How to solve progressions 9. Arithmetic progression (grade 9): formulas, examples. Examples of arithmetic progression

Theme: Arithmetic and geometric progressions

Class: 9

Preparation system: material for preparing the study of the topic in algebra and the preparatory stage for passing the exam

Target: the formation of concepts of arithmetic and geometric progression

Tasks: teach to distinguish between types of progression, teach correctly, use formulas

Arithmetic progression called a sequence of numbers (members of a progression)

in which each subsequent term differs from the previous one by a new term, which is also called a step or a progression difference.

Thus, setting the step of the progression and its first term, you can find any of its elements by the formula

1) Each member of the arithmetic progression, starting from the second number, is the arithmetic mean of the previous and next member of the progression

The converse is also true. If the arithmetic mean of adjacent odd (even) members of the progression is equal to the term between them, then this sequence of numbers is an arithmetic progression. This statement makes it very easy to check any sequence.

Also, by the property of arithmetic progression, the above formula can be generalized to the following

This is easy to verify if we write out the terms to the right of the equal sign

It is often used in practice to simplify computations in problems.

2) The sum of the first n terms of the arithmetic progression is calculated by the formula

Remember well the formula for the sum of an arithmetic progression, it is indispensable for calculations and is quite common in simple life situations.

3) If you need to find not the whole amount, but part of the sequence starting from the k-th member, then the following sum formula will come in handy

4) It is of practical interest to find the sum of n terms of an arithmetic progression starting from the kth number. To do this, use the formula

Find the fortieth term of the arithmetic progression 4; 7; ...

Solution:

According to the condition, we have

Determine the step of the progression

Using the well-known formula, we find the fortieth term of the progression

The arithmetic progression is given by its third and seventh terms. Find the first term of the progression and the sum of ten.

Solution:

Let's write out the given elements of the progression using the formulas

An arithmetic progression is given by the denominator and one of its members. Find the first member of the progression, the sum of its 50 members starting with 50 and the sum of the first 100.

Solution:

Let's write the formula for the hundredth element of the progression

and find the first

Based on the first, we find the 50 term of the progression

Find the sum of the part of the progression

and the sum of the first 100

The sum of the progression is 250. Find the number of members of the arithmetic progression if:

a3-a1 = 8, a2 + a4 = 14, Sn = 111.

Solution:

We write the equations in terms of the first term and the step of the progression and define them

We substitute the obtained values into the sum formula to determine the number of members in the sum

Performing simplifications

and solve the quadratic equation

Of the two values found for the problem condition, only the number 8 is suitable. Thus, the sum of the first eight members of the progression is 111.

Solve the equation

1 + 3 + 5 + ... + x = 307.

Solution:

This equation is the sum of an arithmetic progression. Let's write out its first term and find the difference in progression

We substitute the found values into the formula for the sum of the progression to find the number of terms

As in the previous task, we will simplify and solve the quadratic equation

We choose the more logical of the two values. We have that the sum of 18 members of the progression with given values a1 = 1, d = 2 is equal to Sn = 307.

Examples of problem solving: Arithmetic progression

Task1

The student team contracted to lay ceramic tiles on the floor in the hall of the youth club with an area of 288 m2. Gaining experience, students on each next day, starting from the second, laid out 2 m2 more than the previous one, and they had enough stock of tiles for exactly 11 days of work. Planning for productivity to increase in the same way, the foreman determined that it would take another 5 days to complete the job. How many boxes of tiles should he order if 1 box is enough for 1.2 m2 of floor, and 3 boxes are needed to replace low-quality tiles?

Solution

According to the condition of the problem, it is clear that we are talking about an arithmetic progression in which let

a1 = x, Sn = 288, n = 16

Then we use the formula: Sn = (2а1 + d (n-1)) * n / 0.86 = 200mm Hg. Art.

288 = (2x + 2 * 15) * 16/2

Let's calculate how many m2 students will lay out in 11 days: S11 = (2 * 3 + 2 * 10) * 11.2 = 143m 2

288-143 = 145m2 left after 11 days of work, i.e. for 5 days

145 / 1.2 = 121 (approximate) boxes need to be ordered for 5 days.

121 + 3 = 124 boxes need to be ordered including defect

Answer: 124 boxes

Task2

After each movement of the piston of the vacuum pump, 20% of the air in it is removed from the vessel. Let us determine the air pressure inside the vessel after six piston movements, if the initial pressure was 760 mm Hg. Art.

Solution

Since after each movement of the piston, 20% of the available air is removed from the vessel, 80% of the air remains. To find out the air pressure in the vessel after the successive piston movement, you need to reduce the pressure of the previous piston movement by 0.8.

We have a geometric progression, the first term of which is 760, and the denominator is 0.8. The number expressing the air pressure in the vessel (in mm Hg) after six piston movements is the seventh term in this progression. It is equal to 760 * 0.86 = 200mm Hg. Art.

Answer: 200 mm Hg.

An arithmetic progression is given, where the fifth and tenth terms are 38 and 23, respectively. Find the fifteenth term of the progression and the sum of its first ten terms.

Solution:

Find the number of the term arithmetic progression 5,14,23, ..., if its th term is 239.

Solution:

Find the number of members of the arithmetic progression 9,12,15, ..., if its sum is 306.

Solution:

Find x for which the numbers x-1, 2x-1, x2-5 make up an arithmetic progression

Solution:

Let's find the difference between 1 and 2 terms of the progression:

d = (2x-1) - (x-1) = x

Let's find the difference between 2 and 3 terms of the progression:

d = (x2-5) - (2x-1) = x2-2x-4

Because the difference is the same, then the members of the progression can be equated:

When checking in both cases, an arithmetic progression is obtained

Answer: for x = -1 and x = 4

The arithmetic progression is given by its third and seventh terms a3 = 5; a7 = 13. Find the first term of the progression and the sum of ten.

Solution:

We subtract the first from the second equation, as a result, we find the step of the progression

a1 + 6d- (a1 + 2d) = 4d = 13-5 = 8, so d = 2

We substitute the found value into any of the equations to find the first term of the arithmetic progression

We calculate the sum of the first ten members of the progression

S10 = (2 * 1 + (10-1) * 2) * 10/2 = 100

Answer: a1 = 1; S10 = 100

In an arithmetic progression where the first term is -3.4 and the difference is 3, find the fifth and eleventh terms.

So, we know that a1 = -3.4; d = 3. Find: a5, a11-.

Solution. To find the n-th term of the arithmetic progression, we use the formula: an = a1 + (n - 1) d. We have:

a5 = a1 + (5 - 1) d = -3.4 + 43 = 8.6;

a11 = a1 + (11 - 1) d = -3.4 + 10 * 3 = 26.6.

As you can see, in this case, the solution is not difficult.

The twelfth term of the arithmetic progression is 74, and the difference is -4. Find the thirty-fourth term in this progression.

We are told that a12 = 74; d = -4, but you need to find a34-.

In this problem, it is not possible to immediately apply the formula an = a1 + (n - 1) d, since the first term a1 is not known. This task can be solved in several steps.

1.Using the term a12 and the formula for the nth term, we find a1:

a12 = a1 + (12 - 1) d, now we simplify and substitute d: a12 = a1 + 11 (-4). From this equation we find a1: a1 = a12 - (-44);

We know the twelfth term from the problem statement, so we can easily calculate a1

a1 = 74 + 44 = 118. Go to the second step - calculating a34.

2. Again, using the formula an = a1 + (n - 1) d, since a1 is already known, we will define a34-,

a34 = a1 + (34 - 1) d = 118 + 33 (-4) = 118 - 132 = -14.

Answer: the thirty-fourth term of the arithmetic progression is -14.

As you can see, the solution to the second example is more complicated. The same formula is used twice to get the answer. But everything is so complicated. The solution can be shortened by using additional formulas.

As already noted, if a1 is known in the problem, then the formula for determining the n-th term of an arithmetic progression is very convenient to use. But, if the condition does not specify the first term, then a formula can come to the rescue that connects the n-th term we need and the term ak specified in the problem.

an = ak + (n - k) d.

Let's solve the second example, but using a new formula.

Given: a12 = 74; d = -4. Find: a34-.

We use the formula an = ak + (n - k) d. In our case, it will be:

a34 = a12 + (34 - 12) (-4) = 74 + 22 (-4) = 74 - 88 = -14.

The answer to the problem was received much faster, because there was no need to perform additional actions and look for the first term of the progression.

Using the above formulas, you can solve problems of calculating the difference of an arithmetic progression. So, using the formula an = a1 + (n - 1) d, you can express d:

d = (an - a1) / (n - 1). However, problems with a given first term are not so common, and they can be solved using our formula an = ak + (n - k) d, from which it is seen that d = (an - ak) / (n - k). Let's consider such a task.

Find the difference of the arithmetic progression if it is known that a3 = 36; a8 = 106.

Using the formula we obtained, the solution to the problem can be written in one line:

d = (a8 - a3) / (8 - 3) = (106 - 36) / 5 = 14.

Without this formula in the arsenal, the solution of the problem would have taken much longer, since would have to solve a system of two equations.

Geometric progressions

1. Formula of the th member (common member of the progression).
2. The formula for the sum of the first members of the progression:. When it is customary to talk about a converging geometric progression; in this case, you can calculate the sum of the entire progression using the formula.
3. The formula of the "geometric mean": if,, are three consecutive terms of a geometric progression, then by virtue of the definition we have the ratio: or or .

Understanding of many topics in mathematics and physics is associated with knowledge of the properties of series of numbers. Students in the 9th grade, when studying the subject "Algebra", consider one of the important sequences of numbers - the arithmetic progression. Here are the basic formulas of the arithmetic progression (grade 9), as well as examples of their use for solving problems.

Algebraic or arithmetic progression

The number series, which will be discussed in this article, is called in two different ways, presented in the title of this paragraph. So, in mathematics, an arithmetic progression is understood as a numerical series in which any two numbers standing next to each other differ by the same amount, which is called the difference. Numbers in such a row are usually denoted by letters with a lower integer index, for example, a 1, a 2, a 3, and so on, where the index indicates the number of the row element.

Taking into account the above definition of an arithmetic progression, we can write the following equality: a 2 -a 1 = ... = a n -a n-1 = d, here d is the difference of an algebraic progression and n is any integer. If d> 0, then we can expect that each subsequent member of the series will be larger than the previous one, in this case they speak of an increasing progression. If d<0, тогда предыдущий член будет больше последующего, то есть ряд будет убывать. Частный случай возникает, когда d = 0, то есть ряд представляет собой последовательность, в которой a 1 =a 2 =...=a n .

Arithmetic progression formulas (grade 9 school)

The series of numbers under consideration, since it is ordered and obeys a certain mathematical law, has two properties that are important for its use:

First, knowing only two numbers a 1 and d, you can find any member of the sequence. This is done using the following formula: a n = a 1 + (n-1) * d.
Secondly, to calculate the sum of n terms of the first, it is not necessary to add them in order, since you can use the following formula: S n = n * (a n + a 1) / 2.

The first formula is easy to understand, since it is a direct consequence of the fact that each member of the series under consideration differs from its neighbor by the same difference.

The second formula of the arithmetic progression can be obtained if we pay attention to the fact that the sum a 1 + a n turns out to be equivalent to the sums a 2 + a n-1, a 3 + a n-2, and so on. Indeed, since a 2 = d + a 1, a n-2 = -2 * d + an, a 3 = 2 * d + a 1, and a n-1 = -d + an, then substituting these expressions into the corresponding sums, we get that they will be the same. The factor n / 2 in the 2nd formula (for S n) appears due to the fact that the sums of the type a i + 1 + a ni are exactly n / 2, here i is an integer ranging from 0 to n / 2 -1.

According to the surviving historical evidence, the formula for the sum S n was first obtained by Karl Gauss (the famous German mathematician), when he was asked by a school teacher to add the first 100 numbers.

Example problem # 1: find the difference

Problems in which the question is posed as follows: knowing the arithmetic progression formulas, how to find d (d), are the simplest that can only be for this topic.

Let's give an example: given a numerical sequence -5, -2, 1, 4, ..., it is necessary to determine its difference, that is, d.

To do this is as easy as shelling pears: you need to take two elements and subtract the smaller from the larger one. In this case, we have: d = -2 - (-5) = 3.

To be sure of the answer received, it is recommended to check the remaining differences, since the presented sequence may not satisfy the condition of an algebraic progression. We have: 1 - (- 2) = 3 and 4 - 1 = 3. These data indicate that we got the correct result (d = 3) and proved that a series of numbers in the problem statement is indeed an algebraic progression.

Example problem number 2: find the difference, knowing two terms of the progression

Let's consider another interesting problem, which is posed by the question of how to find the difference. In this case, the arithmetic progression formula must be used for the nth term. So, the problem: given the first and fifth numbers of the series, which corresponds to all the properties of the algebraic progression, for example, these are the numbers a 1 = 8 and a 5 = -10. How to find the difference d?

The solution to this problem should be started by writing the general form of the formula for the nth element: a n = a 1 + d * (- 1 + n). Now you can go in two ways: either substitute the numbers at once and work with them, or express d, and then go on to specific a 1 and a 5. We use the last method, we get: a 5 = a 1 + d * (- 1 + 5) or a 5 = 4 * d + a 1, whence it follows that d = (a 5 -a 1) / 4. Now you can safely substitute the known data from the condition and get the final answer: d = (-10-8) / 4 = -4.5.

Note that in this case the difference in the progression turned out to be negative, that is, there is a decreasing sequence of numbers. It is necessary to pay attention to this fact when solving problems so as not to confuse the signs "+" and "-". All formulas given above are universal, so you should always follow them regardless of the sign of the numbers with which operations are carried out.

An example of solving problem no. 3: find a1, knowing the difference and the element

Let's change a little the condition of the problem. Let there be two numbers: the difference d = 6 and the 9th element of the progression a 9 = 10. How to find a1? The arithmetic progression formulas remain unchanged, we will use them. For the number a 9 we have the following expression: a 1 + d * (9-1) = a 9. From where we easily get the first element of the series: a 1 = a 9 -8 * d = 10 - 8 * 6 = -38.

An example of solving problem number 4: find a1, knowing two elements

This variant of the problem is a complicated version of the previous one. The essence is the same, it is necessary to calculate a 1, but now the difference d is not known, and instead one more element of the progression is given.

An example of this type of problem is the following: find the first number of a sequence for which it is known that it is an arithmetic progression, and that its 15th and 23rd elements are 7 and 12, respectively.

It is necessary to solve this problem by writing the expression for the nth term for each element known from the condition, we have: a 15 = d * (15-1) + a 1 and a 23 = d * (23-1) + a 1. As you can see, we got two linear equations that need to be solved for a 1 and d. Let's do this: subtract the first from the second equation, then we get the following expression: a 23 -a 15 = 22 * d - 14 * d = 8 * d. When deriving the last equation, the values of a 1 were omitted because they cancel out when subtracted. Substituting the known data, we find the difference: d = (a 23 -a 15) / 8 = (12-7) / 8 = 0.625.

The value d must be substituted into any formula for a known element to get the first term of the sequence: a 15 = 14 * d + a 1, whence: a 1 = a 15 -14 * d = 7-14 * 0.625 = -1.75.

Let's check the result, for this we find a 1 through the second expression: a 23 = d * 22 + a 1 or a 1 = a 23 -d * 22 = 12 - 0.625 * 22 = -1.75.

An example of solving problem no. 5: find the sum of n elements

As you can see, up to this point, only one arithmetic progression formula (grade 9) was used for the solution. Now we give a problem, for the solution of which you need to know the second formula, that is, for the sum S n.

There is the following ordered row of numbers -1,1, -2,1, -3,1, ..., you need to calculate the sum of its first 11 elements.

It can be seen from this series that it is decreasing, and a 1 = -1.1. Its difference is: d = -2.1 - (-1.1) = -1. Now let's define the 11th term: a 11 = 10 * d + a 1 = -10 + (-1.1) = -11.1. Having completed the preparatory calculations, you can use the above formula for the sum, we have: S 11 = 11 * (- 1.1 + (- 11.1)) / 2 = -67.1. Since all the terms were negative numbers, then their sum has the corresponding sign.

An example of solving problem number 6: find the sum of elements from n to m

Perhaps this type of problem is the most difficult for most students. Let's give a typical example: given a series of numbers 2, 4, 6, 8 ..., you need to find the sum from 7th to 13th terms.

Formulas arithmetic progression(Grade 9) are used exactly the same as in all the problems earlier. It is recommended to solve this problem in stages:

First, find the sum of 13 terms using the standard formula.
Then calculate this amount for the first 6 items.
After that, subtract the 2nd from the 1st amount.

Let's get down to the solution. As in the previous case, we will carry out preparatory calculations: a 6 = 5 * d + a 1 = 10 + 2 = 12, a 13 = 12 * d + a 1 = 24 + 2 = 26.

Let's calculate two sums: S 13 = 13 * (2 + 26) / 2 = 182, S 6 = 6 * (2 + 12) / 2 = 42. We take the difference and get the desired answer: S 7-13 = S 13 - S 6 = 182-42 = 140. Note that when obtaining this value, it was the sum of 6 elements of the progression that was used as the deducted, since the 7th term is included in the sum S 7-13.

Algebraic or arithmetic progression

The number series, which will be discussed in this article, is called in two different ways, presented in the title of this paragraph. So, in mathematics, an arithmetic progression is understood as a numerical series in which any two numbers standing next to each other differ by the same amount, which is called the difference. Numbers in such a row are usually denoted by letters with a lower integer index, for example, a1, a2, a3, and so on, where the index indicates the number of the row element.

Taking into account the definition of an arithmetic progression given above, we can write the following equality: a2-a1 = ... = an-an-1 = d, here d is the difference of an algebraic progression and n is any integer. If d> 0, then we can expect that each subsequent member of the series will be larger than the previous one, in this case they speak of an increasing progression. If d

Arithmetic progression formulas (grade 9 school)

The series of numbers under consideration, since it is ordered and obeys a certain mathematical law, has two properties that are important for its use:

First, knowing only two numbers a1 and d, you can find any member of the sequence. This is done using this formula: an = a1 + (n-1) * d.

Secondly, to calculate the sum of n terms of the first, it is not necessary to add them in order, since you can use the following formula: Sn = n * (an + a1) / 2.

The first formula is easy to understand, since it is a direct consequence of the fact that each member of the series under consideration differs from its neighbor by the same difference.

The second formula of the arithmetic progression can be obtained if we pay attention to the fact that the sum a1 + an turns out to be equivalent to the sums a2 + an-1, a3 + an-2, and so on. Indeed, since a2 = d + a1, an-2 = -2 * d + an, a3 = 2 * d + a1, and an-1 = -d + an, substituting these expressions into the corresponding sums, we get that they will be the same. The factor n / 2 in the 2nd formula (for Sn) appears due to the fact that there are exactly n / 2 sums like ai + 1 + an-i, here i is an integer ranging from 0 to n / 2- 1.

According to the surviving historical evidence, the formula for the sum of Sn was first obtained by Karl Gauss (the famous German mathematician), when he was tasked by a school teacher to add the first 100 numbers.

Example problem # 1: find the difference

Problems in which the question is posed as follows: knowing the arithmetic progression formulas, how to find d (d), are the simplest that can only be for this topic.

Let's give an example: given a numerical sequence -5, -2, 1, 4, ..., it is necessary to determine its difference, that is, d.

To do this is as easy as shelling pears: you need to take two elements and subtract the smaller from the larger one. In this case, we have: d = -2 - (-5) = 3.

Example problem number 2: find the difference, knowing two terms of the progression

Let's consider another interesting problem, which is posed by the question of how to find the difference. In this case, the arithmetic progression formula must be used for the nth term. So, the problem: given the first and fifth numbers of a series that corresponds to all the properties of an algebraic progression, for example, these are the numbers a1 = 8 and a5 = -10. How to find the difference d?

The solution to this problem should be started by writing the general form of the formula for the nth element: an = a1 + d * (- 1 + n). Now you can go in two ways: either substitute the numbers at once and work with them, or express d, and then go on to specific a1 and a5. Let's use the last method, we get: a5 = a1 + d * (- 1 + 5) or a5 = 4 * d + a1, whence it follows that d = (a5-a1) / 4. Now you can safely substitute the known data from the condition and get the final answer: d = (-10-8) / 4 = -4.5.

An example of solving problem no. 3: find a1, knowing the difference and the element

Let's change a little the condition of the problem. Let there be two numbers: the difference d = 6 and the 9th element of the progression a9 = 10. How to find a1? The arithmetic progression formulas remain unchanged, we will use them. For the number a9 we have the following expression: a1 + d * (9-1) = a9. From where we easily get the first element of the series: a1 = a9-8 * d = 10 - 8 * 6 = -38.

An example of solving problem number 4: find a1, knowing two elements

This variant of the problem is a complicated version of the previous one. The essence is the same, it is necessary to calculate a1, but now the difference d is not known, and instead one more element of the progression is given.

It is necessary to solve this problem by writing the expression for the nth term for each element known from the condition, we have: a15 = d * (15-1) + a1 and a23 = d * (23-1) + a1. As you can see, we got two linear equations that need to be solved for a1 and d. Let's do this: subtract the first from the second equation, then we get the following expression: a23-a15 = 22 * d - 14 * d = 8 * d. When deriving the last equation, the values of a1 were omitted because they cancel out when subtracted. Substituting the known data, we find the difference: d = (a23-a15) / 8 = (12-7) / 8 = 0.625.

The value d must be substituted into any formula for a known element to get the first term in the sequence: a15 = 14 * d + a1, from where: a1 = a15-14 * d = 7-14 * 0.625 = -1.75.

Let's check the result, for this we find a1 through the second expression: a23 = d * 22 + a1 or a1 = a23-d * 22 = 12 - 0.625 * 22 = -1.75.

An example of solving problem no. 5: find the sum of n elements

As you can see, up to this point, only one arithmetic progression formula (grade 9) was used for the solution. Now we present a problem for the solution of which you need to know the second formula, that is, for the sum Sn.

There is the following ordered row of numbers -1,1, -2,1, -3,1, ..., you need to calculate the sum of its first 11 elements.

It can be seen from this series that it is decreasing, and a1 = -1.1. Its difference is: d = -2.1 - (-1.1) = -1. Now let's define the 11th term: a11 = 10 * d + a1 = -10 + (-1.1) = -11.1. Having completed the preparatory calculations, you can use the above formula for the sum, we have: S11 = 11 * (- 1.1 + (- 11.1)) / 2 = -67.1. Since all the terms were negative numbers, then their sum has the corresponding sign.

An example of solving problem number 6: find the sum of elements from n to m

Perhaps this type of problem is the most difficult for most students. Let's give a typical example: given a series of numbers 2, 4, 6, 8 ..., you need to find the sum from 7th to 13th terms.

Arithmetic progression formulas (grade 9) are used exactly the same as in all the problems before. It is recommended to solve this problem in stages:

First, find the sum of 13 terms using the standard formula.

Then calculate this amount for the first 6 items.

After that, subtract the 2nd from the 1st amount.

Let's get down to the solution. As in the previous case, we will carry out preparatory calculations: a6 = 5 * d + a1 = 10 + 2 = 12, a13 = 12 * d + a1 = 24 + 2 = 26.

Let's calculate two sums: S13 = 13 * (2 + 26) / 2 = 182, S6 = 6 * (2 + 12) / 2 = 42. We take the difference and get the desired answer: S7-13 = S13 - S6 = 182-42 = 140. Note that when obtaining this value, it was the sum of 6 elements of the progression that was used as the deducted, since the 7th term is included in the sum S7-13.

Class: 9

Lesson type: lesson in learning new material.

The purpose of the lesson: Formation of the concept of an arithmetic progression as one of the types of sequences, the derivation of the formula for the nth term, acquaintance with the characteristic property of the members of an arithmetic progression. Solving problems.

Lesson Objectives:

Educational- introduce the concept of arithmetic progression; n-th member formulas; characteristic property possessed by members of arithmetic progressions.
Developing- develop the ability to compare mathematical concepts, find similarities and differences, the ability to observe, notice patterns, conduct reasoning by analogy; to form the ability to build and interpret a mathematical model of some real situation.
Educational- to contribute to the fostering of interest in mathematics and its applications, activity, the ability to communicate, arguably defend their views.

Equipment: computer, multimedia projector, presentation (Appendix 1)

Textbooks: Algebra 9, Yu.N. Makarychev, N.G. Mindyuk, K.N. Neshkov, S.B. Suvorov, edited by S.A. Telyakovsky, JSC "Moscow textbooks", 2010

Lesson plan:

Organizational moment, problem statement
Knowledge update, oral work
Learning new material
Primary anchoring
Lesson summary
Homework

In order to increase the clarity and convenience of working with the material, the lesson is accompanied by a presentation. However, this is not required and the same lesson can be taught in classrooms that are not equipped with multimedia equipment. For this, the necessary data can be prepared on a board or in the form of tables and posters.

During the classes

I. Organizational moment, problem statement.

Greetings.

The topic of today's lesson is arithmetic progression. In this lesson, we will learn what an arithmetic progression is, what its general form is, find out how to distinguish an arithmetic progression from other sequences and solve problems where the properties of arithmetic progressions are used.

II. Knowledge update, oral work.

The sequence () is given by the formula: =. What is the number of a member of this sequence if it is equal to 144? 225? 100? Are 48 members of this sequence? 49? 168?

It is known about the sequence () that, ... What is the name of this method of sequencing? Find the first four terms of this sequence.

It is known about the sequence () that. What is the name of this method of sequencing? Find if?

III. Learning new material.

Progression is a sequence of values, each of the following of which is in a certain, common for the entire progression, dependence on the previous one. The term is now largely outdated and is found only in combinations of "arithmetic progression" and "geometric progression".

The term "progression" is of Latin origin (progression, which means "moving forward") and was introduced by the Roman author Boethius (6th century). This term in mathematics used to refer to any sequence of numbers built according to such a law that allows this sequence to be continued indefinitely in one direction. Currently, the term "progression" in the originally broad sense is not used. Two important particular types of progressions - arithmetic and geometric - have retained their names.

Consider a sequence of numbers:

2, 6, 10, 14, 18, :.
11, 8, 5, 2, -1, :.
5, 5, 5, 5, 5, :.

What is the third term in the first sequence? Subsequent member? Previous member? What is the difference between the second and first terms? Third and second members? Fourth and third?

If the sequence is built according to the same law, draw a conclusion, what will be the difference between the sixth and fifth members of the first sequence? Between the seventh and sixth?

Name the next two terms in each sequence. Why do you think so?

(Student Answers)

What property do these sequences have in common? State this property.

(Student Answers)

Numerical sequences with this property are called arithmetic progressions. Invite students to try to formulate a definition for themselves.

Definition of an arithmetic progression: An arithmetic progression is a sequence, each term of which, starting from the second, is equal to the previous one, added to the same number:

(- arithmetic progression, if, where is some number.

Number d, showing how much the next member of the sequence differs from the previous one, is called the difference of the progression:.

Let's take another look at the sequences and talk about the differences. What features does each sequence have and what are they associated with?

If the difference in the arithmetic progression is positive, then the progression is increasing: 2, 6, 10, 14, 18,:. (

If in an arithmetic progression the difference is negative (, then the progression is decreasing: 11, 8, 5, 2, -1,:. (

If the difference is equal to zero () and all members of the progression are equal to the same number, the sequence is called stationary: 5, 5, 5, 5,:.

How to set an arithmetic progression? Consider the following problem.

Task. On the 1st of January, there were 50 tons of coal in the warehouse. Every day for a month, a truck with 3 tons of coal comes to the warehouse. How much coal will be in the warehouse on the 30th, if during this time coal from the warehouse was not consumed.

If we write out the amount of coal in the warehouse of each number, we get an arithmetic progression. How to solve this problem? Do you really have to calculate the amount of coal on each day of the month? Is there any way to do without this? We note that by the 30th, 29 trucks with coal will arrive at the warehouse. Thus, on the 30th, there will be 50 + 329 = 137 tons of coal in the warehouse.

Thus, knowing only the first member of the arithmetic progression and the difference, we can find any member of the sequence. Is this always the case?

Let's analyze how each term of the sequence depends on the first term and the difference:

Thus, we have obtained the formula for the n-th term of the arithmetic progression.

Example 1. Sequence () is an arithmetic progression. Find if and.

Let's use the formula for the nth term ,

Answer: 260.

Consider the following problem:

In the arithmetic progression, even terms were overwritten: 3,:, 7,:, 13: Is it possible to restore the lost numbers?

Students are more likely to compute the difference in the progression first and then find the unknown members of the progression. Then you can ask them to find the relationship between the unknown member of the sequence, the previous and the next.

Solution: Let's take advantage of the fact that in an arithmetic progression, the difference between adjacent terms is constant. Let be the required member of the sequence. Then

Comment. This property of an arithmetic progression is its characteristic property. This means that in any arithmetic progression, each term starting from the second is equal to the arithmetic mean of the previous and subsequent ( ... Conversely, any sequence in which each term, starting with the second is equal to the arithmetic mean of the previous and the next, is an arithmetic progression.

IV. Primary anchoring.

No. 575 ab - orally
No. 576 AVD - orally
No. 577b - independently with verification

The sequence (is an arithmetic progression. Find if and

Let's use the formula for the nth term,

Answer: -24.2.

Find the 23rd and nth terms of the arithmetic progression -8; -6.5; :

Solution: The first term of the arithmetic progression is -8. Let's find the difference of the arithmetic progression, for this it is necessary to subtract the previous one from the next member of the sequence: -6.5 - (- 8) = 1.5.

Let's use the formula for the nth term:

Find the first term of the arithmetic progression () if .

Let's remember the beginning of our lesson, guys. Did you manage to learn something new in today's lesson, make any discoveries? And what were the goals of the lesson we set for ourselves? Do you think we have managed to achieve our goals?

Homework.

Clause 25, No. 578a, No. 580b, No. 582, No. 586a, No. 601a.

Strong Learner Creative Activity: Prove that in arithmetic progression for any numbers such that k the equalities hold and .

Thanks for the tutorial guys. You did a good job today.