Problems on arithmetic progression existed already in ancient times. They appeared and demanded a solution because they had a practical need.

So, in one of the papyri Ancient Egypt", which has a mathematical content - the Rhind papyrus (19th century BC) - contains the following task: divide ten measures of bread among ten people, provided that the difference between each of them is one eighth of the measure."

And in the mathematical works of the ancient Greeks there are elegant theorems related to arithmetic progression. Thus, Hypsicles of Alexandria (2nd century, who composed many interesting problems and added the fourteenth book to Euclid’s Elements), formulated the idea: “In an arithmetic progression having an even number of terms, the sum of the terms of the 2nd half more than the amount members of the 1st on the square of 1/2 the number of members.”

The sequence is denoted by an. The numbers of a sequence are called its members and are usually designated by letters with indices that indicate the serial number of 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? By it we mean the one 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 this progression is considered increasing.

Arithmetic progression is called finite if only its first few terms are taken into account. At very large quantities members is already an endless progression.

Any arithmetic progression is defined by the following formula:

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

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

1. Each term of the progression is the arithmetic mean of the previous term and the subsequent one.
2. Converse: if, starting from the 2nd, each term is the arithmetic mean of the previous term and the subsequent one, i.e. if the condition is met, then this sequence is an arithmetic progression. This equality is at the same time a sign of progression, therefore 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 terms of the sequence, starting with 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 progression numbers).

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

For example: the first term (a1) in an arithmetic progression is given and equal to three, and the difference (d) is equal to 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 an arithmetic progression through any of its kth terms, provided that it is known.

The sum of the terms of an arithmetic progression (meaning the first n terms of a finite 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 problems and the initial data.

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

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

Before we start deciding arithmetic progression problems, let's consider what a number sequence is, since an arithmetic progression is special case number sequence.

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

The first element of the sequence;

The fifth element of the sequence;

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

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

The sequence can be set 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 take up personal time management, and to begin with, count how much time he spends on VKontakte during the week. By recording the time in the table, he will receive a sequence consisting of seven elements:

The first line of the table indicates 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 term formula.

In this case, the dependence of the value of a sequence element on its number is expressed directly in the form of 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 of the nth term.

We do the same thing 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 into the function equation:

If, for example, , That

Let me note once again that in sequence, unlike arbitrary numerical function, the argument can only be a natural number.

3 . The sequence can be specified using a formula that expresses the dependence of the value of the sequence member number n on the values ​​of the previous members.

In this case, it is not enough for us to know only the number of the sequence member 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 sequence members in sequence

, starting from the third: That is, every time, to find the value of the nth term of the sequence, we return to the previous two. This method of specifying a sequence 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 number sequence. is a numerical sequence, each member of which, starting from the second, is equal to the previous one added to the same number.

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

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

For example, 2; 5; 8; eleven;...

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

For example, 2; -1; -4; -7;...

If , then all terms 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 equality by 2:

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

Moreover, since

, and at the same time

, That

, and therefore

Each term of an arithmetic progression, starting with title="k>l">, равен среднему арифметическому двух равноотстоящих. !}

Formula of the th term.

We see that the terms of the arithmetic progression satisfy the following relations:

and finally

We got formula of the nth term.

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

The sum of n terms of an arithmetic progression.

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

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

Let's arrange the terms of the progression first in ascending order of numbers, and then in descending order:

Let's add in pairs:

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

We get:

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

Let's consider solving arithmetic progression problems.

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

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

We found that the difference between 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 31 terms of the progression.

b) Determine whether 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

Many people have heard about arithmetic progression, but not everyone has a good idea of ​​what it is. In this article we will give the corresponding definition, and also consider the question of how to find the difference of an arithmetic progression, and give a number of examples.

Mathematical definition

So if we're talking about about arithmetic or algebraic progression (these concepts define the same thing), this means that there is some number series, satisfying the following law: every two adjacent numbers in a series differ by the same value. Mathematically it is written like this:

Here n means the number of element a n in the sequence, and the number d is the difference of the progression (its name follows from the presented formula).

What does knowing the difference d mean? About how “far” neighboring numbers are from each other. However, knowledge of d is a necessary but not sufficient condition for determining (restoring) the entire progression. You need to know one more number, which can be absolutely any element of the series under consideration, for example, a 4, a10, but, as a rule, they use the first number, that is, a 1.

Formulas for determining progression elements

In general, the information above is already enough to move on to the solution specific tasks. Nevertheless, before the arithmetic progression is given, and it will be necessary to find its difference, we will present a couple of useful formulas, thereby facilitating the subsequent process of solving problems.

It is easy to show that any element of the sequence with number n can be found as follows:

a n = a 1 + (n - 1) * d

Indeed, anyone can check this formula by simple search: if you substitute n = 1, you get the first element, if you substitute n = 2, then the expression gives the sum of the first number and the difference, and so on.

The conditions of many problems are composed in such a way that, given a known pair of numbers, the numbers of which are also given in the sequence, it is necessary to reconstruct the entire number series (find the difference and the first element). Now we will solve this problem in general view.

So, let two elements with numbers n and m be given. Using the formula obtained above, you can create a system of two equations:

a n = a 1 + (n - 1) * d;

a m = a 1 + (m - 1) * d

To find unknown quantities, we use the known simple trick solutions to such a system: subtract the left and right sides in pairs, the equality will remain valid. We have:

a n = a 1 + (n - 1) * d;

a n - a m = (n - 1) * d - (m - 1) * d = d * (n - m)

Thus, we have excluded one unknown (a 1). Now we can write the final expression for determining d:

d = (a n - a m) / (n - m), where n > m

We got very simple formula: to calculate the difference d in accordance with the conditions of the problem, you only need to take the ratio of the differences between the elements themselves and their serial numbers. Should pay attention to one important point attention: the differences are taken between the “senior” and “junior” members, that is, n > m (“senior” means standing further from the beginning of the sequence, its absolute value can be either larger or smaller than the “junior” element).

The expression for the difference d progression should be substituted into any of the equations at the beginning of solving the problem to obtain the value of the first term.

In our age of development computer technology Many schoolchildren try to find solutions for their assignments on the Internet, so questions of this type often arise: find the difference of an arithmetic progression online. For such a request, the search engine will return a number of web pages, by going to which you will need to enter the data known from the condition (this can be either two terms of the progression or the sum of a certain number of them) and instantly receive an answer. However, this approach to solving the problem is unproductive in terms of the student’s development and understanding of the essence of the task assigned to him.

Solution without using formulas

Let's solve the first problem without using any of the given formulas. Let the elements of the series be given: a6 = 3, a9 = 18. Find the difference of the arithmetic progression.

Known elements are close to each other in a row. How many times must the difference d be added to the smallest to get the largest? Three times (the first time adding d, we get the 7th element, the second time - the eighth, and finally, the third time - the ninth). What number must be added to three three times to get 18? This is the number five. Really:

Thus, the unknown difference d = 5.

Of course, the solution could have been carried out using the appropriate formula, but this was not done intentionally. Detailed explanation solution to the problem should become clear and a shining example What is an arithmetic progression?

A task similar to the previous one

Now let's solve a similar problem, but change the input data. So, you should find if a3 = 2, a9 = 19.

Of course, you can again resort to the “head-on” solution method. But since the elements of the series are given, which are relatively far from each other, this method will not be entirely convenient. But using the resulting formula will quickly lead us to the answer:

d = (a 9 - a 3) / (9 - 3) = (19 - 2) / (6) = 17 / 6 ≈ 2.83

Here we have rounded the final number. The extent to which this rounding led to an error can be judged by checking the result obtained:

a 9 = a 3 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 = 18.98

This result differs by only 0.1% from the value given in the condition. Therefore, the rounding used to the nearest hundredths can be considered a successful choice.

Problems involving applying the formula for the an term

Let's consider classic example tasks to determine the unknown d: find the difference of the arithmetic progression if a1 = 12, a5 = 40.

When two numbers of an unknown algebraic sequence are given, and one of them is the element a 1, then you do not need to think long, but should immediately apply the formula for the a n term. IN in this case we have:

a 5 = a 1 + d * (5 - 1) => d = (a 5 - a 1) / 4 = (40 - 12) / 4 = 7

We received the exact number when dividing, so there is no point in checking the accuracy of the calculated result, as was done in the previous paragraph.

Let's solve another similar problem: we need to find the difference of an arithmetic progression if a1 = 16, a8 = 37.

We use an approach similar to the previous one and get:

a 8 = a 1 + d * (8 - 1) => d = (a 8 - a 1) / 7 = (37 - 16) / 7 = 3

What else should you know about arithmetic progression?

In addition to problems of finding an unknown difference or individual elements, it is often necessary to solve problems of the sum of the first terms of a sequence. Consideration of these tasks is beyond the scope of the article, however, for completeness of information we present general formula for the sum of n numbers in a series:

∑ n i = 1 (a i) = n * (a 1 + a n) / 2

Online calculator.
Solving an arithmetic progression.
Given: a n , d, n
Find: a 1

This mathematical program finds \(a_1\) of an arithmetic progression based on user-specified numbers \(a_n, d\) and \(n\).
The numbers \(a_n\) and \(d\) can be specified not only as integers, but also as fractions. Moreover, a fractional number can be entered as a decimal fraction (\(2.5\)) and as common fraction(\(-5\frac(2)(7)\)).

The program not only gives the answer to the problem, but also displays the process of finding a solution.

This online calculator may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework

in mathematics or algebra? In this case, you can also use our programs with detailed solutions. This way you can conduct your own training and/or training of yours. younger brothers

or sisters, while the level of education in the field of problems being solved increases.

If you are not familiar with the rules for entering numbers, we recommend that you familiarize yourself with them.

Rules for entering numbers
The numbers \(a_n\) and \(d\) can be specified not only as integers, but also as fractions.

The number \(n\) can only be a positive integer.
Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a period or a comma. For example, you can enter decimals

so 2.5 or so 2.5
Rules for entering ordinary fractions.

Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative. When entering numerical fraction /
The numerator is separated from the denominator by a division sign:
Input:

Result: \(-\frac(2)(3)\) Whole part &
The numerator is separated from the denominator by a division sign:
separated from the fraction by an ampersand:

Result: \(-1\frac(2)(3)\)

Enter numbers a n , d, n

Find a 1
It was discovered that some scripts necessary to solve this problem were not loaded, and the program may not work.
You may have AdBlock enabled.

In this case, disable it and refresh the page.
JavaScript is disabled in your browser.
For the solution to appear, you need to enable JavaScript.

Here are instructions on how to enable JavaScript in your browser.
Because There are a lot of people willing to solve the problem, your request has been queued.
In a few seconds the solution will appear below. Please wait

sec... If you noticed an error in the solution
, then you can write about this in the Feedback Form. Do not forget indicate which task you decide what.

enter in the fields

Our games, puzzles, emulators:

A little theory.

Number sequence Numbering is often used in everyday practice various items

In a savings bank, using the depositor’s personal account number, you can easily find this account and see what deposit is on it. Let account No. 1 contain a deposit of a1 rubles, account No. 2 contain a deposit of a2 rubles, etc. It turns out number sequence
a 1 , a 2 , a 3 , ..., a N
where N is the number of all accounts. Here, each natural number n from 1 to N is associated with a number a n.

Also studied in mathematics infinite number sequences:
a 1 , a 2 , a 3 , ..., a n , ... .
The number a 1 is called first member of the sequence, number a 2 - second term of the sequence, number a 3 - third term of the sequence etc.
The number a n is called nth (nth) member of the sequence, and the natural number n is its number.

For example, in a sequence of squares natural numbers 1, 4, 9, 16, 25, ..., n 2, (n + 1) 2, ... and 1 = 1 is the first term of the sequence; and n = n 2 is nth term sequences; a n+1 = (n + 1) 2 is the (n + 1)th (n plus first) term of the sequence. Often a sequence can be specified by the formula of its nth term. For example, the formula \(a_n=\frac(1)(n), \; n \in \mathbb(N) \) defines the sequence \(1, \; \frac(1)(2) , \; \frac( 1)(3) , \; \frac(1)(4) , \dots,\frac(1)(n) , \dots \)

Arithmetic progression

The length of the year is approximately 365 days. More exact value is equal to \(365\frac(1)(4)\) days, so every four years an error of one day accumulates.

To account for this error, a day is added to every fourth year, and the extended year is called a leap year.

For example, in the third millennium leap years are the years 2004, 2008, 2012, 2016, ... .

In this sequence, each member, starting from the second, is equal to the previous one, added to the same number 4. Such sequences are called arithmetic progressions.

Definition.
The number sequence a 1, a 2, a 3, ..., a n, ... is called arithmetic progression, if for all natural n the equality
\(a_(n+1) = a_n+d, \)
where d is some number.

From this formula it follows that a n+1 - a n = d. The number d is called the difference arithmetic progression.

By definition of an arithmetic progression we have:
\(a_(n+1)=a_n+d, \quad a_(n-1)=a_n-d, \)
where
\(a_n= \frac(a_(n-1) +a_(n+1))(2) \), where \(n>1 \)

Thus, each term of an arithmetic progression, starting from the second, is equal to the arithmetic mean of its two adjacent terms. This explains the name "arithmetic" progression.

Note that if a 1 and d are given, then the remaining terms of the arithmetic progression can be calculated using the recurrent formula a n+1 = a n + d. In this way it is not difficult to calculate the first few terms of the progression, however, for example, a 100 will already require a lot of calculations. Typically, the nth term formula is used for this. By definition of arithmetic progression
\(a_2=a_1+d, \)
\(a_3=a_2+d=a_1+2d, \)
\(a_4=a_3+d=a_1+3d \)
etc.
At all,
\(a_n=a_1+(n-1)d, \)
since the nth term of an arithmetic progression is obtained from the first term by adding (n-1) times the number d.
This formula is called formula for the nth term of an arithmetic progression.

Sum of the first n terms of an arithmetic progression

Find the sum of all natural numbers from 1 to 100.
Let's write this amount in two ways:
S = l + 2 + 3 + ... + 99 + 100,
S = 100 + 99 + 98 + ... + 2 + 1.
Let's add these equalities term by term:
2S = 101 + 101 + 101 + ... + 101 + 101.
This sum has 100 terms
Therefore, 2S = 101 * 100, hence S = 101 * 50 = 5050.

Let us now consider an arbitrary arithmetic progression
a 1 , a 2 , a 3 , ..., a n , ...
Let S n be the sum of the first n terms of this progression:
S n = a 1 , a 2 , a 3 , ..., a n
Then the sum of the first n terms of an arithmetic progression is equal to
\(S_n = n \cdot \frac(a_1+a_n)(2) \)

Since \(a_n=a_1+(n-1)d\), then replacing a n in this formula we get another formula for finding sum of the first n terms of an arithmetic progression:
\(S_n = n \cdot \frac(2a_1+(n-1)d)(2) \)

Books (textbooks) Abstracts of the Unified State Examination and the Unified State Examination tests online Games, puzzles Plotting graphs of functions Spelling dictionary of the Russian language Dictionary of youth slang Catalog of Russian schools Catalog of secondary educational institutions of Russia Catalog of Russian universities List of tasks

Instructions

An arithmetic progression is a sequence of the form a1, a1+d, a1+2d..., a1+(n-1)d. Number d step progression.It is obvious that the general of an arbitrary n-th term of the arithmetic progression has the form: An = A1+(n-1)d. Then knowing one of the members progression, member progression and step progression, you can, that is, the number of the progress member. Obviously, it will be determined by the formula n = (An-A1+d)/d.

Let now the mth term be known progression and another member progression- nth, but n , as in the previous case, but it is known that n and m do not coincide. Step progression can be calculated using the formula: d = (An-Am)/(n-m). Then n = (An-Am+md)/d.

If the sum of several elements of an arithmetic equation is known progression, as well as its first and last, then the number of these elements can also be determined. The sum of the arithmetic progression will be equal to: S = ((A1+An)/2)n. Then n = 2S/(A1+An) - chdenov progression. Using the fact that An = A1+(n-1)d, this formula can be rewritten as: n = 2S/(2A1+(n-1)d). From this we can express n by solving quadratic equation.

An arithmetic sequence is an ordered set of numbers, each member of which, except the first, differs from the previous one by the same amount. This constant value is called the difference of the progression or its step and can be calculated from the known terms of the arithmetic progression.

Instructions

If the values ​​of the first and second or any other pair of adjacent terms are known from the conditions of the problem, to calculate the difference (d) simply subtract the previous one from the subsequent term. The resulting value can be either positive or negative number- it depends on whether the progression is increasing. IN general form write the solution for an arbitrarily chosen pair (aᵢ and aᵢ₊₁) of neighboring terms of the progression as follows: d = aᵢ₊₁ - aᵢ.

For a pair of terms of such a progression, one of which is the first (a₁), and the other is any other arbitrarily chosen one, it is also possible to create a formula for finding the difference (d). However, in this case, the serial number (i) of an arbitrary selected member of the sequence must be known. To calculate the difference, add both numbers and divide the resulting result by the ordinal number of an arbitrary term reduced by one. In general, write this formula as follows: d = (a₁+ aᵢ)/(i-1).

If, in addition to an arbitrary member of an arithmetic progression with ordinal number i, another member with ordinal number u is known, change the formula from the previous step accordingly. In this case, the difference (d) of the progression will be the sum of these two terms divided by the difference of their ordinal numbers: d = (aᵢ+aᵥ)/(i-v).

The formula for calculating the difference (d) becomes somewhat more complicated if the problem conditions give the value of its first term (a₁) and the sum (Sᵢ) of a given number (i) of the first terms of the arithmetic sequence. To obtain the desired value, divide the sum by the number of terms that make it up, subtract the value of the first number in the sequence, and double the result. Divide the resulting value by the number of terms that make up the sum, reduced by one. In general, write the formula for calculating the discriminant as follows: d = 2*(Sᵢ/i-a₁)/(i-1).