What is the main essence of the formula?
This formula allows you to find any At his number " n " .
Of course, you need to know another first member. A 1. and the difference of progression d.Well, so without these parameters a specific progression and will not write down.
To learn (or saparchable) this formula is not enough. It is necessary to learn its essence and prepare the formula in various tasks. Yes, and do not forget at the right moment, yes ...) how not forget - I do not know. And here how to remember If necessary, I will tell you exactly. For those who are less than the lesson to master.)
So, let's deal with the formula of the N-th member of the arithmetic progression.
What is a formula in general - we imagine.) What is an arithmetic progression, a member number, the difference in progress - is available in the previous lesson. Look, by the way, if not read. Everything is simple there. It remains to figure out what n-th member.
The progression in general can be written in the form of a number of numbers:
a 1, A 2, A 3, A 4, A 5, .....
a 1. - denotes the first term of arithmetic progression, a 3. - the third dick, a 4. - Fourth, and so on. If we are interested in the fifth dick, let's say we work with a 5.if one hundred twenty - with a 120..
And how to designate in general any a member of arithmetic progression, with anyone number? Very simple! Like this:
a N.
That's what it is n-th member of arithmetic progression. Under the letter n, all members of the members are hidden at once: 1, 2, 3, 4, and so on.
And what gives us such a record? Think, instead of the digit, the letters recorded ...
This entry gives us a powerful tool for working with arithmetic progress. Using the designation a N.we can quickly find any member any Arithmetic progression. And also a bunch of tasks on progression to solve. You yourself will see.
In the formula of the N-th member of the arithmetic progression:
| a n \u003d a 1 + (n-1) d |
a 1. - the first term of arithmetic progression;
n. - Member number.
The formula binds the key parameters of any progression: a n; a 1; D. and n.. Around these parameters and all progression tasks are spinning.
The formula of the N-th member can be used to record specific progression. For example, in the task it can be said that the progression is set by the condition:
a n \u003d 5 + (n - 1) · 2.
Such a task may also put in a dead end ... There is no row, no difference ... But, comparing the condition with the formula, it is easy to figure out that in this progression a 1 \u003d 5, and d \u003d 2.
And it happens more angry!) If you take the same condition: a n \u003d 5 + (n-1) · 2,do you reveal brackets and bring similar? We get a new formula:
a n \u003d 3 + 2n.
it Only not overall, but for a specific progression. Here is the underwater stone. Some think that the first member is a triple. Although the first member is a fidder ... just below we will work with such a modified formula.
In the tasks of the progression there is another designation - a n + 1. This, as you guessed, "En Plus the first" member of the progression. Its meaning is simple and harmless.) This is a member of the progression, the number of which is more than N numbers per unit. For example, if we take in any task for a N. fifth dick then a n + 1 It will be a sixth member. Etc.
Most often, the designation a n + 1 It is found in recurrent formulas. Do not scare this terrible word!) It is just a way of expressing a member of arithmetic progression through the previous one. Suppose we are given an arithmetic progression in this form using the recurrent formula:
a n + 1 \u003d a n +3
a 2 \u003d a 1 + 3 \u003d 5 + 3 \u003d 8
a 3 \u003d a 2 + 3 \u003d 8 + 3 \u003d 11
Fourth - through the third, fifth - through the fourth, and so on. And how to calculate immediately, say the twentieth member, a 20. ? But!) While the 19th member do not know, the 20th does not count. This is the fundamental difference between the recurrent formula from the formula of the N-th member. Recurrent works only through previous Member, and the formula of the N-th member - through first and allows immediately find any dick on its number. Without calculating the entire number of numbers in a few.
In arithmetic progression, the recurrent formula is easy to turn into a normal. Calculate a couple of consecutive members, calculate the difference d, Find if necessary, first member a 1., Write the formula in the usual form, and work with it. In GIA, such tasks are often found.
The use of the formula of the N-th member of the arithmetic progression.
To begin with, consider the direct application of the formula. At the end of the previous lesson there was a task:
An arithmetic progression is given (a n). Find a 121 if a 1 \u003d 3, and d \u003d 1/6.
This problem can be solved without any formulas, simply based on the meaning of arithmetic progression. Add, yes add ... AUTOV-OTHER.)
And according to the formula, the decision will take less minute. You can check time.) We decide.
The conditions contain all the data for the use of the formula: a 1 \u003d 3, d \u003d 1/6. It remains to figure out what is equal n. No problem! We need to find a 121.. Here we write:
Please pay attention! Instead of index n. A concrete number appeared: 121. What is quite logical.) We are interested in a member of arithmetic progression. number one hundred twenty one. That will be our n. This is this value n. \u003d 121 We will substitute further in the formula, in brackets. We substitute all the numbers in the formula and believe:
a 121 \u003d 3 + (121-1) · 1/6 \u003d 3 + 20 \u003d 23
That's all things. It could also be possible to find five hundred tenth member, and a thousand third, any. We put instead n. The desired number in the index at the letter " a " And in brackets, and we believe.
I remind you of the essence: this formula allows you to find any Member of arithmetic progression At his number " n " .
I will solve the task of the stiffer. Have we got such a task:
Find the first term of the arithmetic progression (A n) if A 17 \u003d -2; D \u003d -0.5.
If it was difficult, I will tell you the first step. Write down the formula of the N-th member of the arithmetic progression! Yes Yes. Write your hands, right in the notebook:
| a n \u003d a 1 + (n-1) d |
And now, looking at the letters of the formula, we think what the data we have, and what is missing? Available d \u003d -0.5,there is a seventeenth member ... everything? If you think that everything, the task do not decide, yes ...
We still have a room n.! In condition a 17 \u003d -2 Hidden two parameters. This is the value of the seventeenth member (-2), and its number (17). Those. n \u003d 17. This "trifle" often skips past the head, and without it, (without "little things", not head!) The task is not to solve. Although ... and without a head too.)
Now you can simply stupidly substitute our data in the formula:
a 17 \u003d A 1 + (17-1) · (-0,5)
Oh yes, a 17. We know this -2. Well, we substitute:
-2 \u003d a 1 + (17-1) · (-0,5)
Here, in essence, and that's it. It remains to express the first term of the arithmetic progression from the formula, but count. Will the answer: a 1 \u003d 6.
Such a reception is a formula recording and a simple substitution of known data - healthy helps in simple tasks. Well, it is necessary, of course, to be able to express the variable from the formula, and what to do!? Without this skill, mathematics can not be studied at all ...
Another popular task:
Find the difference in arithmetic progression (a n) if A 1 \u003d 2; A 15 \u003d 12.
What do you do? You will be surprised, write the formula!)
| a n \u003d a 1 + (n-1) d |
We think that we know: a 1 \u003d 2; a 15 \u003d 12; And (specially allocated!) n \u003d 15. Boldly substitute in the formula:
12 \u003d 2 + (15-1) D
We consider arithmetic.)
12 \u003d 2 + 14D
d.=10/14 = 5/7
This is the correct answer.
So, the tasks on a N, A 1and D. They praised. It remains to learn the number to find:
The number 99 is a member of the arithmetic progression (A n), where a 1 \u003d 12; d \u003d 3. Find this member.
We substitute in the formula of the N-th member known to us:
a n \u003d 12 + (n-1) · 3
At first glance, there are two unknown values: a n and n. But a N. - This is some member of the number of progression n.... And we know this member of the progression! It's 99. We do not know his number n,so this number is required. We substitute a member of the progression of 99 in the formula:
99 \u003d 12 + (n - 1) · 3
Express from Formula n., believe. We will get the answer: n \u003d 30.
And now the task on the same topic, but more creative):
Determine whether the number 117 will be a member of arithmetic progression (a n):
-3,6; -2,4; -1,2 ...
Again we write a formula. What, no parameters? GM ... And for us why did we mean?) I see the first member of the progression? We see. This is -3.6. You can safely write: a 1 \u003d -3.6. Difference d. Can I define from a number? Easy, if you know what the difference in arithmetic progression is:
d \u003d -2.4 - (-3,6) \u003d 1,2
So, the simplest made. It remains to deal with an unknown number n. And incomprehensible number 117. In the previous problem, at least it was known that it was a member of the progression. And here we do not know ... How to be!? Well, how to be, how to be ... Include creative abilities!)
we suppose That 117 is, after all, a member of our progression. With an unknown number n.. And, exactly as in the previous task, let's try to find this room. Those. We write a formula (yes!)) And we substitute our numbers:
117 \u003d -3,6 + (n-1) · 1.2
Express again from the formulan., believe and get:
Oops! Room happened fractional! One hundred and a half. And fractional numbers in progress can not be. What conclusion will we do? Yes! Number 117. is not Member of our progression. It is somewhere between a hundred first and one hundred second member. If the number turned out to be natural, i.e. Positive whole, the number would be a member of the progression with the found number. And in our case, the answer task will be: not.
Task based on the real version of GIA:
Arithmetic progression is set by condition:
a n \u003d -4 + 6.8N
Find the first and tenth progression members.
Here the progression is not quite familiar here. Some kind of formula ... happens.) However, this formula (as I wrote above) - also, the formula of the N-th member of the arithmetic progression! It also allows find any progression member by its number.
We are looking for the first member. He who thinks That the first member is minus four, fatally mistaken!) Because the formula in the problem is modified. The first member of the arithmetic progression in it hidden. Nothing, now finding.)
Also, as in previous tasks, we substitute n \u003d 1. In this formula:
a 1 \u003d -4 + 6.8 · 1 \u003d 2.8
Here! The first member is 2.8, and not -4!
Similar to looking for a tenth member:
a 10 \u003d -4 + 6.8 · 10 \u003d 64
That's all things.
And now, those who have read up to these lines - the promised bonus.)
Suppose in the complex combat atmosphere of GIA or EGE, you have forgotten the useful formula of the N-th member of arithmetic progression. Something is remembered, but inseparally somehow ... either n. there, then n + 1, then n-1 ... How to be!?
Tranquility! This formula is easy to withdraw. Not very strictly, but for confidence and the right solution is for sure!) To bring it enough to remember the elementary meaning of arithmetic progression and have a couple of time. You just need to draw a picture. For clarity.
We draw a numeric axis and celebrate the first one. second, third, etc. Members. And noting the difference d. between members. Like this:
We look at the picture and we think: what is the second member? Second one d.:
a. 2 \u003d A 1 + 1 · D.
What is the third dick? The third Member equals first Member Plus two d..
a. 3 \u003d A 1 + 2 · D.
Catch? I am not in vain some words allocate bold fonts. Well, okay, one more step).
What is the fourth dick? Fourth Member equals first Member Plus three d..
a. 4 \u003d A 1 + 3 · D.
It's time to figure out that the number of gaps, i.e. d., always one less than the number of the desired member n.. Those., To the number n, number of gapswill be n-1. Therefore, the formula will (without options!):
| a n \u003d a 1 + (n-1) d |
In general, visual pictures are very helpful to solve many tasks in mathematics. Do not neglect pictures. But if the picture is difficult to draw, then ... only formula!) In addition, the formula of the N-th member allows you to connect the entire powerful arsenal of mathematics to the solution - equations, inequalities, systems, etc. The picture is not inserted into the equation ...
Tasks for self solutions.
For workout:
1. In arithmetic progression (a n) a 2 \u003d 3; A 5 \u003d 5.1. Find a 3.
Tip: In the picture, the task is solved seconds for 20 ... by the formula - it turns out more difficult. But to master the formula - it is more useful.) In section 555, this task is solved in the picture, and by the formula. Feel the difference!)
And this is no longer a workout.)
2. In arithmetic progression (A n) A 85 \u003d 19.1; A 236 \u003d 49, 3. Find a 3.
What is reluctant to draw a picture?) Still! It's better in the formula, yes ...
3. Arithmetic progression is given by condition:a 1 \u003d -5.5; a n + 1 \u003d a n +0.5. Find a hundred twenty-fifth member of this progression.
In this task, the progression is set by a recurrent way. But to count up to one hundred and twenty-fifth member ... not all such a feud under power.) But the formula of the N-th member forces to everyone!
4. Dana Arithmetic Progression (A N):
-148; -143,8; -139,6; -135,4, .....
Find the number of the smallest positive member of the progression.
5. Under the condition of task 4, find the amount of the smallest positive and greatest negative members of the progression.
6. The product of the fifth and twelfth members of the increasing arithmetic progression is -2.5, and the sum of the third and eleventh members is zero. Find a 14.
Not the easiest task, yes ...) Here the way "on the fingers" will not roll. Formulas will have to write yes equations to decide.
Answers (in disorder):
3,7; 3,5; 2,2; 37; 2,7; 56,5
Happened? It's nice!)
Not everything works? It happens. By the way, in the last task there is one subtle moment. Care when reading the task will be required. And logic.
The solution of all these tasks is disassembled in detail in section 555. and the element of fantasy for the fourth, and the subtle moment for the sixth, and the general approaches to solve any tasks on the N-th member formula - everything is painted. Recommend.
If you like this site ...
By the way, I have another couple of interesting sites for you.)
It can be accessed in solving examples and find out your level. Testing with instant check. Learn - with interest!)
You can get acquainted with features and derivatives.
When studying algebra in a secondary school (grade 9), one of the important topics are the study of numerical sequences to which the progression is -ometric and arithmetic. In this article, consider arithmetic progression and examples with solutions.
What is the arithmetic progression?
To understand this, it is necessary to define the progression of the progression, as well as to bring the basic formulas that will further be used when solving problems.
Arithmetic or algebraic progression is such a set of ordered rational numbers, each member of which differs from the previous on some permanent value. This value is called a difference. That is, knowing any member of an ordered series of numbers and a difference, one can restore all arithmetic progression.
Let us give an example. The next sequence of numbers will be the progression of arithmetic: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 \u003d 12 - 8 \u003d 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can not be attributed to the type of progression under consideration, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).
Important formulas
We now present the basic formulas that will be needed to solve problems using arithmetic progression. Denote by the symbol a N n-th member of the sequence, where N is an integer. The difference is denoted by the Latin letter d. Then the following expressions are true:
- To determine the value of the N-th member, the formula is suitable: a n \u003d (n - 1) * d + a 1.
- To determine the amount of the first n of the components: S n \u003d (a n + A 1) * N / 2.
To understand any examples of arithmetic progression with a decision in grade 9, it is enough to remember these two formulas, since any tasks of the considered type are being built on their use. We should also not forget that the difference in progression is determined by the formula: d \u003d a n - a n-1.
Example №1: Finding an unknown member
We give a simple example of the progression of the arithmetic and formulas that must be used to solve.
Let the sequence of 10, 8, 6, 4, ..., it is necessary to find five members in it.
From the condition of the problem already follows that the first 4 components are known. The fifth can be defined in two ways:
- Calculate to start a difference. We have: d \u003d 8 - 10 \u003d -2. Similarly, you could take any two other members standing next to each other. For example, D \u003d 4 - 6 \u003d -2. Since it is known that d \u003d a n - a n-1, then d \u003d a 5 - a 4, from where we get: a 5 \u003d a 4 + d. We substitute the known values: a 5 \u003d 4 + (-2) \u003d 2.
- The second method also requires the knowledge of the difference in the progression under consideration, therefore, it is first necessary to determine it, as shown above (D \u003d -2). Knowing that the first term is 1 \u003d 10, we use the formula for n number of sequence. We have: a n \u003d (n - 1) * d + a 1 \u003d (n - 1) * (-2) + 10 \u003d 12 - 2 * n. Substituting in the last expression n \u003d 5, we obtain: a 5 \u003d 12-2 * 5 \u003d 2.
As can be seen, both methods of solving led to the same result. Note that in this example, the difference D of progression is a negative value. Such sequences are called decreasing, as each next term is smaller than the previous one.
Example number 2: The difference of progression
Now complicate a little task, we give an example as
It is known that in some 1st member is 6, and the 7th member is 18. It is necessary to find a difference and restore this sequence up to 7 members.
We use the formula to determine the unknown member: a n \u003d (n - 1) * D + A 1. We substitute the well-known data from the condition, 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 \u003d (18 - 6) / 6 \u003d 2. Thus, they answered the first part of the problem.
To restore a sequence of up to 7 member, it should be used by the definition of algebraic progression, that is, a 2 \u003d a 1 + d, a 3 \u003d a 2 + d and so on. As a result, we restore the entire sequence: a 1 \u003d 6, a 2 \u003d 6 + 2 \u003d 8, a 3 \u003d 8 + 2 \u003d 10, a 4 \u003d 10 + 2 \u003d 12, a 5 \u003d 12 + 2 \u003d 14, a 6 \u003d 14 + 2 \u003d 16, a 7 \u003d 18.
Example number 3: Production of progression
Let's complicate an even stronger the condition of the task. Now it is necessary to answer the question of how to find an arithmetic progression. You can give the following example: two numbers are given, for example, - 4 and 5. It is necessary to make a progression of algebraic so that three more members are placed.
Before starting to solve this task, it is necessary to understand what place will be the given numbers in future progression. Since there will be three more members between them, then a 1 \u003d -4 and a 5 \u003d 5. By installing it, we turn to the task that is similar to the previous one. Again for the N-th member we use the formula, we obtain: a 5 \u003d a 1 + 4 * d. Location: D \u003d (A 5 - A 1) / 4 \u003d (5 - (-4)) / 4 \u003d 2.25. Here we received not a whole value of the difference, however, it is a rational number, so the formulas for algebraic progression remain the same.
Now add the difference found to A 1 and restore the missing member of the progression. We get: a 1 \u003d - 4, a 2 \u003d - 4 + 2.25 \u003d - 1.75, a 3 \u003d -1.75 + 2.25 \u003d 0.5, a 4 \u003d 0.5 + 2.25 \u003d 2.75, a 5 \u003d 5, + 2.25 \u003d 5, which coincided with the condition of the problem.
Example №4: First Member of Progression
We continue to bring examples of arithmetic progression with the solution. In all previous tasks, the first number of algebraic progression was known. Now consider the task of a different type: let two numbers be given, where a 15 \u003d 50 and a 43 \u003d 37. It is necessary to find, from what date this sequence begins.
The formulas used to date suggest knowledge A 1 and D. In the condition of the problem of these numbers, nothing is unknown. Nevertheless, we will write out the expressions for each member, which there is information: a 15 \u003d a 1 + 14 * d and a 43 \u003d a 1 + 42 * d. We received two equations in which 2 unknown values \u200b\u200b(A 1 and D). This means that the task is reduced to solving a system of linear equations.
The specified system is easiest to decide if to express in each equation A 1, and then compare the obtained expressions. The first equation: a 1 \u003d a 15 - 14 * d \u003d 50 - 14 * d; The second equation: a 1 \u003d a 43 - 42 * d \u003d 37 - 42 * d. Equating these expressions, we obtain: 50 - 14 * d \u003d 37 - 42 * d, where D \u003d (37 - 50) / (42 - 14) \u003d - 0.464 (42 - 14) \u003d - 0.464 (only 3 characters of accuracy after the comma) 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 doubts arise in the result resulting, you can check it, for example, to determine 43 member of the progression, which is set in the condition. We obtain: a 43 \u003d a 1 + 42 * d \u003d 56,496 + 42 * (- 0.464) \u003d 37.008. A small error is associated with the fact that when calculations used rounding to thousandth fractions.
Example number 5: Amount
Now consider several examples with solutions for the amount of arithmetic progression.
Let the following progression of the following form: 1, 2, 3, 4, ... ,. How to calculate the amount of 100 of these numbers?
Thanks to the development of computer technologies, you can decide this task, that is, sequentially folded all the numbers that the computing machine will make it immediately as soon as a person presses the ENTER key. However, the task can be solved in mind if you pay attention that the number of numbers presented is the progression of algebraic, and its difference is 1. Using the formula for the amount, we obtain: S n \u003d n * (A 1 + AN) / 2 \u003d 100 * (1 + 100) / 2 \u003d 5050.
It is curious to note that this task is called "Gaussian", since at the beginning of the XVIII century, the famous German still being at the age of 10 years, was able to solve it in the mind in a few seconds. The boy did not know the formula for the amount of algebraic progression, but he noticed that if we fold the numbers in the edges of the sequence, then one result is always obtained, that is, 1 + 100 \u003d 2 + 99 \u003d 3 + 98 \u003d ..., and since These amounts will be exactly 50 (100/2), then it is enough to multiply 50 to 101 to obtain the correct answer.
Example №6: Amount of members from n to m
Another typical example of the sum of arithmetic progression is the following: Dan such numbers range: 3, 7, 11, 15, ..., you need to find what the sum of its members from 8 to 14 will be equal.
The task is solved in two ways. The first one implies the finding of unknown members from 8 to 14, and then their consistent summation. Since the terms are a bit, then this method is not quite laborious. Nevertheless, it is proposed to solve this problem with the second method, which is more versatile.
The idea is to obtain a formula for the sum of algebraic progression among members M and N, where N\u003e M is integers. We drank two expressions for both cases:
- S M \u003d M * (A M + A 1) / 2.
- S n \u003d n * (a n + a 1) / 2.
Since N\u003e M, it is obvious that the amount of the amount includes the first. The last conclusion means that if you take a difference between these sums, and add a member to it (in case of a difference, it is deducted from the sum s), then we obtain the necessary answer to the task. We have: s mn \u003d s n - s m + am \u003d n * (a 1 + an) / 2 - m * (a 1 + am) / 2 + am \u003d a 1 * (n - m) / 2 + an * N / 2 + AM * (1- m / 2). In this expression, it is necessary to substitute the formula for a n and a m. Then we obtain: S Mn \u003d A 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) \u003d a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d * (3 * m - m 2 - 2) / 2.
The resulting formula is somewhat cumbersome, nevertheless the sum S Mn depends only on n, m, a 1 and d. In our case, a 1 \u003d 3, d \u003d 4, n \u003d 14, m \u003d 8. Substituting these numbers, we obtain: S Mn \u003d 301.
As can be seen from the solutions given, all tasks are based on the knowledge of the expression for the N-th member and the formula for the amount of the set of the first components. Before you begin to solve any of these tasks, it is recommended to carefully read the condition, it is clear to understand what is required to be found, and only then proceed to the solution.
Another advice is in the desire for simplicity, that is, if you can answer the question, without applying complex mathematical calculations, it is necessary to act this way, since in this case the probability is less than a mistake. For example, in an example of arithmetic progression with decision No. 6, it would be possible to dwell on the formula S Mn \u003d n * (A 1 + AN) / 2 - M * (A 1 + AM) / 2 + AM, and split the overall task for individual subtasks (In this case, first find members an and AM).
If there are doubts about the result, it is recommended to check it, as was done in some of the examples given. How to find an arithmetic progression, found out. If you figure it out, it is not so difficult.
The theme "Arithmetic progression" is studied in the general course of algebra in schools in grade 9. This topic is important for further in-depth study of the mathematics of numerical rows. In this article, we will get acquainted with the progression of arithmetic, its difference, as well as with typical tasks that schoolchildren may encounter.
The concept of the progression of algebraic
Numerical progression is a sequence of numbers in which each subsequent element can be obtained from the previous one if you use some mathematical law. There are two simple types of progression: geometric and arithmetic, which is also called algebraic. Let us dwell on it in more detail.
Imagine some rational number, we denote by its symbol A 1, where the index indicates its sequence number in the series under consideration. Add to a 1 some other number, denote it d. Then the second element of the series can be reflected as follows: a 2 \u003d a 1 + d. Now add D again, we get: a 3 \u003d a 2 + d. Continuing this mathematical operation, you can get a number of numbers that will be called arithmetic progression.
As can be understood from the above, to find the N-th element of this sequence, it is necessary to use the formula: a n \u003d a 1 + (n - 1) * d. Indeed, substituting n \u003d 1 in the expression, we get a 1 \u003d a 1, if n \u003d 2, then from the formula: a 2 \u003d a 1 + 1 * d, and so on.
For example, if the difference in the progression of arithmetic is 5, and a 1 \u003d 1, then this means that the numeric number of the type under consideration has the appearance: 1, 6, 11, 16, 21, ... as you can see each member more than the previous one on 5 .
Arithmetic progression difference
From the above definition of the number of numbers under consideration, it follows that for its definition, two numbers need to be known: a 1 and d. The latter is called the difference in this progression. It definitely determines the behavior of the entire row. Indeed, if D is positive, the numeric row will constantly increase, on the contrary, in the case of D negative, increasing numbers in a series of only module will occur, their absolute value will decrease with increasing number N.
What is the difference in the progression of arithmetic? Consider the two basic formulas that are used to calculate this value:
- d \u003d a n + 1 -a n, this formula should be directly directly from the definition of the number of the number of numbers.
- d \u003d (-a 1 + a n) / (n-1), this expression is obtained if you express D from the formula shown in the previous paragraph of the article. Note that this expression appeals to uncertainty (0/0), if n \u003d 1. This is due to the fact that the knowledge of at least 2 elements of the row is necessary to determine its difference.
These two basic formulas are used to solve any tasks to find the difference in progression. However, there is another formula that you also need to know.
The sum of the first elements
The formula with which one can determine the amount of any number of members of the progression of algebraic, according to historical evidence, was first obtained by the "prince" of mathematics of the XVIII century Karl Gauss. The German scientist, while still a boy in the elementary class of the village school, noticed that in order to fold natural numbers in a number of 1 to 100, it is necessary to first sum up the first element and the latter (the value obtained will be equal to the sum of the penultimate and second, pre-last and third elements. , and so on), and then this number should be multiplied by the number of these amounts, that is, by 50.
A formula that reflects the outlined result on a private example, can be generalized for an arbitrary case. It will look like: S n \u003d n / 2 * (a n + a 1). Note that to find the specified value, the knowledge of the difference D is not required if two members of the progression (A N and A 1) are known.
Example number 1. Determine the difference, knowing two members of the series A1 and AN
Let us show how to apply the above formula above. We give a simple example: the difference in the progression of arithmetic unknown, it is necessary to determine what it will be equal if A 13 \u003d -5.6 and A 1 \u003d -12.1.
Since we know the values \u200b\u200bof two elements of the numerical sequence, while one of them is the first number, then you can use formula No. 2 to determine the difference d. We have: d \u003d (- 1 * (- 12.1) + (- 5,6)) / 12 \u003d 0.54167. In the expression, we used the value n \u003d 13, since the member is known for this sequential number.
The resulting difference suggests that the progression is increasing, despite the fact that the data in the problem of the problem elements have a negative value. It can be seen that A 13\u003e A 1, although | A 13 |<|a 1 |.
Example number 2. Positive progression members in Example №1
We use the result obtained in the previous example to solve the new task. It is formulated as follows: from which ordinal number of progression elements in Example number 1 will begin to take positive values?
As it was shown, the progression in which A 1 \u003d -12.1 and d \u003d 0.54167 is increasing, so only positive values \u200b\u200bwill be taken from a certain number of numbers. To determine this number n, it is necessary to solve a simple inequality that mathematically writes so: a n\u003e 0 or using the corresponding formula, rewrite the inequality: a 1 + (n - 1) * d\u003e 0. It is necessary to find an unknown N, express it: n\u003e -1 * a 1 / d + 1. Now it remains to substitute the known difference values \u200b\u200band the first member of the sequence. We obtain: n\u003e -1 * (- 12,1) / 0.54167 + 1 \u003d 23,338 or N\u003e 23,338. Since n can only accept integer values, from the resulting inequality it follows that any members of the series that will have a number more than 23 will be positive.
Check the response received by using the above formula to calculate 23 and 24 items of this arithmetic progression. We have: a 23 \u003d -12,1 + 22 * \u200b\u200b0.54167 \u003d -0,18326 (negative number); a 24 \u003d -12,1 + 23 * 0.54167 \u003d 0.3584 (positive value). Thus, the result obtained is true: starting with n \u003d 24, all members of the numerical row will be greater than zero.
Example number 3. How many logs fit?
We give one curious task: during the billet of the forest it was decided to stack on each other as shown in the figure below. How many logs can be laid in this way, knowing that 10 rows fit?
In such a method of folding logs, one interesting thing can be seen: each subsequent row will contain one log less than the previous one, that is, the progression of algebraic, the difference of which is D \u003d 1. Believing that the number of logs of each row is a member of this progression, and also considering that A 1 \u003d 1 (only one log is placed at the very top), we will find the number A 10. We have: a 10 \u003d 1 + 1 * (10-1) \u003d 10. That is, in the 10th row, which lies on Earth, there will be 10 logs.
The total amount of this "pyramidal" design can be obtained if you use the Gauss formula. We obtain: S 10 \u003d 10/2 * (10 + 1) \u003d 55 logs.
Attention!
This topic has additional
Materials in a special section 555.
For those who are strongly "not very ..."
And for those who are "very ...")
Arithmetic progression is a number of numbers in which each number is greater than (or less) the previous one and the same value.
This topic is often complex and incomprehensible. Indices have bukovok, n-th member of progression, progression of difference - all this as something embarrassing, yeah ... We will understand with a sense of an arithmetic progression, and all at once adjusted).
The concept of arithmetic progression.
Arithmetic progression - the concept is very simple and clear. Doubt? In vain.) See ourselves.
I will write unfinished number of numbers:
1, 2, 3, 4, 5, ...
Can you extend this series? What numbers will go further after the top five? Each ... uh-uh ..., briefly, each realized that go beyond numbers 6, 7, 8, 9, etc.
Complete the task. I give an unfinished number of numbers:
2, 5, 8, 11, 14, ...
You can catch regularity, extend a row, and call seventh number of rows?
If you realized that this is the number 20 - I congratulate you! You not only felt key points of arithmetic progression, But and successfully used them into the case! If not realized - we read on.
And now we will transfer key moments from sensations in mathematics.)
First key moment.
Arithmetic progression is dealing with numbers. This is confused at first. We are accustomed to the equation to decide, build graphs and all that ... And then extend a number, find the number of rows ...
Nothing wrong. Just progression is the first acquaintance with the new section of mathematics. The section is called "Rows" and works precisely with the ranks of numbers and expressions. Get used to.)
The second key moment.
In arithmetic progression, any number differs from the previous on the same magnitude.
In the first example, this difference is one. What a number is neither take, it is more than the previous one per unit. In the second - Troika. Any number more than the previous one. Actually, this is the moment and gives us the opportunity to catch pattern and calculate the subsequent numbers.
Third key point.
This moment is not striking, yes ... But very, very important. Here it is: each number of progression is in its place. There is the first number, there is seventh, there are forty fifth, etc. If they are confused as it fell, the pattern will disappear. Arithmetic progression will disappear. There will be just a number of numbers.
That's the whole point.
Of course, new terms and notation appear in the new topic. They need to know. Otherwise, I will not understand the task. For example, you have to decide something, like:
Write the first six members of the arithmetic progression (A n) if a 2 \u003d 5, d \u003d -2.5.
Inspires?) Cooks, some indices ... And the task, by the way - is no easier. You just need to understand the meaning of terms and designations. Now we will master this thing and return to the task.
Terms and designations.
Arithmetic progression - this is a number of numbers in which each number differs from the previous on the same magnitude.
This value is called . Let's discern with this concept in more detail.
The difference of arithmetic progression.
The difference in arithmetic progression - This is the value that any number of progression more previous one.
One important moment. Please pay attention to the word "more". Mathematically, this means that every number of progression is obtained addness The difference in arithmetic progression to the previous number.
To calculate, let's say second number of rows, it is necessary to first Number add This very difference of arithmetic progression. For calculation fifth - The difference is necessary add to fourth Well, etc.
The difference in arithmetic progression may be positive Then every number of rows will actually more than the previous one. Such progression is called increasing. For example:
8; 13; 18; 23; 28; .....
Here every number it turns out addness A positive number, +5 to the previous one.
The difference may be negative Then every number of rows will turn out less than the previous one. Such a progression is called (you will not believe it!) descending.
For example:
8; 3; -2; -7; -12; .....
Here each number is obtained too addness To the previous, but already negative number, -5.
By the way, when working with the progression is very useful to immediately determine its character - it is increasing or decreasing. It helps to orient in a decision to detect the errors and correct them before its too late.
The difference in arithmetic progression denotes, as a rule, the letter d.
How to find d. ? Very simple. It is necessary to take away from any number of any number previous number. Deduct. By the way, the result of subtraction is called "Difference".)
We define, for example, d. For increasing arithmetic progression:
2, 5, 8, 11, 14, ...
Take any number of rows as we want, for example, 11. Take away from it previous number those. eight:
This is the correct answer. For this arithmetic progression, the difference is three.
You can take exactly any number of progression Because For specific progression d -always the same thing. Although somewhere at the beginning of the row, even in the middle, at least anywhere. You can't only take the very first number. Just because at the very first number No previous.)
By the way, knowing that d \u003d 3., It is very simple to find the seventh number of this progression. Add 3 to the fifth number - get six, it will add 17 to the sixth among the top three, received the seventh number - twenty.
Determine d. For decreasing 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 number of progression, for example -7. The previous one has a number -2. Then:
d \u003d -7 - (-2) \u003d -7 + 2 \u003d -5
The difference arithmetic progression can be any number: integer, fractional, irrational sorts.
Other terms and designations.
Each number of rows is called a member of arithmetic progression.
Each member of progression does your number. Rooms go strictly in a few, without any focus. The first, second, third, fourth, etc. For example, in progression 2, 5, 8, 11, 14, ... Two - this is the first member, the five, the second, eleven - the fourth, well, you understood ...) I ask to clearly realize - numbers themselves can be completely any, whole, fractional, negative, which fell, but numbering numbers - Strictly in order!
How to write a progression in general form? No problem! Each number of rows is written in the form of the letter. To indicate arithmetic progression, it is usually the letter a.. The member number is indicated by the index at the bottom right. Members Write through a comma (or a point with a comma), like this:
a 1, A 2, A 3, A 4, A 5, .....
a 1.- this is the first number a 3. - Third, etc. Nothing cunning. Record this series you can briefly like this: (A N.).
Progression is there Finite and endless.
Finite Progression has a limited number of members. Five, thirty eight, as much as you like. But - a finite number.
Infinite Progression - has an infinite number of members as you can guess.)
Record the final progression through a series can be like this, all the members and the point at the end:
a 1, A 2, A 3, A 4, A 5.
Or so, if a lot of members are:
a 1, A 2, ... A 14, A 15.
In a brief record, you will have to additionally specify the number of members. For example (for twenty members), like this:
(a n), n \u003d 20
The infinite progression can be found in the edition at the end of the row, as in the examples of this lesson.
Now you can make up tasks. The tasks are simple, purely to understand the meaning of arithmetic progression.
Examples of tasks for arithmetic progression.
We will analyze the detailed task that is given above:
1. Remove the first six members of the arithmetic progression (A n) if a 2 \u003d 5, d \u003d -2.5.
We translate the task to understandable language. Dana endless arithmetic progression. Known the second number of this progression: a 2 \u003d 5. The difference in progression is known: d \u003d -2.5. It is necessary to find the first, third, fourth, fifth and sixth members of this progression.
For clarity, write a number of the condition of the task. The first six members, where the second member is a fifth:
a 1, 5, A 3, A 4, A 5, A 6, ....
a 3. = a 2. + d.
We substitute in the expression a 2 \u003d 5 and d \u003d -2.5. Do not forget about minus!
a 3.=5+(-2,5)=5 - 2,5 = 2,5
The third member turned out less than the second. Everything is logical. If the number is greater than the previous on negative The amount, then the number itself will turn out to be less than the previous one. Progression is descending. Okay, consider.) 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
Thus, members with the third were calculated on the sixth. It turned out such a series:
a 1, 5, 2.5, 0, -2.5, -5, ....
It remains to find the first member a 1. According to a well-known second. This is a step in the opposite direction, to the left.) So, the difference between the arithmetic progression d. We must not add to a 2., but take away:
a 1. = a 2. - d.
a 1.=5-(-2,5)=5 + 2,5=7,5
That's all things. Quest answer:
7,5, 5, 2,5, 0, -2,5, -5, ...
Alongorally note that we solved this task recurrent way. It is a terrible word means just a member of the progression according to the previous (adjacent) number. Other ways to work with progress We will look at further.
From this simple task you can make one important output.
Remember:
If we are aware of at least one member, and the difference in arithmetic progression, we can find any member of this progression.
Remember? This simple conclusion allows you to solve most of the tasks of the school course on this topic. All tasks are spinning around three main parameters: a member of arithmetic progression, the difference of progression, a member number of progression. Everything.
Of course, the entire previous algebra is not canceled.) In the progression of inequalities, and equations, and other things are trapped. But for the progression itself - everything spins around three parameters.
For example, consider some popular tasks on this topic.
2. Record the final arithmetic progression as a series if n \u003d 5, d \u003d 0,4, and a 1 \u003d 3.6.
Everything is simple here. Everything is already given. It is necessary to remember how members of the arithmetic progression are considered, to calculate, and write. It is advisable not to miss the words in the assignment condition: "final" and " n \u003d 5."So as not to count to complete scoff.) In this progression, only 5 (five) members:
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 \u003d 4.4 + 0.4 \u003d 4.8
a 5. = a 4. + d \u003d 4.8 + 0.4 \u003d 5.2
Left to record the answer:
3,6; 4; 4,4; 4,8; 5,2.
More task:
3. Determine whether the number is 7 member of arithmetic progression (a n) if a 1 \u003d 4.1; d \u003d 1.2.
Hmm ... who knows him? How to determine something?
Like-like ... Yes to record the progression as a series and see Seven will be there or not! We consider:
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 \u003d 6.5 + 1,2 \u003d 7.7
4,1; 5,3; 6,5; 7,7; ...
Now it is clearly seen that we are just slipped Between 6.5 and 7.7! Seven got into our number of numbers, and, it means, the seven will not be a member of a given progression.
Answer: No.
But the task based on the real version of GIA:
4. There are several consecutive members of arithmetic progression:
...; fifteen; x; nine; 6; ...
Here is recorded a row without end and start. There are no member numbers nor the difference d.. Nothing wrong. To solve the task, it suffices to understand the meaning of arithmetic progression. We look and think that you can discover From this series? What are the parameters of the three main?
Members numbers? There is no single number.
But there are three numbers and attention! - Word "Consistent" In condition. This means that the numbers go strictly in order, without skipping. Is there two two in this row neighboring famous numbers? Yes there is! It is 9 and 6. Therefore, we can calculate the difference between the arithmetic progression! From Sixtur Tutter previous number, i.e. Nine:
There were remaining trifles. What number will be the previous one for IKSA? Fifteen. So, X can be easily found easy. To 15 add the difference in arithmetic progression:
That's all. Answer: x \u003d 12.
The following tasks solve themselves. Note: These tasks are not for formulas. Purely on understanding the meaning of arithmetic progression.) Just write down a number of letters to numbers, look and think.
5. Find a first positive member of arithmetic progression, if A 5 \u003d -3; d \u003d 1.1.
6. It is known that the number 5.5 is a member of the arithmetic progression (A n), where a 1 \u003d 1.6; d \u003d 1.3. Determine the number of the N of this member.
7. It is known that in arithmetic progression A 2 \u003d 4; a 5 \u003d 15.1. Find a 3.
8. Several successive members of the arithmetic progression are written:
...; 15.6; x; 3.4; ...
Find a member of the progression indicated by the letter x.
9. The train began moving from the station, evenly increasing the speed of 30 meters per minute. What will be the train speed in five minutes? Answer Give km / h.
10. It is known that in arithmetic progression A 2 \u003d 5; a 6 \u003d -5. Find A 1..
Answers (in disorder): 7.7; 7.5; 9.5; nine; 0.3; four.
Everything worked out? Wonderful! It is possible to develop an arithmetic progression to a higher level in the next lessons.
Not everything happened? No problem. In a special section of 555, all these tasks taken apart to pieces.) And, of course, describes a simple hands-on technique, which immediately displays the solution of these tasks clearly, it is clear at a glance!
By the way, the puzzle about the train has two problemki, which often stumbles people. One - purely for progression, and the second - the total for all applications of mathematics, and physics, too. This is a translation of dimensions from one to another. It is shown how to solve these problems.
In this lesson, we reviewed the elementary meaning of the arithmetic progression and its main parameters. This is enough to solve almost all tasks on this topic. Adjust d. To the numbers, write a row, everything will be decided.
The solution "on the fingers" is well suited for very short pieces of a number, as in the examples of this lesson. If a row is more complete, the calculations are complicated. For example, if in the task 9 to replace "five minutes" on the "thirty five minutes", The task will become essential.)
And there are once simple tasks in essence, but non-valid calculations, for example:
An arithmetic progression is given (a n). Find a 121 if a 1 \u003d 3, and d \u003d 1/6.
And what, we will add a lot more to 1/6? You can kill it !?
You can.) If you do not know a simple formula, according to which such tasks can be solved in a minute. This formula will be in the next lesson. And this task is solved there. In a minute.)
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The sum of arithmetic progression.
The amount of arithmetic progression is simple. And in meaning, and by the formula. But the tasks on this topic are all sorts of. From elementary to quite solid.
First we will deal with the meaning and summary formula. And then they shave. In my pleasure.) The meaning of the amount is simple as soap. To find the amount of arithmetic progression, you just need to gently fold all its members. If these members are small, you can put without any formulas. But if a lot, or very much ... Addition strains.) In this case, the formula saves.
The sum of the amount looks simple:
Let's discern that the beaks are included in the formula. This will clarify much.
S N. - Amount of arithmetic progression. Result of addition all Members, S. first by last. It is important. It is precisely everything Members in a row, without skipping and jumps. And, it is, starting with first. In tasks, such as finding the amount of the third and eighth members, or the amount of members from the fifth on the twentieth - the direct use of the formula will disappoint.)
a 1. - first Member of progression. Everything is clear here, it's just first number of rows.
a N. - last Member of progression. The last number of rows. Not very familiar name, but, in applied to the amount, it is very good. Further you will see.
n. - The number of the last member. It is important to understand that in the formula this number coincides with the number of members folded.
Defend with concept last Member a N.. Backup question: what a member will last If Dana infinite Arithmetic progression?)
For a sure answer, you need to understand the elementary meaning of arithmetic progression and ... carefully read the task!)
In the task of finding the sum of arithmetic progression, always appears (directly or indirectly) the last member who should be limited to. Otherwise the ultimate, concrete amount simply does not exist. To solve, it is important that the progression is set: the ultimate, or endless. It is important that it is asked: near the numbers, or the formula of the N-th member.
The most important thing is to understand that the formula works with the first member of the progression to a member with the number n. Actually, the full name of the formula looks like this: the sum of the first members of the arithmetic progression. The number of these very first members, i.e. n.is determined solely by the task. In the task, all this valuable information is often encrypted, yes ... but nothing, in the examples below, we strip these secrets.)
Examples of tasks for the amount of arithmetic progression.
First of all, useful information:
The main complexity in the tasks on the amount of arithmetic progression is to properly define the elements of the formula.
These very elements of the compilers of tasks are encrypted with an infinite fantasy.) The main thing is not to be afraid. Understanding the essence of the elements, it is enough to decipher them. We analyze several examples in detail. Let's start with a task based on real GIA.
1. Arithmetic progression is given by condition: a n \u003d 2N-3.5. Find the amount of the first 10 of its members.
Good task. Light.) To us to determine the amount by the formula of what you need to know? First Member a 1., last dick a N.yes the number of the last member n.
Where to get the number of the last member n.? Yes, there, in the condition! It says: Find the amount the first 10 members. Well, with what number will be last, Tenth member?) You will not believe his number - the tenth!) It became instead of a N. In the formula we will substitute a 10., and instead n. - dozen. 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 considered by the formula of the N-th member, which is given in the condition of the problem. Do not know how to do it? Visit the previous lesson, without this - in no way.
a 1.\u003d 2 · 1 - 3.5 \u003d -1.5
a 10.\u003d 2 · 10 - 3.5 \u003d 16.5
S N. = S 10..
We found out the value of all elements of the formula of the sum of arithmetic progression. It remains to substitute them, but count:
That's all things. Answer: 75.
Another task based on GIA. A little more complicated:
2. The arithmetic progression (a n) is given, the difference of which is 3.7; a 1 \u003d 2.3. Find the amount of the first 15 of its members.
Immediately write the summary formula:
This formula allows us to find the value of any member by its number. We are looking for a simple substitution:
a 15 \u003d 2.3 + (15-1) · 3,7 \u003d 54.1
It remains to substitute all the elements in the formula sum of arithmetic progression and calculate the answer:
Answer: 423.
By the way, if in the sum of the sum instead a N. Just substitute the formula of the N-th member, we get:
We give the like, we obtain a new formula of the sum of the members of the arithmetic progression:
 |
As you can see, it does not require a N-th member a N.. In some tasks, this formula helps great, yes ... you can remember this formula. And you can simply get it at the right moment, as here. After all, the formula of the sum and the formula of the N-th member should be remembered.)
Now task in the form of a brief encryption):
3. Find the sum of all positive two-digit numbers, multiple three.
In how! Neither your first member nor the last nor progression in general ... how to live!?
You have to think your head and pull out all the elements of the sum of arithmetic progression from the condition. What is two-digit numbers - we know. Of the two tsiferok consist.) What two-digit number will be first? 10, it is necessary to believe.) A last thing Double-digit number? 99, of course! Behind him already three-digit ...
Press three ... um ... these are the numbers that are divided into three aimed, here! A dozen is not divided into three, 11 is not divided ... 12 ... divided! So, something is evaporated. You can already record a number of task condition:
12, 15, 18, 21, ... 96, 99.
Will this range of arithmetic progress? Sure! Each member differs from the previous one strictly on the top three. If you add 2, or 4 to a member, say, the result, i.e. A new number, no longer shares aimed at 3. Before the heap, you can immediately and the difference in arithmetic progression to determine: d \u003d 3. Come true!)
So, you can safely write some progression parameters:
And what will be the number n. last member? The one who thinks is that 99 - fatally mistaken ... rooms - they always go in a row, and we have members - jump over the top three. They do not coincide.
There are two ways to solve. One way - for overhauls. You can paint the progression, the whole range of numbers, and calculate the number of members with your finger.) The second way is for thoughtful. It is necessary to recall the formula of the N-th member. If the formula apply to our task, we get that 99 is a thirtieth member of the progression. Those. n \u003d 30.
We look at the formula sum of arithmetic progression:
We look, and rejoice.) We pulled out the task out of the terms of the task everything you need to calculate the amount:
a 1.= 12.
a 30.= 99.
S N. = S 30..
Elementary arithmetic remains. We substitute the number in the formula and believe:
Answer: 1665.
Another type of popular task:
4. Dana Arithmetic Progression:
-21,5; -20; -18,5; -17; ...
Find the amount of members from the twentieth to thirty fourth.
We look at the sum of the sum and ... are upset.) Formula, remind, considers the amount from the first Member. And the task needs to be considered with twentieth ... The formula does not work.
You can, of course, paint the entire progression in a row, but to post the members from 20 to 34. But ... somehow stupid and long it turns out, right?)
There is a more elegant solution. We break our row into two parts. The first part will be from the first member of the nineteenth. The second part of - from the twentieth to thirty-used. It is clear that if we consider the amount of the members first S 1-19., yes, add up with the sum of the members of the second part S 20-34., I will receive the amount of progression from the first member of the thirty fourth S 1-34.. Like this:
S 1-19. + S 20-34. = S 1-34.
From here it can be seen that finding the amount S 20-34. You can easily subtract
S 20-34. = S 1-34. - S 1-19.
Both amounts in the right part are considered from the first Member, i.e. It is quite applicable to the standard summary formula. Start?
Pull out the problem of the problem of the progression of the problem:
d \u003d 1.5.
a 1.= -21,5.
To calculate the sums of the first 19 and the first 34 members, we will need the 19th and 34th members. We consider them according to the formula of the N-th member, as in the task 2:
a 19.\u003d -21,5 + (19-1) · 1,5 \u003d 5.5
a 34.\u003d -21,5 + (34-1) · 1,5 \u003d 28
There is nothing left. From the amount of 34 members to take out the amount of 19 members:
S 20-34 \u003d S 1-34 - S 1-19 \u003d 110.5 - (-152) \u003d 262.5
Answer: 262.5
One important remark! In solving this task there is a very useful chip. Instead of direct calculation what is needed (s 20-34), We counted what seems to be needed - s 1-19. And then then determined and S 20-34., threading from the full result unnecessary. Such a "fint ears" often saves in evil tasks.)
In this lesson, we reviewed the tasks for which it suffices to understand the meaning of the sum of arithmetic progression. Well, a couple of formulas need to know.)
Practical advice:
When solving any task on the amount of arithmetic progression, I recommend immediately discharge the two main formulas from this topic.
The formula of the N-th member:
These formulas will immediately prompt that you need to look for, in which direction to think to solve the task. Helps.
And now tasks for self-decisions.
5. Find the sum of all two-digit numbers that are not divided by three.
Cool?) Tip is hidden in the comment to the task 4. Well, the task 3 will help.
6. Arithmetic progression is set by condition: A 1 \u003d -5.5; a n + 1 \u003d a n +0.5. Find the amount of the first 24 of its members.
Unusual?) This is a recurrent formula. About it can be read in the previous lesson. Do not ignore the link, such tasks in GIA are often found.
7. Vasya has accumulated for the holiday of money. Whole 4550 rubles! And I decided to give my favorite person myself (myself) for several days of happiness). To live beautifully, without refusing. Spend 500 rubles on the first day, and in every subsequent day spend 50 rubles more than in the previous one! Until the stock of money will end. How many days of happiness did Vasi come?
Difficult?) An additional formula will help from the task 2.
Answers (in disorder): 7, 3240, 6.
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