How to solve arithmetic progression 9. Posts tagged "9th grade arithmetic progression"
Mathematics has its own beauty, just like painting and poetry.
Russian scientist, mechanic N.E. Zhukovsky
Very common problems in entrance examinations in mathematics are problems related to the concept of arithmetic progression. To successfully solve such problems, you must have a good knowledge of the properties of arithmetic progression and have certain skills in their application.
Let us first recall the basic properties of an arithmetic progression and present the most important formulas, related to this concept.
Definition. Number sequence, in which each subsequent term differs from the previous one by the same number, called an arithmetic progression. In this case the numbercalled the progression difference.
For an arithmetic progression, the following formulas are valid:
, (1)
Where . Formula (1) is called the formula of the general term of an arithmetic progression, and formula (2) represents the main property of an arithmetic progression: each term of the progression coincides with the arithmetic mean of its neighboring terms and .
Note that it is precisely because of this property that the progression under consideration is called “arithmetic”.
The above formulas (1) and (2) are generalized as follows:
(3)
To calculate the amount first terms of an arithmetic progressionthe formula is usually used
(5) where and .
If we take into account the formula (1), then from formula (5) it follows
If we denote , then
Where . Since , formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).
In particular , from formula (5) it follows, What
Little known to most students is the property of arithmetic progression, formulated through the following theorem.
Theorem. If , then
Proof. If , then
The theorem is proven.
For example , using the theorem, it can be shown that
Let's move on to consider typical examples of solving problems on the topic “Arithmetic progression”.
Example 1. Let it be. Find .
Solution. Applying formula (6), we obtain . Since and , then or .
Example 2. Let it be three times greater, and when divided by the quotient, the result is 2 and the remainder is 8. Determine and .
Solution. From the conditions of the example, the system of equations follows
Since , , and , then from the system of equations (10) we obtain
The solution to this system of equations is and .
Example 3. Find if and .
Solution. According to formula (5) we have or . However, using property (9), we obtain .
Since and , then from the equality the equation follows or .
Example 4. Find if .
Solution.According to formula (5) we have
However, using the theorem, we can write
From here and from formula (11) we obtain .
Example 5. Given: . Find .
Solution. Since, then. However, therefore.
Example 6. Let , and . Find .
Solution. Using formula (9), we obtain . Therefore, if , then or .
Since and then here we have a system of equations
Solving which, we get and .
Natural root of the equation is .
Example 7. Find if and .
Solution. Since according to formula (3) we have that , then the system of equations follows from the problem conditions
If we substitute the expressioninto the second equation of the system, then we get or .
The roots of a quadratic equation are And .
Let's consider two cases.
1. Let , then . Since and , then .
In this case, according to formula (6), we have
2. If , then , and
Answer: and.
Example 8. It is known that and. Find .
Solution. Taking into account formula (5) and the condition of the example, we write and .
This implies the system of equations
If we multiply the first equation of the system by 2 and then add it to the second equation, we get
According to formula (9) we have. In this regard, it follows from (12) or .
Since and , then .
Answer: .
Example 9. Find if and .
Solution. Since , and by condition , then or .
From formula (5) it is known, What . Since, then.
Hence , here we have a system of linear equations
From here we get and . Taking into account formula (8), we write .
Example 10. Solve the equation.
Solution. From the given equation it follows that . Let us assume that , , and . In this case .
According to formula (1), we can write or .
Since , then equation (13) has the only suitable root .
Example 11. Find the maximum value provided that and .
Solution. Since , then the arithmetic progression under consideration is decreasing. In this regard, the expression takes on its maximum value when it is the number of the minimum positive term of the progression.
Let us use formula (1) and the fact, that and . Then we get that or .
Since , then or . However, in this inequalitylargest natural number, That's why .
If the values of , and are substituted into formula (6), we get .
Answer: .
Example 12. Determine the sum of all two-digit natural numbers that, when divided by the number 6, leave a remainder of 5.
Solution. Let us denote by the set of all two-digit natural numbers, i.e. . Next, we will construct a subset consisting of those elements (numbers) of the set that, when divided by the number 6, give a remainder of 5.
Easy to install, What . Obviously , that the elements of the setform an arithmetic progression, in which and .
To establish the cardinality (number of elements) of the set, we assume that . Since and , it follows from formula (1) or . Taking into account formula (5), we obtain .
The above examples of problem solving can by no means claim to be exhaustive. This article is written based on an analysis of modern methods for solving typical problems on a given topic. For a more in-depth study of methods for solving problems related to arithmetic progression, it is advisable to refer to the list of recommended literature.
1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace and Education, 2013. – 608 p.
2. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. – 216 p.
3. Medynsky M.M. A complete course of elementary mathematics in problems and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. – 208 p.
Still have questions?
To get help from a tutor, register.
website, when copying material in full or in part, a link to the source is required.
Subject: Arithmetic and geometric progressions
Class: 9
Training system: material for preparing the study of algebra topics and the preparatory stage for passing the OGE exam
Target: formation of the concepts of arithmetic and geometric progression
Tasks: teach to distinguish between types of progression, teach correctly, use formulas
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 a step or difference of progression.
Thus, by specifying the progression step and its first term, you can find any of its elements using the formula
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
Find the fortieth term of the arithmetic progression 4;7;...
Solution:
According to the condition we have
Let's determine the progression step
Using a well-known formula, we find the fortieth term of the progression
An 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
An arithmetic progression is given by a 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 sum of the progression is 250. Find the number of terms of the 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.
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
We substitute the found values into the formula for the sum of the progression to find the number of terms
As in the previous task, we will perform simplifications and solve the quadratic equation
We choose the more logical of the two values. We have that the sum of 18 terms of the progression with the given values a1=1, d=2 is equal to Sn=307.
Examples of problem solving: Arithmetic progression
Problem 1
The student team contracted to lay ceramic tiles on the floor in the hall of the youth club with an area of 288 m2. Gaining experience, the students laid out 2 m2 more on each subsequent day, starting from the second, than on the previous day, and their supply of tiles was enough for exactly 11 days of work. Planning that labor productivity would increase in the same way, the foreman determined that it would take another 5 days to complete the work. How many boxes of tiles should he order if 1 box is enough for 1.2 m2 of floor, and 3 boxes are needed to replace low-quality tiles?
Solution
According to the conditions of the problem, it is clear that we are talking about an arithmetic progression in which let
а1=х, Sn=288, n=16
Then we use the formula: Sn= (2a1+d(n-1))*n/0.86=200mmHg. Art.
288=(2x+2*15)*16/2
Let's calculate how many m2 students will lay out in 11 days: S11=(2*3+2*10)*11.2=143m2
288-143=145m2 left after 11 days of work, i.e. for 5 days
145/1.2=121 (approximately) boxes must be ordered for 5 days.
121+3=124 boxes must be ordered taking into account defects
Answer: 124 boxes
Problem 2
After each movement of the vacuum pump piston, 20% of the air in it is removed from the vessel. Let us determine the air pressure inside the vessel after six movements of the piston, if the initial pressure was 760 mm Hg. Art.
Solution
Since after each movement of the piston 20% of the available air is removed from the vessel, 80% of the air remains. To find out the air pressure in the vessel after the next movement of the piston, you need to multiply the pressure of the previous movement of the piston by 0.8.
We have a geometric progression whose first term is 760 and its denominator is 0.8. The number expressing the air pressure in the vessel (in mm Hg) after six movements of the piston is the seventh term of this progression. It is equal to 760*0.86=200mmHg. Art.
Answer: 200 mmHg.
An arithmetic progression is given, where the fifth and tenth terms are equal to 38 and 23, respectively. Find the fifteenth term of the progression and the sum of its first ten terms.
Solution:
Find the number of terms of the arithmetic progression 5,14,23,...,, if its th term is 239.
Solution:
Find the number of terms of an arithmetic progression is 9,12,15,...,, if its sum is 306.
Solution:
Find x for which the numbers x-1, 2x-1, x2-5 form an arithmetic progression
Solution:
Let's find the difference between 1 and 2 terms of the progression:
d=(2x-1)-(x-1)=x
Let's find the difference between 2 and 3 terms of the progression:
d=(x2-5)-(2x-1)=x2-2x-4
Because the difference is the same, then the terms of the progression can be equated:
When checked in both cases, an arithmetic progression is obtained
Answer: at x=-1 and x=4
The arithmetic progression is given by its third and seventh terms a3=5; a7=13. Find the first term of the progression and the sum of ten.
Solution:
We subtract the first from the second equation, as a result we find the progression step
a1+6d-(a1+2d)=4d=13-5=8, then d=2
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
S10=(2*1+(10-1)*2)*10/2=100
Answer: a1=1; S10=100
In an arithmetic progression whose first term is -3.4 and whose difference is 3, find the fifth and eleventh terms.
So we know that a1 = -3.4; d = 3. Find: a5, a11-.
Solution. To find the nth term of an arithmetic progression, we use the formula: an = a1+ (n – 1)d. We have:
a5 = a1 + (5 – 1)d = -3.4 + 4 3 = 8.6;
a11 = a1 + (11 – 1)d = -3.4 + 10 3 = 26.6.
As you can see, in this case, the solution is not difficult.
The twelfth term of an arithmetic progression is 74, and the difference is -4. Find the thirty-fourth term of this progression.
We are told that a12 = 74; d = -4, and we need to find a34-.
In this problem, it is not possible to immediately apply the formula an = a1 + (n – 1)d, because The first term a1 is unknown. This problem can be solved in several steps.
1. Using the term a12 and the formula for the nth term, we find a1:
a12 = a1 + (12 – 1)d, now let’s simplify and substitute d: a12 = a1 + 11 · (-4). From this equation we find a1: a1 = a12 – (-44);
We know the twelfth term from the problem statement, so we can easily calculate a1
a1 = 74 + 44 = 118. Let’s move on to the second step – calculating a34.
2. Again, using the formula an = a1 + (n – 1)d, since a1 is already known, we will determine a34-,
a34 = a1 + (34 – 1)d = 118 + 33 · (-4) = 118 – 132 = -14.
Answer: The thirty-fourth term of the arithmetic progression is -14.
As you can see, the solution to the second example is more complex. The same formula is used twice to obtain the answer. But everything is so complicated. The solution can be shortened by using additional formulas.
As already noted, if a1 is known in the problem, then the formula for determining the nth term of an arithmetic progression is very convenient to use. But, if the condition does not specify the first term, then a formula that connects the nth term we need and the term ak specified in the problem can come to the rescue.
an = ak + (n – k)d.
Let's solve the second example, but using a new formula.
Given: a12 = 74; d = -4. Find: a34-.
We use the formula an = ak + (n – k)d. In our case it will be:
a34 = a12 + (34 – 12) · (-4) = 74 + 22 · (-4) = 74 – 88 = -14.
The answer to the problem was obtained much faster, because there was no need to perform additional actions and look for the first term of the progression.
Using the above formulas, you can solve problems of calculating the difference of an arithmetic progression. So, using the formula an = a1 + (n – 1)d you can express d:
d = (an – a1) / (n – 1). However, problems with a given first term are not encountered so often, and they can be solved using our formula an = ak + (n – k)d, from which it is clear that d = (an – ak) / (n – k). Let's look at this problem.
Find the difference of the arithmetic progression if it is known that a3 = 36; a8 = 106.
Using the formula we obtained, the solution to the problem can be written in one line:
d = (a8 – a3) / (8 – 3) = (106 – 36) / 5 = 14.
Without this formula, solving the problem would have taken much longer, because a system of two equations would have to be solved.
Geometric progressions
1. Formula of the th term (common term of the progression).
2. Formula for the sum of the first terms of the progression: . When it is customary to talk about a convergent geometric progression; in this case, you can calculate the sum of the entire progression using the formula.
3. Formula for the “geometric mean”: if , , are three consecutive terms of a geometric progression, then by definition we have the following relations: either or .
To use presentation previews, create a Google account and log in to it: https://accounts.google.com
Slide captions:
Preview:
Subject
Arithmetic progression
TARGET :
- teach to recognize an arithmetic progression using its definition and sign;
- teach how to solve problems using a definition, a sign, a formula for the general term of a progression.
LESSON OBJECTIVES:
give a definition of an arithmetic progression, prove a sign of an arithmetic progression and teach how to use them in solving problems.
TEACHING METHODS:
updating students' knowledge, independent work, individual work, creating a problem situation.
MODERN TECHNOLOGIES:
ICT, problem-based learning, differentiated learning, health-saving technologies.
LESSON PLAN
Stages of the lesson. | Implementation time. |
|
Organizing time. | 2 minutes |
|
Repetition of what has been covered | 5 minutes |
|
Learning new material | 15 minutes |
|
Physical education minute | 3 minutes |
|
Completing assignments on the topic | 15 minutes |
|
Homework | 2 minutes |
|
Summarizing | 3 minutes |
DURING THE CLASSES:
- In the last lesson we were introduced to the concept of “Sequence”.
Today we will continue to study number sequences, define some of them, and get acquainted with their properties and characteristics.
- Answer the questions: What is a sequence?
What sequences are there?
In what ways can you set the sequence?
What is a number sequence?
What methods of specifying a number sequence do you know? What formula is called recurrent?
- Given numerical sequences:
- 1, 2, 3, 4, 5, …
- 2, 5, 8, 11, 14,…
- 8, 6, 4, 2, 0, - 2, …
- 0,5; 1; 1,5; 2; 2,5; …
Find the pattern of each sequence and name the next three terms of each.
- a n = a n -1 +1
- a n = a n -1 + 3
- a n = a n -1 + (-2)
- a n = a n -1 + 0.5
Give the recurrence formula for each sequence.
Slide 1
A numerical sequence, each member of which, starting from the second, is equal to the previous member added to the same number, is called an arithmetic progression.
The number d is called the difference of an arithmetic progression.
An arithmetic progression is a numerical sequence, so it can be increasing, decreasing, or constant. Give examples of such sequences, name the difference between each progression, and draw a conclusion.
Let us derive the formula for the general term of an arithmetic progression.
On the board: let a 1 is the first term of the progression, d is its difference, then
a 2 =a 1 +d
a 3 =(a 1 +d)+d=a 1 +2d
a 4 =(a 1 +2d)+d=a 1 +3d
a 5 =(a 1 +3d)+d=a 1 +4d
a n =a 1 +d (n-1) - formula of the nth term of an arithmetic progression.
Solve the problem: In an arithmetic progression, the first term is 5 and the difference is 4.
Find the 22nd term of this progression.
The student decides on the board: a n =a 1 +d(n-1)
A 22 =a 1 +21d=5+21*4=89
Physical education minute.
We got up.
Hands on the belt. Tilts left, right, (2 times);
Bend forward, backward (2 times);
Raise your hands up, take a deep breath, lower your hands down, exhale. (2 times)
They shook their hands. Thank you.
We sat down. Let's continue the lesson.
We solve problems using the formula for the general term of an arithmetic progression.
Students are offered the following tasks:
- In an arithmetic progression, the first term is -2, d=3, a n =118.
Find n.
- In an arithmetic progression, the first term is 7, the fifteenth term is –35. Find the difference.
- It is known that in arithmetic progression d=-2, a39=83. Find the first term of the progression.
Students are divided into groups. The task is given for 5 minutes. Next, the first 3 students who solved the problems solve them on the board. The solution is duplicated on the slides.
Let's consider the characteristic properties of an arithmetic progression.
In arithmetic progression
a n -d=a (n-1)
a n +d=a (n+1)
Let's add these two equalities term by term, we get: 2a n =a (n+1) +a (n-1)
A n =(a (n+1) +a (n-1 ))/2
This means that each member of an arithmetic progression, except the first and last, is equal to the arithmetic mean of the preceding and subsequent members.
THEOREM:
A numerical sequence is an arithmetic progression if and only if each of its members, except the first (and the last in the case of a finite sequence), is equal to the arithmetic mean of the preceding and subsequent members (a characteristic property of an arithmetic progression).
Understanding many topics in mathematics and physics is associated with knowledge of the properties of number series. Schoolchildren in the 9th grade, when studying the subject "Algebra", consider one of the important sequences of numbers - an arithmetic progression. We present the basic formulas of arithmetic progression (9th grade), as well as examples of their use for solving problems.
Algebraic or arithmetic progression
The number series that will be discussed in this article is called in two different ways, presented in the title of this paragraph. So, by arithmetic progression in mathematics we mean a number series in which any two adjacent numbers differ by the same amount, called the difference. Numbers in such a series are usually denoted by letters with a lower integer index, for example, a1, a2, a3 and so on, where the index indicates the number of the element of the series.
Taking into account the above definition of arithmetic progression, we can write the following equality: a2-a1 =...=an-an-1=d, here d is the difference of the algebraic progression and n is any integer. If d>0, then we can expect that each subsequent member of the series will be greater than the previous one, in this case we speak of an increasing progression. If d
Arithmetic progression formulas (9th grade school)
The series of numbers in question, since it is ordered and obeys some mathematical law, has two properties that are important for its use:
The first formula is easy to understand, since it is a direct consequence of the fact that each member of the series under consideration differs from its neighbor by the same difference.
The second formula for an arithmetic progression can be obtained by noting that the sum a1+an turns out to be equivalent to the sums a2+an-1, a3+an-2, and so on. Indeed, since a2 = d+a1, an-2 = -2*d+an, a3 = 2*d+a1, and an-1 = -d+an, then substituting these expressions into the corresponding sums, we find that they will be the same. The factor n/2 in the 2nd formula (for Sn) appears due to the fact that sums of type ai+1+an-i turn out to be exactly n/2, here i is an integer ranging from 0 to n/2- 1.
According to surviving historical evidence, the formula for the sum Sn was first obtained by Carl Gauss (the famous German mathematician) when he was given the task by his school teacher to add the first 100 numbers.
Example problem #1: find the difference
Problems in which the question is posed as follows: knowing the formulas of an arithmetic progression, how to find d (d), are the simplest that can only be for this topic.
Let's give an example: given a numerical sequence -5,-2, 1, 4, ..., it is necessary to determine its difference, that is, d.
This can be done as easily as possible: you need to take two elements and subtract the smaller one from the larger one. In this case we have: d = -2 - (-5) = 3.
To be sure of the answer received, it is recommended to check the remaining differences, since the presented sequence may not satisfy the algebraic progression condition. We have: 1-(-2)=3 and 4-1=3. These data indicate that we obtained the correct result (d=3) and proved that the series of numbers in the problem statement really represents an algebraic progression.
Example problem No. 2: find the difference, knowing two terms of the progression
Let's consider another interesting problem, which asks how to find the difference. In this case, the arithmetic progression formula must be used for the nth term. So, the task: given the first and fifth numbers of a series that corresponds to all the properties of an algebraic progression, for example, these are the numbers a1 = 8 and a5 = -10. How to find the difference d?
You should start solving this problem by writing a general formula for the nth element: an = a1+d*(-1+n). Now you can go in two ways: either immediately substitute the numbers and work with them, or express d, and then move on to specific a1 and a5. Using the last method, we get: a5 = a1+d*(-1+5) or a5 = 4*d+a1, which means that d = (a5-a1)/4. Now you can safely substitute the known data from the condition and get the final answer: d = (-10-8)/4 = -4.5.
Note that in this case the progression difference turned out to be negative, that is, there is a decreasing sequence of numbers. It is necessary to pay attention to this fact when solving problems so as not to confuse the signs “+” and “-”. All the formulas given above are universal, so they should always be followed regardless of the sign of the numbers with which the operations are carried out.
An example of solving problem No. 3: find a1, knowing the difference and the element
Let's change the problem statement a little. Let there be two numbers: the difference d=6 and the 9th element of the progression a9 = 10. How to find a1? The formulas for arithmetic progression remain unchanged, let's use them. For the number a9 we have the following expression: a1+d*(9-1) = a9. From where we easily get the first element of the series: a1 = a9-8*d = 10 - 8*6 = -38.
An example of solving problem No. 4: find a1, knowing two elements
This version of the problem is a complicated version of the previous one. The essence is the same, it is necessary to calculate a1, but now the difference d is not known, and instead of it another element of the progression is given.
An example of this type of problem is the following: find the first number of a sequence that is known to be an arithmetic progression and that its 15th and 23rd elements are 7 and 12, respectively.
It is necessary to solve this problem by writing an expression for the nth term for each element known from the condition, we have: a15 = d*(15-1)+a1 and a23 = d*(23-1)+a1. As you can see, we have two linear equations that need to be solved for a1 and d. Let's do this: subtract the first from the second equation, then we get the following expression: a23-a15 = 22*d - 14*d = 8*d. When deriving the last equation, the values of a1 were omitted because they cancel when subtracted. Substituting the known data, we find the difference: d = (a23-a15)/8 = (12-7)/8 = 0.625.
The value of d must be substituted into any formula for a known element to obtain the first term of the sequence: a15 = 14*d+a1, from which: a1=a15-14*d = 7-14*0.625 = -1.75.
Let's check the result obtained; to do this, we find a1 through the second expression: a23 = d*22+a1 or a1 = a23-d*22 = 12 - 0.625*22 = -1.75.
An example of solving problem No. 5: find the sum of n elements
As you can see, up to this point, only one arithmetic progression formula (9th grade) was used for the solution. Now we present a problem, the solutions of which require knowledge of the second formula, that is, for the sum Sn.
There is the following ordered series of numbers -1,1, -2,1, -3,1,..., you need to calculate the sum of its first 11 elements.
From this series it is clear that it is decreasing, and a1 = -1.1. Its difference is equal to: d = -2.1 - (-1.1) = -1. Now let's define the 11th term: a11 = 10*d + a1 = -10 + (-1,1) = -11,1. Having completed the preparatory calculations, you can use the formula noted above for the amount, we have: S11 =11*(-1.1 +(-11.1))/2 = -67.1. Since all the terms were negative numbers, their sum also has the corresponding sign.
An example of solving problem No. 6: find the sum of elements from n to m
Perhaps this type of problem is the most difficult for most schoolchildren. Let's give a typical example: given a series of numbers 2, 4, 6, 8..., you need to find the sum from the 7th to the 13th terms.
Arithmetic progression formulas (grade 9) are used exactly the same as in all previous problems. It is recommended to solve this problem step by step:
Let's get to the solution. Just as in the previous case, we will carry out preparatory calculations: a6 = 5*d+a1 = 10+2 = 12, a13 = 12*d+a1 = 24+2 = 26.
Let's calculate two sums: S13 = 13*(2+26)/2 = 182, S6 = 6*(2+12)/2 = 42. Take the difference and get the desired answer: S7-13 = S13 - S6 = 182-42 = 140. Note that when obtaining this value, it was the sum of 6 elements of the progression that was used as the subtrahend, since the 7th term is included in the sum S7-13.
- The most unusual space objects (6 photos) The most famous space objects
- Interesting facts about the dish
- What is Martini made from: production technology and composition What is the difference between the composition of different types of Martini
- Brodsky Joseph - biography Joseph Brodsky biography and personal