What is non-constant arithmetic progression. Property of members of an arithmetic progression

Arithmetic progression problems have existed since 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 the sequence are called its members and are usually denoted by letters with indices that indicate the serial number of this member (a1, a2, a3 ... it reads: “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. With a very large number of members, this 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:

Each member of the progression is the arithmetic mean of the previous member and the next one.
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 also 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 you to determine the n-th member of an arithmetic progression through any of its k-th member, 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.

The natural series of any numbers such as 1,2,3,...,n,... is the simplest example of an arithmetic progression.

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

Arithmetic and geometric progressions

Theoretical information

Theoretical information

	Arithmetic progression	Geometric progression
Definition	Arithmetic progression a n a sequence is called, each member of which, starting from the second, is equal to the previous member, added with the same number d (d- progression difference)	geometric progression b n a sequence of non-zero numbers is called, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression)
Recurrent formula	For any natural n a n + 1 = a n + d	For any natural n b n + 1 = b n ∙ q, b n ≠ 0
nth term formula	a n = a 1 + d (n - 1)	b n \u003d b 1 ∙ q n - 1, b n ≠ 0
characteristic property
Sum of the first n terms

Examples of tasks with comments

Exercise 1

In arithmetic progression ( a n) a 1 = -6, a 2

According to the formula of the nth term:

a 22 = a 1+ d (22 - 1) = a 1+ 21d

By condition:

a 1= -6, so a 22= -6 + 21d.

It is necessary to find the difference of progressions:

d= a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = - 48.

Answer : a 22 = -48.

Task 2

Find the fifth term of the geometric progression: -3; 6;....

1st way (using n-term formula)

According to the formula of the n-th member of a geometric progression:

b 5 \u003d b 1 ∙ q 5 - 1 = b 1 ∙ q 4.

As b 1 = -3,

2nd way (using recursive formula)

Since the denominator of the progression is -2 (q = -2), then:

b 3 = 6 ∙ (-2) = -12;

b 4 = -12 ∙ (-2) = 24;

b 5 = 24 ∙ (-2) = -48.

Answer : b 5 = -48.

Task 3

In arithmetic progression ( a n) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.

For an arithmetic progression, the characteristic property has the form .

Therefore:

Substitute the data in the formula:

Answer: 95.

Task 4

In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.

To find the sum of the first n terms of an arithmetic progression, two formulas are used:

Which of them is more convenient to apply in this case?

By condition, the formula of the nth member of the original progression is known ( a n) a n= 3n - 4. Can be found immediately and a 1, and a 16 without finding d . Therefore, we use the first formula.

Answer: 368.

Task 5

In arithmetic progression a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.

According to the formula of the nth term:

a 22 = a 1 + d (22 – 1) = a 1+ 21d.

By condition, if a 1= -6, then a 22= -6 + 21d. It is necessary to find the difference of progressions:

d= a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = -48.

Answer : a 22 = -48.

Task 6

Several consecutive terms of a geometric progression are recorded:

Find the term of the progression, denoted by the letter x .

When solving, we use the formula for the nth term b n \u003d b 1 ∙ q n - 1 for geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of these terms of the progression and divide by the previous one. In our example, you can take and divide by. We get that q \u003d 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.

Substituting the found values into the formula, we get:

Answer : .

Task 7

From the arithmetic progressions given by the formula of the nth term, choose the one for which the condition is satisfied a 27 > 9:

Since the specified condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:

Answer: 4.

Task 8

In arithmetic progression a 1= 3, d = -1.5. Specify the largest value of n for which the inequality holds a n > -6.

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

What is this progression?

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

Any sequence of real numbers that is obtained by adding (subtracting) some value from each previous number is called an algebraic (arithmetic) progression. This definition, translated into the language of mathematics, takes the form:

Here i is the ordinal number of the element of the series a i . Thus, knowing only one initial number, you can easily restore the entire series. The parameter d in the formula is called the progression difference.

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

a n \u003d a 1 + d * (n - 1).

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

What is the sum of an arithmetic progression: formula

Before giving the formula for the indicated amount, it is worth considering a simple special case. Given a progression of natural numbers from 1 to 10, you need to find their sum. Since there are few terms in the progression (10), it is possible to solve the problem head-on, that is, sum all the elements in order.

S 10 \u003d 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 \u003d 55.

It is worth considering one interesting thing: since each term differs from the next one by the same value d \u003d 1, then the pairwise summation of the first with the tenth, the second with the ninth, and so on will give the same result. Really:

11 = 1+10 = 2+9 = 3+8 = 4+7 = 5+6.

As you can see, there are only 5 of these sums, that is, exactly two times less than the number of elements in the series. Then multiplying the number of sums (5) by the result of each sum (11), you will come to the result obtained in the first example.

If we generalize these arguments, we can write the following expression:

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

This expression shows that it is not at all necessary to sum all the elements in a row, it is enough to know the value of the first a 1 and the last a n , as well as the total number of terms n.

It is believed that Gauss first thought of this equality when he was looking for a solution to the problem set by his school teacher: to sum the first 100 integers.

Sum of elements from m to n: formula

The formula given in the previous paragraph answers the question of how to find the sum of an arithmetic progression (of the first elements), but often in tasks it is necessary to sum a series of numbers in the middle of the progression. How to do it?

The easiest way to answer this question is by considering the following example: let it be necessary to find the sum of terms from the mth to the nth. To solve the problem, a given segment from m to n of the progression should be represented as a new number series. In this representation, the m-th member a m will be the first, and a n will be numbered n-(m-1). In this case, applying the standard formula for the sum, the following expression will be obtained:

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

Example of using formulas

Knowing how to find the sum of an arithmetic progression, it is worth considering a simple example of using the above formulas.

Below is a numerical sequence, you should find the sum of its members, starting from the 5th and ending with the 12th:

The given numbers indicate that the difference d is equal to 3. Using the expression for the nth element, you can find the values of the 5th and 12th members of the progression. It turns out:

a 5 \u003d a 1 + d * 4 \u003d -4 + 3 * 4 \u003d 8;
a 12 \u003d a 1 + d * 11 \u003d -4 + 3 * 11 \u003d 29.

Knowing the values of the numbers at the ends of the considered algebraic progression, and also knowing what numbers in the series they occupy, you can use the formula for the sum obtained in the previous paragraph. Get:

S 5 12 \u003d (12 - 5 + 1) * (8 + 29) / 2 \u003d 148.

It is worth noting that this value could be obtained differently: first, find the sum of the first 12 elements using the standard formula, then calculate the sum of the first 4 elements using the same formula, and then subtract the second from the first sum.

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

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

The first element of the sequence;

The fifth element of the sequence;

- "nth" element of the sequence, i.e. the element "standing in the queue" at number n.

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

The sequence can be specified in three ways:

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

For example, Someone decided to do personal time management, and to begin with, to calculate how much time he spends on VKontakte during the week. By writing the time in a table, he will get a sequence consisting of seven elements:

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

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

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

For example, if , then

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

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

If, for example, , then

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

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

For example, consider the sequence ,

We can find the values of the members of a sequence in sequence, starting from the third:

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

Now we can define an arithmetic progression. An arithmetic progression is a simple special case of a numerical sequence.

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

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

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

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

If , then each term of the arithmetic progression is less than the previous one, and the progression is waning.

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

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

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

The main property of an arithmetic progression:

Let's look at the picture.

We see that

, and at the same time

Adding these two equalities, we get:

Divide both sides of the equation by 2:

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

Moreover, because

, and at the same time

, then

, and hence

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

th member formula.

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

and finally

We got formula of the nth term.

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

The sum of n members of an arithmetic progression.

In an arbitrary arithmetic progression, the sums of terms equally spaced from the extreme ones are equal to each other:

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

Arrange the terms of the progression first in ascending order of numbers, and then in descending order:

Let's pair it up:

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

We get:

So, the sum of n members of an arithmetic progression can be found using the formulas:

Consider solving arithmetic progression problems.

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

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

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

2 . Given an arithmetic progression -31; -27;...

a) Find the 31 terms of the progression.

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

a) We see that ;

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

In general

In our case , That's why

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.

It is known that in some algebraic progression the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.

Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . We substitute 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. The following example can be given: two numbers are given, for example, 4 and 5. It is necessary to make an algebraic progression so that three more terms are placed 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.