All formulas are in arithmetic progression. Arithmetic progression. Another type of number sequence is geometric
Mathematics has its own beauty, just like painting and poetry.
Russian scientist, mechanic N.E. Zhukovsky
Problems related to the concept of arithmetic progression are very common problems in entrance examinations in mathematics. To successfully solve such problems, it is necessary to know well the properties of the arithmetic progression and have certain skills in their application.
We first recall the main properties of the 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 arithmetic progression. Moreover, the numbercalled the difference in progression.
For an arithmetic progression, the following formulas are valid
, (1)
where . Formula (1) is called the formula for the general term of an arithmetic progression, and formula (2) is 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 considered progression is called "arithmetic".
The above formulas (1) and (2) are generalized as follows:
(3)
To calculate the amount the first members of the arithmetic progressionusually the formula is applied
(5) where and.
Taking into account the formula (1), then formula (5) implies
If we denote, then
where . Since, then formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).
In particular , formula (5) implies, what
The property of the arithmetic progression, formulated by means of the following theorem, is among the little-known to most students.
Theorem. If, then
Proof. If, then
The theorem is proved.
For example , using the theorem, it can be shown that
Let's move on to considering 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 it be three times more, and when dividing by in the quotient, we get 2 and remainder 8. Determine and.
Solution. The condition of the example implies the system of equations
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.By formula (5), we have
However, using the theorem, one can write
From this and 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 the system of equations
Solving which, we get and.
The natural root of the equation is an .
Example 7. Find if and.
Solution. Since by formula (3) we have that, then the problem statement implies the system of equations
If you substitute the expressioninto the second equation of the system, then we get or.
The roots of the 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 down and.
Hence follows 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.
Hence , here we have a system of linear equations
Hence we get and. Taking into account formula (8), we write.
Example 10. Solve the equation.
Solution. From the given equation it follows that. Suppose that,, and. In this case .
According to formula (1), you can write or.
Since, then equation (13) has a single suitable root.
Example 11. Find the maximum value provided that and.
Solution. Since, the considered arithmetic progression is decreasing. In this regard, the expression takes on the maximum value when it is the number of the minimum positive term of the progression.
We use formula (1) and the fact, as. Then we get that or.
Since, then or ... However, in this inequalitygreatest natural number, therefore .
If the values, and are substituted in the formula (6), then we get.
Answer: .
Example 12. Determine the sum of all two-digit natural numbers that, when divided by 6, give a remainder of 5.
Solution. Let us denote by the set of all two-digit 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 the remainder 5.
It is not difficult to establish, what . Obviously , that the elements of the setform an arithmetic progression, in which and.
To establish the cardinality (number of elements) of a set, we assume that. Since and, then from formula (1) it follows or. Taking into account formula (5), we get.
The above examples of solving problems in no way can claim to be exhaustive. This article is written on the basis of an analysis of modern methods for solving typical problems on a given topic. For a deeper study of methods for solving problems associated with arithmetic progression, it is advisable to refer to the list of recommended literature.
1. Collection of problems in mathematics for applicants to technical colleges / Ed. M.I. Skanavi. - 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. Complete course of elementary mathematics in problems and exercises. Book 2: Number sequences and progressions. - M .: Editus, 2015 .-- 208 p.
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Lesson type: learning new material.
Lesson objectives:
- expansion and deepening of students' ideas about the problems solved using arithmetic progression; organization of students' search activity when deriving a 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 set task;
- the development of the desire and need to generalize the facts obtained, the development of independence.
Tasks:
- to generalize and systematize the existing knowledge on the topic “Arithmetic progression”;
- derive formulas for calculating the sum of the first n terms of an arithmetic progression;
- to teach how to apply the obtained formulas in solving various problems;
- to draw the attention of students to the order of actions when finding the value of a numeric expression.
Equipment:
- cards with assignments for working in groups and pairs;
- evaluation paper;
- presentation"Arithmetic progression".
I. Updating basic knowledge.
1. Independent work in pairs.
1st option:
Give a definition of an arithmetic progression. Write down the recurring formula that defines the arithmetic progression. Hello example of arithmetic progression and indicate its difference.
2nd option:
Write down the formula for the nth term of the 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 prepare answers to the same questions.
Students evaluate the partner's work against the board. (The sheets with the answers are handed over).
2. Game moment.
Exercise 1.
Teacher. I have conceived some arithmetic progression. Just ask me 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 ...)
Student questions.
- What is the sixth term in the progression and what is the difference?
- What is the eighth term in the progression and what is the difference?
If there are no more questions, then the teacher can stimulate them - “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 call the number of the number, and the teacher instantly calls the number itself. Explain how I do it?
The teacher remembers the formula for the nth term a n = 3n - 2 and, substituting the given values of n, finds the corresponding values a n.
II. Statement of the educational problem.
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 between 10 people, the difference between each person and his neighbor is equal to 1/8 of the measure”.
- How is this task related to the topic of arithmetic progression? (Each next one gets 1/8 of a measure more, which means the difference d = 1/8, 10 people, which means 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 the barley according to the condition of the task? (The first term in the progression.)
Lesson objective- obtaining the dependence of the sum of the members of the progression on their number, the first term and the difference, and checking whether the problem was solved correctly in ancient times.
Before drawing the conclusion of the formula, let's see 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 the share is the sum of the shares of the 5th and 6th people.
3) 2 measures - 1/8 measures = 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 to the problem.
1. Working 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 the Little Gauss).
S 100 = (1 + 100) ∙ 50 = 5050
Output:
III group: Find the sum of natural numbers from 1 to 21.
Solution: 1 + 21 = 2 + 20 = 3 + 19 = 4 + 18 ...
Output:
IV group: Find the sum of natural numbers from 1 to 101.
Output:
This method for solving the considered problems is called the “Gauss Method”.
2. Each group presents a 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 us find this sum by reasoning in a similar way:
4. Have we solved the task at hand?(Yes.)
IV. Primary comprehension and application of the obtained formulas in solving problems.
1. Checking the solution to an old problem using the formula.
2. Application of the formula in solving various problems.
3. Exercises to form the ability to apply the formula when solving problems.
A) No. 613
Given: ( a n) - arithmetic progression;
(a n): 1, 2, 3, ..., 1500
Find: S 1500
Solution: , a 1 = 1, a 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 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: ( a n) - arithmetic progression;
a 1 = 200, d = 30, n = 12
Find: S 12
Solution:
Answer: Denis received 4380 rubles in a year.
Vi. Homework briefing.
- p. 4.3 - learn the derivation of the formula.
- №№ 585, 623 .
- Create 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. Evaluation sheet
2. Continue sentences
- Today in the lesson 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. Textbook for educational institutions. Ed. G.V. Dorofeeva. M .: "Education", 2009.
Yes, yes: the arithmetic progression is not a toy for you :) Well, friends, if you are reading this text, then the internal cap-evidence tells me that you do not yet know what an arithmetic progression is, but you really (no, like this: SOOOOO!) Want to know. Therefore, I will not torment you with long introductions and get straight to the point.
Let's start with a couple of examples. Consider several sets of numbers:
- 1; 2; 3; 4; ...
- 15; 20; 25; 30; ...
- $ \ sqrt (2); \ 2 \ sqrt (2); \ 3 \ sqrt (2); ... $
What do all these sets have in common? At first glance, nothing. But actually there is something. Namely: each next element differs from the previous one by the same number.
Judge for yourself. The first set is simply consecutive numbers, each next one more than the previous one. In the second case, the difference between the adjacent numbers is already equal to five, but this difference is still constant. In the third case, roots in general. However, $ 2 \ sqrt (2) = \ sqrt (2) + \ sqrt (2) $, and $ 3 \ sqrt (2) = 2 \ sqrt (2) + \ sqrt (2) $, i.e. and in this case, each next element simply increases by $ \ sqrt (2) $ (and don't be afraid that this number is irrational).
So: all such sequences are called arithmetic progressions. Let's give a strict definition:
Definition. A sequence of numbers in which each next differs from the previous by exactly the same amount is called an arithmetic progression. The very amount by which the numbers differ is called the difference of the progression and is most often denoted by the letter $ d $.
Designation: $ \ left (((a) _ (n)) \ right) $ - the progression itself, $ d $ - its difference.
And just a couple of important remarks. First, only orderly sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. You cannot rearrange or swap numbers.
Secondly, the sequence itself can be either finite or infinite. For example, the set (1; 2; 3) is obviously a finite arithmetic progression. But if you write something in the spirit (1; 2; 3; 4; ...) - this is already an endless progression. The ellipsis after the four, as it were, hints that there are still quite a few numbers going on. Infinitely many, for example. :)
I would also like to note that progressions are increasing and decreasing. We have already seen the increasing ones - the same set (1; 2; 3; 4; ...). And here are examples of decreasing progressions:
- 49; 41; 33; 25; 17; ...
- 17,5; 12; 6,5; 1; −4,5; −10; ...
- $ \ sqrt (5); \ \ sqrt (5) -1; \ \ sqrt (5) -2; \ \ sqrt (5) -3; ... $
Okay, okay: this last example may seem overly complicated. But the rest, I think you understand. Therefore, we will introduce new definitions:
Definition. An arithmetic progression is called:
- increasing if each next element is larger than the previous one;
- decreasing if, on the contrary, each subsequent element is less than the previous one.
In addition, there are so-called "stationary" sequences - they consist of the same repeating number. For example, (3; 3; 3; ...).
There remains only one question: how to distinguish an increasing progression from a decreasing one? Fortunately, it all depends on the sign of the number $ d $, i.e. difference progression:
- If $ d \ gt 0 $, then the progression is increasing;
- If $ d \ lt 0 $, then the progression is obviously decreasing;
- Finally, there is the case $ d = 0 $ - in this case the whole progression is reduced to a stationary sequence of identical numbers: (1; 1; 1; 1; ...), etc.
Let's try to calculate the difference $ d $ for the three decreasing progressions given above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract the number on the left from the number on the right. It will look like this:
- 41−49=−8;
- 12−17,5=−5,5;
- $ \ sqrt (5) -1- \ sqrt (5) = - 1 $.
As you can see, in all three cases the difference really turned out to be negative. And now that we have more or less figured out the definitions, it's time to figure out how progressions are described and what their properties are.
Progression members and recurrent formula
Since the elements of our sequences cannot be swapped, they can be numbered:
\ [\ left (((a) _ (n)) \ right) = \ left \ (((a) _ (1)), \ ((a) _ (2)), ((a) _ (3 )), ... \ right \) \]
The individual elements of this set are called members of the progression. They are indicated by a number: the first term, the second term, etc.
In addition, as we already know, the adjacent members of the progression are related by the formula:
\ [((a) _ (n)) - ((a) _ (n-1)) = d \ Rightarrow ((a) _ (n)) = ((a) _ (n-1)) + d \]
In short, to find the $ n $ th term in the progression, you need to know the $ n-1 $ th term and the $ d $ difference. Such a formula is called recurrent, since with its help you can find any number, only knowing the previous one (and in fact - all the previous ones). This is very inconvenient, so there is a more tricky formula that reduces any calculations to the first term and the difference:
\ [((a) _ (n)) = ((a) _ (1)) + \ left (n-1 \ right) d \]
Surely you have already met this formula. They love to give it in all sorts of reference books and reshebniks. And in any sensible textbook on mathematics, she goes one of the first.
Nevertheless, I suggest we practice a little.
Problem number 1. Write out the first three terms of the arithmetic progression $ \ left (((a) _ (n)) \ right) $, if $ ((a) _ (1)) = 8, d = -5 $.
Solution. So, we know the first term $ ((a) _ (1)) = 8 $ and the difference of the progression $ d = -5 $. Let's use the formula just given and substitute $ n = 1 $, $ n = 2 $ and $ n = 3 $:
\ [\ begin (align) & ((a) _ (n)) = ((a) _ (1)) + \ left (n-1 \ right) d; \\ & ((a) _ (1)) = ((a) _ (1)) + \ left (1-1 \ right) d = ((a) _ (1)) = 8; \\ & ((a) _ (2)) = ((a) _ (1)) + \ left (2-1 \ right) d = ((a) _ (1)) + d = 8-5 = 3; \\ & ((a) _ (3)) = ((a) _ (1)) + \ left (3-1 \ right) d = ((a) _ (1)) + 2d = 8-10 = -2. \\ \ end (align) \]
Answer: (8; 3; −2)
That's all! Please note: our progression is decreasing.
Of course, $ n = 1 $ could not have been substituted - the first term is already known to us. However, substituting one, we made sure that our formula works even for the first term. In other cases, it all boiled down to trivial arithmetic.
Problem number 2. Write out the first three terms of the arithmetic progression if its seventh term is −40 and the seventeenth term is −50.
Solution. Let's write down the condition of the problem in the usual terms:
\ [((a) _ (7)) = - 40; \ quad ((a) _ (17)) = - 50. \]
\ [\ left \ (\ begin (align) & ((a) _ (7)) = ((a) _ (1)) + 6d \\ & ((a) _ (17)) = ((a) _ (1)) + 16d \\ \ end (align) \ right. \]
\ [\ left \ (\ begin (align) & ((a) _ (1)) + 6d = -40 \\ & ((a) _ (1)) + 16d = -50 \\ \ end (align) \ right. \]
I put the sign of the system because these requirements must be fulfilled simultaneously. And now note that if we subtract the first from the second equation (we have the right to do this, since we have a system), we get this:
\ [\ begin (align) & ((a) _ (1)) + 16d- \ left (((a) _ (1)) + 6d \ right) = - 50- \ left (-40 \ right); \\ & ((a) _ (1)) + 16d - ((a) _ (1)) - 6d = -50 + 40; \\ & 10d = -10; \\ & d = -1. \\ \ end (align) \]
That's how easy we found the difference in the progression! It remains to substitute the found number into any of the equations of the system. For example, in the first:
\ [\ begin (matrix) ((a) _ (1)) + 6d = -40; \ quad d = -1 \\ \ Downarrow \\ ((a) _ (1)) - 6 = -40; \\ ((a) _ (1)) = - 40 + 6 = -34. \\ \ end (matrix) \]
Now, knowing the first term and the difference, it remains to find the second and third terms:
\ [\ begin (align) & ((a) _ (2)) = ((a) _ (1)) + d = -34-1 = -35; \\ & ((a) _ (3)) = ((a) _ (1)) + 2d = -34-2 = -36. \\ \ end (align) \]
Ready! The problem has been solved.
Answer: (-34; -35; -36)
Pay attention to an interesting property of the progression that we discovered: if we take the $ n $ th and $ m $ th terms and subtract them from each other, we get the difference of the progression multiplied by the number $ n-m $:
\ [((a) _ (n)) - ((a) _ (m)) = d \ cdot \ left (n-m \ right) \]
A simple but very useful property that you should definitely know - with its help, you can significantly speed up the solution of many problems in progressions. Here's a prime example:
Problem number 3. The fifth term of the arithmetic progression is 8.4, and its tenth term is 14.4. Find the fifteenth term of this progression.
Solution. Since $ ((a) _ (5)) = 8.4 $, $ ((a) _ (10)) = 14.4 $, and you need to find $ ((a) _ (15)) $, then we note following:
\ [\ begin (align) & ((a) _ (15)) - ((a) _ (10)) = 5d; \\ & ((a) _ (10)) - ((a) _ (5)) = 5d. \\ \ end (align) \]
But by condition $ ((a) _ (10)) - ((a) _ (5)) = 14.4-8.4 = $ 6, therefore $ 5d = $ 6, whence we have:
\ [\ begin (align) & ((a) _ (15)) - 14.4 = 6; \\ & ((a) _ (15)) = 6 + 14.4 = 20.4. \\ \ end (align) \]
Answer: 20.4
That's all! We did not need to compose some systems of equations and calculate the first term and the difference - everything was solved in just a couple of lines.
Now let's consider another type of tasks - to find negative and positive members of the progression. It is no secret that if the progression increases, while the first term is negative, then sooner or later positive terms will appear in it. And on the contrary: the members of the decreasing progression will sooner or later become negative.
At the same time, it is far from always possible to grope this moment "head-on", sequentially going through the elements. Often, problems are designed in such a way that without knowing the formulas, the calculations would take several sheets - we would just fall asleep while we found the answer. Therefore, we will try to solve these problems in a faster way.
Problem number 4. How many negative terms are in the arithmetic progression -38.5; −35.8; ...?
Solution. So, $ ((a) _ (1)) = - 38.5 $, $ ((a) _ (2)) = - 35.8 $, from where we immediately find the difference:
Note that the difference is positive, so the progression increases. The first term is negative, so at some point we really will stumble upon positive numbers. The only question is when it will happen.
Let's try to find out: how long (i.e. up to what natural number $ n $) the negativity of the terms is preserved:
\ [\ begin (align) & ((a) _ (n)) \ lt 0 \ Rightarrow ((a) _ (1)) + \ left (n-1 \ right) d \ lt 0; \\ & -38.5+ \ left (n-1 \ right) \ cdot 2.7 \ lt 0; \ quad \ left | \ cdot 10 \ right. \\ & -385 + 27 \ cdot \ left (n-1 \ right) \ lt 0; \\ & -385 + 27n-27 \ lt 0; \\ & 27n \ lt 412; \\ & n \ lt 15 \ frac (7) (27) \ Rightarrow ((n) _ (\ max)) = 15. \\ \ end (align) \]
The last line requires clarification. So, we know that $ n \ lt 15 \ frac (7) (27) $. On the other hand, we will only be satisfied with integer values of the number (moreover: $ n \ in \ mathbb (N) $), so the largest allowed number is exactly $ n = 15 $, and in no case is 16.
Problem number 5. In arithmetic progression $ (() _ (5)) = - 150, (() _ (6)) = - 147 $. Find the number of the first positive term of this progression.
It would be exactly the same problem as the previous one, but we don't know $ ((a) _ (1)) $. But the neighboring terms are known: $ ((a) _ (5)) $ and $ ((a) _ (6)) $, so we can easily find the difference of the progression:
In addition, we will try to express the fifth term in terms of the first and the difference according to the standard formula:
\ [\ begin (align) & ((a) _ (n)) = ((a) _ (1)) + \ left (n-1 \ right) \ cdot d; \\ & ((a) _ (5)) = ((a) _ (1)) + 4d; \\ & -150 = ((a) _ (1)) + 4 \ cdot 3; \\ & ((a) _ (1)) = - 150-12 = -162. \\ \ end (align) \]
Now we proceed by analogy with the previous task. We find out at what point in our sequence there will be positive numbers:
\ [\ begin (align) & ((a) _ (n)) = - 162+ \ left (n-1 \ right) \ cdot 3 \ gt 0; \\ & -162 + 3n-3 \ gt 0; \\ & 3n \ gt 165; \\ & n \ gt 55 \ Rightarrow ((n) _ (\ min)) = 56. \\ \ end (align) \]
The smallest integer solution to this inequality is 56.
Please note: in the last task, everything was reduced to a strict inequality, so the $ n = 55 $ option will not suit us.
Now that we have learned how to solve simple problems, let's move on to more complex ones. But first, let's study another very useful property of arithmetic progressions, which in the future will save us a lot of time and unequal cells. :)
Arithmetic mean and equal indents
Consider several consecutive members of the increasing arithmetic progression $ \ left (((a) _ (n)) \ right) $. Let's try to mark them on the number line:
Members of an arithmetic progression on a number lineI specifically noted arbitrary terms $ ((a) _ (n-3)), ..., ((a) _ (n + 3)) $, not any $ ((a) _ (1)) , \ ((a) _ (2)), \ ((a) _ (3)) $, etc. Because the rule, which I will now talk about, works the same for any "segments".
And the rule is very simple. Let's remember the recurrence formula and write it down for all marked members:
\ [\ begin (align) & ((a) _ (n-2)) = ((a) _ (n-3)) + d; \\ & ((a) _ (n-1)) = ((a) _ (n-2)) + d; \\ & ((a) _ (n)) = ((a) _ (n-1)) + d; \\ & ((a) _ (n + 1)) = ((a) _ (n)) + d; \\ & ((a) _ (n + 2)) = ((a) _ (n + 1)) + d; \\ \ end (align) \]
However, these equalities can be rewritten differently:
\ [\ begin (align) & ((a) _ (n-1)) = ((a) _ (n)) - d; \\ & ((a) _ (n-2)) = ((a) _ (n)) - 2d; \\ & ((a) _ (n-3)) = ((a) _ (n)) - 3d; \\ & ((a) _ (n + 1)) = ((a) _ (n)) + d; \\ & ((a) _ (n + 2)) = ((a) _ (n)) + 2d; \\ & ((a) _ (n + 3)) = ((a) _ (n)) + 3d; \\ \ end (align) \]
Well, so what? And the fact that the terms $ ((a) _ (n-1)) $ and $ ((a) _ (n + 1)) $ lie at the same distance from $ ((a) _ (n)) $. And this distance is equal to $ d $. The same can be said about the members $ ((a) _ (n-2)) $ and $ ((a) _ (n + 2)) $ - they are also removed from $ ((a) _ (n)) $ the same distance equal to $ 2d $. You can continue indefinitely, but the meaning is well illustrated by the picture.
The members of the progression lie at the same distance from the center What does this mean for us? This means that you can find $ ((a) _ (n)) $ if the neighboring numbers are known:
\ [((a) _ (n)) = \ frac (((a) _ (n-1)) + ((a) _ (n + 1))) (2) \]
We came up with an excellent statement: every member of the arithmetic progression is equal to the arithmetic mean of the neighboring terms! Moreover: we can deviate from our $ ((a) _ (n)) $ left and right not one step, but $ k $ steps - and still the formula will be correct:
\ [((a) _ (n)) = \ frac (((a) _ (n-k)) + ((a) _ (n + k))) (2) \]
Those. we can easily find some $ ((a) _ (150)) $ if we know $ ((a) _ (100)) $ and $ ((a) _ (200)) $, because $ (( a) _ (150)) = \ frac (((a) _ (100)) + ((a) _ (200))) (2) $. At first glance, it may seem that this fact does not give us anything useful. However, in practice, many problems are specially "sharpened" for the use of the arithmetic mean. Take a look:
Problem number 6. Find all values of $ x $ for which the numbers $ -6 ((x) ^ (2)) $, $ x + 1 $ and $ 14 + 4 ((x) ^ (2)) $ are consecutive members of the arithmetic progression (in order).
Solution. Since the indicated numbers are members of the progression, the condition of the arithmetic mean is satisfied for them: the central element $ x + 1 $ can be expressed in terms of adjacent elements:
\ [\ begin (align) & x + 1 = \ frac (-6 ((x) ^ (2)) + 14 + 4 ((x) ^ (2))) (2); \\ & x + 1 = \ frac (14-2 ((x) ^ (2))) (2); \\ & x + 1 = 7 - ((x) ^ (2)); \\ & ((x) ^ (2)) + x-6 = 0. \\ \ end (align) \]
The result is a classic quadratic equation. Its roots: $ x = 2 $ and $ x = -3 $ - these are the answers.
Answer: −3; 2.
Problem number 7. Find the $$ values for which the numbers $ -1; 4-3; (() ^ (2)) + 1 $ make an arithmetic progression (in that order).
Solution. Again, we express the middle term in terms of the arithmetic mean of the neighboring terms:
\ [\ begin (align) & 4x-3 = \ frac (x-1 + ((x) ^ (2)) + 1) (2); \\ & 4x-3 = \ frac (((x) ^ (2)) + x) (2); \ quad \ left | \ cdot 2 \ right .; \\ & 8x-6 = ((x) ^ (2)) + x; \\ & ((x) ^ (2)) - 7x + 6 = 0. \\ \ end (align) \]
Again the quadratic equation. And again there are two roots: $ x = 6 $ and $ x = 1 $.
Answer: 1; 6.
If in the process of solving a problem you get out some brutal numbers, or you are not completely sure of the correctness of the answers found, then there is a wonderful technique that allows you to check: did we solve the problem correctly?
For example, in problem no. 6 we received answers -3 and 2. How to check that these answers are correct? Let's just plug them in and see what happens. Let me remind you that we have three numbers ($ -6 (() ^ (2)) $, $ + 1 $ and $ 14 + 4 (() ^ (2)) $), which must form an arithmetic progression. Substitute $ x = -3 $:
\ [\ begin (align) & x = -3 \ Rightarrow \\ & -6 ((x) ^ (2)) = - 54; \\ & x + 1 = -2; \\ & 14 + 4 ((x) ^ (2)) = 50. \ end (align) \]
Received numbers -54; −2; 50, which differ by 52, is undoubtedly an arithmetic progression. The same thing happens for $ x = 2 $:
\ [\ begin (align) & x = 2 \ Rightarrow \\ & -6 ((x) ^ (2)) = - 24; \\ & x + 1 = 3; \\ & 14 + 4 ((x) ^ (2)) = 30. \ end (align) \]
Again a progression, but with a difference of 27. Thus, the problem is solved correctly. Those interested can check the second problem on their own, but I'll say right away: everything is correct there too.
In general, while solving the last problems, we came across another interesting fact, which also needs to be remembered:
If three numbers are such that the second is the arithmetic mean of the first and the last, then these numbers form an arithmetic progression.
In the future, understanding this statement will allow us to literally "construct" the necessary progressions, based on the condition of the problem. But before we get down to such "construction", we should pay attention to one more fact, which directly follows from what has already been considered.
Grouping and sum of elements
Let's go back to the number axis again. Let us note there several members of the progression, between which, perhaps. there are a lot of other members:
The number line has 6 elements markedLet's try to express "left tail" in terms of $ ((a) _ (n)) $ and $ d $, and "right tail" in terms of $ ((a) _ (k)) $ and $ d $. It's very simple:
\ [\ begin (align) & ((a) _ (n + 1)) = ((a) _ (n)) + d; \\ & ((a) _ (n + 2)) = ((a) _ (n)) + 2d; \\ & ((a) _ (k-1)) = ((a) _ (k)) - d; \\ & ((a) _ (k-2)) = ((a) _ (k)) - 2d. \\ \ end (align) \]
Now, note that the following sums are equal:
\ [\ begin (align) & ((a) _ (n)) + ((a) _ (k)) = S; \\ & ((a) _ (n + 1)) + ((a) _ (k-1)) = ((a) _ (n)) + d + ((a) _ (k)) - d = S; \\ & ((a) _ (n + 2)) + ((a) _ (k-2)) = ((a) _ (n)) + 2d + ((a) _ (k)) - 2d = S. \ end (align) \]
Simply put, if we consider as a start two elements of the progression, which in total are equal to some number $ S $, and then we begin to walk from these elements in opposite directions (towards each other or vice versa to move away), then the sums of the elements that we will stumble upon will also be equal$ S $. This can be most clearly represented graphically:
Equal indentation gives equal amounts Understanding this fact will allow us to solve problems of a fundamentally higher level of complexity than those that we considered above. For example, such:
Problem number 8. Determine the difference of the arithmetic progression in which the first term is 66, and the product of the second and twelfth terms is the smallest possible.
Solution. Let's write down everything we know:
\ [\ begin (align) & ((a) _ (1)) = 66; \\ & d =? \\ & ((a) _ (2)) \ cdot ((a) _ (12)) = \ min. \ end (align) \]
So, we do not know the difference of the progression $ d $. Actually, the whole solution will be built around the difference, since the product $ ((a) _ (2)) \ cdot ((a) _ (12)) $ can be rewritten as follows:
\ [\ begin (align) & ((a) _ (2)) = ((a) _ (1)) + d = 66 + d; \\ & ((a) _ (12)) = ((a) _ (1)) + 11d = 66 + 11d; \\ & ((a) _ (2)) \ cdot ((a) _ (12)) = \ left (66 + d \ right) \ cdot \ left (66 + 11d \ right) = \\ & = 11 \ cdot \ left (d + 66 \ right) \ cdot \ left (d + 6 \ right). \ end (align) \]
For those in the tank: I took out the common factor of 11 from the second parenthesis. Thus, the sought product is a quadratic function with respect to the variable $ d $. Therefore, consider the function $ f \ left (d \ right) = 11 \ left (d + 66 \ right) \ left (d + 6 \ right) $ - its graph will be a parabola with branches up, since if we expand the brackets, then we get:
\ [\ begin (align) & f \ left (d \ right) = 11 \ left (((d) ^ (2)) + 66d + 6d + 66 \ cdot 6 \ right) = \\ & = 11 (( d) ^ (2)) + 11 \ cdot 72d + 11 \ cdot 66 \ cdot 6 \ end (align) \]
As you can see, the coefficient at the leading term is 11 - this is a positive number, so we really are dealing with a parabola with branches up:
quadratic function graph - parabola
Please note: this parabola takes its minimum value at its vertex with the abscissa $ ((d) _ (0)) $. Of course, we can calculate this abscissa according to the standard scheme (there is also the formula $ ((d) _ (0)) = (- b) / (2a) \; $), but it would be much more reasonable to notice that the desired vertex lies on the axis symmetry of the parabola, so the point $ ((d) _ (0)) $ is equidistant from the roots of the equation $ f \ left (d \ right) = 0 $:
\ [\ begin (align) & f \ left (d \ right) = 0; \\ & 11 \ cdot \ left (d + 66 \ right) \ cdot \ left (d + 6 \ right) = 0; \\ & ((d) _ (1)) = - 66; \ quad ((d) _ (2)) = - 6. \\ \ end (align) \]
That is why I was in no hurry to open the brackets: in the original form, the roots were very, very easy to find. Therefore, the abscissa is equal to the arithmetic mean of the numbers −66 and −6:
\ [((d) _ (0)) = \ frac (-66-6) (2) = - 36 \]
What does the discovered number give us? With it, the required product takes on the smallest value (by the way, we haven't counted $ ((y) _ (\ min)) $ - we don't need this). At the same time, this number is the difference between the initial progression, i.e. we found the answer. :)
Answer: −36
Problem number 9. Insert three numbers between the numbers $ - \ frac (1) (2) $ and $ - \ frac (1) (6) $ so that they together with the given numbers form an arithmetic progression.
Solution. Basically, we need to make a sequence of five numbers, with the first and last numbers already known. Let's denote the missing numbers by the variables $ x $, $ y $ and $ z $:
\ [\ left (((a) _ (n)) \ right) = \ left \ (- \ frac (1) (2); x; y; z; - \ frac (1) (6) \ right \ ) \]
Note that the number $ y $ is the "middle" of our sequence - it is equidistant from both the numbers $ x $ and $ z $, and from the numbers $ - \ frac (1) (2) $ and $ - \ frac (1) ( 6) $. And if at the moment we cannot get $ y $ from the numbers $ x $ and $ z $, then the situation is different with the ends of the progression. Remembering the arithmetic mean:
Now, knowing $ y $, we will find the remaining numbers. Note that $ x $ lies between the numbers $ - \ frac (1) (2) $ and the $ y = - \ frac (1) (3) $ just found. That's why
Reasoning similarly, we find the remaining number:
Ready! We found all three numbers. Let's write them down in the answer in the order in which they should be inserted between the original numbers.
Answer: $ - \ frac (5) (12); \ - \ frac (1) (3); \ - \ frac (1) (4) $
Problem number 10. Insert several numbers between the numbers 2 and 42, which together with these numbers form an arithmetic progression, if you know that the sum of the first, second and last of the inserted numbers is 56.
Solution. An even more difficult task, which, however, is solved according to the same scheme as the previous ones - through the arithmetic mean. The problem is that we don't know exactly how many numbers to insert. Therefore, for definiteness, let us assume that after inserting everything there will be exactly $ n $ numbers, and the first of them is 2, and the last is 42. In this case, the sought arithmetic progression can be represented as:
\ [\ left (((a) _ (n)) \ right) = \ left \ (2; ((a) _ (2)); ((a) _ (3)); ...; (( a) _ (n-1)); 42 \ right \) \]
\ [((a) _ (2)) + ((a) _ (3)) + ((a) _ (n-1)) = 56 \]
Note, however, that the numbers $ ((a) _ (2)) $ and $ ((a) _ (n-1)) $ are obtained from the numbers 2 and 42 at the edges by one step towards each other, i.e. ... to the center of the sequence. This means that
\ [((a) _ (2)) + ((a) _ (n-1)) = 2 + 42 = 44 \]
But then the expression written above can be rewritten as follows:
\ [\ begin (align) & ((a) _ (2)) + ((a) _ (3)) + ((a) _ (n-1)) = 56; \\ & \ left (((a) _ (2)) + ((a) _ (n-1)) \ right) + ((a) _ (3)) = 56; \\ & 44 + ((a) _ (3)) = 56; \\ & ((a) _ (3)) = 56-44 = 12. \\ \ end (align) \]
Knowing $ ((a) _ (3)) $ and $ ((a) _ (1)) $, we can easily find the difference of the progression:
\ [\ begin (align) & ((a) _ (3)) - ((a) _ (1)) = 12 - 2 = 10; \\ & ((a) _ (3)) - ((a) _ (1)) = \ left (3-1 \ right) \ cdot d = 2d; \\ & 2d = 10 \ Rightarrow d = 5. \\ \ end (align) \]
It remains only to find the rest of the members:
\ [\ begin (align) & ((a) _ (1)) = 2; \\ & ((a) _ (2)) = 2 + 5 = 7; \\ & ((a) _ (3)) = 12; \\ & ((a) _ (4)) = 2 + 3 \ cdot 5 = 17; \\ & ((a) _ (5)) = 2 + 4 \ cdot 5 = 22; \\ & ((a) _ (6)) = 2 + 5 \ cdot 5 = 27; \\ & ((a) _ (7)) = 2 + 6 \ cdot 5 = 32; \\ & ((a) _ (8)) = 2 + 7 \ cdot 5 = 37; \\ & ((a) _ (9)) = 2 + 8 \ cdot 5 = 42; \\ \ end (align) \]
Thus, already at the 9th step we will come to the left end of the sequence - the number 42. In total, it was necessary to insert only 7 numbers: 7; 12; 17; 22; 27; 32; 37.
Answer: 7; 12; 17; 22; 27; 32; 37
Word problems with progressions
In conclusion, I would like to consider a couple of relatively simple tasks. Well, how simple: for most students who study mathematics at school and have not read what is written above, these tasks may seem like a tin. Nevertheless, it is precisely such problems that come across in the OGE and USE in mathematics, so I recommend that you familiarize yourself with them.
Problem number 11. The brigade produced 62 parts in January, and in each next month it produced 14 more parts than in the previous one. How many parts did the team make in November?
Solution. Obviously, the number of parts, scheduled by month, will represent an increasing arithmetic progression. Moreover:
\ [\ begin (align) & ((a) _ (1)) = 62; \ quad d = 14; \\ & ((a) _ (n)) = 62+ \ left (n-1 \ right) \ cdot 14. \\ \ end (align) \]
November is the 11th month of the year, so we need to find $ ((a) _ (11)) $:
\ [((a) _ (11)) = 62 + 10 \ cdot 14 = 202 \]
Consequently, 202 parts will be manufactured in November.
Problem number 12. The bookbinding workshop bound 216 books in January, and each next month it bound 4 more books than the previous one. How many books did the workshop bind in December?
Solution. All the same:
$ \ begin (align) & ((a) _ (1)) = 216; \ quad d = 4; \\ & ((a) _ (n)) = 216+ \ left (n-1 \ right) \ cdot 4. \\ \ end (align) $
December is the last, 12th month of the year, so we are looking for $ ((a) _ (12)) $:
\ [((a) _ (12)) = 216 + 11 \ cdot 4 = 260 \]
This is the answer - 260 books will be bound in December.
Well, if you have read this far, I hasten to congratulate you: you have successfully completed the "Young Fighter Course" in arithmetic progressions. You can safely proceed to the next lesson, where we will study the formula for the sum of a progression, as well as important and very useful consequences from it.
For example, the sequence \ (2 \); \(5\); \(eight\); \(eleven\); \ (14 \) ... is an arithmetic progression, because each next element differs from the previous one by three (can be obtained from the previous one by adding a triplet):
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 negative. For example, in arithmetic progression \ (16 \); \(ten\); \(4\); \ (- 2 \); \ (- 8 \) ... the difference of the progression \ (d \) is equal to minus six.
And in this case, each next element will be less than the previous one. These progressions are called decreasing.
Arithmetic progression notation
Progression is indicated by a small Latin letter.
The numbers forming the progression call it members of(or elements).
They are denoted by the same letter as the arithmetic progression, but with a numerical index equal to the number of the element in order.
For example, the 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 \) \)
Problem solving for arithmetic progression
In principle, the above information is already enough to solve almost any problem for an arithmetic progression (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 the arithmetic progression are given: \ (62; 49; 36 ... \) Find the value of the first negative term of this progression ..
Solution:
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We are given the first elements of the sequence and we know that it is an arithmetic progression. That is, each element differs from the neighboring one by the same number. Find out which one, subtracting the previous one from the next element: \ (d = 49-62 = -13 \). |
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Now we can restore our progression to the (first negative) element we need. |
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Ready. You can write an answer. |
Answer: \(-3\)
Example (OGE).
Several consecutive elements of the arithmetic progression are given: \ (… 5; x; 10; 12,5 ... \) Find the value of the element indicated by the letter \ (x \).
Solution:
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To find \ (x \), we need to know how much the next element differs from the previous one, in other words, the difference of the progression. Let's find it from two known neighboring elements: \ (d = 12.5-10 = 2.5 \). |
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And now we find the desired one without any problems: \ (x = 5 + 2.5 = 7.5 \). |
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Ready. You can write an answer. |
Answer: \(7,5\).
Example (OGE).
The arithmetic progression is specified 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:
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We need to find the sum of the first six terms of the progression. But we do not know their meanings, we are only given the first element. Therefore, first we calculate the values in turn, using the given to us: \ (n = 1 \); \ (a_ (1 + 1) = a_1 + 5 = -11 + 5 = -6 \) |
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\ (S_6 = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = \) |
The amount you are looking for has been found. |
Answer: \ (S_6 = 9 \).
Example (OGE).
In arithmetic progression \ (a_ (12) = 23 \); \ (a_ (16) = 51 \). Find the difference between this progression.
Solution:
Answer: \ (d = 7 \).
Important Arithmetic Progression Formulas
As you can see, many arithmetic progression problems can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each next 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) \). What is it, we \ (385 \) times add four? Or imagine that in the penultimate example, you need to find the sum of the first seventy-three elements. You will be tortured to count ...
Therefore, in such cases, they do not solve "head-on", but use special formulas derived for the arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum \ (n \) of the first terms.
Formula \ (n \) - th member: \ (a_n = a_1 + (n-1) d \), where \ (a_1 \) is the first term of the progression;
\ (n \) - number of the element being searched for;
\ (a_n \) is a member of the progression with the number \ (n \).
This formula allows us to quickly find at least the three hundredth, even 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 \).
The 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 \) members of this progression.
Solution:
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\ (S_ (25) = \) \ (\ frac (a_1 + a_ (25)) (2) \) \ (\ cdot 25 \) |
To calculate the sum of the first twenty-five elements, we need to know the value of the first and twenty-fifth terms. |
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\ (n = 1; \) \ (a_1 = 3.4 1-0.6 = 2.8 \) |
Now we find the twenty-fifth term, substituting twenty-five instead of \ (n \). |
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\ (n = 25; \) \ (a_ (25) = 3.4 25-0.6 = 84.4 \) |
Well, now we can calculate the required amount without any problems. |
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\ (S_ (25) = \) \ (\ frac (a_1 + a_ (25)) (2) \) \ (\ cdot 25 = \) |
The answer is ready. |
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:
The 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 \ (n \) of the first elements;
\ (a_1 \) - the first summed term;
\ (d \) - progression difference;
\ (n \) - the number of items in the sum.
Example.
Find the sum of the first \ (33 \) - ex members of the arithmetic progression: \ (17 \); \ (15.5 \); \(fourteen\)…
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. We conclude the topic by considering problems in which you need not only to apply formulas, but also to 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:
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\ (S_n = \) \ (\ frac (2a_1 + (n-1) d) (2) \) \ (\ cdot n \) |
The task is very similar to the previous one. We start to solve also: first we find \ (d \). |
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\ (d = a_2-a_1 = -19 - (- 19.3) = 0.3 \) |
Now I would substitute \ (d \) in the formula for the sum ... and here a small nuance emerges - we do not know \ (n \). In other words, we do not know how many terms will need to be added. How to find out? Let's think. We'll stop adding elements when we get to the first positive element. That is, you need to find out the number of this element. How? Let's write down the formula for calculating any element of the arithmetic progression: \ (a_n = a_1 + (n-1) d \) for our case. |
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\ (a_n = a_1 + (n-1) d \) |
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\ (a_n = -19.3 + (n-1) 0.3 \) |
We need \ (a_n \) to be greater than zero. Let's find out at what \ (n \) this will happen. |
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\ (- 19.3+ (n-1) 0.3> 0 \) |
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\ ((n-1) 0.3> 19.3 \) \ (|: 0.3 \) |
We divide both sides of the inequality by \ (0,3 \). |
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\ (n-1> \) \ (\ frac (19,3) (0,3) \) |
Move minus one, remembering to change signs |
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\ (n> \) \ (\ frac (19,3) (0,3) \) \ (+ 1 \) |
We calculate ... |
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\ (n> 65,333 ... \) |
... and it turns out that the first positive element will have the number \ (66 \). Accordingly, the last negative has \ (n = 65 \). Let's check it out just in case. |
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\ (n = 65; \) \ (a_ (65) = - 19.3+ (65-1) 0.3 = -0.1 \) |
Thus, we need to add the first \ (65 \) elements. |
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\ (S_ (65) = \) \ (\ frac (2 \ cdot (-19.3) + (65-1) 0.3) (2) \)\ (\ cdot 65 \) |
The answer is ready. |
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 \ (26 \) th to \ (42 \) element inclusive.
Solution:
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\ (a_1 = -33; \) \ (a_ (n + 1) = a_n + 4 \) |
In this problem, you also need to find the sum of the elements, but starting not from the first, but from \ (26 \) - th. For such a case, we have no formula. How to decide? |
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For our progression \ (a_1 = -33 \), and the difference \ (d = 4 \) (after all, it is the four that we add to the previous element to find the next one). Knowing this, we find the sum of the first \ (42 \) - yh elements. |
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\ (S_ (42) = \) \ (\ frac (2 \ cdot (-33) + (42-1) 4) (2) \)\ (\ cdot 42 = \) |
Now the sum of the first \ (25 \) - ty elements. |
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\ (S_ (25) = \) \ (\ frac (2 \ cdot (-33) + (25-1) 4) (2) \)\ (\ cdot 25 = \) |
Finally, we calculate the answer. |
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\ (S = S_ (42) -S_ (25) = 2058-375 = 1683 \) |
Answer: \ (S = 1683 \).
For the 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.
Someone is wary of the word "progression", 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 remain). And to understand the essence (and in mathematics there is nothing more important than "understanding the essence") of the arithmetic sequence is not so difficult, having analyzed several elementary concepts.
Mathematical number sequence
It is customary to name a series of numbers by a numerical sequence, each of which has its own number.
a 1 - 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 n-th member of the sequence;
However, we are not interested in any arbitrary set of numbers and numbers. We will focus our attention on the numerical sequence, in which the value of the nth term is associated with its ordinal number by a dependence that can be clearly formulated mathematically. In other words: the numerical value of the n-th number is some function of n.
a - value of a member of a numerical sequence;
n is its serial number;
f (n) is a function where the ordinal in the numerical sequence n is an argument.
Definition
It is customary to call an arithmetic progression 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 for the next number;
d - difference (a certain number).
It is easy to determine that if the difference is positive (d> 0), then each subsequent term of the series under consideration will be larger 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 “ascending”.
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 any arbitrary member a n of an arithmetic progression. You can do this by calculating sequentially 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 meaning of the five-thousandth or eight-millionth member. The traditional calculation will take a long time. However, a specific arithmetic progression can be investigated using specific formulas. There is also a formula for the nth term: the value of any member of an arithmetic progression can be defined as the sum of the first term of the progression with the difference of the progression, multiplied by the number of the sought term, decreased by one.
The formula is universal for both 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 nth term of an arithmetic progression.
Condition: there is an arithmetic progression with parameters:
The first term in the sequence is 3;
The difference in the number series is 1.2.
Assignment: you need to find the value of 214 members
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 in the sequence is 258.6.
The advantages of this calculation method are obvious - the whole 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 a certain segment of it. This also does not require calculating the values of each term and then summing. This method is applicable if the number of terms to be found is small. In other cases, it is more convenient to use the following formula.
The sum of the members of the arithmetic progression from 1 to n is equal to the sum of the first and nth members, multiplied by the number of the member n and divided by half. 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 in the sequence is zero;
The difference is 0.5.
In the problem, you need to determine the sum of the members of the series from 56th 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, substituting the data of their 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 members of the progression from 56th to 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:
s 101 - s 55 = 2,525 - 742.5 = 1,782.5
An example of practical application of arithmetic progression
At the end of the article, let's go back to the example of the arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider an example.
Boarding a taxi (which includes 3 km of run) 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 price.
30 - 3 = 27 km.
2. Further calculation is nothing more than an analysis of an arithmetic number series.
Member number - the number of kilometers traveled (minus the first three).
The member value is the sum.
The first term in this problem will be equal to a 1 = 50 p.
Difference in progression d = 22 p.
the number we are interested in is the value of the (27 + 1) -th term of the arithmetic progression - the counter reading at the end of the 27th kilometer is 27.999… = 28 km.
a 28 = 50 + 22 ∙ (28 - 1) = 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 type of number sequence is geometric
Geometric progression is characterized by large, in comparison with arithmetic, rates of change. It is no coincidence that in politics, sociology, medicine, they often say that the process develops exponentially in order to show the high rate of spread of a phenomenon, for example, a disease during an epidemic.
The Nth term of the geometric numerical 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, respectively, is 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 member of the geometric progression;
b n + 1 - the formula of the next term of the geometric progression;
q is the denominator of a geometric progression (constant number).
If the graph of the arithmetic progression is a straight line, then the geometric one paints a slightly different picture:
As in the case of arithmetic, a geometric progression has a formula for the value of an arbitrary term. Any n-th term of the geometric progression is equal to the product of the first term by 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 = b 1 ∙ q (5-1) = 3 ∙ 1.5 4 = 15.1875
The sum of a given number of members is calculated in the same way 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 considered above, the value of the sum of the first n terms of the considered numerical series will take the form:
Example. The geometric progression starts with the first term equal to 1. The denominator is set equal to 3. Find the sum of the first eight terms.
s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280
