For example, the sequence \(2\); \(5\); \(eight\); \(eleven\); \(14\)… is an arithmetic progression, because each next element differs from the previous one by three (can be obtained from the previous one by adding three):
In this progression, the difference \(d\) is positive (equal to \(3\)), and therefore each next term is greater than the previous one. Such progressions are called increasing.
However, \(d\) can also be a negative number. For example, in arithmetic progression \(16\); \(ten\); \(four\); \(-2\); \(-8\)… the progression difference \(d\) is equal to minus six.
And in this case, each next element will be less than the previous one. These progressions are called decreasing.
Arithmetic progression notation
Progression is denoted by a small Latin letter.
The numbers that form a progression are called it members(or elements).
They are denoted by the same letter as the arithmetic progression, but with a numerical index equal to the element number in order.
For example, the arithmetic progression \(a_n = \left\( 2; 5; 8; 11; 14…\right\)\) consists of the elements \(a_1=2\); \(a_2=5\); \(a_3=8\) and so on.
In other words, for the progression \(a_n = \left\(2; 5; 8; 11; 14…\right\)\)
Solving problems on an arithmetic progression
In principle, the above information is already enough to solve almost any problem on an arithmetic progression (including those offered at the OGE).
Example (OGE).
The arithmetic progression is given by the conditions \(b_1=7; d=4\). Find \(b_5\).
Solution:
Answer: \(b_5=23\)
Example (OGE).
The first three terms of an arithmetic progression are given: \(62; 49; 36…\) Find the value of the first negative term of this progression..
Solution:
|
We are given the first elements of the sequence and know that it is an arithmetic progression. That is, each element differs from the neighboring one by the same number. Find out which one by subtracting the previous one from the next element: \(d=49-62=-13\). |
|
|
Now we can restore our progression to the desired (first negative) element. |
|
|
Ready. You can write an answer. |
Answer: \(-3\)
Example (OGE).
Several successive elements of an arithmetic progression are given: \(...5; x; 10; 12.5...\) Find the value of the element denoted by the letter \(x\).
Solution:
|
|
To find \(x\), we need to know how much the next element differs from the previous one, in other words, the progression difference. Let's find it from two known neighboring elements: \(d=12.5-10=2.5\). |
|
|
And now we find what we are looking for without any problems: \(x=5+2.5=7.5\). |
|
|
Ready. You can write an answer. |
Answer: \(7,5\).
Example (OGE).
The arithmetic progression is given by the following conditions: \(a_1=-11\); \(a_(n+1)=a_n+5\) Find the sum of the first six terms of this progression.
Solution:
|
We need to find the sum of the first six terms of the progression. But we do not know their meanings, we are given only the first element. Therefore, we first calculate the values in turn, using the given to us: \(n=1\); \(a_(1+1)=a_1+5=-11+5=-6\) |
|
|
\(S_6=a_1+a_2+a_3+a_4+a_5+a_6=\) |
The requested amount has been found. |
Answer: \(S_6=9\).
Example (OGE).
In arithmetic progression \(a_(12)=23\); \(a_(16)=51\). Find the difference of this progression.
Solution:
Answer: \(d=7\).
Important Arithmetic Progression Formulas
As you can see, many arithmetic progression problems can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each next element in this chain is obtained by adding the same number to the previous one (the difference of the progression).
However, sometimes there are situations when it is very inconvenient to solve "on the forehead". For example, imagine that in the very first example, we need to find not the fifth element \(b_5\), but the three hundred and eighty-sixth \(b_(386)\). What is it, we \ (385 \) times to add four? Or imagine that in the penultimate example, you need to find the sum of the first seventy-three elements. Counting is confusing...
Therefore, in such cases, they do not solve “on the forehead”, but use special formulas derived for arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum \(n\) of the first terms.
Formula for the \(n\)th member: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first member of the progression;
\(n\) – number of the required element;
\(a_n\) is a member of the progression with the number \(n\).
This formula allows us to quickly find at least the three hundredth, even the millionth element, knowing only the first and the progression difference.
Example.
The arithmetic progression is given by the conditions: \(b_1=-159\); \(d=8,2\). Find \(b_(246)\).
Solution:
Answer: \(b_(246)=1850\).
The formula for the sum of the first n terms is: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where
\(a_n\) is the last summed term;
Example (OGE).
The arithmetic progression is given by the conditions \(a_n=3.4n-0.6\). Find the sum of the first \(25\) terms of this progression.
Solution:
|
\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\) |
To calculate the sum of the first twenty-five elements, we need to know the value of the first and twenty-fifth term. |
|
|
\(n=1;\) \(a_1=3.4 1-0.6=2.8\) |
Now let's find the twenty-fifth term by substituting twenty-five instead of \(n\). |
|
|
\(n=25;\) \(a_(25)=3.4 25-0.6=84.4\) |
Well, now we calculate the required amount without any problems. |
|
|
\(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25=\) |
The answer is ready. |
Answer: \(S_(25)=1090\).
For the sum \(n\) of the first terms, you can get another formula: you just need to \(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25\ ) instead of \(a_n\) substitute the formula for it \(a_n=a_1+(n-1)d\). We get:
The formula for the sum of the first n terms is: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where
\(S_n\) – the required sum \(n\) of the first elements;
\(a_1\) is the first term to be summed;
\(d\) – progression difference;
\(n\) - the number of elements in the sum.
Example.
Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15,5\); \(fourteen\)…
Solution:
Answer: \(S_(33)=-231\).
More complex arithmetic progression problems
Now you have all the information you need to solve almost any arithmetic progression problem. Let's finish the topic by considering problems in which you need to not only apply formulas, but also think a little (in mathematics, this can be useful ☺)
Example (OGE).
Find the sum of all negative terms of the progression: \(-19.3\); \(-19\); \(-18.7\)…
Solution:
|
\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\) |
The task is very similar to the previous one. We start solving the same way: first we find \(d\). |
|
|
\(d=a_2-a_1=-19-(-19.3)=0.3\) |
Now we would substitute \(d\) into the formula for the sum ... and here a small nuance pops up - we don't know \(n\). In other words, we do not know how many terms will need to be added. How to find out? Let's think. We will stop adding elements when we get to the first positive element. That is, you need to find out the number of this element. How? Let's write down the formula for calculating any element of an arithmetic progression: \(a_n=a_1+(n-1)d\) for our case. |
|
|
\(a_n=a_1+(n-1)d\) |
||
|
\(a_n=-19.3+(n-1) 0.3\) |
We need \(a_n\) to be greater than zero. Let's find out for what \(n\) this will happen. |
|
|
\(-19.3+(n-1) 0.3>0\) |
||
|
\((n-1) 0.3>19.3\) \(|:0.3\) |
We divide both sides of the inequality by \(0,3\). |
|
|
\(n-1>\)\(\frac(19,3)(0,3)\) |
We transfer minus one, not forgetting to change signs |
|
|
\(n>\)\(\frac(19,3)(0,3)\) \(+1\) |
Computing... |
|
|
\(n>65,333…\) |
…and it turns out that the first positive element will have the number \(66\). Accordingly, the last negative has \(n=65\). Just in case, let's check it out. |
|
|
\(n=65;\) \(a_(65)=-19.3+(65-1) 0.3=-0.1\) |
Thus, we need to add the first \(65\) elements. |
|
|
\(S_(65)=\) \(\frac(2 \cdot (-19,3)+(65-1)0,3)(2)\)\(\cdot 65\) |
The answer is ready. |
Answer: \(S_(65)=-630.5\).
Example (OGE).
The arithmetic progression is given by the conditions: \(a_1=-33\); \(a_(n+1)=a_n+4\). Find the sum from \(26\)th to \(42\) element inclusive.
Solution:
|
\(a_1=-33;\) \(a_(n+1)=a_n+4\) |
In this problem, you also need to find the sum of elements, but starting not from the first, but from the \(26\)th. We don't have a formula for this. How to decide? |
|
|
For our progression \(a_1=-33\), and the difference \(d=4\) (after all, we add four to the previous element to find the next one). Knowing this, we find the sum of the first \(42\)-uh elements. |
|
\(S_(42)=\) \(\frac(2 \cdot (-33)+(42-1)4)(2)\)\(\cdot 42=\) |
Now the sum of the first \(25\)-th elements. |
|
\(S_(25)=\) \(\frac(2 \cdot (-33)+(25-1)4)(2)\)\(\cdot 25=\) |
And finally, we calculate the answer. |
|
\(S=S_(42)-S_(25)=2058-375=1683\) |
Answer: \(S=1683\).
For an arithmetic progression, there are several more formulas that we have not considered in this article due to their low practical usefulness. However, you can easily find them.
Attention!
There are additional
material in Special Section 555.
For those who strongly "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 is often difficult and incomprehensible. Letter indices, 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 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 of an 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 is 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, new terms and notation appear in the new topic. 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 conclusion allows us 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 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 learn arithmetic progression at a higher level in the following 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.)
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
When studying algebra in a secondary school (grade 9), one of the important topics is the study of numerical sequences, which include progressions - geometric and arithmetic. In this article, we will consider an arithmetic progression and examples with solutions.
What is an arithmetic progression?
To understand this, it is necessary to give a definition of the progression under consideration, as well as to give the basic formulas that will be further used 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 . We substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 \u003d 6 + 6 * d. From this expression, you can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, the first part of the problem was answered.
To restore the sequence to the 7th member, 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 and 7 = 18.
Example #3: making a progression
Let us complicate the condition of the problem even more. Now you need to answer the question of how to find an arithmetic progression. We can give the following example: two numbers are given, for example, 4 and 5. It is necessary to make an algebraic progression so that three more terms fit between these.
Before starting to solve this problem, it is necessary to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 \u003d -4 and a 5 \u003d 5. Having established this, we proceed to a task that is similar to the previous one. Again, for the nth term, we use the formula, we get: a 5 \u003d a 1 + 4 * d. From: d \u003d (a 5 - a 1) / 4 \u003d (5 - (-4)) / 4 \u003d 2.25. Here, the difference is not an integer value, but it is a 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 members of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 \u003d 2.75 + 2.25 \u003d 5, which coincided with the condition of the problem.
Example #4: The first member of the progression
We continue to give examples of an arithmetic progression with a solution. In all previous problems, the first number of the algebraic progression was known. Now 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 from what number this sequence begins.
The formulas that have been used so far assume knowledge of a 1 and d. Nothing is known about these numbers in the condition of the problem. Nevertheless, let's write out the expressions for each term about which we have information: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We got 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 specified system is easiest to solve if you 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 \u003d a 43 - 42 * d \u003d 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d \u003d 37 - 42 * d, whence the difference d \u003d (37 - 50) / (42 - 14) \u003d - 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 \u003d 50 - 14 * d \u003d 50 - 14 * (- 0.464) \u003d 56.496.
If there are doubts about the result, you can check it, for example, determine the 43rd member of the progression, which is specified in the condition. We get: a 43 \u003d a 1 + 42 * d \u003d 56.496 + 42 * (- 0.464) \u003d 37.008. A small error is due to the fact that rounding to thousandths was used in the calculations.
Example #5: Sum
Now let's look at some examples with solutions for the sum of an arithmetic progression.
Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?
Thanks to the development of computer technology, this problem can be solved, that is, sequentially add up all the numbers, 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 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.
It is curious to note that this problem is called "Gaussian", since at the beginning of the 18th century the famous German, still at the age of only 10 years old, was able to solve it in his mind 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 pairs of numbers located at the edges of the sequence, 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 #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.
The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them up sequentially. Since there are few terms, this method is not laborious enough. Nevertheless, it is proposed to solve this problem by the second method, which is more universal.
The idea is to get a formula for the sum of an algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:
- S m \u003d m * (a m + a 1) / 2.
- S n \u003d n * (a n + a 1) / 2.
Since n > m, it is obvious that the 2 sum includes the first one. 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), then we get the necessary answer to the problem. We have: S mn \u003d S n - S m + a m \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m \u003d 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 the knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before you start solving any of these problems, it is recommended that you carefully read the condition, clearly understand what you want to find, and only then proceed with the solution.
Another tip is to strive for simplicity, that is, if you can answer the question without using complex mathematical calculations, then you need to do just that, since in this case the probability of making a mistake is less. For example, in the example of an arithmetic progression with solution 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 break the general task into separate subtasks (in this case, first find the terms a n and a m).
If there are doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. How to find an arithmetic progression, found out. Once you figure it out, it's not that hard.
Arithmetic progression name a sequence of numbers (members of a progression)
In which each subsequent term differs from the previous one by a steel term, which is also called step or progression difference.
Thus, by setting the step of the progression and its first term, you can find any of its elements using the formula
Properties of an arithmetic progression
1) Each member of the arithmetic progression, starting from the second number, is the arithmetic mean of the previous and next member of the progression
The converse is also true. If the arithmetic mean of neighboring odd (even) members of the progression is equal to the member that stands between them, then this sequence of numbers is an arithmetic progression. By this assertion 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 we 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 by the formula
Remember well the formula for the sum of an arithmetic progression, it is indispensable in calculations and is quite common in simple life situations.
3) If you need to find not the entire sum, but a part of the sequence starting from its k -th member, then the following sum formula will come in handy in you
4) It is of practical interest to find the sum of n members of an arithmetic progression starting from the kth number. To do this, use the formula
This is where the theoretical material ends and we move on to solving problems that are common in practice.
Example 1. Find the fortieth term of the arithmetic progression 4;7;...
Solution:
According to the condition, we have
Define the progression step
According to the well-known formula, we find the fortieth term of the progression
Example2. The arithmetic progression is given by its third and seventh members. Find the first term of the progression and the sum of ten.
Solution:
We write the given elements of the progression according to the formulas
We subtract the first equation from the second equation, as a result we find the progression step
The found value is substituted into any of the equations to find the first term of the arithmetic progression
Calculate the sum of the first ten terms of the progression
Without applying complex calculations, we found all the required values.
Example 3. An arithmetic progression is given by the denominator and one of its members. 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 the formula for the hundredth element of the progression
and find the first
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 sum of the progression is 250.
Example 4
Find the number of members of an arithmetic progression if:
a3-a1=8, a2+a4=14, Sn=111.
Solution:
We write the equations in terms of the first term and the step of the progression and define them
We substitute the obtained values into the sum formula to determine the number of terms in the sum
Making simplifications
and solve the quadratic equation
Of the two values found, only the number 8 is suitable for the condition of the problem. 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. We write out its first term and find the difference of the progression
Before we start to decide arithmetic progression problems, consider what a number sequence is, since an arithmetic progression is a special case of a number sequence.
A numerical sequence is a numerical set, each element of which has its own serial number. The elements of this set are called members of the sequence. The ordinal number of a sequence element is indicated by an index:
The first element of the sequence;
The fifth element of the sequence;
- "nth" element of the sequence, i.e. the element "standing in the queue" at number n.
There is a dependency between the value of a sequence element and its ordinal number. Therefore, we can consider a sequence as a function whose argument is the ordinal number of an element of the sequence. In other words, one can say that the sequence is a function of the natural argument:
The sequence can be specified in three ways:
1 . The sequence can be specified using a table. In this case, we simply set the value of each member of the sequence.
For example, Someone decided to do personal time management, and to begin with, to calculate how much time he spends on VKontakte during the week. By writing the time in a table, he will get a sequence consisting of seven elements:
The first line of the table contains the number of the day of the week, the second - the time in minutes. We see that, that is, on Monday Someone spent 125 minutes on VKontakte, that is, on Thursday - 248 minutes, and, that is, on Friday, only 15.
2 . The sequence can be specified using the nth member formula.
In this case, the dependence of the value of a sequence element on its number is expressed directly as a formula.
For example, if , then
To find the value of a sequence element with a given number, we substitute the element number into the formula for the nth member.
We do the same if we need to find the value of a function if the value of the argument is known. We substitute the value of the argument instead in the equation of the function:
If, for example, 
, then
Once again, I note that in a sequence, in contrast to an arbitrary numeric function, only a natural number can be an argument.
3 . The sequence can be specified using a formula that expresses the dependence of the value of the member of the sequence with number n on the value of the previous members. In this case, it is not enough for us to know only the number of a sequence member in order to find its value. We need to specify the first member or first few members of the sequence.
For example, consider the sequence 
, 
We can find the values of the members of a sequence in sequence, starting from the third:
That is, each time to find the value of the nth member of the sequence, we return to the previous two. This way of sequencing is called recurrent, from the Latin word recurro- come back.
Now we can define an arithmetic progression. An arithmetic progression is a simple special case of a numerical sequence.
Arithmetic progression is called a numerical sequence, each member of which, starting from the second, is equal to the previous one, added with the same number.
The number is called the difference of an arithmetic progression. The difference of an arithmetic progression can be positive, negative, or zero.
If title="(!LANG:d>0">, то каждый член арифметической прогрессии больше предыдущего, и прогрессия является !} increasing.
For example, 2; 5; eight; eleven;...
If , then each term of the arithmetic progression is less than the previous one, and the progression is waning.
For example, 2; -one; -four; -7;...
If , then all members of the progression are equal to the same number, and the progression is stationary.
For example, 2;2;2;2;...
The main property of an arithmetic progression:
Let's look at the picture.
We see that
, and at the same time
Adding these two equalities, we get:
.
Divide both sides of the equation by 2:
So, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of two neighboring ones:
Moreover, since
, and at the same time
, then
, and hence
Each member of the arithmetic progression starting with title="(!LANG:k>l">, равен среднему арифметическому двух равноотстоящих.
!}
th member formula.
We see that for the members of the arithmetic progression, the following relations hold:
and finally
We got formula of the nth term.
IMPORTANT! Any member of an arithmetic progression can be expressed in terms of and . Knowing the first term and the difference of an arithmetic progression, you can find any of its members.
The sum of n members of an arithmetic progression.
In an arbitrary arithmetic progression, the sums of terms equally spaced from the extreme ones are equal to each other:
Consider an arithmetic progression with n members. Let the sum of n members of this progression be equal to .
Arrange the terms of the progression first in ascending order of numbers, and then in descending order:
Let's pair it up:
The sum in each parenthesis is , the number of pairs is n.
We get:
So, the sum of n members of an arithmetic progression can be found using the formulas:
Consider solving arithmetic progression problems.
1 . The sequence is given by the formula of the nth term: . Prove that this sequence is an arithmetic progression.
Let us prove that the difference between two adjacent members of the sequence is equal to the same number.
We have obtained that the difference of two adjacent members of the sequence does not depend on their number and is a constant. Therefore, by definition, this sequence is an arithmetic progression.
2 . Given an arithmetic progression -31; -27;...
a) Find the 31 terms of the progression.
b) Determine if the number 41 is included in this progression.
a) We see that ;
Let's write down the formula for the nth term for our progression.
In general 
In our case 
, that's why 
