Tasks for arithmetic progression already existed in ancient times. They appeared and demanded a solution, because they had a practical need.

So, in one of the papyri of Ancient Egypt, which has a mathematical content - the Rhind papyrus (XIX century BC) - contains the following task: divide ten measures of bread into ten people, provided that the difference between each of them is one eighth of a measure.

And in the mathematical works of the ancient Greeks there are elegant theorems related to arithmetic progression. So, Hypsicles of Alexandria (2nd century, who compiled many interesting problems and added the fourteenth book to Euclid's "Elements", formulated the idea: "In an arithmetic progression with an even number of members, the sum of the members of the 2nd half is greater than the sum of the members of the 1st by the square 1 / 2 members.

The sequence an is denoted. The numbers of a sequence are called its members and are usually denoted by letters with indices that indicate serial number this member (a1, a2, a3 ... read: "a 1st", "a 2nd", "a 3rd" and so on).

The sequence can be infinite or finite.

What is an arithmetic progression? It is understood as obtained by adding the previous term (n) with the same number d, which is the difference of the progression.

If d<0, то мы имеем убывающую прогрессию. Если d>0, then such a progression is considered to be increasing.

An arithmetic progression is said to be finite if only a few of its first terms are taken into account. At very in large numbers members is already an infinite progression.

Any arithmetic progression is given by the following formula:

an =kn+b, while b and k are some numbers.

The statement, which is the opposite, is absolutely true: if the sequence is given by a similar formula, then this is exactly an arithmetic progression, which has the properties:

1. Each member of the progression is the arithmetic mean of the previous member and the next one.
2. The opposite: if, starting from the 2nd, each term is the arithmetic mean of the previous term and the next, i.e. if the condition is met, then the given sequence is an arithmetic progression. This equality is at the same time a sign of progression, so it is usually called a characteristic property of progression.
   In the same way, the theorem that reflects this property is true: a sequence is an arithmetic progression only if this equality is true for any of the members of the sequence, starting from the 2nd.

The characteristic property for any four numbers of an arithmetic progression can be expressed by the formula an + am = ak + al if n + m = k + l (m, n, k are the numbers of the progression).

In an arithmetic progression, any necessary (Nth) term can be found by applying the following formula:

For example: the first term (a1) in an arithmetic progression is given and equals three, and the difference (d) equals four. You need to find the forty-fifth term of this progression. a45 = 1+4(45-1)=177

The formula an = ak + d(n - k) allows us to determine nth term arithmetic progression through any of its k-th term, provided that it is known.

The sum of the members of an arithmetic progression (assuming the 1st n members of the final progression) is calculated as follows:

Sn = (a1+an) n/2.

If the 1st term is also known, then another formula is convenient for calculation:

Sn = ((2a1+d(n-1))/2)*n.

The sum of an arithmetic progression that contains n terms is calculated as follows:

The choice of formulas for calculations depends on the conditions of the tasks and the initial data.

Natural series of any numbers like 1,2,3,...,n,...- the simplest example arithmetic progression.

In addition to the arithmetic progression, there is also a geometric one, which has its own properties and characteristics.

The concept of a numerical sequence implies that each natural number corresponds to some real value. Such a series of numbers can be both arbitrary and have certain properties - a progression. In the latter case, each subsequent element (member) of the sequence can be calculated using the previous one.

An arithmetic progression is a sequence of numerical values ​​in which its adjacent members differ by the same number(all elements of the series, starting from the 2nd one, have a similar property). This number - the difference between the previous and subsequent member - is constant and is called the progression difference.

Progression Difference: Definition

Consider a sequence consisting of j values ​​A = a(1), a(2), a(3), a(4) … a(j), j belongs to the set natural numbers N. An arithmetic progression, according to its definition, is a sequence in which a(3) - a(2) = a(4) - a(3) = a(5) - a(4) = ... = a(j) – a(j-1) = d. The value of d is the desired difference of this progression.

d = a(j) - a(j-1).

Allocate:

• An increasing progression, in which case d > 0. Example: 4, 8, 12, 16, 20, …
• decreasing progression, then d< 0. Пример: 18, 13, 8, 3, -2, …

Difference of progression and its arbitrary elements

If 2 arbitrary members of the progression (i-th, k-th) are known, then the difference for this sequence can be established based on the relation:

a(i) = a(k) + (i - k)*d, so d = (a(i) - a(k))/(i-k).

The progression difference and its first term

This expression will help determine the unknown value only in cases where the number of the sequence element is known.

Progression difference and its sum

The sum of a progression is the sum of its terms. To calculate the total value of its first j elements, use the corresponding formula:

S(j) =((a(1) + a(j))/2)*j, but since a(j) = a(1) + d(j – 1), then S(j) = ((a(1) + a(1) + d(j – 1))/2)*j=(( 2a(1) + d(– 1))/2)*j.

First level

Arithmetic progression. Detailed theory with examples (2019)

Numeric sequence

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

Numeric sequence
For example, for our sequence:

The assigned number is specific to only one sequence number. 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 the number is called the -th member of the sequence.

We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .

In our case:

Let's say we have a numerical sequence in which the difference between adjacent numbers is the same and equal.
For example:

etc.
Such a numerical sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius as early as the 6th century and was understood in a more broad sense, as an infinite number sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which the ancient Greeks were engaged in.

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

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

a)
b)
c)
d)

Got it? 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 member. Exists two way to find it.

1. Method

We can add to the previous value of the progression number 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 member of the described arithmetic progression is equal to.

2. Way

What if we needed to find the value of the th term of the progression? The summation would have taken us more than one hour, and it is not a fact that we would not have made mistakes when adding the numbers.
Of course, mathematicians have come up with a way in which you do not need to add the difference of an arithmetic progression to the previous value. Look closely at the drawn picture ... Surely you have already noticed a certain pattern, namely:

For example, let's see what makes up the value of the -th member of this arithmetic progression:


In other words:

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

Calculated? Compare your entries with the answer:

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

Arithmetic progression equation.

Arithmetic progressions are either 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 it out in practice.
We are given an arithmetic progression consisting of the following numbers:

Since then:

Thus, we were convinced that the formula works both in decreasing and in increasing arithmetic progression.
Try to find the -th and -th members of this arithmetic progression on your own.

Let's compare the results:

Arithmetic progression property

Let's complicate the task - we derive the property of an arithmetic progression.
Suppose we are given the following condition:
- arithmetic progression, find the value.
It's easy, you say, and start counting according to the formula you already know:

Let, a, then:

Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but what if we are given numbers in the condition? Agree, there is a possibility of making mistakes in the calculations.
Now think, is it possible to solve this problem in one step using any formula? Of course, yes, and we will try to bring it out now.

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

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

Let's sum the previous and next members of the progression:

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

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

Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, one of the greatest mathematicians of all time, the "king of mathematicians" - Karl Gauss, easily deduced for himself ...

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

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

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


Tried? What did you notice? Correctly! Their sums are equal

Now answer, how many such pairs will there be in the progression given to us? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two members of an arithmetic progression is equal, and similar equal pairs, we get that 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 progression difference. Try to substitute in the sum formula, the formula of the th member.
What did you get?

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

How much did you get?
Gauss turned out that the sum of the terms is equal, and the sum of the terms. Is that how you decided?

In fact, the formula for the sum of members of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people used the properties of an arithmetic progression with might and main.
For example, imagine Ancient Egypt and the most large-scale construction of that time - the construction of the pyramid ... The figure 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? Count how many blocks are needed to build one wall if block bricks are placed in the base. I hope you will not count by moving your finger across the monitor, do you remember the last formula and everything we said about arithmetic progression?

AT this case the progression looks like this:
Arithmetic progression difference.
The number of members of an arithmetic progression.
Let's substitute our data into the last formulas (we count the number of blocks in 2 ways).

Method 1.

Method 2.

And now you can also calculate on the monitor: compare the obtained values ​​​​with the number of blocks that are in our pyramid. Did it agree? Well done, you have mastered the sum of the th terms of an arithmetic progression.
Of course, you can’t build a pyramid from the blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:

Workout

Tasks:

1. Masha is getting in shape for the summer. Every day she increases the number of squats by. How many times will Masha squat in weeks if she did squats at the first workout.
2. What is the sum of all odd numbers contained in.
3. When storing logs, lumberjacks stack them in such a way that each upper layer contains one log less than the previous one. How many logs are in one masonry, if the base 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 squat once a day.

2. First odd number, last number.
   Arithmetic progression difference.
   The number of odd numbers in - half, however, check this fact using the formula for finding the -th member of an arithmetic progression:

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

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

3. Recall the problem about the pyramids. For our case, a , since each top layer is reduced by one log, there are only a bunch of layers, that is.
   Substitute the data in the formula:

   Answer: There are logs in the masonry.

Summing up

1. - a numerical sequence in which the difference between adjacent numbers is the same and equal. It is increasing and decreasing.
2. Finding formula th member 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 - the number of numbers in the progression.
4. The sum of the members of an arithmetic progression can be found in two ways:
   
   , where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Numeric sequence

Let's sit down and start writing some numbers. For example:

You can write any numbers, and there can be as many as you like. But you can always tell which of them is the first, which is the second, and so on, that is, we can number them. This is an example of a number sequence.

Numeric 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 only one. And we will not assign this number to any other number from this set.

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

We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .

It is very convenient if the -th member of the sequence can be given 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). Or (, difference).

nth term formula

We call recurrent a formula 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 such a formula, we have to calculate the previous nine. For example, let. Then:

Well, now it's clear what the formula is?

In each line, we add to, multiplied by some number. For what? Very simple: this is the number of the current member minus:

Much more comfortable 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 member is equal. And what is the difference? And here's what:

(after all, it is called the difference because it is equal to the difference of successive members of the progression).

So the formula is:

Then the hundredth term is:

What is the sum of all natural numbers from to?

According to legend, the great mathematician Carl Gauss, being a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last number is equal, the sum of the second and 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? 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 two-digit numbers, multiples.

Solution:

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

The formula for the th term for this progression is:

How many terms are in the progression if they must all be two digits?

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 1m more than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
2. A cyclist rides more miles each day than the previous one. On the first day he traveled km. How many days does he have to drive to cover a kilometer? How many kilometers will he travel on the last day of the journey?
3. The price of a refrigerator in the store is reduced by the same amount every year. Determine how much the price of a refrigerator decreased every year if, put up for sale for rubles, six years later it was sold for rubles.

Answers:

1. The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
   .
   Answer:
2. Here it is given:, it is necessary to find.
   Obviously, you need to use the same sum formula as in the previous problem:
   .
   Substitute the values:

   The root obviously doesn't fit, so the answer.
   Let's calculate the distance traveled over the last day using the formula of the -th member:
   (km).
   Answer:

3. Given: . Find: .
   It doesn't get easier:
   (rub).
   Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN

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

Arithmetic progression is increasing () and decreasing ().

For example:

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

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

Property of members of an arithmetic progression

It makes it easy to find a member of the progression if its neighboring members are known - where is the number of numbers in the progression.

The sum of the members of an arithmetic progression

There are two ways to find the sum:

Where is the number of values.

Where is the number of values.

First level

Arithmetic progression. Detailed theory with examples (2019)

Numeric sequence

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

Numeric sequence
For example, for our sequence:

The assigned number is specific to only one sequence number. 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 the number is called the -th member of the sequence.

We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .

In our case:

Let's say we have a numerical sequence in which the difference between adjacent numbers is the same and equal.
For example:

etc.
Such a numerical sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius as early as the 6th century and was understood in a broader sense as an endless numerical sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which the ancient Greeks were engaged in.

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

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

a)
b)
c)
d)

Got it? 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 member. Exists two way to find it.

1. Method

We can add to the previous value of the progression number 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 member of the described arithmetic progression is equal to.

2. Way

What if we needed to find the value of the th term of the progression? The summation would have taken us more than one hour, and it is not a fact that we would not have made mistakes when adding the numbers.
Of course, mathematicians have come up with a way in which you do not need to add the difference of an arithmetic progression to the previous value. Look closely at the drawn picture ... Surely you have already noticed a certain pattern, namely:

For example, let's see what makes up the value of the -th member of this arithmetic progression:


In other words:

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

Calculated? Compare your entries with the answer:

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

Arithmetic progression equation.

Arithmetic progressions are either 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 it out in practice.
We are given an arithmetic progression consisting of the following numbers:

Since then:

Thus, we were convinced that the formula works both in decreasing and in increasing arithmetic progression.
Try to find the -th and -th members of this arithmetic progression on your own.

Let's compare the results:

Arithmetic progression property

Let's complicate the task - we derive the property of an arithmetic progression.
Suppose we are given the following condition:
- arithmetic progression, find the value.
It's easy, you say, and start counting according to the formula you already know:

Let, a, then:

Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but what if we are given numbers in the condition? Agree, there is a possibility of making mistakes in the calculations.
Now think, is it possible to solve this problem in one step using any formula? Of course, yes, and we will try to bring it out now.

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

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

Let's sum the previous and next members of the progression:

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

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

Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, one of the greatest mathematicians of all time, the "king of mathematicians" - Karl Gauss, easily deduced for himself ...

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

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

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


Tried? What did you notice? Correctly! Their sums are equal

Now answer, how many such pairs will there be 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 equal pairs, we get that the total sum 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 progression difference. Try to substitute in the sum formula, the formula of the th member.
What did you get?

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

How much did you get?
Gauss turned out that the sum of the terms is equal, and the sum of the terms. Is that how you decided?

In fact, the formula for the sum of members of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people used the properties of an arithmetic progression with might and main.
For example, imagine Ancient Egypt and the largest construction site of that time - the construction of a pyramid ... The figure 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? Count how many blocks are needed to build one wall if block bricks are placed in the base. I hope you will not count by moving your finger across the monitor, do you remember the last formula and everything we said about arithmetic progression?

In this case, the progression looks like this:
Arithmetic progression difference.
The number of members of an arithmetic progression.
Let's substitute our data into the last formulas (we count the number of blocks in 2 ways).

Method 1.

Method 2.

And now you can also calculate on the monitor: compare the obtained values ​​​​with the number of blocks that are in our pyramid. Did it agree? Well done, you have mastered the sum of the th terms of an arithmetic progression.
Of course, you can’t build a pyramid from the blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:

Workout

Tasks:

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

2. First odd number, last number.
   Arithmetic progression difference.
   The number of odd numbers in - half, however, check this fact using the formula for finding the -th member of an arithmetic progression:

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

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

3. Recall the problem about the pyramids. For our case, a , since each top layer is reduced by one log, there are only a bunch of layers, that is.
   Substitute the data in the formula:

   Answer: There are logs in the masonry.

Summing up

1. - a numerical sequence in which the difference between adjacent numbers is the same and equal. It is increasing and decreasing.
2. Finding formula th member 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 - the number of numbers in the progression.
4. The sum of the members of an arithmetic progression can be found in two ways:
   
   , where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Numeric sequence

Let's sit down and start writing some numbers. For example:

You can write any numbers, and there can be as many as you like. But you can always tell which of them is the first, which is the second, and so on, that is, we can number them. This is an example of a number sequence.

Numeric 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 only one. And we will not assign this number to any other number from this set.

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

We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .

It is very convenient if the -th member of the sequence can be given 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). Or (, difference).

nth term formula

We call recurrent a formula 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 such a formula, we have to calculate the previous nine. For example, let. Then:

Well, now it's clear what the formula is?

In each line, we add to, multiplied by some number. For what? Very simple: this is the number of the current member minus:

Much more comfortable 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 member is equal. And what is the difference? And here's what:

(after all, it is called the difference because it is equal to the difference of successive members of the progression).

So the formula is:

Then the hundredth term is:

What is the sum of all natural numbers from to?

According to legend, the great mathematician Carl Gauss, being a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last number is equal, the sum of the second and 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? 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 two-digit multiples.

Solution:

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

The formula for the th term for this progression is:

How many terms are in the progression if they must all be two digits?

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 1m more than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
2. A cyclist rides more miles each day than the previous one. On the first day he traveled km. How many days does he have to drive to cover a kilometer? How many kilometers will he travel on the last day of the journey?
3. The price of a refrigerator in the store is reduced by the same amount every year. Determine how much the price of a refrigerator decreased every year if, put up for sale for rubles, six years later it was sold for rubles.

Answers:

1. The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
   .
   Answer:
2. Here it is given:, it is necessary to find.
   Obviously, you need to use the same sum formula as in the previous problem:
   .
   Substitute the values:

   The root obviously doesn't fit, so the answer.
   Let's calculate the distance traveled over the last day using the formula of the -th member:
   (km).
   Answer:

3. Given: . Find: .
   It doesn't get easier:
   (rub).
   Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN

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

Arithmetic progression is increasing () and decreasing ().

For example:

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

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

Property of members of an arithmetic progression

It makes it easy to find a member of the progression if its neighboring members are known - where is the number of numbers in the progression.

The sum of the members of an arithmetic progression

There are two ways to find the sum:

Where is the number of values.

Where is the number of values.

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 when solving various tasks;
• 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.

1. What is the sixth term of the progression and what is the difference?
2. 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.

A 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

Find: S 1500

Solution: , and 1 = 1, and 1500 = 1500,

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

Find: n
Solution:

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
Find: S 12
Solution:

Answer: Denis received 4380 rubles for the year.

VI. Homework instruction.

1. p. 4.3 - learn the derivation of the formula.
2. №№ 585, 623 .
3. 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 general educational institutions. Ed. G.V. Dorofeeva. Moscow: Enlightenment, 2009.