The sum of the first 100 numbers of an arithmetic progression. The formula for the sum of members of a finite arithmetic progression

In this lesson, we will derive the formula for the sum of terms of a finite arithmetic progression and solve some problems using this formula.

Theme: Progressions

Lesson: The formula for the sum of members of a finite arithmetic progression

1. Introduction

Consider the problem: find the sum of natural numbers from 1 to 100 inclusive.

Given: 1, 2, 3, ..., 98, 99, 100.

Find: S100=1+2+3 … +98 + 99 + 100.

Solution: S100=(1+100)+(2+99)+(3+98)+…+(50+51)=101+101+101+…+101=101 x 50=5050.

Answer: 5050.

The sequence of natural numbers 1, 2, 3, ..., 98, 99, 100 is arithmetic progression: a1=1, d=1.

We have found the sum of the first hundred natural numbers, i.e. the sum of the first n members of an arithmetic progression.

The considered solution was proposed by the great mathematician Carl Friedrich Gauss, who lived in the 19th century. The problem was solved by him at the age of 5 years.

History reference: Johann Carl Friedrich Gauss (1777 - 1855) - German mathematician, mechanic, physicist and astronomer. Considered one of the greatest mathematicians of all time, "the king of mathematicians". Laureate of the Copley medal (1838), foreign member of the Swedish (1821) and Russian (1824) Academies of Sciences, of the English Royal Society. According to legend, a school mathematics teacher, in order to keep children busy for a long time, suggested that they calculate the sum of numbers from 1 to 100. Young Gauss noticed that pairwise sums from opposites to opposites are the same: 1+100=101, 2+99=101, etc. and instantly got the result: 101x50=5050.

2. Derivation of the formula for the sum of the first n terms of an arithmetic progression

Consider a similar problem for an arbitrary arithmetic progression.

Find: the sum of the first n members of an arithmetic progression.

Let us show that all expressions in brackets are equal to each other, namely, to the expression . Let d be the difference of an arithmetic progression. Then:

And so on. Therefore, we can write:

Where do we get the formula for the sum of the first n members of an arithmetic progression:

3. Solving problems on the application of the formula for the sum of the first n terms of an arithmetic progression

1. Solve the problem of the sum of natural numbers from 1 to 100 using the formula for the sum of the first n members of an arithmetic progression:

Solution: a1=1, d=1, n=100.

General formula:

In our case: .

Answer: 5050.

General formula:

. Let's find by the formula of the n-th member of the arithmetic progression: .

In our case: .

To find , you first need to find .

This can be done using the general formula .First, apply this formula to find the difference of an arithmetic progression.

i.e. . Means .

Now we can find .

Using the formula for the sum of the first n terms of an arithmetic progression

, let's find .

4. Derivation of the second formula for the sum of the first n terms of an arithmetic progression

We obtain the second formula for the sum of the first n terms of an arithmetic progression, namely: we prove that .

Proof:

In the formula for the sum of the first n terms of an arithmetic progression let us substitute the expression for , namely . We get: , i.e. . Q.E.D.

Let's analyze the obtained formulas. For calculations by the first formula you need to know the first term, the last term and n by the second formula - you need to know the first term, the difference and n.

Finally, note that in any case Sn is a quadratic function of n, because .

5. Solving problems on the application of the second formula for the sum of the first n terms of an arithmetic progression

General formula:

In our case:.

Answer: 403.

2. Find the sum of all two-digit numbers that are multiples of 4.

(12; 16; 20; ...; 96) - a set of numbers that satisfy the condition of the problem.

So we have an arithmetic progression.

n find from the formula for:.

i.e. . Means .

Using the second formula for the sum of the first n terms of an arithmetic progression

, let's find .

It is required to find the sum of all terms from the 10th to the 25th inclusive.

One way to solve it is as follows:

Hence, .

6. Summary of the lesson

So, we have derived formulas for the sum of members of a finite arithmetic progression. These formulas have been used to solve some problems.

In the next lesson, we will get acquainted with the characteristic property of an arithmetic progression.

1. Makarychev Yu. N. et al. Algebra grade 9 (textbook for secondary school).-M.: Education, 1992.

2. Makarychev Yu. N., Mindyuk N. G., Neshkov, K. I. Algebra for Grade 9 with deepening. study mathematics.-M.: Mnemozina, 2003.

3. Makarychev Yu. N., Mindyuk N. G. Additional chapters to the school textbook of algebra grade 9.-M .: Education, 2002.

4. Galitsky M. L., Goldman A. M., Zvavich L. I. Collection of problems in algebra for grades 8-9 (textbook for students of schools and classes with in-depth study of mathematics). - M .: Education, 1996.

5. Mordkovich A. G. Algebra grade 9, textbook for general education institutions. - M.: Mnemosyne, 2002.

6. Mordkovich A. G., Mishutina T. N., Tulchinskaya E. E. Algebra grade 9, problem book for educational institutions. - M.: Mnemosyne, 2002.

7. Glazer G. I. History of mathematics at school. Grades 7-8 (a guide for teachers).-M.: Enlightenment, 1983.

1. College section. ru in mathematics.

2. Portal of Natural Sciences.

3. Exponential. ru Educational mathematical site.

1. No. 362, 371, 377, 382 (Makarychev Yu. N. et al. Algebra Grade 9).

2. No. 12.96 (Galitsky M. L., Goldman A. M., Zvavich L. I. Collection of problems in algebra for grades 8-9).

When studying algebra in a secondary 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 an arithmetic progression and examples with solutions.

What is an arithmetic progression?

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

An arithmetic or algebraic progression is such a set of ordered rational numbers, each member of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of an ordered series of numbers and the difference, you can restore the entire arithmetic progression.

Let's take an example. The next sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the considered type of progression, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).

Important Formulas

We now give the basic formulas that will be needed to solve problems using an arithmetic progression. Let a n denote the nth member of the sequence, where n is an integer. The difference is denoted by the Latin letter d. Then the following expressions are true:

To determine the value of the nth term, the formula is suitable: a n \u003d (n-1) * d + a 1.
To determine the sum of the first n terms: S n = (a n + a 1)*n/2.

To understand any examples of an arithmetic progression with a solution in grade 9, it is enough to remember these two formulas, since any problems of the type in question are built on their use. Also, do not forget that the progression difference is determined by the formula: d = a n - a n-1 .

Example #1: Finding an Unknown Member

We give a simple example of an arithmetic progression and the formulas that must be used to solve.

Let the sequence 10, 8, 6, 4, ... be given, it is necessary to find five terms in it.

It already follows from the conditions of the problem that the first 4 terms are known. The fifth can be defined in two ways:

Let's calculate the difference first. We have: d = 8 - 10 = -2. Similarly, one could take any two other terms standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d \u003d a n - a n-1, then d \u003d a 5 - a 4, from where we get: a 5 \u003d a 4 + d. We substitute the known values: a 5 = 4 + (-2) = 2.
The second method also requires knowledge of the difference of the progression in question, so you first need to determine it, as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for the n number of the sequence. We have: a n \u003d (n - 1) * d + a 1 \u003d (n - 1) * (-2) + 10 \u003d 12 - 2 * n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.

As you can see, both solutions lead to the same result. Note that in this example the difference d of the progression is negative. Such sequences are called decreasing because each successive term is less than the previous one.

Example #2: progression difference

Now let's complicate the task a little, give an example of how

It is known that in some 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 the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 \u003d 6 + 6 * d. From this expression, you can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, the first part of the problem was answered.

To restore the sequence to the 7th member, 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 and 7 = 18.

Example #3: making a progression

Let us complicate the condition of the problem even more. Now you need to answer the question of how to find an arithmetic progression. We can give the following example: two numbers are given, 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 \u003d -4 and a 5 \u003d 5. Having established this, we proceed to a task that is similar to the previous one. Again, for the nth term, we use the formula, we get: a 5 \u003d a 1 + 4 * d. From: d \u003d (a 5 - a 1) / 4 \u003d (5 - (-4)) / 4 \u003d 2.25. Here, the difference is not an integer value, but it is a rational number, so the formulas for the algebraic progression remain the same.

Now let's 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 \u003d 2.75 + 2.25 \u003d 5, which coincided with the condition of the problem.

Example #4: The first member of the progression

We continue to give examples of an arithmetic progression with a solution. In all 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 from what number this sequence begins.

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

The specified system is easiest to solve if you express a 1 in each equation, and then compare the resulting expressions. First equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 \u003d a 43 - 42 * d \u003d 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d \u003d 37 - 42 * d, whence the difference d \u003d (37 - 50) / (42 - 14) \u003d - 0.464 (only 3 decimal places are given).

Knowing d, you can use any of the 2 expressions above for a 1 . For example, first: a 1 \u003d 50 - 14 * d \u003d 50 - 14 * (- 0.464) \u003d 56.496.

If there are doubts about the result, you can check it, for example, determine the 43rd member of the progression, which is specified in the condition. We get: a 43 \u003d a 1 + 42 * d \u003d 56.496 + 42 * (- 0.464) \u003d 37.008. A small error is due to the fact that rounding to thousandths was used in the calculations.

Example #5: Sum

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 to calculate the sum of 100 of these numbers?

Thanks to the development of computer technology, this problem can be solved, that is, sequentially add up all the numbers, which the computer will do as soon as a person presses the Enter key. However, the problem can be solved mentally if you 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 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.

It is curious to note that this problem is called "Gaussian", since at the beginning of the 18th century the famous German, still at the age of only 10 years old, was able to solve it in his mind 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 pairs of numbers located at the edges of the sequence, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer, it is enough to multiply 50 by 101.

Example #6: sum of terms 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 terms from 8 to 14 will be.

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

The idea is to get a formula for the sum of an algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:

S m \u003d m * (a m + a 1) / 2.
S n \u003d n * (a n + a 1) / 2.

Since n > m, it is obvious that the 2 sum includes the first one. The last conclusion means that if we take the difference between these sums, and add the term a m to it (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 \u003d S n - S m + a m \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m \u003d a 1 * (n - m) / 2 + a n * n / 2 + a m * (1- m / 2). It is necessary to substitute formulas for a n and a m into this expression. 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, however, the sum 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 above solutions, all problems are based on the knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before you start solving any of these problems, it is recommended that you carefully read the condition, clearly understand what you want to find, and only then proceed with the solution.

Another tip is to strive for simplicity, that is, if you can answer the 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 the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and break the general task into separate subtasks (in this case, first find the terms a n and a m).

If there are doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. How to find an arithmetic progression, found out. Once you figure it out, it's not that hard.

The sum of an arithmetic progression.

The sum of an arithmetic progression is a simple thing. Both in meaning and in formula. But there are all sorts of tasks on this topic. From elementary to quite solid.

First, let's deal with the meaning and formula of the sum. And then we'll decide. For your own pleasure.) The meaning of the sum is as simple as lowing. To find the sum of an arithmetic progression, you just need to carefully add all its members. If these terms are few, you can add without any formulas. But if there is a lot, or a lot ... addition is annoying.) In this case, the formula saves.

The sum formula is simple:

Let's figure out what kind of letters are included in the formula. This will clear up a lot.

S n is the sum of an arithmetic progression. Addition result all members, with first on last. It is important. Add up exactly all members in a row, without gaps and jumps. And, exactly, starting from first. In problems like finding the sum of the third and eighth terms, or the sum of terms five through twentieth, direct application of the formula will be disappointing.)

a 1 - first member of the progression. Everything is clear here, it's simple first row number.

a n- last member of the progression. The last number of the row. Not a very familiar name, but, when applied to the amount, it is very suitable. Then you will see for yourself.

n is the number of the last member. It is important to understand that in the formula this number coincides with the number of added terms.

Let's define the concept last member a n. Filling question: what kind of member will last, if given endless arithmetic progression?

For a confident answer, you need to understand the elementary meaning of an arithmetic progression and ... read the assignment carefully!)

In the task of finding the sum of an arithmetic progression, the last term always appears (directly or indirectly), which should be limited. Otherwise, a finite, specific amount just doesn't exist. For the solution, it does not matter what kind of progression is given: finite or infinite. It doesn't matter how it is given: by a series of numbers, or by the formula of the nth member.

The most important thing is to understand that the formula works from the first term of the progression to the term with the number n. Actually, the full name of the formula looks like this: the sum of the first n terms of an arithmetic progression. The number of these very first members, i.e. n, is determined solely by the task. In the task, all this valuable information is often encrypted, yes ... But nothing, in the examples below we will reveal these secrets.)

Examples of tasks for the sum of an arithmetic progression.

First of all, useful information:

The main difficulty in tasks for the sum of an arithmetic progression is the correct determination of the elements of the formula.

The authors of the assignments encrypt these very elements with boundless imagination.) The main thing here is not to be afraid. Understanding the essence of the elements, it is enough just to decipher them. Let's take a look at a few examples in detail. Let's start with a task based on a real GIA.

1. The arithmetic progression is given by the condition: a n = 2n-3.5. Find the sum of the first 10 terms.

Good job. Easy.) To determine the amount according to the formula, what do we need to know? First member a 1, last term a n, yes the number of the last term n.

Where to get the last member number n? Yes, in the same place, in the condition! It says find the sum first 10 members. Well, what number will it be last, tenth member?) You won’t believe it, his number is tenth!) Therefore, instead of a n we will substitute into the formula a 10, but instead n- ten. Again, the number of the last member is the same as the number of members.

It remains to be determined a 1 and a 10. This is easily calculated by the formula of the nth term, which is given in the problem statement. Don't know how to do it? Visit the previous lesson, without this - nothing.

a 1= 2 1 - 3.5 = -1.5

a 10\u003d 2 10 - 3.5 \u003d 16.5

S n = S 10.

We found out the meaning of all elements of the formula for the sum of an arithmetic progression. It remains to substitute them, and count:

That's all there is to it. Answer: 75.

Another task based on the GIA. A little more complicated:

2. Given an arithmetic progression (a n), the difference of which is 3.7; a 1 \u003d 2.3. Find the sum of the first 15 terms.

We immediately write the sum formula:

This formula allows us to find the value of any member by its number. We are looking for a simple substitution:

a 15 \u003d 2.3 + (15-1) 3.7 \u003d 54.1

It remains to substitute all the elements in the formula for the sum of an arithmetic progression and calculate the answer:

Answer: 423.

By the way, if in the sum formula instead of a n just substitute the formula of the nth term, we get:

We give similar ones, we get a new formula for the sum of members of an arithmetic progression:

As you can see, the nth term is not required here. a n. In some tasks, this formula helps out a lot, yes ... You can remember this formula. And you can simply withdraw it at the right time, as here. After all, the formula for the sum and the formula for the nth term must be remembered in every way.)

Now the task in the form of a short encryption):

3. Find the sum of all positive two-digit numbers that are multiples of three.

How! No first member, no last, no progression at all... How to live!?

You will have to think with your head and pull out from the condition all the elements of the sum of an arithmetic progression. What are two-digit numbers - we know. They consist of two numbers.) What two-digit number will first? 10, presumably.) last thing two digit number? 99, of course! The three-digit ones will follow him ...

Multiples of three... Hm... These are numbers that are evenly divisible by three, here! Ten is not divisible by three, 11 is not divisible... 12... is divisible! So, something is emerging. You can already write a series according to the condition of the problem:

12, 15, 18, 21, ... 96, 99.

Will this series be an arithmetic progression? Certainly! Each term differs from the previous one strictly by three. If 2, or 4, is added to the term, say, the result, i.e. a new number will no longer be divided by 3. You can immediately determine the difference of the arithmetic progression to the heap: d = 3. Useful!)

So, we can safely write down some progression parameters:

What will be the number n last member? Anyone who thinks that 99 is fatally mistaken ... Numbers - they always go in a row, and our members jump over the top three. They don't match.

There are two solutions here. One way is for the super hardworking. You can paint the progression, the whole series of numbers, and count the number of terms with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If the formula is applied to our problem, we get that 99 is the thirtieth member of the progression. Those. n = 30.

We look at the formula for the sum of an arithmetic progression:

We look and rejoice.) We pulled out everything necessary for calculating the amount from the condition of the problem:

a 1= 12.

a 30= 99.

S n = S 30.

What remains is elementary arithmetic. Substitute the numbers in the formula and calculate:

Answer: 1665

Another type of popular puzzles:

4. An arithmetic progression is given:

-21,5; -20; -18,5; -17; ...

Find the sum of terms from the twentieth to thirty-fourth.

We look at the sum formula and ... we are upset.) The formula, let me remind you, calculates the sum from the first member. And in the problem you need to calculate the sum since the twentieth... The formula won't work.

You can, of course, paint the entire progression in a row, and put the members from 20 to 34. But ... somehow it turns out stupidly and for a long time, right?)

There is a more elegant solution. Let's break our series into two parts. The first part will from the first term to the nineteenth. Second part - twenty to thirty-four. It is clear that if we calculate the sum of the terms of the first part S 1-19, let's add it to the sum of the members of the second part S 20-34, we get the sum of the progression from the first term to the thirty-fourth S 1-34. Like this:

S 1-19 + S 20-34 = S 1-34

This shows that to find the sum S 20-34 can be done by simple subtraction

S 20-34 = S 1-34 - S 1-19

Both sums on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Are we getting started?

We extract the progression parameters from the task condition:

d = 1.5.

a 1= -21,5.

To calculate the sums of the first 19 and the first 34 terms, we will need the 19th and 34th terms. We count them according to the formula of the nth term, as in problem 2:

a 19\u003d -21.5 + (19-1) 1.5 \u003d 5.5

a 34\u003d -21.5 + (34-1) 1.5 \u003d 28

There is nothing left. Subtract the sum of 19 terms from the sum of 34 terms:

S 20-34 = S 1-34 - S 1-19 = 110.5 - (-152) = 262.5

Answer: 262.5

One important note! There is a very useful feature in solving this problem. Instead of direct calculation what you need (S 20-34), we counted what, it would seem, is not needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the full result. Such a "feint with the ears" often saves in evil puzzles.)

In this lesson, we examined problems for which it is enough to understand the meaning of the sum of an arithmetic progression. Well, you need to know a couple of formulas.)

Practical advice:

When solving any problem for the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.

Formula of the nth term:

These formulas will immediately tell you what to look for, in which direction to think in order to solve the problem. Helps.

And now the tasks for independent solution.

5. Find the sum of all two-digit numbers that are not divisible by three.

Cool?) The hint is hidden in the note to problem 4. Well, problem 3 will help.

6. Arithmetic progression is given by the condition: a 1 =-5.5; a n+1 = a n +0.5. Find the sum of the first 24 terms.

Unusual?) This is a recurrent formula. You can read about it in the previous lesson. Do not ignore the link, such puzzles are often found in the GIA.

7. Vasya saved up money for the Holiday. As much as 4550 rubles! And I decided to give the most beloved person (myself) a few days of happiness). Live beautifully without denying yourself anything. Spend 500 rubles on the first day, and spend 50 rubles more on each subsequent day than on the previous one! Until the money runs out. How many days of happiness did Vasya have?

Is it difficult?) An additional formula from task 2 will help.

Answers (in disarray): 7, 3240, 6.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

Lesson type: learning new material.

Lesson Objectives:

expansion and deepening of students' ideas about tasks solved using arithmetic progression; organization of search activity of students when deriving the formula for the sum of the first n members of an arithmetic progression;
development of skills to independently acquire new knowledge, use already acquired knowledge to achieve the task;
development of the desire and need to generalize the facts obtained, the development of independence.

Tasks:

generalize and systematize the existing knowledge on the topic “Arithmetic progression”;
derive formulas for calculating the sum of the first n members of an arithmetic progression;
teach how to apply the obtained formulas in solving various problems;
draw students' attention to the procedure for finding the value of a numerical expression.

Equipment:

cards with tasks for work in groups and pairs;
evaluation paper;
presentation"Arithmetic progression".

I. Actualization of basic knowledge.

1. Independent work in pairs.

1st option:

Define an arithmetic progression. Write down a recursive formula that defines an arithmetic progression. Give an example of an arithmetic progression and indicate its difference.

2nd option:

Write down the formula for the nth term of an arithmetic progression. Find the 100th term of an arithmetic progression ( a n}: 2, 5, 8 …
At this time, two students on the back of the board are preparing answers to the same questions.
Students evaluate the partner's work by comparing it with the board. (Leaflets with answers are handed over).

2. Game moment.

Exercise 1.

Teacher. I conceived some arithmetic progression. Ask me only two questions so that after the answers you can quickly name the 7th member of this progression. (1, 3, 5, 7, 9, 11, 13, 15…)

Questions from students.

What is the sixth term of the progression and what is the difference?
What is the eighth term of the progression and what is the difference?

If there are no more questions, then the teacher can stimulate them - a “ban” on d (difference), that is, it is not allowed to ask what the difference is. You can ask questions: what is the 6th term of the progression and what is the 8th term of the progression?

Task 2.

There are 20 numbers written on the board: 1, 4, 7 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58.

The teacher stands with his back to the blackboard. The students say the number of the number, and the teacher immediately calls the number itself. Explain how I can do it?

The teacher remembers the formula of the nth term a n \u003d 3n - 2 and, substituting the given values of n, finds the corresponding values a n .

II. Statement of the educational task.

I propose to solve an old problem dating back to the 2nd millennium BC, found in Egyptian papyri.

Task:“Let it be said to you: divide 10 measures of barley between 10 people, the difference between each person and his neighbor is 1/8 of the measure.”

How does this problem relate to the topic of arithmetic progression? (Each next person gets 1/8 of the measure more, so the difference is d=1/8, 10 people, so n=10.)
What do you think the number 10 means? (The sum of all members of the progression.)
What else do you need to know to make it easy and simple to divide barley according to the condition of the problem? (The first term of the progression.)

Lesson objective- obtaining the dependence of the sum of the terms of the progression on their number, the first term and the difference, and checking whether the problem was solved correctly in ancient times.

Before deriving the formula, let's see how the ancient Egyptians solved the problem.

And they solved it like this:

1) 10 measures: 10 = 1 measure - average share;
2) 1 measure ∙ = 2 measures - doubled average share.
doubled average the share is the sum of the shares of the 5th and 6th person.
3) 2 measures - 1/8 measure = 1 7/8 measures - twice the share of the fifth person.
4) 1 7/8: 2 = 5/16 - the share of the fifth; and so on, you can find the share of each previous and subsequent person.

We get the sequence:

III. The solution of the task.

1. Work in groups

1st group: Find the sum of 20 consecutive natural numbers: S 20 \u003d (20 + 1) ∙ 10 \u003d 210.

In general

II group: Find the sum of natural numbers from 1 to 100 (Legend of Little Gauss).

S 100 \u003d (1 + 100) ∙ 50 \u003d 5050

Conclusion:

III group: Find the sum of natural numbers from 1 to 21.

Solution: 1+21=2+20=3+19=4+18…

Conclusion:

IV group: Find the sum of natural numbers from 1 to 101.

Conclusion:

This method of solving the considered problems is called the “Gauss method”.

2. Each group presents the solution to the problem on the board.

3. Generalization of the proposed solutions for an arbitrary arithmetic progression:

a 1 , a 2 , a 3 ,…, a n-2 , a n-1 , a n .
S n \u003d a 1 + a 2 + a 3 + a 4 + ... + a n-3 + a n-2 + a n-1 + a n.

We find this sum by arguing similarly:

4. Have we solved the task?(Yes.)

IV. Primary comprehension and application of the obtained formulas in solving problems.

1. Checking the solution of an old problem by the formula.

2. Application of the formula in solving various problems.

3. Exercises for the formation of the ability to apply the formula in solving problems.

A) No. 613

Given :( and n) - arithmetic progression;

(a n): 1, 2, 3, ..., 1500

To find: S 1500

Decision: , and 1 = 1, and 1500 = 1500,

B) Given: ( and n) - arithmetic progression;
(and n): 1, 2, 3, ...
S n = 210

To find: n
Decision:

V. Independent work with mutual verification.

Denis went to work as a courier. In the first month, his salary was 200 rubles, in each subsequent month it increased by 30 rubles. How much did he earn in a year?

Given :( and n) - arithmetic progression;
a 1 = 200, d=30, n=12
To find: S 12
Decision:

Answer: Denis received 4380 rubles for the year.

VI. Homework instruction.

p. 4.3 - learn the derivation of the formula.
№№ 585, 623 .
Compose a problem that would be solved using the formula for the sum of the first n terms of an arithmetic progression.

VII. Summing up the lesson.

1. Score sheet

2. Continue the sentences

Today in class I learned...
Learned Formulas...
I think that …

3. Can you find the sum of numbers from 1 to 500? What method will you use to solve this problem?

Bibliography.

1. Algebra, 9th grade. Textbook for educational institutions. Ed. G.V. Dorofeeva. Moscow: Enlightenment, 2009.

NUMERICAL SEQUENCES VI

§ 144. The sum of the members of an arithmetic progression

They say that once an elementary school teacher, wanting to occupy the class for a long time with independent work, gave the children a “difficult” task - to calculate the sum of all natural numbers from 1 to 100:

1 + 2 + 3 + 4 + ... + 100.

One of the students immediately suggested a solution. Here it is.:

1+2 +3+... + 98 +99+ 100 = (1 + 100) + (2 + 99) + (3 + 98) + ... +(49 + 52)+ (50 + 51) =
= 101 + 101 + . . . + 101 + 101 = 101 50 = 5050.
50 times

It was Carl Gauss, who later became one of the most famous mathematicians in the world*.

*A similar case with Gauss actually took place. However, here it is greatly simplified. The numbers suggested by the teacher were five-digit and made up an arithmetic progression with a three-digit difference.

The idea of such a solution can be used to find the sum of the terms of any arithmetic progression.

Lemma. The sum of two terms of a finite arithmetic progression, equidistant from the ends, is equal to the sum of the extreme terms.

For example, in a finite arithmetic progression

1, 2, 3.....98, 99, 100

terms 2 and 99, 3 and 98, 4 and 97, etc. are equidistant from the ends of this progression. Therefore, their sums 2 + 99, 3 + 98, 4 + 97 are equal to the sum of the extreme terms 1 + 100.

Proof of the lemma. Let in a finite arithmetic progression

a 1 , a 2 , ..., a n - 1 , a n

any two members are equally distant from the ends. Let's assume that one of them is k -th term from the left, that is a k , and the other - k th term from the right, i.e. a n -k+ one . Then

a k + a n -k+ 1 =[a 1 + (k - 1)d ] + [a 1 + (n - k )d ] = 2a 1 + (n - 1)d .

The sum of the extreme terms of this progression is equal to

a 1 + a n = a 1 + [a 1 + (n - 1)d ] = 2a 1 + (n - 1)d .

Thus,

a k + a n -k+ 1 = a 1 + a n

Q.E.D.

Using the lemma just proved, it is easy to obtain a general formula for the sum P members of any arithmetic progression.

S n = a 1 +a 2 + ...+ a n - 1 + a n

S n = a n + a n - 1 + ... + a 2 + a 1 .

Adding these two equalities term by term, we get:

2S n = (a 1 +a n ) + (a 2 +a n - 1)+...+(a n - 1 +a 2) + (a n +a 1)

a 1 +a n = a 2 +a n - 1 = a 3 +a n - 2 =... .

2S n = n (a 1 +a n ),

The sum of the members of a finite arithmetic progression is equal to the product of half the sum of the extreme members and the number of all members.

In particular,

Exercises

971. Find the sum of all odd three-digit numbers.

972. How many strokes will a clock make during a day if it strikes only the number of whole hours?

973. What is the sum of the first P natural numbers?

974. Derive the formula for the length of the path traveled by the body during uniformly accelerated motion:

where v 0 - initial speed in m/sec , a - acceleration in m/sec 2 , t - travel time sec.

975. Find the sum of all irreducible fractions with a denominator of 3 between positive integers t and P (t< п ).

976. A worker maintains 16 looms that operate automatically. Performance per machine a m/h. The worker turned on the first machine at 7 h, and each next by 5 min later than the previous one. Find out the output in meters for the first 2 h work.

977. Solve equations:

a) 1 + 7 + 13 + ... + X = 280;

b) ( X + 1) + (X + 4) + (X + 7) +...+ (X + 28) = 155

978. From July 1 to July 12, inclusive, the air temperature rose daily by an average of 1/2 degree. Knowing that the average temperature during this time turned out to be 18 3/4 degrees, determine what the air temperature was on July 1.

979. Find an arithmetic progression whose arithmetic mean P the first terms for any P equal to their number.

980. Find the sum of the first twenty terms of an arithmetic progression in which

a 6 + a 9 + a 12 + a 15 = 20.