What is the main essence of the formula?
This formula allows you to find any BY HIS NUMBER " n " .
Of course, you also need to know the first term. a 1 and the difference of the progression d, well, without these parameters, you cannot record a specific progression.
Memorizing (or sparging) this formula is not enough. It is necessary to assimilate its essence and apply the formula in various tasks. Moreover, do not forget at the right time, yes ...) How not forget- I do not know. But how to remember if necessary, I will tell you exactly. Those who master the lesson to the end.)
So, let's deal with the formula for the nth term arithmetic progression.
What is a formula in general - we can imagine.) What is an arithmetic progression, number of a member, the difference in progression - is available in the previous lesson. Take a look, by the way, if you haven't read it. Everything is simple there. It remains to figure out what is nth term.
Progression into general view can be written as a series of numbers:
a 1, a 2, a 3, a 4, a 5, .....
a 1- denotes the first member of an arithmetic progression, a 3- third term, a 4- the fourth, and so on. If we are interested in the fifth term, say we are working with a 5 if one hundred and twentieth - from a 120.
And how to designate in general terms any member of arithmetic progression, s any number? Very simple! Like this:
a n
That's what it is the nth term of the arithmetic progression. The letter n hides all the numbers of the members at once: 1, 2, 3, 4, and so on.
And what does such a recording give us? Just think, instead of a number they wrote down a letter ...
This entry gives us a powerful tool for working with arithmetic progression. Using the notation a n, we can quickly find any member any arithmetic progression. And to solve a bunch of problems in progression. You will see for yourself.
In the formula for the nth term of the arithmetic progression:
| a n = a 1 + (n-1) d |
a 1- the first member of the arithmetic progression;
n- member number.
The formula connects the key parameters of any progression: a n; a 1; d and n. All the problems in the progression revolve around these parameters.
The nth term formula can also be used to record a specific progression. For example, the problem may say that the progression is specified by the condition:
a n = 5 + (n-1) 2.
Such a problem can even confuse ... There is no row, no difference ... But, comparing the condition with the formula, it is easy to figure out that in this progression a 1 = 5 and d = 2.
And it happens even angrier!) If we take the same condition: a n = 5 + (n-1) 2, yes to open the brackets and bring similar ones? Let's get a new formula:
a n = 3 + 2n.
This Only not general, but for a specific progression. This is where the pitfall lurks. Some people think that the first term is a triplet. Although in reality the first term is a five ... A little later we will work with such a modified formula.
In the tasks for the progression, one more designation is found - a n + 1... This is, you guessed it, the "en plus first" term in the progression. Its meaning is simple and harmless.) It is a member of the progression whose number is greater than n by one. For example, if in some problem we take for a n fifth term then a n + 1 will be the sixth member. Etc.
Most often the designation a n + 1 occurs in recursive formulas. Do not be intimidated by this terrible word!) This is just a way of expressing a member of an arithmetic progression through the previous one. Suppose we are given an arithmetic progression like this, using a recurrent formula:
a n + 1 = a n +3
a 2 = a 1 + 3 = 5 + 3 = 8
a 3 = a 2 + 3 = 8 + 3 = 11
The fourth - through the third, the fifth - through the fourth, and so on. And how to count right away, say, the twentieth term, a 20? But no way!) Until the 19th term is recognized, the 20th cannot be counted. This is the fundamental difference between the recurrent formula and the formula for the nth term. Recurrent only works through previous term, and the formula of the nth term is through first and allows straightaway find any member by its number. Without counting the whole series of numbers in order.
In an arithmetic progression, a recurring formula can be easily turned into an ordinary one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write down the formula in its usual form, and work with it. In the GIA, such tasks are often encountered.
Application of the formula for the n-th member of an arithmetic progression.
First, let's look at the direct application of the formula. At the end of the previous lesson, there was a problem:
You are given an arithmetic progression (a n). Find a 121 if a 1 = 3 and d = 1/6.
This problem can be solved without any formulas, simply proceeding from the meaning of the arithmetic progression. Add, yes add ... an hour or two.)
And according to the formula, the solution will take less than a minute. You can time it.) We decide.
The conditions provide all the data for using the formula: a 1 = 3, d = 1/6. It remains to figure out what it is equal to n. No problem! We need to find a 121... So we write:
Please pay attention! Instead of an index n a specific number appeared: 121. Which is quite logical.) We are interested in a member of the arithmetic progression number one hundred twenty one. This will be ours n. It is this meaning n= 121 we will substitute further into the formula, in parentheses. We substitute all the numbers in the formula and calculate:
a 121 = 3 + (121-1) 1/6 = 3 + 20 = 23
That's all there is to it. Just as quickly one could find the five hundred and tenth term, and the thousand and three, any. We put instead n the desired index number for the letter " a " and in brackets, and we count.
Let me remind you the point: this formula allows you to find any term of arithmetic progression BY HIS NUMBER " n " .
Let's solve the task more cunning. Let us have such a problem:
Find the first term of the arithmetic progression (a n) if a 17 = -2; d = -0.5.
If you have any difficulties, I will tell you the first step. Write down the formula for the nth term of the arithmetic progression! Yes Yes. Write with your hands, right in your notebook:
| a n = a 1 + (n-1) d |
And now, looking at the letters of the formula, we figure out what data we have and what is missing? There is d = -0.5, there is a seventeenth member ... Is that all? If you think that's all, then you won't solve the problem, yes ...
We still have a number n! In condition a 17 = -2 hidden two parameters. This is both the value of the seventeenth term (-2) and its number (17). Those. n = 17. This "trifle" often slips past the head, and without it, (without the "trifle", and not the head!) The problem cannot be solved. Although ... without a head, too.)
Now you can just stupidly substitute our data into the formula:
a 17 = a 1 + (17-1) (-0.5)
Oh yes, a 17 we know it's -2. Okay, let's substitute:
-2 = a 1 + (17-1) (-0.5)
That, in essence, is all. It remains to express the first term of the arithmetic progression from the formula, and calculate. The answer will be: a 1 = 6.
This technique - writing a formula and a simple substitution of known data - helps a lot in simple tasks. Well, you must, of course, be able to express a variable from a formula, but what to do !? Without this skill, mathematics can be avoided at all ...
Another popular puzzle:
Find the difference of the arithmetic progression (a n) if a 1 = 2; a 15 = 12.
What are we doing? You will be surprised, we are writing the formula!)
| a n = a 1 + (n-1) d |
Consider what we know: a 1 = 2; a 15 = 12; and (I will specially highlight it!) n = 15. Feel free to substitute in the formula:
12 = 2 + (15-1) d
We count arithmetic.)
12 = 2 + 14d
d=10/14 = 5/7
This is the correct answer.
So, tasks for a n, a 1 and d solved. It remains to learn how to find the number:
The number 99 is a member of the arithmetic progression (a n), where a 1 = 12; d = 3. Find the number of this member.
We substitute the quantities known to us in the formula for the nth term:
a n = 12 + (n-1) 3
At first glance, there are two unknowns: a n and n. But a n is some member of the progression with a number n... And we know this member of the progression! It's 99. We don't know his number. n, so this number is required to be found. We substitute the term of the progression 99 into the formula:
99 = 12 + (n-1) 3
We express from the formula n, consider. We get the answer: n = 30.
And now a puzzle on the same topic, but more creative):
Determine if the number 117 is a member of the arithmetic progression (a n):
-3,6; -2,4; -1,2 ...
We write the formula again. What, there are no parameters? Hm ... Why are we given eyes?) See the first member of the progression? We see. This is -3.6. You can safely write: a 1 = -3.6. Difference d can be determined from a number? It's easy if you know what the difference of an arithmetic progression is:
d = -2.4 - (-3.6) = 1.2
So, we did the simplest thing. It remains to deal with an unknown number n and an incomprehensible number 117. In the previous problem, at least it was known that it was a member of the progression that was given. And here we don't even know ... How to be !? Well, how to be, how to be ... Turn on creativity!)
We suppose that 117 is, after all, a member of our progression. With an unknown number n... And, just like in the previous task, let's try to find this number. Those. we write the formula (yes, yes!)) and substitute our numbers:
117 = -3.6 + (n-1) 1.2
Again we express from the formulan, we count and get:
Oops! Number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion can we draw? Yes! Number 117 is not a member of our progression. It is somewhere between the one hundred and first and one hundred and second members. If the number turned out to be natural, i.e. a positive integer, then the number would be a member of the progression with the found number. And in our case, the answer to the problem will be: no.
The task based on the real version of the GIA:
The arithmetic progression is specified by the condition:
a n = -4 + 6.8n
Find the first and tenth members of the progression.
Here the progression is not set in a completely familiar way. Some kind of formula ... It happens.) However, this formula (as I wrote above) - is also a formula for the nth term of an arithmetic progression! She also allows find any member of the progression by its number.
We are looking for the first member. The one who thinks. that the first term is minus four, is fatally mistaken!) Because the formula in the problem is modified. The first term of the arithmetic progression in it hidden. Nothing, we'll find it now.)
Just as in the previous tasks, we substitute n = 1 into this formula:
a 1 = -4 + 6.8 1 = 2.8
Here! The first term is 2.8, not -4!
Similarly, we look for the tenth term:
a 10 = -4 + 6.8 10 = 64
That's all there is to it.
And now, for those who have read to these lines - the promised bonus.)
Suppose, in a difficult combat situation of the GIA or USE, you have forgotten a useful formula for the n-th term of the arithmetic progression. Something is recalled, but somehow uncertainly ... Either n there or n + 1, or n-1 ... How to be !?
Calm! This formula is easy to deduce. Not very strict, but for certainty and the correct solution, it will definitely be enough!) For a conclusion, it is enough to remember the elementary meaning of the arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.
Draw a number axis and mark the first one on it. second, third, etc. members. And note the difference d between members. Like this:
We look at the picture and figure out: what is the second term equal to? Second one thing d:
a 2 = a 1 + 1 D
What is the third term? Third term equals first term plus two d.
a 3 = a 1 + 2 D
Do you get it? It's not for nothing that I highlight some words in bold. Okay, one more step).
What is the fourth term? Fourth term equals first term plus three d.
a 4 = a 1 + 3 D
It's time to figure out that the number of gaps, i.e. d, always one less than the number of the required term n. That is, to the number n, number of intervals will n-1. Therefore, the formula will be (no options!):
| a n = a 1 + (n-1) d |
In general, pictorial pictures are very helpful in solving many problems in mathematics. Don't neglect pictures. But if it’s difficult to draw a picture, then ... only the formula!) In addition, the formula of the nth term allows you to connect to the solution the entire powerful arsenal of mathematics - equations, inequalities, systems, etc. You can't put a picture into an equation ...
Tasks for independent solution.
To warm up:
1. In an arithmetic progression (a n) a 2 = 3; a 5 = 5.1. Find a 3.
Hint: according to the picture, the problem is solved in 20 seconds ... According to the formula, it turns out more difficult. But for mastering the formula it is more useful.) Section 555 solved this problem both by the picture and by the formula. Feel the difference!)
And this is no longer a warm-up.)
2. In arithmetic progression (a n) a 85 = 19.1; a 236 = 49, 3. Find a 3.
What, do you feel reluctant to draw a picture?) Of course! Better by the formula, yes ...
3. The arithmetic progression is specified by the condition:a 1 = -5.5; a n + 1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.
In this task, the progression is given in a recurrent way. But counting up to one hundred and twenty-fifth term ... Not everyone can do such a feat.) But the formula of the n-th term is within everyone's power!
4. Given an arithmetic progression (a n):
-148; -143,8; -139,6; -135,4, .....
Find the number of the smallest positive term in the progression.
5. According to the condition of task 4, find the sum of the smallest positive and largest negative members of the progression.
6. The product of the fifth and twelfth terms of the increasing arithmetic progression is -2.5, and the sum of the third and eleventh terms is zero. Find a 14.
Not the easiest task, yes ...) Here, the "on the fingers" method will not work. We'll have to write formulas and solve equations.
Answers (in disarray):
3,7; 3,5; 2,2; 37; 2,7; 56,5
Happened? It's nice!)
Not everything works out? It happens. By the way, there is one subtle point in the last task. Carefulness when reading the problem will be required. And logic.
The solution of all these problems is discussed in detail in Section 555. And the element of fantasy for the fourth, and the delicate moment for the sixth, and general approaches for solving any problems on the formula of the nth term - everything is written out. Recommend.
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. Instant validation testing. 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 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, to 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 work 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 back side the boards prepare answers to these same questions.
Students evaluate the partner's work against the board. (The answer sheets are handed over).
2. Game moment.
Exercise 1.
Teacher. I have in mind 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 told 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 antiquity.
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
Conclusion:
III group: Find the sum of natural numbers from 1 to 21.
Solution: 1 + 21 = 2 + 20 = 3 + 19 = 4 + 18 ...
Conclusion:
IV-th group: Find the sum of natural numbers from 1 to 101.
Conclusion:
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 a 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 will immediately get down to business.
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 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 just 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. Firstly, 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 might seem overly complicated. But the rest, I think, is clear to you. Therefore, we will introduce new definitions:
Definition. An arithmetic progression is called:
- ascending 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, because 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.
However, 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 definitely need to 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 $, from where we immediately find the difference:
Note that the difference is positive, so the progression increases. The first term is negative, so at some point we really will stumble upon positive numbers. The only question is when it will happen.
Let's try to find out: how long (i.e. until what natural number$ n $) the negativity of the members 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 needs some explanation. So, we know that $ n \ lt 15 \ frac (7) (27) $. On the other hand, we will be satisfied with only 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 recursion 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 terms $ ((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 have deduced 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) \]
It turned out to be 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 $ form 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 $.
Answer: 1; 6.
If in the process of solving a problem you get out some brutal numbers, or you are not completely sure of the correctness of the answers found, then there is a wonderful technique that allows you to check: did we solve the problem correctly?
For example, in problem no. 6 we received answers -3 and 2. How to check that these answers are correct? Let's just plug them into the initial condition and see what happens. Let me remind you that we have three numbers ($ -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 stumbled upon one more 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 $ S $ number, 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, 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 for 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
Pay attention: this parabola takes its minimum value at its vertex with 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 smallest value(we, by the way, have not counted $ ((y) _ (\ min)) $ - this is not required from us). At the same time, this number is the difference between the original 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. So
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 desired 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 problems. 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 binding workshop bound 216 books in January, and each 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 look 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 the progression, as well as important and very useful consequences from it.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very ..."
And for those who are "very even ...")
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. Indices for letters, the n-th term of the progression, the difference in the progression - all this is somehow confusing, yes ... Let's figure out the meaning of the arithmetic progression and everything will work out right away.)
Arithmetic progression concept.
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 row? What numbers will go next, after the five? Everyone ... uh-uh ..., in short, everyone will realize 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 will be able to catch the pattern, extend the series, and name seventh row number?
If you figured out that this number is 20 - I congratulate you! Not only did you feel key points of arithmetic progression, but also successfully used them in business! If you haven't figured it out, read on.
Now let's translate the key points from sensation to mathematics.)
First key point.
Arithmetic progression deals with series of numbers. This is confusing at first. We are used to solving equations, plotting graphs and all that ... And then extend the series, find the number of the series ...
Nothing wrong. It's just that progressions are the first acquaintance with a new branch of mathematics. The section is called "Rows" and it works with series of numbers and expressions. Get used to it.)
Second key point.
In an arithmetic progression, any number is different 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 greater than the previous one by three. Actually, it is this moment that gives us the opportunity to catch the pattern and calculate the subsequent numbers.
The third key point.
This moment is not striking, yes ... But it is very, very important. Here it is: each number in the progression stands in its place. There is the first number, there is the seventh, there is the forty-fifth, etc. If you mix them up at random, the pattern will disappear. The arithmetic progression will also disappear. Just a row of numbers will remain.
That's the whole point.
Of course, new terms and designations appear in the new topic. You need to know them. Otherwise, you will not understand the task. For example, you have to decide something like:
Write out 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 terms and designations. Now we will master this business 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 quantity is called ... Let's deal with this concept in more detail.
Difference of arithmetic progression.
Difference of arithmetic progression is the amount by which any number of the progression more the previous one.
One important point. Please pay attention to the word "more". Mathematically, this means that each number in the progression is obtained adding the difference of the arithmetic progression to the previous number.
For calculation, let's say second number of the series, it is necessary to the first the number add this same difference of arithmetic progression. For calculation fifth- the difference is necessary add To fourth, well, etc.
Difference of arithmetic progression may be positive, then each number of the row will turn out really more than the previous one. This progression is called increasing. For instance:
8; 13; 18; 23; 28; .....
Here every number is obtained adding positive number, +5 to the previous one.
The difference can be negative, then each number in the row will be less than the previous one. Such a progression is called (you won't believe it!) decreasing.
For instance:
8; 3; -2; -7; -12; .....
Here every number is obtained too adding to the previous, but already 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 navigate the solution, to detect your mistakes and fix them before it's too late.
Difference of arithmetic progression denoted, as a rule, 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 the subtraction is called the "difference".)
Let us define, for example, d for increasing arithmetic progression:
2, 5, 8, 11, 14, ...
We take any number of the row that we want, for example, 11. Subtract from it previous number, those. eight:
This is the correct answer. For this arithmetic progression, the difference is three.
You can take exactly any number of progression, since for a specific progression d -always the same. At least somewhere in the beginning of the row, at least in the middle, at least anywhere. You cannot take only the very first number. Just because the very first number there is no previous one.)
By the way, knowing that d = 3, it is very easy to find the seventh number of this progression. Add 3 to the fifth number - we get the sixth, it will be 17. Add three to the sixth number, we get the seventh number - twenty.
We define d for a decreasing arithmetic progression:
8; 3; -2; -7; -12; .....
I remind you that, regardless of the signs, to determine d it is necessary from any number take away the previous one. We choose any number of the progression, for example -7. The previous one is -2. Then:
d = -7 - (-2) = -7 + 2 = -5
The difference of the arithmetic progression can be any number: whole, fractional, irrational, whatever.
Other terms and designations.
Each number in the series is called a member of an arithmetic progression.
Each member of the progression has its own 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 term, five is the second, eleven is the fourth, well, you understand ...) Please understand clearly - the numbers themselves can be absolutely any, whole, fractional, negative, whatever, but numbering of numbers- strictly in order!
How to record a general progression? No problem! Each number in the row is written as a letter. As a rule, the letter is used to denote an arithmetic progression a... The member number is indicated by an index at the bottom right. We write members 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 briefly write this series like this: (a n).
Progressions are finite and endless.
The ultimate the progression has a limited number of members. Five, thirty-eight, whatever. But - a finite number.
Endless progression - has an infinite number of members, as you might guess.)
You can write the final progression through a series like this, all the members and a dot at the end:
a 1, a 2, a 3, a 4, a 5.
Or so, 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 endless progression can be recognized by the ellipsis at the end of the row, as in the examples in this lesson.
Now you can solve tasks. The tasks are simple, purely for understanding the meaning of the arithmetic progression.
Examples of tasks on arithmetic progression.
Let's analyze the task in detail, which is given above:
1. Write out the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.
We translate the task into an understandable language. An infinite arithmetic progression is given. The second number of this progression is known: a 2 = 5. The difference in progression is known: d = -2.5. It is necessary 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 terms, where the second term is a five:
a 1, 5, a 3, a 4, a 5, a 6, ....
a 3 = a 2 + d
Substitute into expression a 2 = 5 and d = -2.5... Do not forget about the minus!
a 3=5+(-2,5)=5 - 2,5 = 2,5
The third term is smaller than the second. Everything is logical. If the number is greater than the previous one by negative value, then the number itself will turn out to be less than the previous one. The 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 are calculated. The result is such 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 need not add 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 answer:
7,5, 5, 2,5, 0, -2,5, -5, ...
Along the way, I will note that we solved this task recurrent way. This scary word only means searching for a member of the progression. by the previous (adjacent) number. We will consider other ways of working with progression later.
One important conclusion can be drawn from this simple task.
Remember:
If we know at least one term and the difference of an arithmetic progression, we can find any member of this progression.
Do you remember? This simple conclusion allows you to solve most of the tasks of the school course on this topic. All tasks revolve around three main parameters: member of arithmetic progression, difference of progression, number of member of the progression. Everything.
Of course, all the previous algebra is not canceled.) Inequalities, equations, and other things are attached to the progression. But by the very progression- everything revolves around three parameters.
Let's take a look at some of the popular assignments on this topic as an example.
2. Write down the final arithmetic progression as a series, if n = 5, d = 0.4, and a 1 = 3.6.
Everything is simple here. Everything has already been given. You need to remember how the members of an arithmetic progression are counted, count, and write them down. It is advisable not to miss the words in the condition of the assignment: "final" and " n = 5". Not to count until completely blue in the face.) There are only 5 (five) members in this progression:
a 2 = a 1 + d = 3.6 + 0.4 = 4
a 3 = a 2 + d = 4 + 0.4 = 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 whether the number 7 is a member of the arithmetic progression (a n), if a 1 = 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 there, or not! We consider:
a 2 = a 1 + d = 4.1 + 1.2 = 5.3
a 3 = a 2 + d = 5.3 + 1.2 = 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 a 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.
The answer is no.
And here is a task based on a real version of the GIA:
4. Several consecutive members of the arithmetic progression are written out:
...; 15; X; 9; 6; ...
A row is written here without end and beginning. No member numbers, no difference d... Nothing wrong. To solve the problem, it is enough to understand the meaning of the arithmetic progression. We look and think about what is possible discover from this series? What are the three main parameters?
Member numbers? There is not a single number here.
But there are three numbers and - attention! - word "consecutive" in the condition. This means that the numbers are strictly in order, without gaps. Are there two in this row neighboring known numbers? Yes, I have! These are 9 and 6. So we can calculate the difference of the arithmetic progression! We subtract from the six previous number, i.e. nine:
There are mere trifles left. What is the previous number for the X? Fifteen. This means that x can be easily found by simple addition. Add the difference of the arithmetic progression to 15:
That's all. Answer: x = 12
We solve the following problems ourselves. Note: these problems are not about 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 the arithmetic progression a 2 = 4; a 5 = 15.1. Find a 3.
8. Written out several consecutive members of the arithmetic progression:
...; 15.6; X; 3.4; ...
Find the term in the progression denoted by the letter x.
9. The train started moving from the station, steadily increasing its speed by 30 meters per minute. What will the train speed be in five minutes? Give your answer in km / h.
10. It is known that in the arithmetic progression a 2 = 5; a 6 = -5. Find a 1.
Answers (in disarray): 7.7; 7.5; 9.5; 9; 0.3; 4.
Everything worked out? Wonderful! You can master the arithmetic progression at a higher level in the following lessons.
Not everything worked out? No problem. In Special Section 555, all these tasks are sorted out to pieces.) And, of course, a simple practical technique is described that immediately highlights the solution of such tasks clearly, clearly, as if 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 is purely in progression, and the second is common for any problems in mathematics, and physics too. This is a translation of dimensions from one to another. In it is shown how these problems should be solved.
In this lesson, we examined the elementary meaning of the 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 a row, as in the examples in this lesson. If the row 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 significantly angrier.)
And there are also tasks that are simple in essence, but incredible in terms of calculations, for example:
You are given an arithmetic progression (a n). Find a 121 if a 1 = 3 and d = 1/6.
And what, we will add many, many times by 1/6 ?! You can kill it !?
You can.) If you do not know the simple formula by which such tasks can be solved in a minute. This formula will be in the next lesson. And this 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. Instant validation testing. Learning - with interest!)
you can get acquainted with functions and derivatives.
When studying algebra in comprehensive school(Grade 9) one of the important topics is the study of numerical sequences, which include progressions - geometric and arithmetic. In this article, we will consider the arithmetic progression and examples with solutions.
What is an arithmetic progression?
To understand this, it is necessary to give a definition of the considered progression, as well as to give the basic formulas that will be further used in solving problems.
An arithmetic or algebraic progression is a set of ordered rational numbers, each term of which differs from the previous one by some constant amount. This value is called the difference. That is, knowing any member of the ordered series of numbers and the difference, you can restore the entire arithmetic progression.
Let's give an example. The following sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the considered type of progression, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).
Important formulas
Let us now give the basic formulas that will be needed to solve problems using an arithmetic progression. Let us denote by a n the nth term of the sequence, where n is an integer. The difference is denoted by the Latin letter d. Then the following expressions are valid:
- To determine the value of the nth term, the formula is suitable: a n = (n-1) * d + a 1.
- To determine the sum of the first n terms: S n = (a n + a 1) * n / 2.
To understand any examples of arithmetic progression with a solution in grade 9, it is enough to remember these two formulas, since any problems of the type under consideration are built on their use. You should also remember that the difference in progression is determined by the formula: d = a n - a n-1.
Example # 1: finding an unknown member
Let's give a simple example of an arithmetic progression and formulas that must be used to solve.
Let the sequence 10, 8, 6, 4, ... be given, it is necessary to find five terms in it.
It already follows from the problem statement that the first 4 terms are known. The fifth can be defined in two ways:
- Let's calculate the difference first. We have: d = 8 - 10 = -2. Likewise, one could take any two other members standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d = a n - a n-1, then d = a 5 - a 4, whence we obtain: a 5 = a 4 + d. Substitute the known values: a 5 = 4 + (-2) = 2.
- The second method also requires knowing the difference of the considered progression, so first you need to determine it, as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for n number of the sequence. We have: a n = (n - 1) * d + a 1 = (n - 1) * (-2) + 10 = 12 - 2 * n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.
As you can see, both methods of solution led to the same result. Note that in this example, the difference d of the progression is negative. Such sequences are called decreasing, since each next term is less than the previous one.
Example # 2: Progression Difference
Now let's complicate the task a little, let's give an example how
It is known that in some the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.
Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1. We substitute in it the known data from the condition, that is, the numbers a 1 and a 7, we have: 18 = 6 + 6 * d. From this expression, you can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, the answer to the first part of the problem.
To restore a sequence up to 7 terms, you should use the definition of an algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d, and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2 = 8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16, a 7 = 18.
Example # 3: making a progression
Let us complicate the condition of the problem even more. Now it is necessary to answer the question of how to find the arithmetic progression. You can give the following example: given two numbers, for example, - 4 and 5. It is necessary to compose an algebraic progression so that three more terms fit between these.
Before starting to solve this problem, it is necessary to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 = -4 and a 5 = 5. Having established this, we proceed to the problem, which is similar to the previous one. Again, for the n-th term, we use the formula, we get: a 5 = a 1 + 4 * d. From where: d = (a 5 - a 1) / 4 = (5 - (-4)) / 4 = 2.25. Here we received not an integer value of the difference, but it is a rational number, so the formulas for the algebraic progression remain the same.
Now add the found difference to a 1 and restore the missing members of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 = 2.75 + 2.25 = 5, which coincided with the condition of the problem.
Example # 4: the first term of the progression
Let's continue to give examples of arithmetic progression with a solution. In all the previous problems, the first number of the algebraic progression was known. Now consider a problem of a different type: let there be given two numbers, where a 15 = 50 and a 43 = 37. It is necessary to find the number from which this sequence begins.
The formulas used so far assume knowledge of a 1 and d. Nothing is known about these numbers in the problem statement. Nevertheless, we write out expressions for each member about which there is information: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. Received two equations, in which 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.
This system is easiest to solve if you express a 1 in each equation, and then compare the resulting expressions. The first equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 = a 43 - 42 * d = 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d = 37 - 42 * d, whence the difference d = (37 - 50) / (42 - 14) = - 0.464 (only 3 decimal places are given).
Knowing d, you can use any of the 2 above expressions for a 1. For example, the first one: a 1 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.
If you have doubts about the result, you can check it, for example, determine the 43 term of the progression, which is specified in the condition. We get: a 43 = a 1 + 42 * d = 56.496 + 42 * (- 0.464) = 37.008. A small error is due to the fact that the calculations used rounding to thousandths.
Example # 5: amount
Now let's look at some examples with solutions for the sum of an arithmetic progression.
Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How do you calculate the sum of these 100 numbers?
Thanks to the development of computer technology, it is possible to solve this problem, that is, to add up all the numbers sequentially, which the computer will do as soon as a person presses the Enter key. However, the problem can be solved in the mind, if we pay attention that the presented series of numbers is an algebraic progression, and its difference is 1. Applying the formula for the sum, we get: S n = n * (a 1 + an) / 2 = 100 * (1 + 100) / 2 = 5050.
It is curious to note that this problem is called "Gaussian", since at the beginning of the 18th century the famous German, while still only 10 years old, was able to solve it in his head in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add in pairs the numbers on the edges of the sequence, you always get one result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since of these amounts will be exactly 50 (100/2), then to get the correct answer, it is enough to multiply 50 by 101.
Example # 6: sum of members from n to m
Another typical example of the sum of an arithmetic progression is the following: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its members from 8 to 14 will equal.
The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then adding them sequentially. Since there are few terms, this method is not laborious enough. Nevertheless, it is proposed to solve this problem by the second method, which is more universal.
The idea is to obtain a formula for the sum of the algebraic progression between the terms m and n, where n> m are integers. Let us write out two expressions for the sum for both cases:
- S m = m * (a m + a 1) / 2.
- S n = n * (a n + a 1) / 2.
Since n> m, it is obvious that the 2 sum includes the first. The last conclusion means that if we take the difference between these sums, and add to it the term a m (in the case of taking the difference, it is subtracted from the sum S n), then we get the necessary answer to the problem. We have: S mn = S n - S m + am = n * (a 1 + an) / 2 - m * (a 1 + am) / 2 + am = a 1 * (n - m) / 2 + an * n / 2 + am * (1- m / 2). In this expression it is necessary to substitute the formulas for a n and a m. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d * (3 * m - m 2 - 2) / 2.
The resulting formula is somewhat cumbersome; nevertheless, the sum of S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.
As can be seen from the solutions presented, all problems are based on knowledge of the expression for the nth term and the formula for the sum of the set of the first terms. Before proceeding with the solution of any of these problems, it is recommended to carefully read the condition, clearly understand what needs to be found, and only then proceed to the solution.
Another tip is to strive for simplicity, that is, if you can answer a question without using complex mathematical calculations, then you need to do just that, since in this case the probability of making a mistake is less. For example, in the example of an arithmetic progression with solution # 6, one could stop at the formula S mn = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and split common task into separate subtasks (in this case, first find the terms a n and a m).
If there are doubts about the result obtained, it is recommended to check it, as was done in some of the above examples. We figured out how to find the arithmetic progression. If you figure it out, it's not that difficult.