INSTRUCTIONS AND PROPHECIES OF THE Blessed MOTHER ALIPIA GOLOSEEVSKY, Kyiv...
Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")
Arithmetic progression- this is a series of numbers in which each number is greater (or less) than the previous one by the same amount.
This topic is often difficult and incomprehensible. Letter indexes, nth member progressions, the difference in progression - all this is somehow confusing, yes ... Let's figure out the meaning of the arithmetic progression and everything will work out right away.)
The concept of arithmetic progression.
Arithmetic progression is a very simple and clear concept. Doubt? In vain.) See for yourself.
I'll write an unfinished series of numbers:
1, 2, 3, 4, 5, ...
Can you extend this line? What numbers will go next, after the five? Everyone ... uh ..., in short, everyone will figure out that the numbers 6, 7, 8, 9, etc. will go further.
Let's complicate the task. I give an unfinished series of numbers:
2, 5, 8, 11, 14, ...
You can catch the pattern, extend the series, and name seventh row number?
If you figured out that this number is 20 - I congratulate you! You not only felt key points arithmetic progression, but also successfully used them in business! If you don't understand, read on.
Now let's translate the key points from sensations into mathematics.)
First key point.
Arithmetic progression deals with series of numbers. This is confusing at first. We are used to solving equations, building graphs and all that ... And then extend the series, find the number of the series ...
It's OK. It's just that progressions are the first acquaintance with a new branch of mathematics. The section is called "Series" and works with series of numbers and expressions. Get used to it.)
Second key point.
In an arithmetic progression, any number differs from the previous one by the same amount.
In the first example, this difference is one. Whatever number you take, it is one more than the previous one. In the second - three. Any number is three times greater than the previous one. Actually, it is this moment that gives us the opportunity to catch the pattern and calculate the subsequent numbers.
Third key point.
This moment is not striking, yes ... But very, very important. Here he is: each progression number stands in its place. There is the first number, there is the seventh, there is the forty-fifth, and so on. If you confuse them haphazardly, the pattern will disappear. The arithmetic progression will also disappear. It's just a series of numbers.
That's the whole point.
Of course, in new topic new terms and notation appear. They need to know. Otherwise, you won't understand the task. For example, you have to decide something like:
Write down the first six terms of the arithmetic progression (a n) if a 2 = 5, d = -2.5.
Does it inspire?) Letters, some indexes... And the task, by the way, couldn't be easier. You just need to understand the meaning of the terms and notation. Now we will master this matter and return to the task.
Terms and designations.
Arithmetic progression is a series of numbers in which each number is different from the previous one by the same amount.
This value is called . Let's deal with this concept in more detail.
Arithmetic progression difference.
Arithmetic progression difference is the amount by which any progression number more the previous one.
One important point. Please pay attention to the word "more". Mathematically, this means that each progression number is obtained adding the difference of an arithmetic progression to the previous number.
To calculate, let's say second numbers of the row, it is necessary to first number add this very difference of an arithmetic progression. For calculation fifth- the difference is necessary add to fourth well, etc.
Arithmetic progression difference may be positive then each number of the series will turn out to be real more than the previous one. This progression is called increasing. For example:
8; 13; 18; 23; 28; .....
Here each number is adding positive number, +5 to the previous one.
The difference can be negative then each number in the series will be less than the previous one. This progression is called (you won't believe it!) decreasing.
For example:
8; 3; -2; -7; -12; .....
Here every number is obtained too adding to the previous, 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 find your bearings in the decision, to detect your mistakes and correct them before it's too late.
Arithmetic progression difference usually denoted by the letter d.
How to find d? Very simple. It is necessary to subtract from any number of the series previous number. Subtract. By the way, the result of subtraction is called "difference".)
Let's define, for example, d for an increasing arithmetic progression:
2, 5, 8, 11, 14, ...
We take any number of the row that we want, for example, 11. Subtract from it the previous number those. eight:
This is the correct answer. For this arithmetic progression, the difference is three.
You can just take any number of progressions, because for a specific progression d-always the same. At least somewhere at the beginning of the row, at least in the middle, at least anywhere. You can not take only the very first number. Just because the very first number no previous.)
By the way, knowing that d=3, finding the seventh number of this progression is very simple. We add 3 to the fifth number - we get the sixth, it will be 17. We add three to the sixth number, we get the seventh number - twenty.
Let's define d for a decreasing arithmetic progression:
8; 3; -2; -7; -12; .....
I remind you that, regardless of the signs, to determine d needed from any number take away the previous one. We choose any number of progression, for example -7. His previous number is -2. Then:
d = -7 - (-2) = -7 + 2 = -5
The difference of an arithmetic progression can be any number: integer, fractional, irrational, any.
Other terms and designations.
Each number in the series is called member of an arithmetic progression.
Each member of the progression has his number. The numbers are strictly in order, without any tricks. First, second, third, fourth, etc. For example, in the progression 2, 5, 8, 11, 14, ... two is the first member, five is the second, eleven is the fourth, well, you understand ...) Please clearly understand - the numbers themselves can be absolutely any, whole, fractional, negative, whatever, but numbering- strictly in order!
How to write a progression in general form? No problem! Each number in the series is written as a letter. To denote an arithmetic progression, as a rule, the letter is used a. The member number is indicated by the index at the bottom right. Members are written separated by commas (or semicolons), like this:
a 1 , a 2 , a 3 , a 4 , a 5 , .....
a 1 is the first number a 3- third, etc. Nothing tricky. You can write this series briefly like this: (a n).
There are progressions finite and infinite.
ultimate the progression has a limited number of members. Five, thirty-eight, whatever. But it's a finite number.
Endless progression - has an infinite number of members, as you might guess.)
You can write a final progression through a series like this, all members and a dot at the end:
a 1 , a 2 , a 3 , a 4 , a 5 .
Or like this, if there are many members:
a 1 , a 2 , ... a 14 , a 15 .
In a short entry, you will have to additionally indicate the number of members. For example (for twenty members), like this:
(a n), n = 20
An infinite progression can be recognized by the ellipsis at the end of the row, as in the examples in this lesson.
Now you can already solve tasks. The tasks are simple, purely for understanding the meaning of the arithmetic progression.
Examples of tasks for arithmetic progression.
Let's take a closer look at the task above:
1. Write down the first six members of the arithmetic progression (a n), if a 2 = 5, d = -2.5.
We translate the task into understandable language. Given an infinite arithmetic progression. The second number of this progression is known: a 2 = 5. Known progression difference: d = -2.5. We need to find the first, third, fourth, fifth and sixth members of this progression.
For clarity, I will write down a series according to the condition of the problem. The first six members, where the second member is five:
a 1 , 5 , a 3 , a 4 , a 5 , a 6 ,....
a 3 = a 2 + d
We substitute in the expression a 2 = 5 and d=-2.5. Don't forget the minus!
a 3=5+(-2,5)=5 - 2,5 = 2,5
The third term is less than the second. Everything is logical. If the number is greater than the previous one negative value, so the number itself will be less than the previous one. Progression is decreasing. Okay, let's take it into account.) We consider the fourth member of our series:
a 4 = a 3 + d
a 4=2,5+(-2,5)=2,5 - 2,5 = 0
a 5 = a 4 + d
a 5=0+(-2,5)= - 2,5
a 6 = a 5 + d
a 6=-2,5+(-2,5)=-2,5 - 2,5 = -5
So, the terms from the third to the sixth have been calculated. This resulted in a series:
a 1 , 5 , 2.5 , 0 , -2.5 , -5 , ....
It remains to find the first term a 1 according to the well-known second. This is a step in the other direction, to the left.) Hence, the difference of the arithmetic progression d should not be added to a 2, a take away:
a 1 = a 2 - d
a 1=5-(-2,5)=5 + 2,5=7,5
That's all there is to it. Task response:
7,5, 5, 2,5, 0, -2,5, -5, ...
In passing, I note that we solved this task recurrent way. This terrible word means, only, the search for a member of the progression by the previous (adjacent) number. Other ways to work with progression will be discussed later.
One important conclusion can be drawn from this simple task.
Remember:
If we know at least one member and the difference of an arithmetic progression, we can find any member of this progression.
Remember? This simple derivation allows us to solve most problems school course on this topic. All tasks revolve around three main parameters: member of an arithmetic progression, difference of a progression, number of a member of a progression. Everything.
Of course, all previous algebra is not cancelled.) Inequalities, equations, and other things are attached to the progression. But according to the progression- everything revolves around three parameters.
For example, consider some popular tasks on this topic.
2. Write the final arithmetic progression as a series if n=5, d=0.4, and a 1=3.6.
Everything is simple here. Everything is already given. You need to remember how the members of an arithmetic progression are calculated, count, and write down. It is advisable not to skip the words in the task condition: "final" and " n=5". In order not to count until you are completely blue in the face.) There are only 5 (five) members in this progression:
a 2 \u003d a 1 + d \u003d 3.6 + 0.4 \u003d 4
a 3 \u003d a 2 + d \u003d 4 + 0.4 \u003d 4.4
a 4 = a 3 + d = 4.4 + 0.4 = 4.8
a 5 = a 4 + d = 4.8 + 0.4 = 5.2
It remains to write down the answer:
3,6; 4; 4,4; 4,8; 5,2.
Another task:
3. Determine if the number 7 will be a member of an arithmetic progression (a n) if a 1 \u003d 4.1; d = 1.2.
Hmm... Who knows? How to define something?
How-how ... Yes, write down the progression in the form of a series and see if there will be a seven or not! We believe:
a 2 \u003d a 1 + d \u003d 4.1 + 1.2 \u003d 5.3
a 3 \u003d a 2 + d \u003d 5.3 + 1.2 \u003d 6.5
a 4 = a 3 + d = 6.5 + 1.2 = 7.7
4,1; 5,3; 6,5; 7,7; ...
Now it is clearly seen that we are just seven slipped through between 6.5 and 7.7! The seven did not get into our series of numbers, and, therefore, the seven will not be a member of the given progression.
Answer: no.
And here is a task based on a real version of the GIA:
4. Several consecutive members of the arithmetic progression are written out:
...; fifteen; X; 9; 6; ...
Here is a series without end and beginning. No member numbers, no difference d. It's OK. To solve the problem, it is enough to understand the meaning of an arithmetic progression. Let's see and see what we can to know from this line? What are the parameters of the three main ones?
Member numbers? There is not a single number here.
But there are three numbers and - attention! - word "consecutive" in condition. This means that the numbers are strictly in order, without gaps. Are there two in this row? neighboring known numbers? Yes there is! These are 9 and 6. So we can calculate the difference of an arithmetic progression! We subtract from the six previous number, i.e. nine:
There are empty spaces left. What number will be the previous one for x? Fifteen. So x can be easily found by simple addition. To 15 add the difference of an arithmetic progression:
That's all. Answer: x=12
We solve the following problems ourselves. Note: these puzzles are not for formulas. Purely for understanding the meaning of an arithmetic progression.) We just write down a series of numbers-letters, look and think.
5. Find the first positive term of the arithmetic progression if a 5 = -3; d = 1.1.
6. It is known that the number 5.5 is a member of the arithmetic progression (a n), where a 1 = 1.6; d = 1.3. Determine the number n of this member.
7. It is known that in an arithmetic progression a 2 = 4; a 5 \u003d 15.1. Find a 3 .
8. Several consecutive members of the arithmetic progression are written out:
...; 15.6; X; 3.4; ...
Find the term of the progression, denoted by the letter x.
9. The train started moving from the station, gradually increasing its speed by 30 meters per minute. What will be the speed of the train in five minutes? Give your answer in km/h.
10. It is known that in an arithmetic progression a 2 = 5; a 6 = -5. Find a 1.
Answers (in disarray): 7.7; 7.5; 9.5; 9; 0.3; four.
Everything worked out? Wonderful! You can master the arithmetic progression for more high level, in the next lessons.
Didn't everything work out? No problem. In Special Section 555, all these puzzles are broken down piece by piece.) And, of course, a simple practical technique is described that immediately highlights the solution of such tasks clearly, clearly, as in the palm of your hand!
By the way, in the puzzle about the train there are two problems on which people often stumble. One - purely by progression, and the second - common to any tasks in mathematics, and physics too. This is a translation of dimensions from one to another. It shows how these problems should be solved.
In this lesson, we examined the elementary meaning of an arithmetic progression and its main parameters. This is enough to solve almost all problems on this topic. Add d to the numbers, write a series, everything will be decided.
The finger solution works well for very short pieces of the series, as in the examples in this lesson. If the series is longer, the calculations become more complicated. For example, if in problem 9 in the question, replace "five minutes" on the "thirty-five minutes" the problem will become much worse.)
And there are also tasks that are simple in essence, but utterly absurd in terms of calculations, for example:
Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.
And what, we will add 1/6 many, many times?! Is it possible to kill yourself!?
You can.) If you do not know a simple formula by which you can solve such tasks in a minute. This formula will be in the next lesson. And that problem is solved there. In a minute.)
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By the way, I have a couple more interesting sites for you.)
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you can get acquainted with functions and derivatives.
First level
Arithmetic progression. Detailed theory with examples (2019)
Numeric sequence
So let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many as you like (in our case, them). No matter how many numbers we write, we can always say which of them is the first, which is the second, and so on to the last, that is, we can number them. This is an example of a number sequence:
Numeric sequence
For example, for our sequence:
The assigned number is specific to only one sequence number. In other words, there are no three second numbers in the sequence. The second number (like the -th number) is always the same.
The number with the number is called the -th member of the sequence.
We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .
In our case: 
Let's say we have a numerical sequence in which the difference between adjacent numbers is the same and equal.
For example: 
etc.
Such a numerical sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius as early as the 6th century and was understood in a more broad sense, as an infinite number sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which the ancient Greeks were engaged in.
This is a numerical sequence, each member of which is equal to the previous one, added with the same number. This number is called the difference of an arithmetic progression and is denoted.
Try to determine which number sequences are an arithmetic progression and which are not:
a)
b)
c)
d)
Got it? Compare our answers:
Is arithmetic progression - b, c.
Is not arithmetic progression - a, d.
Let's return to the given progression () and try to find the value of its th member. Exists two way to find it.
1. Method
We can add to the previous value of the progression number until we reach the th term of the progression. It’s good that we don’t have much to summarize - only three values:
So, the -th member of the described arithmetic progression is equal to.
2. Method
What if we needed to find the value of the th term of the progression? The summation would have taken us more than one hour, and it is not a fact that we would not have made mistakes when adding the numbers.
Of course, mathematicians have come up with a way in which you do not need to add the difference of an arithmetic progression to the previous value. Look closely at the drawn picture ... Surely you have already noticed a certain pattern, namely:
For example, let's see what makes up the value of the -th member of this arithmetic progression: 
In other words:
Try to independently find in this way the value of a member of this arithmetic progression.
Calculated? Compare your entries with the answer:
Pay attention that you got exactly the same number as in the previous method, when we successively added the members of an arithmetic progression to the previous value.
Let's try to "depersonalize" this formula - let's bring it into general form and get:
|
Arithmetic progression equation. |
Arithmetic progressions are either increasing or decreasing.
Increasing- progressions in which each subsequent value of the terms is greater than the previous one.
For example:
Descending- progressions in which each subsequent value of the terms is less than the previous one.
For example:
The derived formula is used in the calculation of terms in both increasing and decreasing terms of an arithmetic progression.
Let's check it out in practice.
We are given an arithmetic progression consisting of the following numbers:
Since then:
Thus, we were convinced that the formula works both in decreasing and in increasing arithmetic progression.
Try to find the -th and -th members of this arithmetic progression on your own.
Let's compare the results:
Arithmetic progression property
Let's complicate the task - we derive the property of an arithmetic progression.
Suppose we are given the following condition:
- arithmetic progression, find the value.
It's easy, you say, and start counting according to the formula you already know:
Let, a, then:
Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but what if we are given numbers in the condition? Agree, there is a possibility of making mistakes in the calculations.
Now think, is it possible to solve this problem in one step using any formula? Of course, yes, and we will try to bring it out now.
Let's denote the desired term of the arithmetic progression as, we know the formula for finding it - this is the same formula that we derived at the beginning:
, then:
- the previous member of the progression is:
- the next term of the progression is:
Let's sum the previous and next members of the progression:
It turns out that the sum of the previous and subsequent members of the progression is twice the value of the member of the progression located between them. In other words, in order to find the value of a progression member with known previous and successive values, it is necessary to add them and divide by.
That's right, we got the same number. Let's fix the material. Calculate the value for the progression yourself, because it is not difficult at all.
Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, one of the greatest mathematicians of all time, the "king of mathematicians" - Karl Gauss, easily deduced for himself ...
When Carl Gauss was 9 years old, the teacher, busy checking the work of students in other classes, asked the following task at the lesson: “Calculate the sum of all natural numbers from to (according to other sources up to) inclusive. What was the surprise of the teacher when one of his students (it was Karl Gauss) after a minute gave the correct answer to the task, while most of the classmates of the daredevil after long calculations received the wrong result ...
Young Carl Gauss noticed a pattern that you can easily notice.
Let's say we have an arithmetic progression consisting of -ti members: We need to find the sum of the given members of the arithmetic progression. Of course, we can manually sum all the values, but what if we need to find the sum of its terms in the task, as Gauss was looking for?
Let's depict the progression given to us. Look closely at the highlighted numbers and try to perform various mathematical operations with them. 
Tried? What did you notice? Correctly! Their sums are equal
Now answer, how many such pairs will there be in the progression given to us? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two members of an arithmetic progression is equal, and similar equal pairs, we get that total amount is equal to:
.
Thus, the formula for the sum of the first terms of any arithmetic progression will be:
In some problems, we do not know the th term, but we know the progression difference. Try to substitute in the sum formula, the formula of the th member.
What did you get?
Well done! Now let's return to the problem that was given to Carl Gauss: calculate for yourself what the sum of numbers starting from the -th is, and the sum of the numbers starting from the -th.
How much did you get?
Gauss turned out that the sum of the terms is equal, and the sum of the terms. Is that how you decided?
In fact, the formula for the sum of members of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people used the properties of an arithmetic progression with might and main.
For example, imagine Ancient Egypt and the most large-scale construction of that time - the construction of the pyramid ... The figure shows one side of it. 
Where is the progression here you say? Look carefully and find a pattern in the number of sand blocks in each row of the pyramid wall. 
Why not an arithmetic progression? Count how many blocks are needed to build one wall if block bricks are placed in the base. I hope you will not count by moving your finger across the monitor, do you remember the last formula and everything we said about arithmetic progression?
AT this case the progression looks like this:
Arithmetic progression difference.
The number of members of an arithmetic progression.
Let's substitute our data into the last formulas (we count the number of blocks in 2 ways).
Method 1.
Method 2.
And now you can also calculate on the monitor: compare the obtained values with the number of blocks that are in our pyramid. Did it agree? Well done, you have mastered the sum of the th terms of an arithmetic progression.
Of course, you can’t build a pyramid from the blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:
Workout
Tasks:
- Masha is getting in shape for the summer. Every day she increases the number of squats by. How many times will Masha squat in weeks if she did squats at the first workout.
- What is the sum of all odd numbers contained in.
- When storing logs, lumberjacks stack them in such a way that each upper layer contains one log less than the previous one. How many logs are in one masonry, if the base of the masonry is logs.
Answers:
- Let us define the parameters of the arithmetic progression. In this case
(weeks = days).Answer: In two weeks, Masha should squat once a day.
- First odd number, last number.
Arithmetic progression difference.
The number of odd numbers in - half, however, check this fact using the formula for finding the -th member of an arithmetic progression:The numbers do contain odd numbers.
We substitute the available data into the formula:Answer: The sum of all odd numbers contained in is equal to.
- Recall the problem about the pyramids. For our case, a , since each top layer is reduced by one log, there are only a bunch of layers, that is.
Substitute the data in the formula:Answer: There are logs in the masonry.
Summing up
- - a numerical sequence in which the difference between adjacent numbers is the same and equal. It is increasing and decreasing.
- Finding formula th member of an arithmetic progression is written by the formula - , where is the number of numbers in the progression.
- Property of members of an arithmetic progression- - where - the number of numbers in the progression.
- The sum of the members of an arithmetic progression can be found in two ways:
, where is the number of values.
ARITHMETIC PROGRESSION. AVERAGE LEVEL
Numeric sequence
Let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many as you like. But you can always tell which of them is the first, which is the second, and so on, that is, we can number them. This is an example of a number sequence.
Numeric sequence is a set of numbers, each of which can be assigned a unique number.
In other words, each number can be associated with a certain natural number, and only one. And we will not assign this number to any other number from this set.
The number with the number is called the -th member of the sequence.
We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .
It is very convenient if the -th member of the sequence can be given by some formula. For example, the formula
sets the sequence:
And the formula is the following sequence:
For example, an arithmetic progression is a sequence (the first term here is equal, and the difference). Or (, difference).
nth term formula
We call recurrent a formula in which, in order to find out the -th term, you need to know the previous or several previous ones:
To find, for example, the th term of the progression using such a formula, we have to calculate the previous nine. For example, let. Then:
Well, now it's clear what the formula is?
In each line, we add to, multiplied by some number. For what? Very simple: this is the number of the current member minus:
Much more comfortable now, right? We check:
Decide for yourself:
In an arithmetic progression, find the formula for the nth term and find the hundredth term.
Solution:
The first term is equal. And what is the difference? And here's what:
(after all, it is called the difference because it is equal to the difference of successive members of the progression).
So the formula is:
Then the hundredth term is:
What is the sum of all natural numbers from to?
According to legend, the great mathematician Carl Gauss, being a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last number is equal, the sum of the second and penultimate is the same, the sum of the third and the 3rd from the end is the same, and so on. How many such pairs are there? That's right, exactly half the number of all numbers, that is. So,
The general formula for the sum of the first terms of any arithmetic progression will be:
Example:
Find the sum of all two-digit numbers, multiples.
Solution:
The first such number is this. Each next is obtained by adding a number to the previous one. Thus, the numbers of interest to us form an arithmetic progression with the first term and the difference.
The formula for the th term for this progression is:
How many terms are in the progression if they must all be two digits?
Very easy: .
The last term of the progression will be equal. Then the sum:
Answer: .
Now decide for yourself:
- Every day the athlete runs 1m more than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
- A cyclist rides more miles each day than the previous one. On the first day he traveled km. How many days does he have to drive to cover a kilometer? How many kilometers will he travel on the last day of the journey?
- The price of a refrigerator in the store is reduced by the same amount every year. Determine how much the price of a refrigerator decreased every year if, put up for sale for rubles, six years later it was sold for rubles.
Answers:
- The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
.
Answer: - Here it is given:, it is necessary to find.
Obviously, you need to use the same sum formula as in the previous problem:
.
Substitute the values:The root obviously doesn't fit, so the answer.
Let's calculate the distance traveled over the last day using the formula of the -th member:
(km).
Answer: - Given: . Find: .
It doesn't get easier:
(rub).
Answer:
ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN
This is a numerical sequence in which the difference between adjacent numbers is the same and equal.
Arithmetic progression is increasing () and decreasing ().
For example:
The formula for finding the n-th member of an arithmetic progression
is written as a formula, where is the number of numbers in the progression.
Property of members of an arithmetic progression
It makes it easy to find a member of the progression if its neighboring members are known - where is the number of numbers in the progression.
The sum of the members of an arithmetic progression
There are two ways to find the sum:
Where is the number of values.
Where is the number of values.
Instruction
An arithmetic progression is a sequence of the form a1, a1+d, a1+2d..., a1+(n-1)d. Number d step progressions.Obviously, the total of an arbitrary nth term of the arithmetic progressions has the form: An = A1+(n-1)d. Then knowing one of the members progressions, member progressions and step progressions, can be , that is, the number of the progression term. Obviously, it will be determined by the formula n = (An-A1+d)/d.
Let the mth term be known now progressions and some other member progressions- n-th, but n , as in the previous case, but it is known that n and m do not match.Step progressions can be calculated by the formula: d = (An-Am)/(n-m). Then n = (An-Am+md)/d.
If the sum of several elements of an arithmetic progressions, as well as its first and last , then the number of these elements can also be determined. The sum of the arithmetic progressions will be equal to: S = ((A1+An)/2)n. Then n = 2S/(A1+An) are chdenov progressions. Using the fact that An = A1+(n-1)d, this formula can be rewritten as: n = 2S/(2A1+(n-1)d). From this one can express n by solving quadratic equation.
An arithmetic sequence is such an ordered set of numbers, each member of which, except for the first, differs from the previous one by the same amount. This constant is called the difference of the progression or its step and can be calculated from the known members of the arithmetic progression.
Instruction
If the values of the first and second or any other pair of neighboring terms are known from the conditions of the problem, to calculate the difference (d), simply subtract the previous term from the next term. The resulting value can be either positive or negative number- it depends on whether the progression is increasing. AT general form write the solution for an arbitrary pair (aᵢ and aᵢ₊₁) of neighboring members of the progression as follows: d = aᵢ₊₁ - aᵢ.
For a pair of members of such a progression, one of which is the first (a₁), and the other is any other arbitrarily chosen one, one can also make a formula for finding the difference (d). However, in this case, the serial number (i) of an arbitrary chosen member of the sequence must be known. To calculate the difference, add both numbers, and divide the result by the ordinal number of an arbitrary term reduced by one. In general, write this formula as follows: d = (a₁+ aᵢ)/(i-1).
If, in addition to an arbitrary member of the arithmetic progression with ordinal number i, another member with ordinal number u is known, change the formula from the previous step accordingly. In this case, the difference (d) of the progression will be the sum of these two terms divided by their difference serial numbers: d = (aᵢ+aᵥ)/(i-v).
The formula for calculating the difference (d) becomes somewhat more complicated if the value of its first member (a₁) and the sum (Sᵢ) of a given number (i) of the first members of the arithmetic sequence are given in the conditions of the problem. To get the desired value, divide the sum by the number of terms that made it up, subtract the value of the first number in the sequence, and double the result. Divide the resulting value by the number of terms that made up the sum reduced by one. In general, write down the formula for calculating the discriminant as follows: d = 2*(Sᵢ/i-a₁)/(i-1).
First level
Arithmetic progression. Detailed theory with examples (2019)
Numeric sequence
So let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many as you like (in our case, them). No matter how many numbers we write, we can always say which of them is the first, which is the second, and so on to the last, that is, we can number them. This is an example of a number sequence:
Numeric sequence
For example, for our sequence:
The assigned number is specific to only one sequence number. In other words, there are no three second numbers in the sequence. The second number (like the -th number) is always the same.
The number with the number is called the -th member of the sequence.
We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .
In our case:
Let's say we have a numerical sequence in which the difference between adjacent numbers is the same and equal.
For example:
etc.
Such a numerical sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius as early as the 6th century and was understood in a broader sense as an endless numerical sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which the ancient Greeks were engaged in.
This is a numerical sequence, each member of which is equal to the previous one, added with the same number. This number is called the difference of an arithmetic progression and is denoted.
Try to determine which number sequences are an arithmetic progression and which are not:
a)
b)
c)
d)
Got it? Compare our answers:
Is arithmetic progression - b, c.
Is not arithmetic progression - a, d.
Let's return to the given progression () and try to find the value of its th member. Exists two way to find it.
1. Method
We can add to the previous value of the progression number until we reach the th term of the progression. It’s good that we don’t have much to summarize - only three values:
So, the -th member of the described arithmetic progression is equal to.
2. Method
What if we needed to find the value of the th term of the progression? The summation would have taken us more than one hour, and it is not a fact that we would not have made mistakes when adding the numbers.
Of course, mathematicians have come up with a way in which you do not need to add the difference of an arithmetic progression to the previous value. Look closely at the drawn picture ... Surely you have already noticed a certain pattern, namely:
For example, let's see what makes up the value of the -th member of this arithmetic progression:
In other words:
Try to independently find in this way the value of a member of this arithmetic progression.
Calculated? Compare your entries with the answer:
Pay attention that you got exactly the same number as in the previous method, when we successively added the members of an arithmetic progression to the previous value.
Let's try to "depersonalize" this formula - we bring it into a general form and get:
|
Arithmetic progression equation. |
Arithmetic progressions are either increasing or decreasing.
Increasing- progressions in which each subsequent value of the terms is greater than the previous one.
For example:
Descending- progressions in which each subsequent value of the terms is less than the previous one.
For example:
The derived formula is used in the calculation of terms in both increasing and decreasing terms of an arithmetic progression.
Let's check it out in practice.
We are given an arithmetic progression consisting of the following numbers:
Since then:
Thus, we were convinced that the formula works both in decreasing and in increasing arithmetic progression.
Try to find the -th and -th members of this arithmetic progression on your own.
Let's compare the results:
Arithmetic progression property
Let's complicate the task - we derive the property of an arithmetic progression.
Suppose we are given the following condition:
- arithmetic progression, find the value.
It's easy, you say, and start counting according to the formula you already know:
Let, a, then:
Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but what if we are given numbers in the condition? Agree, there is a possibility of making mistakes in the calculations.
Now think, is it possible to solve this problem in one step using any formula? Of course, yes, and we will try to bring it out now.
Let's denote the desired term of the arithmetic progression as, we know the formula for finding it - this is the same formula that we derived at the beginning:
, then:
- the previous member of the progression is:
- the next term of the progression is:
Let's sum the previous and next members of the progression:
It turns out that the sum of the previous and subsequent members of the progression is twice the value of the member of the progression located between them. In other words, in order to find the value of a progression member with known previous and successive values, it is necessary to add them and divide by.
That's right, we got the same number. Let's fix the material. Calculate the value for the progression yourself, because it is not difficult at all.
Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, one of the greatest mathematicians of all time, the "king of mathematicians" - Karl Gauss, easily deduced for himself ...
When Carl Gauss was 9 years old, the teacher, busy checking the work of students from other classes, asked the following task at the lesson: "Calculate the sum of all natural numbers from up to (according to other sources up to) inclusive." What was the surprise of the teacher when one of his students (it was Karl Gauss) after a minute gave the correct answer to the task, while most of the classmates of the daredevil after long calculations received the wrong result ...
Young Carl Gauss noticed a pattern that you can easily notice.
Let's say we have an arithmetic progression consisting of -ti members: We need to find the sum of the given members of the arithmetic progression. Of course, we can manually sum all the values, but what if we need to find the sum of its terms in the task, as Gauss was looking for?
Let's depict the progression given to us. Look closely at the highlighted numbers and try to perform various mathematical operations with them.
Tried? What did you notice? Correctly! Their sums are equal
Now answer, how many such pairs will there be in the progression given to us? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two terms of an arithmetic progression is equal, and similar equal pairs, we get that the total sum is equal to:
.
Thus, the formula for the sum of the first terms of any arithmetic progression will be:
In some problems, we do not know the th term, but we know the progression difference. Try to substitute in the sum formula, the formula of the th member.
What did you get?
Well done! Now let's return to the problem that was given to Carl Gauss: calculate for yourself what the sum of numbers starting from the -th is, and the sum of the numbers starting from the -th.
How much did you get?
Gauss turned out that the sum of the terms is equal, and the sum of the terms. Is that how you decided?
In fact, the formula for the sum of members of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people used the properties of an arithmetic progression with might and main.
For example, imagine Ancient Egypt and the largest construction site of that time - the construction of a pyramid ... The figure shows one side of it.
Where is the progression here you say? Look carefully and find a pattern in the number of sand blocks in each row of the pyramid wall.
Why not an arithmetic progression? Count how many blocks are needed to build one wall if block bricks are placed in the base. I hope you will not count by moving your finger across the monitor, do you remember the last formula and everything we said about arithmetic progression?
In this case, the progression looks like this:
Arithmetic progression difference.
The number of members of an arithmetic progression.
Let's substitute our data into the last formulas (we count the number of blocks in 2 ways).
Method 1.
Method 2.
And now you can also calculate on the monitor: compare the obtained values with the number of blocks that are in our pyramid. Did it agree? Well done, you have mastered the sum of the th terms of an arithmetic progression.
Of course, you can’t build a pyramid from the blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:
Workout
Tasks:
- Masha is getting in shape for the summer. Every day she increases the number of squats by. How many times will Masha squat in weeks if she did squats at the first workout.
- What is the sum of all odd numbers contained in.
- When storing logs, lumberjacks stack them in such a way that each top layer contains one less log than the previous one. How many logs are in one masonry, if the base of the masonry is logs.
Answers:
- Let us define the parameters of the arithmetic progression. In this case
(weeks = days).Answer: In two weeks, Masha should squat once a day.
- First odd number, last number.
Arithmetic progression difference.
The number of odd numbers in - half, however, check this fact using the formula for finding the -th member of an arithmetic progression:The numbers do contain odd numbers.
We substitute the available data into the formula:Answer: The sum of all odd numbers contained in is equal to.
- Recall the problem about the pyramids. For our case, a , since each top layer is reduced by one log, there are only a bunch of layers, that is.
Substitute the data in the formula:Answer: There are logs in the masonry.
Summing up
- - a numerical sequence in which the difference between adjacent numbers is the same and equal. It is increasing and decreasing.
- Finding formula th member of an arithmetic progression is written by the formula - , where is the number of numbers in the progression.
- Property of members of an arithmetic progression- - where - the number of numbers in the progression.
- The sum of the members of an arithmetic progression can be found in two ways:
, where is the number of values.
ARITHMETIC PROGRESSION. AVERAGE LEVEL
Numeric sequence
Let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many as you like. But you can always tell which of them is the first, which is the second, and so on, that is, we can number them. This is an example of a number sequence.
Numeric sequence is a set of numbers, each of which can be assigned a unique number.
In other words, each number can be associated with a certain natural number, and only one. And we will not assign this number to any other number from this set.
The number with the number is called the -th member of the sequence.
We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .
It is very convenient if the -th member of the sequence can be given by some formula. For example, the formula
sets the sequence:
And the formula is the following sequence:
For example, an arithmetic progression is a sequence (the first term here is equal, and the difference). Or (, difference).
nth term formula
We call recurrent a formula in which, in order to find out the -th term, you need to know the previous or several previous ones:
To find, for example, the th term of the progression using such a formula, we have to calculate the previous nine. For example, let. Then:
Well, now it's clear what the formula is?
In each line, we add to, multiplied by some number. For what? Very simple: this is the number of the current member minus:
Much more comfortable now, right? We check:
Decide for yourself:
In an arithmetic progression, find the formula for the nth term and find the hundredth term.
Solution:
The first term is equal. And what is the difference? And here's what:
(after all, it is called the difference because it is equal to the difference of successive members of the progression).
So the formula is:
Then the hundredth term is:
What is the sum of all natural numbers from to?
According to legend, the great mathematician Carl Gauss, being a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last number is equal, the sum of the second and penultimate is the same, the sum of the third and the 3rd from the end is the same, and so on. How many such pairs are there? That's right, exactly half the number of all numbers, that is. So,
The general formula for the sum of the first terms of any arithmetic progression will be:
Example:
Find the sum of all two-digit multiples.
Solution:
The first such number is this. Each next is obtained by adding a number to the previous one. Thus, the numbers of interest to us form an arithmetic progression with the first term and the difference.
The formula for the th term for this progression is:
How many terms are in the progression if they must all be two digits?
Very easy: .
The last term of the progression will be equal. Then the sum:
Answer: .
Now decide for yourself:
- Every day the athlete runs 1m more than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
- A cyclist rides more miles each day than the previous one. On the first day he traveled km. How many days does he have to drive to cover a kilometer? How many kilometers will he travel on the last day of the journey?
- The price of a refrigerator in the store is reduced by the same amount every year. Determine how much the price of a refrigerator decreased every year if, put up for sale for rubles, six years later it was sold for rubles.
Answers:
- The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
.
Answer: - Here it is given:, it is necessary to find.
Obviously, you need to use the same sum formula as in the previous problem:
.
Substitute the values:The root obviously doesn't fit, so the answer.
Let's calculate the distance traveled over the last day using the formula of the -th member:
(km).
Answer: - Given: . Find: .
It doesn't get easier:
(rub).
Answer:
ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN
This is a numerical sequence in which the difference between adjacent numbers is the same and equal.
Arithmetic progression is increasing () and decreasing ().
For example:
The formula for finding the n-th member of an arithmetic progression
is written as a formula, where is the number of numbers in the progression.
Property of members of an arithmetic progression
It makes it easy to find a member of the progression if its neighboring members are known - where is the number of numbers in the progression.
The sum of the members of an arithmetic progression
There are two ways to find the sum:
Where is the number of values.
Where is the number of values.
The sum of an arithmetic progression.
The sum of an arithmetic progression is a simple thing. Both in meaning and in formula. But there are all sorts of tasks on this topic. From elementary to quite solid.
First, let's deal with the meaning and formula of the sum. And then we'll decide. For your own pleasure.) The meaning of the sum is as simple as lowing. To find the sum of an arithmetic progression, you just need to carefully add all its members. If these terms are few, you can add without any formulas. But if there is a lot, or a lot ... addition is annoying.) In this case, the formula saves.
The sum formula is simple:
Let's figure out what kind of letters are included in the formula. This will clear up a lot.
S n is the sum of an arithmetic progression. Addition result all members, with first on last. It is important. Add up exactly all members in a row, without gaps and jumps. And, exactly, starting from first. In puzzles, such as finding the sum of the third and eighth terms, or the sum of the terms from the fifth to the twentieth - direct application formulas are disappointing.)
a 1 - the first member of the progression. Everything is clear here, it's simple first row number.
a n- last member of the progression. The last number of the row. Not a very familiar name, but, when applied to the amount, it is very suitable. Then you will see for yourself.
n is the number of the last member. It is important to understand that in the formula this number coincides with the number of added terms.
Let's define the concept last member a n. Filling question: what kind of member will last, if given endless arithmetic progression?
For a confident answer, you need to understand the elementary meaning of an arithmetic progression and ... read the assignment carefully!)
In the task of finding the sum of an arithmetic progression, the last term always appears (directly or indirectly), which should be limited. Otherwise, a finite, specific amount just doesn't exist. For the solution, it does not matter what kind of progression is given: finite or infinite. It doesn't matter how it is given: by a series of numbers, or by the formula of the nth member.
The most important thing is to understand that the formula works from the first term of the progression to the term with the number n. Actually, the full name of the formula looks like this: the sum of the first n terms of an arithmetic progression. The number of these very first members, i.e. n, is determined solely by the task. In the task, all this valuable information is often encrypted, yes ... But nothing, in the examples below we will reveal these secrets.)
Examples of tasks for the sum of an arithmetic progression.
Primarily, useful information:
The main difficulty in tasks for the sum of an arithmetic progression is the correct determination of the elements of the formula.
The authors of the assignments encrypt these very elements with boundless imagination.) The main thing here is not to be afraid. Understanding the essence of the elements, it is enough just to decipher them. Let's take a look at a few examples in detail. Let's start with a task based on a real GIA.
1. The arithmetic progression is given by the condition: a n = 2n-3.5. Find the sum of the first 10 terms.
Good job. Easy.) To determine the amount according to the formula, what do we need to know? First Member a 1, last term a n, yes the number of the last term n.
Where to get the last member number n? Yes, in the same place, in the condition! It says find the sum first 10 members. Well, what number will it be last, tenth member?) You won’t believe it, his number is tenth!) Therefore, instead of a n we will substitute into the formula a 10, but instead n- ten. Again, the number of the last member is the same as the number of members.
It remains to be determined a 1 and a 10. This is easily calculated by the formula of the nth term, which is given in the problem statement. Don't know how to do it? Visit the previous lesson, without this - nothing.
a 1= 2 1 - 3.5 = -1.5
a 10\u003d 2 10 - 3.5 \u003d 16.5
S n = S 10.
We found out the meaning of all elements of the formula for the sum of an arithmetic progression. It remains to substitute them, and count:
That's all there is to it. Answer: 75.
Another task based on the GIA. A little more complicated:
2. Given an arithmetic progression (a n), the difference of which is 3.7; a 1 \u003d 2.3. Find the sum of the first 15 terms.
We immediately write the sum formula:
This formula allows us to find the value of any member by its number. We are looking for a simple substitution:
a 15 \u003d 2.3 + (15-1) 3.7 \u003d 54.1
It remains to substitute all the elements in the formula for the sum of an arithmetic progression and calculate the answer:
Answer: 423.
By the way, if in the sum formula instead of a n just substitute the formula of the nth term, we get:
We give similar ones, we get a new formula for the sum of members of an arithmetic progression:
 |
As you can see, the nth term is not required here. a n. In some tasks, this formula helps out a lot, yes ... You can remember this formula. And you can simply withdraw it at the right time, as here. After all, the formula for the sum and the formula for the nth term must be remembered in every way.)
Now the task in the form of a short encryption):
3. Find the sum of all positive two-digit numbers that are multiples of three.
How! No first member, no last, no progression at all... How to live!?
You will have to think with your head and pull out from the condition all the elements of the sum of an arithmetic progression. What are two-digit numbers - we know. They consist of two numbers.) What two-digit number will first? 10, presumably.) last thing two digit number? 99, of course! The three-digit ones will follow him ...
Multiples of three... Hm... These are numbers that are evenly divisible by three, here! Ten is not divisible by three, 11 is not divisible... 12... is divisible! So, something is emerging. You can already write a series according to the condition of the problem:
12, 15, 18, 21, ... 96, 99.
Will this series be an arithmetic progression? Of course! Each term differs from the previous one strictly by three. If 2, or 4, is added to the term, say, the result, i.e. a new number will no longer be divided by 3. You can immediately determine the difference of the arithmetic progression to the heap: d = 3. Useful!)
So, we can safely write down some progression parameters:
What will be the number n last member? Anyone who thinks that 99 is fatally mistaken ... Numbers - they always go in a row, and our members jump over the top three. They don't match.
There are two solutions here. One way is for the super hardworking. You can paint the progression, the whole series of numbers, and count the number of terms with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If the formula is applied to our problem, we get that 99 is the thirtieth member of the progression. Those. n = 30.
We look at the formula for the sum of an arithmetic progression:
We look and rejoice.) We pulled out everything necessary for calculating the amount from the condition of the problem:
a 1= 12.
a 30= 99.
S n = S 30.
What remains is elementary arithmetic. Substitute the numbers in the formula and calculate:
Answer: 1665
Another type of popular puzzles:
4. An arithmetic progression is given:
-21,5; -20; -18,5; -17; ...
Find the sum of terms from the twentieth to thirty-fourth.
We look at the sum formula and ... we are upset.) The formula, let me remind you, calculates the sum from the first member. And in the problem you need to calculate the sum since the twentieth... The formula won't work.
You can, of course, paint the entire progression in a row, and put the members from 20 to 34. But ... somehow it turns out stupidly and for a long time, right?)
There is a more elegant solution. Let's break our series into two parts. The first part will from the first term to the nineteenth. The second part - twenty to thirty-four. It is clear that if we calculate the sum of the terms of the first part S 1-19, let's add it to the sum of the members of the second part S 20-34, we get the sum of the progression from the first term to the thirty-fourth S 1-34. Like this:
S 1-19 + S 20-34 = S 1-34
This shows that to find the sum S 20-34 can be done by simple subtraction
S 20-34 = S 1-34 - S 1-19
Both sums on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Are we getting started?
We extract the progression parameters from the task condition:
d = 1.5.
a 1= -21,5.
To calculate the sums of the first 19 and the first 34 terms, we will need the 19th and 34th terms. We count them according to the formula of the nth term, as in problem 2:
a 19\u003d -21.5 + (19-1) 1.5 \u003d 5.5
a 34\u003d -21.5 + (34-1) 1.5 \u003d 28
There is nothing left. Subtract the sum of 19 terms from the sum of 34 terms:
S 20-34 = S 1-34 - S 1-19 = 110.5 - (-152) = 262.5
Answer: 262.5
One important note! In solving this problem, there is useful feature. Instead of direct calculation what you need (S 20-34), we counted what, it would seem, is not needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the full result. Such a "feint with the ears" often saves in evil puzzles.)
In this lesson, we examined problems for which it is enough to understand the meaning of the sum of an arithmetic progression. Well, you need to know a couple of formulas.)
When solving any problem for the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.
Formula of the nth term:
These formulas will immediately tell you what to look for, in which direction to think in order to solve the problem. Helps.
And now the tasks for independent solution.
5. Find the sum of all two-digit numbers that are not divisible by three.
Cool?) The hint is hidden in the note to problem 4. Well, problem 3 will help.
6. Arithmetic progression is given by the condition: a 1 =-5.5; a n+1 = a n +0.5. Find the sum of the first 24 terms.
Unusual?) This is a recurrent formula. You can read about it in the previous lesson. Do not ignore the link, such puzzles are often found in the GIA.
7. Vasya saved up money for the Holiday. As much as 4550 rubles! And I decided to give the most beloved person (myself) a few days of happiness). Live beautifully without denying yourself anything. Spend 500 rubles on the first day, and spend 50 rubles more on each subsequent day than on the previous one! Until the money runs out. How many days of happiness did Vasya have?
Is it difficult?) An additional formula from task 2 will help.
Answers (in disarray): 7, 3240, 6.
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By the way, I have a couple more interesting sites for you.)
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