Mathematics has its own beauty, just like painting and poetry.
Russian scientist, mechanic N.E. Zhukovsky
Very common tasks in entrance examinations in mathematics are problems related to the concept 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 has been 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 .
Roots 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 given equation 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, which when divided by 6 leaves 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 the analysis modern methods solutions typical tasks 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 school curriculum. – M.: Lenand / URSS, 2014. – 216 p.
3. Medynsky M.M. Full course elementary mathematics in problems and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. – 208 p.
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Concept 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– 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 terms differ from each other by 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.
Lesson type: learning new material.
Lesson objectives:
- expanding and deepening students’ understanding of problems solved using arithmetic progression; organization search activity students when deriving the formula for the sum of the first n terms of an arithmetic progression;
- developing the ability to independently acquire new knowledge and use already acquired knowledge to achieve a given task;
- developing the desire and need to generalize the facts obtained, developing independence.
Tasks:
- summarize and systematize existing knowledge on the topic “Arithmetic progression”;
- derive formulas for calculating the sum of the first n terms of an arithmetic progression;
- teach how to apply the obtained formulas when solving various problems;
- draw students' attention to the procedure for finding the value of a numerical expression.
Equipment:
- cards with tasks for working in groups and pairs;
- evaluation paper;
- presentation“Arithmetic progression.”
I. Updating of basic knowledge.
1. Independent work in pairs.
1st option:
Define arithmetic progression. Write it down recurrence formula, which is used to define an arithmetic progression. Please provide 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 the 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 their partner's work by checking them on the board. (Sheets with answers are handed in.)
2. Game moment.
Exercise 1.
Teacher. I thought of some arithmetic progression. Ask me only two questions so that after the answers you can quickly name the 7th term 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 equal to. You can ask questions: what is the 6th term of the progression equal to and what is the 8th term of the progression equal to?
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 board. Students call out the number, and the teacher instantly calls out the number itself. Explain how I can do this?
The teacher remembers the formula for the nth term a n = 3n – 2 and, substituting the specified values n, finds the corresponding values a n.
II. Setting a learning task.
I propose to solve an ancient problem dating back to the 2nd millennium BC, found in Egyptian papyri.
Task:“Let it be said to you: divide 10 measures of barley among 10 people, the difference between each person and his neighbor is 1/8 of the measure.”
- How is this problem related to the topic arithmetic progression? (Each next person receives 1/8 of the measure more, which means the difference is d=1/8, 10 people, which means n=10.)
- What do you think the number 10 measures means? (Sum of all terms of the progression.)
- What else do you need to know to make it easy and simple to divide the barley according to the conditions of the problem? (First term of 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 we deduce the formula, let's look at how the ancient Egyptians solved the problem.
And they solved it as follows:
1) 10 measures: 10 = 1 measure – average share;
2) 1 measure ∙ = 2 measures – doubled average share.
Doubled average share is the sum of the shares of the 5th and 6th person.
3) 2 measures – 1/8 measures = 1 7/8 measures – double the share of the fifth person.
4) 1 7/8: 2 = 5/16 – fraction of a fifth; and so on, you can find the share of each previous and subsequent person.
We get the sequence:
III. Solving the problem.
1. Work in groups
Group I: Find the sum of 20 consecutive natural numbers: S 20 =(20+1)∙10 =210.
In general 
II group: Find the sum of natural numbers from 1 to 100 (The Legend of Little Gauss).
S 100 = (1+100)∙50 = 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 problems considered 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 =a 1 + a 2 + a 3 + a 4 +…+ a n-3 + a n-2 + a n-1 + a n.
Let's find this sum using similar reasoning:
4. Have we solved the problem?(Yes.)
IV. Primary understanding and application of the obtained formulas when solving problems.
1. Checking the solution to an ancient problem using the formula.
2. Application of the formula in solving various problems.
3. Exercises to develop the ability to apply formulas when solving problems.
A) No. 613
Given: ( a n) – arithmetic progression;
(a n): 1, 2, 3, …, 1500
Find: S 1500
Solution: , a 1 = 1, and 1500 = 1500,
B) Given: ( a n) – arithmetic progression;
(a n): 1, 2, 3, …
S n = 210
Find: n
Solution:
V. Independent work with mutual verification.
Denis started working 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 total in a year?
Given: ( a 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.
- Section 4.3 – learn the derivation of the formula.
- №№ 585, 623 .
- Create a problem that can 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...
- Formulas learned...
- I believe 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. M.: “Enlightenment”, 2009.