When studying algebra in a secondary school (grade 9) one of important topics is the study of number sequences, which include progressions - geometric and arithmetic. In this article we will look at an arithmetic progression and examples with solutions.

What is an arithmetic progression?

To understand this, it is necessary to define the progression in question, as well as give basic formulas, which 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 . Let's substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 = 6 + 6 * d. From this expression you can easily calculate the difference: d = (18 - 6) /6 = 2. Thus, we have answered the first part of the problem.

To restore the sequence to the 7th term, 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, a 7 = 18.

Example No. 3: drawing up a progression

Let's complicate it further stronger condition tasks. Now we need to answer the question of how to find an arithmetic progression. You can cite next example: two numbers are given, for example, - 4 and 5. It is necessary to create 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 will be occupied given numbers in future progression. Since there will be three more terms between them, then a 1 = -4 and a 5 = 5. Having established this, we move on to the problem, which is similar to the previous one. Again, for the nth term we use the formula, we get: a 5 = a 1 + 4 * d. From: d = (a 5 - a 1)/4 = (5 - (-4)) / 4 = 2.25. What we got here is not an integer value of the difference, but it is 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 terms 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 = 2.75 + 2.25 = 5, which coincided with the conditions of the problem.

Example No. 4: first term of progression

Let's continue to give examples of arithmetic progression with solutions. In all previous tasks the first number of the algebraic progression was known. Now let's 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 which number this sequence begins with.

The formulas used so far assume knowledge of a 1 and d. In the problem statement, nothing is known about these numbers. Nevertheless, we will write down expressions for each term about which information is available: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We received 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 easiest way to solve this system is to 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 = a 43 - 42 * d = 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d = 37 - 42 * d, whence the difference d = (37 - 50) / (42 - 14) = - 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 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.

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

Example No. 5: amount

Now let's look at several examples with solutions for the sum of an arithmetic progression.

Let it be given numerical progression of the following form: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?

Thanks to development computer technology you can solve this problem, that is, add all the numbers sequentially, which computer will do as soon as the 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 equal to 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.

It is interesting to note that this problem is called "Gaussian" because in early XVIII century, the famous German, while still only 10 years old, was able to solve it in his head 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 the numbers at the ends of the sequence in pairs, 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 No. 6: sum of terms from n to m

One more typical example the sum of an arithmetic progression is as follows: 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 equal to.

The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them sequentially. Since there are few terms, this method is not quite labor-intensive. Nevertheless, it is proposed to solve this problem using a second method, which is more universal.

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

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

Since n > m, it is obvious that the 2nd sum includes the first. 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), we will obtain the necessary answer to the problem. We have: S mn = S n - S m + a m =n * (a 1 + a n) / 2 - m *(a 1 + a m)/2 + a m = 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 knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before starting to solve any of these problems, it is recommended that you carefully read the condition, clearly understand what you need to find, and only then proceed with the solution.

Another tip is to strive for simplicity, that is, if you can answer a question without using complex mathematical calculations, then you need to do just that, since in this case the likelihood 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 = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and break common task into separate subtasks (in in this case first find the terms a n and a m).

If you have doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. We found out how to find an arithmetic progression. If you figure it out, it's not that difficult.

Arithmetic progression name a sequence of numbers (terms of a progression)

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

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

Properties of an arithmetic progression

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

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

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

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

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

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

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

3) If you need to find not the whole sum, but part of the sequence starting from its kth term, then the following sum formula will be useful to you

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

On this theoretical material ends and we move on to solving common problems in practice.

Example 1. Find the fortieth term of the arithmetic progression 4;7;...

Solution:

According to the condition we have

Let's determine the progression step

By well-known formula find the fortieth term of the progression

Example 2. Arithmetic progression is given by its third and seventh terms. Find the first term of the progression and the sum of ten.

Solution:

Let us write down the given elements of the progression using the formulas

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

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

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

Without applying complex calculations We found all the required quantities.

Example 3. An arithmetic progression is given by the denominator and one of its terms. Find the first term of the progression, the sum of its 50 terms starting from 50 and the sum of the first 100.

Solution:

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

and find the first one

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

Finding the sum of the part of the progression

and the sum of the first 100

The progression amount is 250.

Example 4.

Find the number of terms of an arithmetic progression if:

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

Solution:

Let's write the equations in terms of the first term and the progression step and determine them

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

We carry out simplifications

and solve the quadratic equation

Of the two values ​​found, only the number 8 fits the problem conditions. Thus, the sum of the first eight terms of the progression is 111.

Example 5.

Solve the equation

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

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

Before we start deciding arithmetic progression problems, let's look at what it is number sequence, since arithmetic progression is special case number sequence.

The number sequence is 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 one by one, 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 Latin word recurro- come back.

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

Arithmetic progression 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; 11;...

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 drawing.

We see that

, and at the same time

Adding these two equalities, we get:

.

Let's 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

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

This math 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, 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.
Whole and fractional part 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 is separated from the fraction by an ampersand: &
Input:
Result: \(-1\frac(2)(3)\)

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A little theory.

Number sequence

Numbering is often used in everyday practice various items to indicate the order in which they appear. For example, the houses on each street are numbered. In the library, reader's subscriptions are numbered and then arranged in the order of assigned numbers in special card files.

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 term 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,
\(a_n=a_1+(n-1)d, \)
because 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:
\(S_n = n \cdot \frac(2a_1+(n-1)d)(2) \)

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

What kind of progression is this?

Before moving on to the question (how to find the sum of an arithmetic progression), it is worth understanding what we are talking about.

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, when translated into mathematical language, takes the form:

Here i is the serial number of the element of the row a i. Thus, knowing just one starting 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 for the series of numbers under consideration the following equality holds:

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

That is, to find the value of the nth element in order, you should 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 = 1+2+3+4+5+6+7+8+9+10 = 55.

One thing worth considering interesting thing: since each term differs from the next one by the same value d = 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 of the series. Then multiplying the number of sums (5) by the result of each sum (11), you will arrive at the result obtained in the first example.

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

S n = 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 total number n terms.

It is believed that Gauss was the first to think of this equality when he was looking for a solution to a given problem. school teacher task: 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 (the first elements), but often in problems it is necessary to sum a series of numbers in the middle of the progression. How to do this?

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, you should represent the given segment from m to n of the progression as a new number series. In this m-th representation the term 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 = (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 terms, 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 terms of the progression. It turns out:

a 5 = a 1 + d * 4 = -4 + 3 * 4 = 8;

a 12 = a 1 + d * 11 = -4 + 3 * 11 = 29.

Knowing the values ​​of the numbers at the ends of the algebraic progression under consideration, as well as knowing what numbers in the series they occupy, you can use the formula for the sum obtained in the previous paragraph. It will turn out:

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

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