) Can S equal 1?

S.

Solution.

a) The number 8 is the sum of four consecutive terms of an arithmetic progression. For example, 8 = − 1 + 1 + 3 + 5.

b) Let the number 1 be the sum of the first k terms of an arithmetic progression with the first term A and the difference d. Then

means the number k- divisor 2, which contradicts the condition

c) Any natural number is the sum of an arithmetic progression consisting of terms. If we replace all terms of this progression with their opposites, we get an arithmetic progression consisting of 2 n members, the sum of which is − n.

In the previous paragraph we showed that S cannot be equal to 1. Similarly, it can be shown that S cannot equal −1. Number S may be equal to 0, for example, for progression −1; 0; 1. Thus, S can take any integer value except −1 and 1.

Answer: a) yes; b) no; c) any integer values ​​except −1 and 1.

Source: Unified State Exam in Mathematics 05/08/2014. Early wave, reserve day. Option 1.

a) Can S equal 9?

b) Can S equal 2?

c) Find all the values ​​that it can take S.

Solution.

a) The number is the sum of six consecutive terms of an arithmetic progression. For example,

b) Let the number be the sum of the first terms of an arithmetic progression with the first term and the difference Then

this means the number is a divisor, which contradicts the condition

c) Any natural number is the sum of an arithmetic progression consisting of terms. If we replace all the terms of this progression with their opposites, we get an arithmetic progression consisting of terms whose sum is equal to

In the previous paragraph, we showed that it cannot equal. Similarly, we can show that it cannot equal. A number can equal, for example, for a progression. Thus, it can take any integer values ​​except and

Answer: a) yes; b) no; c) any integer values ​​except and

Source: Unified State Exam in Mathematics 05/08/2014. Early wave, reserve day. Option 2.

· Task prototype ·

a) Is there an arithmetic progression of length 5 composed of terms of this sequence?

b) Is it possible to create an arithmetic progression? infinite length from these numbers?

c) Can the progression have 2013 members?

Solution.

a) Consider the sequence: It is easy to see that this is a progression with a difference

b) Let there be an infinite arithmetic progression, all of whose terms are members of a given sequence. Let, for definiteness, the first term of this progression be equal and the difference of this progression be equal to Then we take a natural such that Then we get that This means that the th term of our progression is negative, but this cannot be.

c) Consider the following arithmetic progression: ...; It is clear that each of these fractions is a member of this sequence.

Answer: a) yes; b) no; c) yes.

Source: A. Larin: Training option No. 22.

a) Are there any progressions for which and are distinct natural numbers?

b) Are there any progressions for which and are distinct natural numbers?

c) Which smallest value can take a fraction if it is known that and are distinct natural numbers.

Solution.

a) A suitable example is the progressions and respectively. For these progressions we have and

b) Let us assume that such progressions exist. Then one of the numbers or is not less than 1, and the second is greater than 1. This means either and , and, therefore, Hence, using the properties of an arithmetic progression, we obtain:

We have arrived at a contradiction.

c) Let us denote by and the differences of arithmetic progressions and, respectively. It follows from the condition that numbers are both natural and integers and not equal to zero. We have:

The denominators of the fractions and are positive, and the numerators of these fractions have same sign. This means that the numbers and have the same sign, that is, either , or In both cases, we obtain that

If the progressions and are progressions and respectively, then and This example shows that the smallest possible value of the fraction is

Answer: a) yes, for example, and accordingly; b) no; c) 2.

· Task prototype ·

Increasing arithmetic progressions consist of natural numbers.

a) Are there such progressions for which ?

b) Are there such progressions for which ?

c) Which highest value can accept a work if ?

Solution.

a) A suitable example is the progressions and respectively. For these progressions we have

b) Let us denote by and the differences of arithmetic progressions and, respectively. Then

If , then we came to a contradiction, because according to the condition and

c) As before, we denote by and the differences of arithmetic progressions and, respectively. Then, by condition and by what was proved in point b, we have: So,

If progressions and are progressions and respectively, then

This example shows that the largest possible value of the product is

Answer: a) yes, for example, and accordingly; b) no; c) 98.

Dans n various natural numbers that make up an arithmetic progression

a) Can the sum of all these numbers be equal to 14?

b) What is the largest value n, if the sum of all given numbers is less than 900?

c) Find everything possible values n, if the sum of all given numbers is 123.

Solution.

a) Yes, it can. The numbers 2, 3, 4, 5 form an arithmetic progression, their sum is 14.

b) Let a- first member, d- difference, n- the number of terms of the progression, then their sum is equal. For the number of terms to be greatest, the first term and the difference must be the smallest. Let them be equal to 1, then by condition the Greatest natural solution this inequality n= 41. This result is obtained with progression

c) For the sum of terms of an arithmetic progression we have:

Thus, the number of terms of the progression n is a divisor of the number 246. If then left side greater than 246: therefore, Since we find that or Progressions of three and six terms with a sum of 123 exist: for example, 40, 41, 42 and 3, 10, 17, 24, 31, 38.

Job type: 11

Condition

Natasha needs to make 300 paper cranes. Every day she makes the same number of cranes more than the previous day. On the first day, Natasha made 6 cranes. How many cranes were made on the last day if the whole job took 15 days?

Solution

It follows from the condition that the number of paper “cranes” increased by the same number every day. The number of paper “cranes” made daily forms an arithmetic progression, with the first term of the progression being equal to 6. According to the formula for the sum of the first terms of an arithmetic progression, we have

a_1+a_2+a_3+...+a_(15)= \frac(a_1+a_(15))(2)\cdot15= 300,

6+a_(15)=40,

a_(15)=40-6=34.

On the last day Natasha made 34 paper “cranes”

Answer

Job type: 11
Topic: Arithmetic and geometric progressions

Condition

Kolya needs to plant 350 rose bushes. Every day he plants the same number of bushes more than the previous day. On the first day he planted 8 rose bushes. How many bushes were planted on the last day if all the work took 20 days?

Solution

It follows from the condition that the number of planted rose bushes increased by the same number every day. The number of roses planted daily forms an arithmetic progression, with the first term being 8. Using the formula for the sum of the first terms of an arithmetic progression, we obtain a_1+a_2+a_3+...+a_(20)= \frac(a_1+a_(20))(2)\cdot20= 350,

8+a_(20)=35,

a_(20)=35-8=27.

On the last day Kolya planted 27 rose bushes.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level" Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 11
Topic: Arithmetic and geometric progressions

Condition

The tiler must lay 320 m2 of tiles. If he lays 6 m2 more per day than planned, the work will be completed 12 days earlier. Determine how much square meters tiles per day the tiler plans to lay.

Entry 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 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 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 more in a broad sense, like an infinite number 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 bring her to general view and we 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 true. 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 the progression term given the known previous and consecutive values, you need to add them up and divide them 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 for himself 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 close 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 total amount 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 on your own what the sum of the 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 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 an arithmetic progression.
For example, imagine Ancient Egypt and the most large-scale construction 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 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 less log 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. MIDDLE 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, great mathematician Karl Gauss, as a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last date is equal, the sum of the second and the penultimate is the same, the sum of the third and the 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 double digit numbers, 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 on the first day he ran km m?
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 previous task:
   .
   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.