Arithmetic progression name a sequence of numbers (terms of a progression)
In which each subsequent term differs from the previous one by a new term, which is also called step or progression difference.
Thus, by specifying the progression step and its first term, you can find any of its elements using the formula
Properties of an arithmetic progression
1) Each member of an arithmetic progression, starting from the second number, is the arithmetic mean of the previous and next members of the progression
The converse is also true. If the arithmetic mean of adjacent odd (even) terms of a progression is equal to the term that stands between them, then this sequence of numbers is an arithmetic progression. Using this statement, it is very easy to check any sequence.
Also, by the property of arithmetic progression, the above formula can be generalized to the following
This is easy to verify if you write the terms to the right of the equal sign
It is often used in practice to simplify calculations in problems.
2) The sum of the first n terms of an arithmetic progression is calculated using the formula
Remember well the formula for the sum of an arithmetic progression; it is indispensable in calculations and is quite often found in simple life situations.
3) If you need to find not the whole sum, but part of the sequence starting from its kth term, then the following sum formula will be useful to you
4) Of practical interest is finding the sum of n terms of an arithmetic progression starting from the kth number. To do this, use the formula
On this theoretical material ends and we move on to solving common problems in practice.
Example 1. Find the fortieth term of the arithmetic progression 4;7;...
Solution:
According to the condition we have
Let's determine the progression step
Using a well-known formula, we find the fortieth term of the progression
Example 2. An arithmetic progression is given by its third and seventh terms. Find the first term of the progression and the sum of ten.
Solution:
Let us write down the given elements of the progression using the formulas
We subtract the first from the second equation, as a result we find the progression step
We substitute the found value into any of the equations to find the first term of the arithmetic progression
We calculate the sum of the first ten terms of the progression
Without using complex calculations, we found all the required quantities.
Example 3. An arithmetic progression is given by the denominator and one of its terms. Find the first term of the progression, the sum of its 50 terms starting from 50 and the sum of the first 100.
Solution:
Let's write down the formula for the hundredth element of the progression
and find the first one
Based on the first, we find the 50th term of the progression
Finding the sum of the part of the progression
and the sum of the first 100
The progression amount is 250.
Example 4.
Find the number of terms of an arithmetic progression if:
a3-a1=8, a2+a4=14, Sn=111.
Solution:
Let's write the equations in terms of the first term and the progression step and determine them
We substitute the obtained values into the sum formula to determine the number of terms in the sum
We carry out simplifications
and solve the quadratic equation
Of the two values found, only the number 8 fits the problem conditions. Thus, the sum of the first eight terms of the progression is 111.
Example 5.
Solve the equation
1+3+5+...+x=307.
Solution: This equation is the sum of an arithmetic progression. Let's write out its first term and find the difference in progression
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 basic to quite solid.
First, let's understand the meaning and formula of the amount. And then we'll decide. For your own pleasure.) The meaning of the amount is as simple as a moo. To find the sum of an arithmetic progression, you just need to carefully add all its terms. 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 comes to the rescue.
The formula for the amount is simple:
Let's figure out what kind of letters are included in the formula. This will clear things up a lot.
S n - the sum of an arithmetic progression. Addition result everyone members, with first By last. It is important. They add up exactly All members in a row, without skipping or skipping. And, precisely, starting from first. In problems like finding the sum of the third and eighth terms, or the sum of the fifth to twentieth terms, direct application of the formula will disappoint.)
a 1 - 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 series. Not a very familiar name, but when applied to the amount, it’s very suitable. Then you will see for yourself.
n - 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. Tricky question: which member will be the last one if given endless arithmetic progression?)
To answer confidently, you need to understand the elementary meaning of arithmetic progression and... read the task 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 final, specific amount simply doesn't exist. For the solution, it does not matter whether the progression is given: finite or infinite. It doesn’t matter how it is given: a series of numbers, or a formula for the nth term.
The most important thing is to understand that the formula works from the first term of the progression to the term with 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 a task, all this valuable information is often encrypted, yes... But never mind, in the examples below we reveal these secrets.)
Examples of tasks on the sum of an arithmetic progression.
First of all, helpful information:
The main difficulty in tasks involving the sum of an arithmetic progression lies in the correct determination of the elements of the formula.
The task writers encrypt these same elements with boundless imagination.) The main thing here is not to be afraid. Understanding the essence of the elements, it is enough to simply decipher them. Let's 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 its first 10 terms.
Good job. Easy.) To determine the amount using the formula, what do we need to know? First member a 1, last term a n, yes the number of the last member n.
Where can I get the last member's number? n? Yes, right there, on condition! It says: find the sum first 10 members. Well, what number will it be with? 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, and instead n- ten. I repeat, the number of the last member coincides with the number of members.
It remains to determine a 1 And a 10. This is easily calculated using the formula for the nth term, which is given in the problem statement. Don't know how to do this? Attend the previous lesson, without this there is no way.
a 1= 2 1 - 3.5 = -1.5
a 10=2·10 - 3.5 =16.5
S n = S 10.
We have found out the meaning of all elements of the formula for the sum of an arithmetic progression. All that remains is to substitute them and count:
That's 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 =2.3. Find the sum of its first 15 terms.
We immediately write the sum formula:
This formula allows us to find the value of any term by its number. We look for a simple substitution:
a 15 = 2.3 + (15-1) 3.7 = 54.1
It remains to substitute all the elements into 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 We simply substitute the formula for the nth term and get:
Let us present similar ones and obtain a new formula for the sum of terms of an arithmetic progression:
 |
As you can see, it is not required here nth term a n. In some problems this formula helps a lot, yes... You can remember this formula. Or you can simply display it at the right time, like here. After all, you always need to remember the formula for the sum and the formula for the nth term.)
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.
Wow! Neither your first member, nor your last, nor progression at all... How to live!?
You will have to think with your head and pull out all the elements of the sum of the arithmetic progression from the condition. We know what two-digit numbers are. They consist of two numbers.) What two-digit number will be first? 10, presumably.) A last thing double digit number? 99, of course! The three-digit ones will follow him...
Multiples of three... Hm... These are numbers that are 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 down a series according to the conditions of the problem:
12, 15, 18, 21, ... 96, 99.
Will this series be an arithmetic progression? Certainly! Each term differs from the previous one by strictly three. If you add 2 or 4 to a term, say, the result, i.e. the new number is no longer divisible by 3. You can immediately determine the difference of the arithmetic progression: d = 3. It will come in handy!)
So, we can safely write down some progression parameters:
What will the number be? n last member? Anyone who thinks that 99 is fatally mistaken... The numbers always go in a row, but our members jump over three. They don't match.
There are two solutions here. One way is for the super hardworking. You can write down the progression, the entire series of numbers, and count the number of members with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If we apply the formula to our problem, we find that 99 is the thirtieth term of the progression. Those. n = 30.
Let's look at the formula for the sum of an arithmetic progression:
We look and rejoice.) We pulled out from the problem statement everything necessary to calculate the amount:
a 1= 12.
a 30= 99.
S n = S 30.
All that remains is elementary arithmetic. We substitute the numbers into the formula and calculate:
Answer: 1665
Another type of popular puzzle:
4. Given an arithmetic progression:
-21,5; -20; -18,5; -17; ...
Find the sum of terms from twentieth to thirty-four.
We look at the formula for the amount and... we get upset.) The formula, let me remind you, calculates the amount 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, write out the entire progression in a series, and add terms from 20 to 34. But... it’s somehow stupid and takes a long time, right?)
There is a more elegant solution. Let's divide our series into two parts. The first part will be from the first term to the nineteenth. Second part - from 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 with the sum of the terms 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
From this we can see that find the sum S 20-34 can be done by simple subtraction
S 20-34 = S 1-34 - S 1-19
Both amounts on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Let's get started?
We extract the progression parameters from the problem statement:
d = 1.5.
a 1= -21,5.
To calculate the sums of the first 19 and first 34 terms, we will need the 19th and 34th terms. We calculate them using the formula for the nth term, as in problem 2:
a 19= -21.5 +(19-1) 1.5 = 5.5
a 34= -21.5 +(34-1) 1.5 = 28
There's nothing left. From the sum of 34 terms subtract the sum of 19 terms:
S 20-34 = S 1-34 - S 1-19 = 110.5 - (-152) = 262.5
Answer: 262.5
One important note! There is a very useful trick in solving this problem. Instead of direct calculation what you need (S 20-34), we counted something that would seem not to be needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the complete result. This kind of “feint with your ears” often saves you in wicked problems.)
In this lesson we looked at 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 involving the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.
Formula for the nth term:
These formulas will immediately tell you what to look for and in what 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. The arithmetic progression is given by the condition: a 1 = -5.5; a n+1 = a n +0.5. Find the sum of its first 24 terms.
Unusual?) This recurrence formula. You can read about it in the previous lesson. Don’t ignore the link, such problems are often found in the State Academy of Sciences.
7. Vasya saved up money for the holiday. As much as 4550 rubles! And I decided to give my favorite person (myself) a few days of happiness). Live beautifully without denying yourself anything. Spend 500 rubles on the first day, and on each subsequent day spend 50 rubles more than the previous one! Until the money runs out. How many days of happiness did Vasya have?
Is it difficult?) The additional formula from problem 2 will help.
Answers (in disarray): 7, 3240, 6.
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Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
An arithmetic progression is a series of numbers in which each number is greater (or less) than the previous one by the same amount.
This topic often seems complex and incomprehensible. The indices of the letters, the nth term of the progression, the difference of the progression - all this is somehow confusing, yes... Let's figure out the meaning of the arithmetic progression and everything will get better right away.)
The concept of arithmetic progression.
Arithmetic progression is a very simple and clear concept. Do you have any doubts? In vain.) See for yourself.
I'll write an unfinished series of numbers:
1, 2, 3, 4, 5, ...
Can you extend this series? What numbers will come next, after the five? Everyone... uh..., in short, everyone will realize that the numbers 6, 7, 8, 9, etc. will come next.
Let's complicate the task. I give you 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 realized that this number is 20, congratulations! Not only did you feel key points 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 sensations into mathematics.)
First key point.
Arithmetic progression deals with series of numbers. This is confusing at first. We are used to solving equations, drawing graphs and all that... But here we 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 specifically 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 is three more than the previous one. Actually, it is this moment that gives us the opportunity to grasp the pattern and calculate subsequent numbers.
Third key point.
This moment is not striking, yes... But it is 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, etc. If you mix them up at random, the pattern will disappear. Arithmetic progression will also disappear. What's left is just a series of numbers.
That's the whole point.
Of course, in new topic new terms and designations appear. You need to know them. Otherwise you won’t understand the task. For example, you will have to decide something like:
Write down the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.
Inspiring?) Letters, some indexes... And the task, by the way, couldn’t be simpler. You just need to understand the meaning of the terms and designations. 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 quantity is called . Let's look at this concept in more detail.
Arithmetic progression difference.
Arithmetic progression difference is the amount by which any progression number more previous one.
One important point. Please pay attention to the word "more". Mathematically, this means that each progression number is by adding difference of arithmetic progression to the previous number.
To calculate, let's say second numbers of the series, you need 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 in 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 obtained by adding positive number, +5 to the previous one.
The difference may 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 each number is also obtained by adding to the previous one, but already negative number, -5.
By the way, when working with progression, it is very useful to immediately determine its nature - whether it is increasing or decreasing. This helps a lot to navigate the decision, spot 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 in the series previous number. Subtract. By the way, the result of subtraction is called "difference".)
Let us define, for example, d for increasing arithmetic progression:
2, 5, 8, 11, 14, ...
We take any number in the series that we want, for example, 11. We subtract from it previous number those. 8:
This is the correct answer. For this arithmetic progression, the difference is three.
You can take it any progression number, 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 cannot take only the very first number. Simply because the very first number no previous one.)
By the way, knowing that d=3, finding the seventh number of this progression is very simple. Let's add 3 to the fifth number - we get the sixth, it will be 17. Let's add three to the sixth number, we get the seventh number - twenty.
Let's define d for descending arithmetic progression:
8; 3; -2; -7; -12; .....
I remind you that, regardless of the signs, to determine d need from any number take away the previous one. Choose any progression number, 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 number.
Other terms and designations.
Each number in the series is called 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 clearly understand - the numbers themselves can be absolutely anything, whole, fractional, negative, whatever, but numbering of numbers- strictly in order!
How to write a progression in general form? No problem! Each number in a series is written as a letter. To denote an arithmetic progression, the letter is usually used a. The member number is indicated by an index at the bottom right. We write terms separated by commas (or semicolons), like this:
a 1, a 2, a 3, a 4, a 5, .....
a 1- this is the first number, a 3- third, etc. Nothing fancy. This series can be written briefly like this: (a n).
Progressions happen finite and infinite.
Ultimate the progression has a limited number of members. Five, thirty-eight, whatever. But it's a finite number.
Infinite progression - has an infinite number of members, as you might guess.)
Write down finite progression you can go through a series like this, all the terms 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 short note 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 solve the tasks. The tasks are simple, purely for understanding the meaning of an arithmetic progression.
Examples of tasks on arithmetic progression.
Let's look at the task given above in detail:
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 understandable language. An infinite arithmetic progression is given. The second number of this progression is known: a 2 = 5. The progression difference is known: d = -2.5. We need to find the first, third, fourth, fifth and sixth terms of this progression.
For clarity, I will write down a series according to the conditions of the problem. The first six terms, where the second term is 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. Don't forget about the minus!
a 3=5+(-2,5)=5 - 2,5 = 2,5
The third term turned out less than two. Everything is logical. If the number is greater than the previous one negative value, which means the number itself will be less than the previous one. Progression is decreasing. Okay, let's take it into account.) We count the fourth term 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, terms from the third to the sixth were calculated. The result is the following 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.) So, 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 it. Assignment answer:
7,5, 5, 2,5, 0, -2,5, -5, ...
In passing, I would like to note that we solved this task recurrent way. This terrible word means only the search for a member of the progression according to the previous (adjacent) number. We'll look at other ways to work with progression below.
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 term of this progression.
Do you remember? This simple conclusion allows you to solve most of the problems of the 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 the progression. All.
Of course, all previous algebra is not canceled.) Inequalities, equations, and other things are attached to progression. But according to the progression itself- everything revolves around three parameters.
As an example, let's look at some popular tasks on this topic.
2. Write the finite 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 them, and write them down. It is advisable not to miss the words in the task conditions: “final” and “ n=5". So as not to count until you are 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 will be a member of the arithmetic progression (a n), if a 1 = 4.1; d = 1.2.
Hmm... Who knows? How to determine something?
How-how... Write down the progression in the form of a series and see whether there will be a seven there or not! We count:
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 visible that we are just seven slipped through between 6.5 and 7.7! Seven did not fall into our series of numbers, and, therefore, seven will not be a member of the given progression.
Answer: no.
Here's a problem based on real option GIA:
4. Several consecutive terms of the arithmetic progression are written out:
...; 15; X; 9; 6; ...
Here is a series written 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 look and see what's possible to know 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 "consistent" in 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. Therefore, we can calculate the difference of the arithmetic progression! Subtract from six previous number, i.e. nine:
There are mere trifles left. What number will be the previous one for 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 based on formulas. Purely to understand the meaning of an arithmetic progression.) We just write down a series of numbers and letters, look and figure it out.
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 term.
7. It is known that in arithmetic progression a 2 = 4; a 5 = 15.1. Find a 3 .
8. Several consecutive terms of the arithmetic progression are written out:
...; 15.6; X; 3.4; ...
Find the term of the progression indicated by the letter x.
9. The train began moving from the station, uniformly increasing speed by 30 meters per minute. What will be the speed of the train in five minutes? Give your answer in km/hour.
10. It is known that in 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? Amazing! You can master arithmetic progression for more high level, in the following lessons.
Didn't everything work out? No problem. In Special Section 555, all these problems are sorted out piece by piece.) And, of course, a simple practical technique is described that immediately highlights the solution to such tasks clearly, clearly, at a glance!
By the way, in the train puzzle there are two problems that people often stumble over. One is purely in terms of progression, and the second is general for any problems 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 looked at 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 solved.
The finger solution works well for very short pieces of a row, as in the examples in this tutorial. If the series is longer, the calculations become more complicated. For example, if in problem 9 in the question we replace "five minutes" on "thirty-five minutes" the problem will become significantly worse.)
And there are also tasks that are simple in essence, but absurd in terms of calculations, for example:
An arithmetic progression (a n) is given. Find a 121 if a 1 =3 and d=1/6.
So what, are we going to add 1/6 many, many times?! You can kill yourself!?
You can.) If you don’t know simple formula, which allows you to solve such tasks in a minute. This formula will be in the next lesson. And this problem is solved there. In a minute.)
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When studying algebra in secondary school(9th grade) one of important topics is the study of number sequences, which include progressions - geometric and arithmetic. In this article we will look at an arithmetic progression and examples with solutions.
What is an arithmetic progression?
To understand this, it is necessary to define the progression in question, as well as provide the basic formulas that will be used later in solving problems.
It is known that in some algebraic progression 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 . Let's substitute the known data from the condition into it, 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, we have answered the first part of the problem.
To restore the sequence to the 7th term, 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 No. 3: drawing up a progression
Let's complicate it further stronger condition tasks. Now we need to answer the question of how to find an arithmetic progression. The following example can be given: two numbers are given, for example - 4 and 5. It is necessary to create an algebraic progression so that three more terms are placed between these.
Before you start solving this problem, you need 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 move on to the problem, which is similar to the previous one. Again, for the nth term we use the formula, we get: a 5 = a 1 + 4 * d. From: d = (a 5 - a 1)/4 = (5 - (-4)) / 4 = 2.25. What we got here is not an integer value of the difference, but it is rational number, so the formulas for the algebraic progression remain the same.
Now let's add the found difference to a 1 and restore the missing terms 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 conditions of the problem.
Example No. 4: first term of progression
Let's continue to give examples of arithmetic progression with solutions. In all previous problems, the first number of the algebraic progression was known. Now let's consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find which number this sequence begins with.
The formulas used so far assume knowledge of a 1 and d. In the problem statement, nothing is known about these numbers. Nevertheless, we will write down expressions for each term about which information is available: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We received two equations in which there are 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.
The easiest way to solve this system is to express a 1 in each equation and then compare the resulting expressions. 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 expressions above for a 1. For example, first: a 1 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.
If you have doubts about the result obtained, you can check it, for example, determine the 43rd 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. The small error is due to the fact that rounding to thousandths was used in the calculations.
Example No. 5: amount
Now let's look at several examples with solutions for the sum of an arithmetic progression.
Let a numerical progression be given the following type: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?
Thanks to the development of computer technology, it is possible to solve this problem, that is, add all the numbers sequentially, which the computer will do as soon as a person presses the Enter key. However, the problem can be solved mentally if you pay attention that the presented series of numbers is an algebraic progression, and its difference is equal to 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.
It is interesting to note that this problem is called “Gaussian” because at the beginning of the 18th century the famous German, 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 the numbers at the ends of the sequence in pairs, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer it is enough to multiply 50 by 101.
Example No. 6: sum of terms 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 terms from 8 to 14 will be equal to.
The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them sequentially. Since there are few terms, this method is not quite labor-intensive. Nevertheless, it is proposed to solve this problem using a second method, which is more universal.
The idea is to obtain a formula for the sum of the algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:
- S m = m * (a m + a 1) / 2.
- S n = n * (a n + a 1) / 2.
Since n > m, it is obvious that the 2nd sum includes the first. The last conclusion means that if we take the difference between these sums and add the term a m to it (in the case of taking the difference, it is subtracted from the sum S n), we will obtain the necessary answer to the problem. We have: S mn = S n - S m + a m =n * (a 1 + a n) / 2 - m *(a 1 + a m)/2 + a m = a 1 * (n - m) / 2 + a n * n/2 + a m * (1- m/2). It is necessary to substitute formulas for a n and a m into this expression. 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, however, the sum 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 above solutions, all problems are based on knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before starting to solve any of these problems, it is recommended that you carefully read the condition, clearly understand what you need to find, and only then proceed with 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 likelihood of making a mistake is less. For example, in the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and divide the overall problem into separate subtasks (V in this case first find the terms a n and a m).
If you have doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. We found out how to find an arithmetic progression. If you figure it out, it's not that difficult.
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Number sequence
So, let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many of them as you like (in our case, there are them). No matter how many numbers we write, we can always say which one is first, which one is second, and so on until the last, that is, we can number them. This is an example of a number sequence:
Number sequence
For example, for our sequence:
The assigned number is specific to only one number in the sequence. 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 number is called the th term of the sequence.
We usually call the entire sequence by some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member: .
In our case: 
Let's say we have number sequence, in which the difference between adjacent numbers is the same and equal.
For example: 
etc.
This number sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius back in the 6th century and was understood in a broader sense as an infinite numerical sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which was studied by the ancient Greeks.
This is a number sequence, each member of which is equal to the previous one added to the same number. This number is called the difference of an arithmetic progression and is designated.
Try to determine which number sequences are an arithmetic progression and which are not:
a)
b)
c)
d)
Got it? Let's 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 term. Exists two way to find it.
1. Method
We can add the progression number to the previous value 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 term 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 take us more than one hour, and it is not a fact that we would not make mistakes when adding numbers.
Of course, mathematicians have come up with a way in which it is not necessary to add the difference of an arithmetic progression to the previous value. Take a closer look at the drawn picture... Surely you have already noticed a certain pattern, namely:
For example, let’s see what the value of the th term of this arithmetic progression consists of: 
In other words:
Try to find the value of a member of a given arithmetic progression yourself in this way.
Did you calculate? Compare your notes with the answer:
Please note that you got exactly the same number as in the previous method, when we sequentially added the terms of the arithmetic progression to the previous value.
Let's try to "depersonalize" this formula- let's bring her to general form and we get:
|
Arithmetic progression equation. |
Arithmetic progressions can be 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 this in practice.
We are given an arithmetic progression consisting of the following numbers: Let's check what the th number of this arithmetic progression will be if we use our formula to calculate it:
Since then:
Thus, we are convinced that the formula operates in both decreasing and increasing arithmetic progression.
Try to find the th and th terms of this arithmetic progression yourself.
Let's compare the results:
Arithmetic progression property
Let's complicate the problem - we will derive the property of arithmetic progression.
Let's say we are given the following condition:
- arithmetic progression, find the value.
Easy, you say and start counting according to the formula you already know:
Let, ah, 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 a mistake in the calculations.
Now think about whether it is possible to solve this problem in one step using any formula? Of course yes, and that’s what we’ll try to bring out now.
Let us denote the required term of the arithmetic progression as, the formula for finding it is known to us - this is the same formula we derived at the beginning:
, Then:
- the previous term of the progression is:
- the next term of the progression is:
Let's sum up the previous and subsequent terms of the progression:
It turns out that the sum of the previous and subsequent terms of the progression is the double value of the progression term located between them. In other words, to find the value of a progression term with known previous and successive values, you need to add them and divide by.
That's right, we got the same number. Let's secure the material. Calculate the value for the progression yourself, it’s not at all difficult.
Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, was easily deduced by one of the greatest mathematicians of all time, the “king of mathematicians” - Karl Gauss...
When Carl Gauss was 9 years old, a teacher, busy checking the work of students in other classes, asked the following problem in class: “Calculate the sum of all natural numbers from to (according to other sources up to) inclusive.” Imagine the teacher’s surprise when one of his students (this was Karl Gauss) a minute later gave the correct answer to the task, while most of the daredevil’s classmates, after long calculations, received the wrong result...
Young Carl Gauss noticed a certain pattern that you can easily notice too.
Let's say we have an arithmetic progression consisting of -th terms: We need to find the sum of these terms of the arithmetic progression. Of course, we can manually sum all the values, but what if the task requires finding the sum of its terms, as Gauss was looking for?
Let us depict the progression given to us. Take a closer look at the highlighted numbers and try to perform various mathematical operations with them. 
Have you tried it? What did you notice? Right! Their sums are equal
Now tell me, how many such pairs are there in total 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 pairs are equal, we obtain 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 difference of the progression. Try to substitute the formula of the th term into the sum formula.
What did you get?
Well done! Now let's return to the problem that was asked to Carl Gauss: calculate for yourself what the sum of numbers starting from the th is equal to and the sum of the numbers starting from the th.
How much did you get?
Gauss found that the sum of the terms is equal, and the sum of the terms. Is that what you decided?
In fact, the formula for the sum of the terms of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people made full use of the properties of the arithmetic progression.
For example, imagine Ancient Egypt and the largest construction project of that time - the construction of a pyramid... The picture 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? Calculate how many blocks are needed to build one wall if the base is block bricks. I hope you won’t count while moving your finger across the monitor, 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 terms of an arithmetic progression.
Let's substitute our data into the last formulas (calculate the number of blocks in 2 ways).
Method 1.
Method 2.
And now you can calculate on the monitor: compare the obtained values with the number of blocks that are in our pyramid. Got it? Well done, you have mastered the sum of the nth terms of an arithmetic progression.
Of course, you can’t build a pyramid from 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:
Training
Tasks:
- Masha is getting in shape for summer. Every day she increases the number of squats by. How many times will Masha do squats in a week if she did squats at the first training session?
- What is the sum of all odd numbers contained in.
- When storing logs, loggers stack them in such a way that each upper layer contains one less log than the previous one. How many logs are in one masonry, if the foundation 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 do squats once a day.
- First odd number, last number.
Arithmetic progression difference.
The number of odd numbers in is half, however, let’s check this fact using the formula for finding the th term of an arithmetic progression:Numbers do contain odd numbers.
Let's substitute the available data into the formula:Answer: The sum of all odd numbers contained in is equal.
- Let's remember the problem about pyramids. For our case, a , since each top layer is reduced by one log, then in total there are a bunch of layers, that is.
Let's substitute the data into the formula:Answer: There are logs in the masonry.
Let's sum it up
- - a number sequence in which the difference between adjacent numbers is the same and equal. It can be increasing or decreasing.
- Finding formula The th term 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 is the number of numbers in progression.
- The sum of the terms of an arithmetic progression can be found in two ways:
, where is the number of values.
ARITHMETIC PROGRESSION. AVERAGE LEVEL
Number sequence
Let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many of them as you like. But we can always say which one is first, which one is second, and so on, that is, we can number them. This is an example of a number sequence.
Number 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 a unique one. And we will not assign this number to any other number from this set.
The number with number is called the th member of the sequence.
We usually call the entire sequence by some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member: .
It is very convenient if the th term of the sequence can be specified 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 is). Or (, difference).
nth term formula
We call a formula recurrent 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 this formula, we will have to calculate the previous nine. For example, let it. Then:
Well, is it clear now what the formula is?
In each line we add to, multiplied by some number. Which one? Very simple: this is the number of the current member minus:
Much more convenient 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. What is the difference? Here's what:
(This is why it is called difference because it is equal to the difference of successive terms of the progression).
So, the formula:
Then the hundredth term is equal to:
What is the sum of all natural numbers from to?
According to legend, the great mathematician Carl Gauss, as a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last date is equal, the sum of the second and the 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 in total? 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 subsequent number is obtained by adding to the previous number. Thus, the numbers we are interested in form an arithmetic progression with the first term and the difference.
Formula of the th term for this progression:
How many terms are there in the progression if they all have to be two-digit?
Very easy: .
The last term of the progression will be equal. Then the sum:
Answer: .
Now decide for yourself:
- Every day the athlete runs more meters than the previous day. How many total kilometers will he run in a week if he ran km m on the first day?
- A cyclist travels more kilometers every day than the previous day. On the first day he traveled km. How many days does he need to travel to cover a kilometer? How many kilometers will he travel during the last day of his journey?
- The price of a refrigerator in a store decreases by the same amount every year. Determine how much the price of a refrigerator decreased each 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: , must be found.
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 is.
Let's calculate the path traveled over the last day using the formula of the th term:
(km).
Answer: - Given: . Find: .
It couldn't be simpler:
(rub).
Answer:
ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS
This is a number sequence in which the difference between adjacent numbers is the same and equal.
Arithmetic progression can be increasing () and decreasing ().
For example:
Formula for finding the nth term of an arithmetic progression
is written by the formula, where is the number of numbers in progression.
Property of members of an arithmetic progression
It allows you to easily find a term of a progression if its neighboring terms are known - where is the number of numbers in the progression.
Sum of terms of an arithmetic progression
There are two ways to find the amount:
Where is the number of values.
Where is the number of values.
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
Now the most important thing.
You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough...
For what?
For successful passing the Unified State Exam, for admission to college on a budget and, MOST IMPORTANTLY, for life.
I won’t convince you of anything, I’ll just say one thing...
People who have received a good education earn much more than those who have not received it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?
GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
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You will need solve problems against time.
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It's like in sports - you need to repeat it many times to win for sure.
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