Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")
An arithmetic progression is a series of numbers in which each number is greater (or less) than the previous one by the same amount.
This topic is often difficult and incomprehensible. Letter indexes, nth member progressions, the difference in progression - all this is somehow confusing, yes ... Let's figure out the meaning of the arithmetic progression and everything will work out right away.)
The concept of arithmetic progression.
Arithmetic progression is a very simple and clear concept. Doubt? In vain.) See for yourself.
I'll write an unfinished series of numbers:
1, 2, 3, 4, 5, ...
Can you extend this line? What numbers will go next, after the five? Everyone ... uh ..., in short, everyone will figure out that the numbers 6, 7, 8, 9, etc. will go further.
Let's complicate the task. I give an unfinished series of numbers:
2, 5, 8, 11, 14, ...
You can catch the pattern, extend the series, and name seventh row number?
If you figured out that this number is 20 - I congratulate you! You not only felt key points of an arithmetic progression, but also successfully used them in business! If you don't understand, read on.
Now let's translate the key points from sensations into mathematics.)
First key point.
Arithmetic progression deals with series of numbers. This is confusing at first. We are used to solving equations, building graphs and all that ... And then extend the series, find the number of the series ...
It's OK. It's just that progressions are the first acquaintance with a new branch of mathematics. The section is called "Series" and works with series of numbers and expressions. Get used to it.)
Second key point.
In an arithmetic progression, any number differs from the previous one by the same amount.
In the first example, this difference is one. Whatever number you take, it is one more than the previous one. In the second - three. Any number is three times greater than the previous one. Actually, it is this moment that gives us the opportunity to catch the pattern and calculate the subsequent numbers.
Third key point.
This moment is not striking, yes ... But very, very important. Here he is: each progression number is in its place. There is the first number, there is the seventh, there is the forty-fifth, and so on. If you confuse them haphazardly, the pattern will disappear. will disappear and arithmetic progression. It's just a series of numbers.
That's the whole point.
Of course, in new topic new terms and notation appear. They need to know. Otherwise, you won't understand the task. For example, you have to decide something like:
Write down the first six terms of the arithmetic progression (a n) if a 2 = 5, d = -2.5.
Does it inspire?) Letters, some indexes... And the task, by the way, couldn't be easier. You just need to understand the meaning of the terms and notation. Now we will master this matter and return to the task.
Terms and designations.
Arithmetic progression is a series of numbers in which each number is different from the previous one by the same amount.
This value is called . Let's deal with this concept in more detail.
Arithmetic progression difference.
Arithmetic progression difference is the amount by which any progression number more the previous one.
One important point. Please pay attention to the word "more". Mathematically, this means that each progression number is obtained adding the difference of an arithmetic progression to the previous number.
To calculate, let's say second numbers of the row, it is necessary to first number add this very difference of an arithmetic progression. For calculation fifth- the difference is necessary add to fourth well, etc.
Arithmetic progression difference may be positive then each number of the series will turn out to be real more than the previous one. This progression is called increasing. For example:
8; 13; 18; 23; 28; .....
Here each number is adding positive number, +5 to the previous one.
The difference can be negative then each number in the series will be less than the previous one. This progression is called (you won't believe it!) decreasing.
For example:
8; 3; -2; -7; -12; .....
Here every number is obtained too adding to the previous one, but negative number, -5.
By the way, when working with a progression, it is very useful to immediately determine its nature - whether it is increasing or decreasing. It helps a lot to find your bearings in the decision, to detect your mistakes and correct them before it's too late.
Arithmetic progression difference usually denoted by the letter d.
How to find d? Very simple. It is necessary to subtract from any number of the series previous number. Subtract. By the way, the result of subtraction is called "difference".)
Let's define, for example, d for an increasing arithmetic progression:
2, 5, 8, 11, 14, ...
We take any number of the row that we want, for example, 11. Subtract from it the previous number those. eight:
This is the correct answer. For this arithmetic progression, the difference is three.
You can just take any number of progressions, because for a specific progression d-always the same. At least somewhere at the beginning of the row, at least in the middle, at least anywhere. You can not take only the very first number. Just because the very first number no previous.)
By the way, knowing that d=3, finding the seventh number of this progression is very simple. We add 3 to the fifth number - we get the sixth, it will be 17. We add three to the sixth number, we get the seventh number - twenty.
Let's define d for a decreasing arithmetic progression:
8; 3; -2; -7; -12; .....
I remind you that, regardless of the signs, to determine d needed from any number take away the previous one. We choose any number of progression, for example -7. His previous number is -2. Then:
d = -7 - (-2) = -7 + 2 = -5
The difference of an arithmetic progression can be any number: integer, fractional, irrational, any.
Other terms and designations.
Each number in the series is called member of an arithmetic progression.
Each member of the progression has his number. The numbers are strictly in order, without any tricks. First, second, third, fourth, etc. For example, in the progression 2, 5, 8, 11, 14, ... two is the first member, five is the second, eleven is the fourth, well, you understand ...) Please clearly understand - the numbers themselves can be absolutely any, whole, fractional, negative, whatever, but numbering- strictly in order!
How to record a progression in general view? No problem! Each number in the series is written as a letter. To denote an arithmetic progression, as a rule, the letter is used a. The member number is indicated by the index at the bottom right. Members are written separated by commas (or semicolons), like this:
a 1 , a 2 , a 3 , a 4 , a 5 , .....
a 1 is the first number a 3- third, etc. Nothing tricky. You can write this series briefly like this: (a n).
There are progressions finite and infinite.
Ultimate the progression has a limited number of members. Five, thirty-eight, whatever. But it's a finite number.
Endless progression - has an infinite number of members, as you might guess.)
You can write a final progression through a series like this, all members and a dot at the end:
a 1 , a 2 , a 3 , a 4 , a 5 .
Or like this, if there are many members:
a 1 , a 2 , ... a 14 , a 15 .
In a short entry, you will have to additionally indicate the number of members. For example (for twenty members), like this:
(a n), n = 20
An infinite progression can be recognized by the ellipsis at the end of the row, as in the examples in this lesson.
Now you can already solve tasks. The tasks are simple, purely for understanding the meaning of the arithmetic progression.
Examples of tasks for arithmetic progression.
Let's take a closer look at the task above:
1. Write down the first six members of the arithmetic progression (a n), if a 2 = 5, d = -2.5.
We translate the task into understandable language. Given an infinite arithmetic progression. The second number of this progression is known: a 2 = 5. Known progression difference: d = -2.5. We need to find the first, third, fourth, fifth and sixth members of this progression.
For clarity, I will write down a series according to the condition of the problem. The first six members, where the second member is five:
a 1 , 5 , a 3 , a 4 , a 5 , a 6 ,....
a 3 = a 2 + d
We substitute in the expression a 2 = 5 and d=-2.5. Don't forget the minus!
a 3=5+(-2,5)=5 - 2,5 = 2,5
The third term is less than the second. Everything is logical. If the number is greater than the previous one negative value, so the number itself will be less than the previous one. Progression is decreasing. Okay, let's take it into account.) We consider the fourth member of our series:
a 4 = a 3 + d
a 4=2,5+(-2,5)=2,5 - 2,5 = 0
a 5 = a 4 + d
a 5=0+(-2,5)= - 2,5
a 6 = a 5 + d
a 6=-2,5+(-2,5)=-2,5 - 2,5 = -5
So, the terms from the third to the sixth have been calculated. This resulted in a series:
a 1 , 5 , 2.5 , 0 , -2.5 , -5 , ....
It remains to find the first term a 1 according to the well-known second. This is a step in the other direction, to the left.) Hence, the difference of the arithmetic progression d should not be added to a 2, a take away:
a 1 = a 2 - d
a 1=5-(-2,5)=5 + 2,5=7,5
That's all there is to it. Task response:
7,5, 5, 2,5, 0, -2,5, -5, ...
In passing, I note that we solved this task recurrent way. This terrible word means, only, the search for a member of the progression by the previous (adjacent) number. Other ways to work with progression will be discussed later.
One important conclusion can be drawn from this simple task.
Remember:
If we know at least one member and the difference of an arithmetic progression, we can find any member of this progression.
Remember? This simple derivation allows us to solve most problems school course on this topic. All tasks revolve around three main parameters: member of an arithmetic progression, difference of a progression, number of a member of a progression. Everything.
Of course, all previous algebra is not cancelled.) Inequalities, equations, and other things are attached to the progression. But according to the progression- everything revolves around three parameters.
For example, consider some popular tasks on this topic.
2. Write the final arithmetic progression as a series if n=5, d=0.4, and a 1=3.6.
Everything is simple here. Everything is already given. You need to remember how the members of an arithmetic progression are calculated, count, and write down. It is advisable not to skip the words in the task condition: "final" and " n=5". In order not to count until you are completely blue in the face.) There are only 5 (five) members in this progression:
a 2 \u003d a 1 + d \u003d 3.6 + 0.4 \u003d 4
a 3 \u003d a 2 + d \u003d 4 + 0.4 \u003d 4.4
a 4 = a 3 + d = 4.4 + 0.4 = 4.8
a 5 = a 4 + d = 4.8 + 0.4 = 5.2
It remains to write down the answer:
3,6; 4; 4,4; 4,8; 5,2.
Another task:
3. Determine if the number 7 will be a member of an arithmetic progression (a n) if a 1 \u003d 4.1; d = 1.2.
Hmm... Who knows? How to define something?
How-how ... Yes, write down the progression in the form of a series and see if there will be a seven or not! We believe:
a 2 \u003d a 1 + d \u003d 4.1 + 1.2 \u003d 5.3
a 3 \u003d a 2 + d \u003d 5.3 + 1.2 \u003d 6.5
a 4 = a 3 + d = 6.5 + 1.2 = 7.7
4,1; 5,3; 6,5; 7,7; ...
Now it is clearly seen that we are just seven slipped through between 6.5 and 7.7! The seven did not get into our series of numbers, and, therefore, the seven will not be a member of the given progression.
Answer: no.
And here is a problem based on real version GIA:
4. Several consecutive members of the arithmetic progression are written out:
...; fifteen; X; 9; 6; ...
Here is a series without end and beginning. No member numbers, no difference d. It's OK. To solve the problem, it is enough to understand the meaning of an arithmetic progression. Let's see and see what we can to know from this line? What are the parameters of the three main ones?
Member numbers? There is not a single number here.
But there are three numbers and - attention! - word "consecutive" in condition. This means that the numbers are strictly in order, without gaps. Are there two in this row? neighboring known numbers? Yes there is! These are 9 and 6. So we can calculate the difference of an arithmetic progression! We subtract from the six previous number, i.e. nine:
There are empty spaces left. What number will be the previous one for x? Fifteen. So x can be easily found by simple addition. To 15 add the difference of an arithmetic progression:
That's all. Answer: x=12
We solve the following problems ourselves. Note: these puzzles are not for formulas. Purely for understanding the meaning of an arithmetic progression.) We just write down a series of numbers-letters, look and think.
5. Find the first positive term of the arithmetic progression if a 5 = -3; d = 1.1.
6. It is known that the number 5.5 is a member of the arithmetic progression (a n), where a 1 = 1.6; d = 1.3. Determine the number n of this member.
7. It is known that in an arithmetic progression a 2 = 4; a 5 \u003d 15.1. Find a 3 .
8. Several consecutive members of the arithmetic progression are written out:
...; 15.6; X; 3.4; ...
Find the term of the progression, denoted by the letter x.
9. The train started moving from the station, gradually increasing its speed by 30 meters per minute. What will be the speed of the train in five minutes? Give your answer in km/h.
10. It is known that in an arithmetic progression a 2 = 5; a 6 = -5. Find a 1.
Answers (in disarray): 7.7; 7.5; 9.5; 9; 0.3; four.
Everything worked out? Wonderful! You can master the arithmetic progression for more high level, in the next lessons.
Didn't everything work out? No problem. In Special Section 555, all these puzzles are broken down piece by piece.) And, of course, a simple practical technique is described that immediately highlights the solution of such tasks clearly, clearly, as in the palm of your hand!
By the way, in the puzzle about the train there are two problems on which people often stumble. One - purely by progression, and the second - common to any tasks in mathematics, and physics too. This is a translation of dimensions from one to another. It shows how these problems should be solved.
In this lesson, we examined the elementary meaning of an arithmetic progression and its main parameters. This is enough to solve almost all problems on this topic. Add d to the numbers, write a series, everything will be decided.
The finger solution works well for very short pieces of the series, as in the examples in this lesson. If the series is longer, the calculations become more complicated. For example, if in problem 9 in the question, replace "five minutes" on the "thirty-five minutes" the problem will become much worse.)
And there are also tasks that are simple in essence, but utterly absurd in terms of calculations, for example:
Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.
And what, we will add 1/6 many, many times?! Is it possible to kill yourself!?
You can.) If you do not know a simple formula by which you can solve such tasks in a minute. This formula will be in the next lesson. And that problem is solved there. In a minute.)
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
Lesson type: learning new material.
Lesson Objectives:
- expansion and deepening of students' ideas about tasks solved using arithmetic progression; organization search activity students when deriving the formula for the sum of the first n members of an arithmetic progression;
- development of skills to independently acquire new knowledge, use already acquired knowledge to achieve the task;
- development of the desire and need to generalize the facts obtained, the development of independence.
Tasks:
- generalize and systematize the existing knowledge on the topic “Arithmetic progression”;
- derive formulas for calculating the sum of the first n members of an arithmetic progression;
- teach how to apply the obtained formulas in solving various problems;
- draw students' attention to the procedure for finding the value of a numerical expression.
Equipment:
- cards with tasks for work in groups and pairs;
- evaluation paper;
- presentation"Arithmetic progression".
I. Actualization of basic knowledge.
1. Independent work in pairs.
1st option:
Define an arithmetic progression. Write down a recursive formula that defines an arithmetic progression. Give an example of an arithmetic progression and indicate its difference.
2nd option:
Write down the formula for the nth term of an arithmetic progression. Find the 100th term of an arithmetic progression ( a n}: 2, 5, 8 …
At this time, two students on the back of the board are preparing answers to the same questions.
Students evaluate the partner's work by comparing it with the board. (Leaflets with answers are handed over).
2. Game moment.
Exercise 1.
Teacher. I conceived some arithmetic progression. Ask me only two questions so that after the answers you can quickly name the 7th member of this progression. (1, 3, 5, 7, 9, 11, 13, 15…)
Questions from students.
- What is the sixth term of the progression and what is the difference?
- What is the eighth term of the progression and what is the difference?
If there are no more questions, then the teacher can stimulate them - a “ban” on d (difference), that is, it is not allowed to ask what the difference is. You can ask questions: what is the 6th term of the progression and what is the 8th term of the progression?
Task 2.
There are 20 numbers written on the board: 1, 4, 7 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58.
The teacher stands with his back to the blackboard. The students say the number of the number, and the teacher immediately calls the number itself. Explain how I can do it?
The teacher remembers the formula of the nth term a n \u003d 3n - 2 and, substituting the given values of n, finds the corresponding values a n .
II. Statement of the educational task.
I propose to solve an old problem dating back to the 2nd millennium BC, found in Egyptian papyri.
A task:“Let it be said to you: divide 10 measures of barley between 10 people, the difference between each person and his neighbor is 1/8 of the measure.”
- How does this problem relate to the topic of arithmetic progression? (Each next person gets 1/8 of the measure more, so the difference is d=1/8, 10 people, so n=10.)
- What do you think the number 10 means? (The sum of all members of the progression.)
- What else do you need to know to make it easy and simple to divide barley according to the condition of the problem? (The first term of the progression.)
Lesson objective- obtaining the dependence of the sum of the terms of the progression on their number, the first term and the difference, and checking whether the problem was solved correctly in ancient times.
Before deriving the formula, let's see how the ancient Egyptians solved the problem.
And they solved it like this:
1) 10 measures: 10 = 1 measure - average share;
2) 1 measure ∙ = 2 measures - doubled average share.
doubled average the share is the sum of the shares of the 5th and 6th person.
3) 2 measures - 1/8 measure = 1 7/8 measures - twice the share of the fifth person.
4) 1 7/8: 2 = 5/16 - the share of the fifth; and so on, you can find the share of each previous and subsequent person.
We get the sequence:
III. The solution of the task.
1. Work in groups
1st group: Find the sum of 20 consecutive natural numbers: S 20 \u003d (20 + 1) ∙ 10 \u003d 210.
In general 
II group: Find the sum of natural numbers from 1 to 100 (Legend of Little Gauss).
S 100 \u003d (1 + 100) ∙ 50 \u003d 5050
Conclusion:
III group: Find the sum of natural numbers from 1 to 21.
Solution: 1+21=2+20=3+19=4+18…
Conclusion:
IV group: Find the sum of natural numbers from 1 to 101.
Conclusion:
This method of solving the considered problems is called the “Gauss method”.
2. Each group presents the solution to the problem on the board.
3. Generalization of the proposed solutions for an arbitrary arithmetic progression:
a 1 , a 2 , a 3 ,…, a n-2 , a n-1 , a n .
S n \u003d a 1 + a 2 + a 3 + a 4 + ... + a n-3 + a n-2 + a n-1 + a n.
We find this sum by arguing similarly:
4. Have we solved the task?(Yes.)
IV. Primary comprehension and application of the obtained formulas in solving problems.
1. Checking the solution of an old problem by the formula.
2. Application of the formula in solving various problems.
3. Exercises for the formation of the ability to apply the formula in solving problems.
A) No. 613
Given :( and n) - arithmetic progression;
(a n): 1, 2, 3, ..., 1500
Find: S 1500
Solution: , and 1 = 1, and 1500 = 1500,
B) Given: ( and n) - arithmetic progression;
(and n): 1, 2, 3, ...
S n = 210
Find: n
Solution:
V. Independent work with mutual verification.
Denis went to work as a courier. In the first month, his salary was 200 rubles, in each subsequent month it increased by 30 rubles. How much did he earn in a year?
Given :( and n) - arithmetic progression;
a 1 = 200, d=30, n=12
Find: S 12
Solution:
Answer: Denis received 4380 rubles for the year.
VI. Homework instruction.
- p. 4.3 - learn the derivation of the formula.
- №№ 585, 623 .
- Compose a problem that would be solved using the formula for the sum of the first n terms of an arithmetic progression.
VII. Summing up the lesson.
1. Score sheet
2. Continue the sentences
- Today in class I learned...
- Learned Formulas...
- I think that …
3. Can you find the sum of numbers from 1 to 500? What method will you use to solve this problem?
Bibliography.
1. Algebra, 9th grade. Tutorial for educational institutions. Ed. G.V. Dorofeeva. Moscow: Enlightenment, 2009.
The concept of a numerical sequence implies that each natural number corresponds to some real value. Such a series of numbers can be both arbitrary and have certain properties - a progression. In the latter case, each subsequent element (member) of the sequence can be calculated using the previous one.
An arithmetic progression is a sequence of 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). Given number- the difference between the previous and subsequent member 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 of d is the desired difference of this progression.
d = a(j) - a(j-1).
Allocate:
- 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 of progression and its arbitrary elements
If 2 arbitrary members of the progression (i-th, k-th) are known, then the difference for this sequence can be established based on the relation:
a(i) = a(k) + (i - k)*d, so d = (a(i) - a(k))/(i-k).
The progression difference and its first term
This expression will help determine the 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 corresponding 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.
Mathematics has its own beauty, as does painting and poetry.
Russian scientist, mechanic N.E. Zhukovsky
Very common tasks entrance examinations in mathematics are tasks related to the concept of an arithmetic progression. To successfully solve such problems, it is necessary to know the properties of an arithmetic progression well and have certain skills in their application.
Let us first recall the main properties of an arithmetic progression and present the most important formulas, associated with this concept.
Definition. Numeric sequence, in which each subsequent term differs from the previous one by the same number, called an arithmetic progression. At the same time, the numberis called the progression difference.
For an arithmetic progression, the formulas are valid
, (1)
where . Formula (1) is called the formula of the common term of an arithmetic progression, and formula (2) is the main property of an arithmetic progression: each member of the progression coincides with the arithmetic mean of its neighboring members and .
Note that it is precisely because of this property that the progression under consideration is called "arithmetic".
Formulas (1) and (2) above are summarized as follows:
(3)
To calculate the sum first members of an arithmetic progressionthe formula is usually used
(5) where and .
If we take into account the formula (1), then formula (5) implies
If we designate
where . Since , then formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).
In particular , from formula (5) it follows, what
Among the little-known to most students is the property of an arithmetic progression, formulated by means of the following theorem.
Theorem. If , then
Proof. If , then
The theorem has been proven.
For example , using the theorem, it can be shown that
Let's move on to the consideration of typical examples of solving problems on the topic "Arithmetic progression".
Example 1 Let and . Find .
Solution. Applying formula (6), we obtain . Since and , then or .
Example 2 Let three times more, and when dividing by in the quotient, it turns out 2 and the remainder is 8. Determine and.
Solution. The system of equations follows from the condition of the example
Since , , and , then from the system of equations (10) we obtain
The solution of this system of equations are 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.By formula (5) we have
However, using the theorem, one 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 condition of the problem
If we substitute the expressioninto the second equation of the system, then we get or .
Rooted 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 connection, from (12) it follows 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 .
Consequently , 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. It follows from the given equation that . Let's assume that , , and . In this case .
According to formula (1), we can write or .
Since , equation (13) has a unique suitable root .
Example 11. Find the maximum value provided that and .
Solution. Since , then the considered arithmetic progression is decreasing. In this regard, the expression takes on a maximum value when it is the number of the minimum positive member of the progression.
We use formula (1) and the fact, which and . Then we get that or .
Because , then or . However, in this inequalitylargest natural number, that's why .
If the values , and are substituted into formula (6), then we get .
Answer: .
Example 12. Find the sum of all two-digit natural numbers that, when divided by 6, have a remainder of 5.
Solution. Denote by the set of all two-valued natural numbers, i.e. . Next, we 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 determine the cardinality (number of elements) of the set, we assume that . Since and , then formula (1) implies or . Taking into account formula (5), we obtain .
The above examples of solving problems can by no means claim to be exhaustive. This article is based on the analysis modern methods solving typical problems on a given topic. For a deeper study of methods for solving problems related to arithmetic progression, it is advisable to refer to the list of recommended literature.
1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.
2. Suprun V.P. Mathematics for high school students: additional sections school curriculum. – M.: Lenand / URSS, 2014. - 216 p.
3. Medynsky M.M. Full course elementary mathematics in tasks and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. - 208 p.
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Someone treats the word "progression" with caution, as a very complex term from the sections of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi counter (where they still remain). And to understand the essence (and in mathematics there is nothing more important than “to understand the essence”) of an arithmetic sequence is not so difficult, having analyzed a few elementary concepts.
Mathematical number sequence
It is customary to call a numerical sequence a series of numbers, each of which has its own number.
and 1 is the first member of the sequence;
and 2 is the second member 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 figures and numbers interests us. We will focus our attention on a numerical sequence in which the value of the n-th member is related to its ordinal number by a dependence that can be clearly formulated mathematically. In other words: the numerical value of the nth number is some function of n.
a - value of a member of the numerical sequence;
n is its serial number;
f(n) is a function where the ordinal in the numeric 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 member of an arithmetic sequence is as follows:
a n - the value of the current member of the arithmetic progression;
a n+1 - the formula of the next number;
d - difference (a 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 numerical sequence 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.
The value of the specified member
Sometimes it is necessary to determine the value of some arbitrary term a n of an arithmetic progression. You can do this by calculating successively the values of all members of the arithmetic progression, from the first to the desired one. However, this way is not always acceptable if, for example, it is necessary to find the value of the five thousandth or eight millionth term. The traditional calculation will take a long time. However, a specific arithmetic progression can be investigated using certain formulas. There is also a formula for the nth term: the value of any member of an arithmetic progression can be determined as the sum of the first member of the progression with the difference of the progression, multiplied by the number of the desired member, minus one.
The formula is universal for increasing and decreasing progression.
An example of calculating the value of a given member
Let's solve the following problem of finding the value of the n-th member of an arithmetic progression.
Condition: there is an arithmetic progression with parameters:
The first member of the sequence is 3;
The difference in the number series is 1.2.
Task: it is necessary to find the value of 214 terms
Solution: to determine the value of a given member, 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 member of the sequence is equal to 258.6.
The advantages of this calculation method are obvious - the entire solution takes no more than 2 lines.
Sum of a given number of members
Very often, in a given arithmetic series, it is required to determine the sum of the values of some of its segments. It also doesn't need to calculate the values of each term and then sum them up. This method is applicable if the number of terms whose sum must be found is small. In other cases, it is more convenient to use the following formula.
The sum of the members of an arithmetic progression from 1 to n is equal to the sum of the first and nth members, multiplied by the member number n and divided by two. If in the formula the value of the n-th member 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.
In the problem, it is required to determine the sum of the terms of the series from 56 to 101.
Solution. Let's use the formula for determining the sum of the progression:
s(n) = (2∙a1 + d∙(n-1))∙n/2
First, we determine the sum of the values of 101 members 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 the 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
So the sum of the arithmetic progression for this example is:
s 101 - s 55 \u003d 2,525 - 742.5 \u003d 1,782.5
Example of practical application of arithmetic progression
At the end of the article, let's return to the example of the arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider such an example.
Getting into a taxi (which includes 3 km) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles / km. Travel distance 30 km. Calculate the cost of the trip.
1. Let's discard the first 3 km, the price of which is included in the landing cost.
30 - 3 = 27 km.
2. Further calculation is nothing more than parsing an arithmetic number series.
The member number is 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 p.
the number of interest to us - the value of the (27 + 1)th member of the arithmetic progression - the meter reading at the end of the 27th kilometer - 27.999 ... = 28 km.
a 28 \u003d 50 + 22 ∙ (28 - 1) \u003d 644
Calculations of calendar data 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 luminary. In addition, various numerical series are successfully used in statistics and other applied branches of mathematics.
Another kind of number sequence is geometric
A geometric progression is characterized by a large, compared with an arithmetic, rate of change. It is no coincidence that in politics, sociology, medicine, often, in order to show the high speed of the spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops exponentially.
The N-th member 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 member is 1, the denominator is 2, respectively, 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 member of the geometric progression;
b n+1 - the formula of the next member of the geometric progression;
q is the denominator of a geometric progression (constant number).
If the graph of an arithmetic progression is a straight line, then the geometric one draws a slightly different picture:
As in the case of arithmetic, a geometric progression has a formula for the value of an arbitrary member. Any n-th 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. Find the 5th term of the progression
b 5 \u003d b 1 ∙ q (5-1) \u003d 3 ∙ 1.5 4 \u003d 15.1875
The sum of a given number of members is also calculated using a special formula. The sum of the first n members of a geometric progression is equal to the difference between the product of the nth member of the progression and its denominator and the first member 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 members of the considered number series will take the form:
Example. The geometric progression starts with the first term equal to 1. The denominator is set equal to 3. Let's find the sum of the first eight terms.
s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280