For example, the sequence \(2\); \(5\); \(8\); \(11\); \(14\)... is an arithmetic progression, because each subsequent element differs from the previous one by three (can be obtained from the previous one by adding three):

In this progression, the difference \(d\) is positive (equal to \(3\)), and therefore each next term is greater than the previous one. Such progressions are called increasing.

However, \(d\) can also be a negative number. For example, in arithmetic progression \(16\); \(10\); \(4\); \(-2\); \(-8\)... the progression difference \(d\) is equal to minus six.

And in this case, each next element will be smaller than the previous one. These progressions are called decreasing.

Arithmetic progression notation

Progression is indicated by a small Latin letter.

Numbers that form a progression are called members(or elements).

They are denoted by the same letter as an arithmetic progression, but with a numerical index equal to the number of the element in order.

For example, arithmetic progression\(a_n = \left\( 2; 5; 8; 11; 14…\right\)\) consists of the elements \(a_1=2\); \(a_2=5\); \(a_3=8\) and so on.

In other words, for the progression \(a_n = \left\(2; 5; 8; 11; 14…\right\)\)

Solving arithmetic progression problems

In principle, the information presented above is already enough to solve almost any arithmetic progression problem (including those offered at the OGE).

Example (OGE). The arithmetic progression is specified by the conditions \(b_1=7; d=4\). Find \(b_5\).
Solution:

Answer: \(b_5=23\)

Example (OGE). The first three terms of an arithmetic progression are given: \(62; 49; 36…\) Find the value of the first negative term of this progression..
Solution:

Answer: \(-3\)

Example (OGE). Given several consecutive elements of an arithmetic progression: \(…5; x; 10; 12.5...\) Find the value of the element designated by the letter \(x\).
Solution:

Answer: \(7,5\).

Example (OGE). The arithmetic progression is defined by the following conditions: \(a_1=-11\); \(a_(n+1)=a_n+5\) Find the sum of the first six terms of this progression.
Solution:

Answer: \(S_6=9\).

Example (OGE). In arithmetic progression \(a_(12)=23\); \(a_(16)=51\). Find the difference of this progression.
Solution:

Answer: \(d=7\).

Important formulas for arithmetic progression

As you can see, many problems on arithmetic progression can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each subsequent element in this chain is obtained by adding the same number to the previous one (the difference of the progression).

However, sometimes there are situations when it is very inconvenient to decide “head-on”. For example, imagine that in the very first example we need to find not the fifth element \(b_5\), but the three hundred and eighty-sixth \(b_(386)\). Should we add four \(385\) times? Or imagine that in the penultimate example you need to find the sum of the first seventy-three elements. You'll be tired of counting...

Therefore, in such cases they do not solve things “head-on”, but use special formulas derived for arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum of \(n\) first terms.

Formula of the \(n\)th term: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first term of the progression;
\(n\) – number of the required element;
\(a_n\) – term of the progression with number \(n\).

This formula allows us to quickly find even the three-hundredth or the millionth element, knowing only the first and the difference of the progression.

Example. The arithmetic progression is specified by the conditions: \(b_1=-159\); \(d=8.2\). Find \(b_(246)\).
Solution:

Answer: \(b_(246)=1850\).

Formula for the sum of the first n terms: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where

\(a_n\) – the last summed term;

Example (OGE). The arithmetic progression is specified by the conditions \(a_n=3.4n-0.6\). Find the sum of the first \(25\) terms of this progression.
Solution:

Answer: \(S_(25)=1090\).

For the sum \(n\) of the first terms, you can get another formula: you just need to \(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25\ ) instead of \(a_n\) substitute the formula for it \(a_n=a_1+(n-1)d\). We get:

Formula for the sum of the first n terms: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where

\(S_n\) – the required sum of \(n\) first elements;
\(a_1\) – the first summed term;
\(d\) – progression difference;
\(n\) – number of elements in total.

Example. Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15.5\); \(14\)…
Solution:

Answer: \(S_(33)=-231\).

More complex arithmetic progression problems

Now you have all the information you need to solve almost any arithmetic progression problem. Let’s finish the topic by considering problems in which you not only need to apply formulas, but also think a little (in mathematics this can be useful ☺)

Example (OGE). Find the sum of all negative terms of the progression: \(-19.3\); \(-19\); \(-18.7\)…
Solution:

Answer: \(S_(65)=-630.5\).

Example (OGE). The arithmetic progression is specified by the conditions: \(a_1=-33\); \(a_(n+1)=a_n+4\). Find the sum from the \(26\)th to the \(42\) element inclusive.
Solution:

Answer: \(S=1683\).

For arithmetic progression, there are several more formulas that we did not consider in this article due to their low practical usefulness. However, you can easily find them.

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, which amounted to a lot interesting tasks and who added the fourteenth book to Euclid’s Elements, formulated the thought: “In an arithmetic progression, which has even number terms, the sum of the terms of the 2nd half is greater than the sum of the terms of the 1st by the square of 1/2 the number of terms.”

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.

An 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 also a sign of progression, which is why 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 1st 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.

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, the fractional number can be entered in the form of a decimal fraction (\(2.5\)) and in the form 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 in preparing 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 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.

At the savings bank by number personal account The depositor 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. A more accurate value is \(365\frac(1)(4)\) 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:
\(S_n = n \cdot \frac(2a_1+(n-1)d)(2) \)

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

Arithmetic progression - sequence numerical values, in which its neighboring members differ from each other by the same number (all elements of the series, starting from the 2nd, have a similar property). This number– the difference between the previous and subsequent terms 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 of 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 d is the desired difference of this progression.

d = a(j) – a(j-1).

Highlight:

• 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 progression and its arbitrary elements

If 2 arbitrary terms of the progression are known (i-th, k-th), then the difference for a given sequence can be determined based on the relationship:

a(i) = a(k) + (i – k)*d, which means d = (a(i) – a(k))/(i-k).

Difference of progression and its first term

This expression will help determine an 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 appropriate 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.

Some people treat the word “progression” with caution, as a very complex term from the branches of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi meter (where they still exist). And understand the essence (and in mathematics there is nothing more important than “getting the essence”) arithmetic sequence It's not that difficult once you understand a few basic concepts.

Mathematical number sequence

A numerical sequence is usually called a series of numbers, each of which has its own number.

a 1 is the first member of the sequence;

and 2 is the second term of the sequence;

and 7 is the seventh member of the sequence;

and n is the nth member of the sequence;

However, not any arbitrary set of numbers and numbers interests us. We will focus our attention on a numerical sequence in which the value of the nth term is related to its ordinal number by a relationship that can be clearly formulated mathematically. In other words: numerical value The nth number is some function of n.

a is the value of a member of a numerical sequence;

n is its serial number;

f(n) is a function, where the ordinal number in the numerical sequence n is the argument.

Definition

An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth term of an arithmetic sequence is as follows:

a n - the value of the current member of the arithmetic progression;

a n+1 - formula of the next number;

d - difference (certain number).

It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one and such an arithmetic progression will be increasing.

In the graph below it is easy to see why the number sequence is called “increasing”.

In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.

Specified member value

Sometimes it is necessary to determine the value of any arbitrary term a n of an arithmetic progression. This can be done by sequentially calculating the values ​​of all members of the arithmetic progression, starting from the first to the desired one. However, this path is not always acceptable if, for example, it is necessary to find the value of the five thousandth or eight millionth term. Traditional calculations will take a lot of time. However, a specific arithmetic progression can be studied using certain formulas. There is also a formula for the nth term: the value of any term of an arithmetic progression can be determined as the sum of the first term of the progression with the difference of the progression, multiplied by the number of the desired term, reduced by one.

The formula is universal for increasing and decreasing progression.

An example of calculating the value of a given term

Let's solve the following problem of finding the value of the nth term of an arithmetic progression.

Condition: there is an arithmetic progression with parameters:

The first term of the sequence is 3;

The difference in the number series is 1.2.

Task: you need to find the value of 214 terms

Solution: to determine the value of a given term, we use the formula:

a(n) = a1 + d(n-1)

Substituting the data from the problem statement into the expression, we have:

a(214) = a1 + d(n-1)

a(214) = 3 + 1.2 (214-1) = 258.6

Answer: The 214th term of the sequence is equal to 258.6.

The advantages of this method of calculation are obvious - the entire solution takes no more than 2 lines.

Sum of a given number of terms

Very often, in a given arithmetic series, it is necessary to determine the sum of the values ​​of some of its segments. To do this, there is also no need to calculate the values ​​of each term and then add them up. This method is applicable if the number of terms whose sum needs to be found is small. In other cases, it is more convenient to use the following formula.

The sum of the terms of an arithmetic progression from 1 to n is equal to the sum of the first and nth terms, multiplied by the number of the term n and divided by two. If in the formula the value of the nth term is replaced by the expression from the previous paragraph of the article, we get:

Calculation example

For example, let’s solve a problem with the following conditions:

The first term of the sequence is zero;

The difference is 0.5.

The problem requires determining the sum of the terms of the series from 56 to 101.

Solution. Let's use the formula for determining the amount of progression:

s(n) = (2∙a1 + d∙(n-1))∙n/2

First, we determine the sum of the values ​​of 101 terms of the progression by substituting the given conditions of our problem into the formula:

s 101 = (2∙0 + 0.5∙(101-1))∙101/2 = 2,525

Obviously, in order to find out the sum of terms of the progression from the 56th to the 101st, it is necessary to subtract S 55 from S 101.

s 55 = (2∙0 + 0.5∙(55-1))∙55/2 = 742.5

Thus, the sum of the arithmetic progression for this example is:

s 101 - s 55 = 2,525 - 742.5 = 1,782.5

Example of practical application of arithmetic progression

At the end of the article, let's return to the example of an arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider this example.

Boarding a taxi (which includes 3 km of travel) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles/km. Travel distance is 30 km. Calculate the cost of the trip.

1. Let’s discard the first 3 km, the price of which is included in the cost of landing.

30 - 3 = 27 km.

2. Further calculation is nothing more than parsing an arithmetic number series.

Member number - the number of kilometers traveled (minus the first three).

The value of the member is the sum.

The first term in this problem will be equal to a 1 = 50 rubles.

Progression difference d = 22 r.

the number we are interested in is the value of the (27+1)th term of the arithmetic progression - the meter reading at the end of the 27th kilometer is 27.999... = 28 km.

a 28 = 50 + 22 ∙ (28 - 1) = 644

Calendar data calculations for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of the celestial body to the star. In addition, various number series are successfully used in statistics and other applied areas of mathematics.

Another type of number sequence is geometric

Geometric progression is characterized by greater rates of change compared to arithmetic progression. It is no coincidence that in politics, sociology, and medicine, in order to show the high speed of spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops in geometric progression.

The Nth term of the geometric number series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first term is 1, the denominator is correspondingly equal to 2, then:

n=1: 1 ∙ 2 = 2

n=2: 2 ∙ 2 = 4

n=3: 4 ∙ 2 = 8

n=4: 8 ∙ 2 = 16

n=5: 16 ∙ 2 = 32,

b n - the value of the current term of the geometric progression;

b n+1 - formula of the next term of the geometric progression;

q is the denominator of the geometric progression (a constant number).

If the graph of an arithmetic progression is a straight line, then a geometric progression paints a slightly different picture:

As in the case of arithmetic, geometric progression has a formula for the value of an arbitrary term. Any nth term of a geometric progression is equal to the product of the first term and the denominator of the progression to the power of n reduced by one:

Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Let's find the 5th term of the progression

b 5 = b 1 ∙ q (5-1) = 3 ∙ 1.5 4 = 15.1875

The sum of a given number of terms is also calculated using a special formula. The sum of the first n terms of a geometric progression is equal to the difference between the product of the nth term of the progression and its denominator and the first term of the progression, divided by the denominator reduced by one:

If b n is replaced using the formula discussed above, the value of the sum of the first n terms of the number series under consideration will take the form:

Example. The geometric progression starts with the first term equal to 1. The denominator is set to 3. Let's find the sum of the first eight terms.

s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280