The formula for the difference of an arithmetic progression. The sum of the first n-terms of an arithmetic progression
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
On this theoretical material ends and we move on to solving common practical problems.
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. 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 members 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
The sum of an arithmetic progression.
The sum of an arithmetic progression is a simple thing. Both in meaning and in formula. But there are all sorts of tasks on this topic. From elementary to quite solid.
First, let's deal with the meaning and formula of the sum. And then we'll decide. For your own pleasure.) The meaning of the sum is as simple as lowing. To find the sum of an arithmetic progression, you just need to carefully add all its members. If these terms are few, you can add without any formulas. But if there is a lot, or a lot ... addition is annoying.) In this case, the formula saves.
The sum formula is simple:
Let's figure out what kind of letters are included in the formula. This will clear up a lot.
S n is the sum of an arithmetic progression. Addition result all members, with first By last. It is important. Add up exactly All members in a row, without gaps and jumps. And, exactly, starting from first. In problems like finding the sum of the third and eighth terms, or the sum of terms five through twentieth, direct application of the formula will be disappointing.)
a 1 - first member of the progression. Everything is clear here, it's simple first row number.
a n- last member of the progression. last number row. Not a very familiar name, but, when applied to the amount, it is very suitable. Then you will see for yourself.
n is the number of the last member. It is important to understand that in the formula this number coincides with the number of added terms.
Let's define the concept last member a n. Filling question: what kind of member will last, if given endless arithmetic progression?
For a confident answer, you need to understand the elementary meaning of an arithmetic progression and ... read the assignment carefully!)
In the task of finding the sum of an arithmetic progression, the last term always appears (directly or indirectly), which should be limited. Otherwise, a finite, specific amount just doesn't exist. For the solution, it does not matter what kind of progression is given: finite or infinite. It doesn't matter how it is given: by a series of numbers, or by the formula of the nth member.
The most important thing is to understand that the formula works from the first term of the progression to the term with the number n. Actually, the full name of the formula looks like this: the sum of the first n terms of an arithmetic progression. The number of these very first members, i.e. n, is determined solely by the task. In the task, all this valuable information is often encrypted, yes ... But nothing, in the examples below we will reveal these secrets.)
Examples of tasks for the sum of an arithmetic progression.
First of all, helpful information:
The main difficulty in tasks for the sum of an arithmetic progression is the correct determination of the elements of the formula.
The authors of the assignments encrypt these very elements with boundless imagination.) The main thing here is not to be afraid. Understanding the essence of the elements, it is enough just to decipher them. Let's take a look at a few examples in detail. Let's start with a task based on a real GIA.
1. The arithmetic progression is given by the condition: a n = 2n-3.5. Find the sum of the first 10 terms.
Good job. Easy.) To determine the amount according to the formula, what do we need to know? First member a 1, last term a n, yes the number of the last term n.
Where to get the last member number n? Yes, in the same place, in the condition! It says find the sum first 10 members. Well, what number will it be last, tenth member?) You won’t believe it, his number is tenth!) Therefore, instead of a n we will substitute into the formula a 10, but instead n- ten. Again, the number of the last member is the same as the number of members.
It remains to be determined a 1 And a 10. This is easily calculated by the formula of the nth term, which is given in the problem statement. Don't know how to do it? Visit the previous lesson, without this - nothing.
a 1= 2 1 - 3.5 = -1.5
a 10\u003d 2 10 - 3.5 \u003d 16.5
S n = S 10.
We found out the meaning of all elements of the formula for the sum of an arithmetic progression. It remains to substitute them, and count:
That's all there is to it. Answer: 75.
Another task based on the GIA. A little more complicated:
2. Given an arithmetic progression (a n), the difference of which is 3.7; a 1 \u003d 2.3. Find the sum of the first 15 terms.
We immediately write the sum formula:
This formula allows us to find the value of any member by its number. We are looking for a simple substitution:
a 15 \u003d 2.3 + (15-1) 3.7 \u003d 54.1
It remains to substitute all the elements in the formula for the sum of an arithmetic progression and calculate the answer:
Answer: 423.
By the way, if in the sum formula instead of a n just substitute the formula of the nth term, we get:
We give similar ones, we get a new formula for the sum of members of an arithmetic progression:
 |
As you can see, there is no need nth member a n. In some tasks, this formula helps out a lot, yes ... You can remember this formula. And you can simply withdraw it at the right time, as here. After all, the formula for the sum and the formula for the nth term must be remembered in every way.)
Now the task in the form of a short encryption):
3. Find the sum of all positive two-digit numbers that are multiples of three.
How! No first member, no last, no progression at all... How to live!?
You will have to think with your head and pull out from the condition all the elements of the sum of an arithmetic progression. What are two-digit numbers - we know. They consist of two numbers.) What two-digit number will first? 10, presumably.) last thing two digit number? 99, of course! The three-digit ones will follow him ...
Multiples of three... Hm... These are numbers that are evenly divisible by three, here! Ten is not divisible by three, 11 is not divisible... 12... is divisible! So, something is emerging. You can already write a series according to the condition of the problem:
12, 15, 18, 21, ... 96, 99.
Will this series be an arithmetic progression? Certainly! Each term differs from the previous one strictly by three. If 2, or 4, is added to the term, say, the result, i.e. a new number will no longer be divided by 3. You can immediately determine the difference of the arithmetic progression to the heap: d = 3. Useful!)
So, we can safely write down some progression parameters:
What will be the number n last member? Anyone who thinks that 99 is fatally mistaken ... Numbers - they always go in a row, and our members jump over the top three. They don't match.
There are two solutions here. One way is for the super hardworking. You can paint the progression, the whole series of numbers, and count the number of terms with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If the formula is applied to our problem, we get that 99 is the thirtieth member of the progression. Those. n = 30.
We look at the formula for the sum of an arithmetic progression:
We look and rejoice.) We pulled out everything necessary for calculating the amount from the condition of the problem:
a 1= 12.
a 30= 99.
S n = S 30.
What remains is elementary arithmetic. Substitute the numbers in the formula and calculate:
Answer: 1665
Another type of popular puzzles:
4. An arithmetic progression is given:
-21,5; -20; -18,5; -17; ...
Find the sum of terms from the twentieth to thirty-fourth.
We look at the sum formula and ... we are upset.) The formula, let me remind you, calculates the sum from the first member. And in the problem you need to calculate the sum since the twentieth... The formula won't work.
You can, of course, paint the entire progression in a row, and put the members from 20 to 34. But ... somehow it turns out stupidly and for a long time, right?)
There is a more elegant solution. Let's break our series into two parts. The first part will from the first term to the nineteenth. Second part - twenty to thirty-four. It is clear that if we calculate the sum of the terms of the first part S 1-19, let's add it to the sum of the members of the second part S 20-34, we get the sum of the progression from the first term to the thirty-fourth S 1-34. Like this:
S 1-19 + S 20-34 = S 1-34
This shows that to find the sum S 20-34 can be done by simple subtraction
S 20-34 = S 1-34 - S 1-19
Both sums on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Are we getting started?
We extract the progression parameters from the task condition:
d = 1.5.
a 1= -21,5.
To calculate the sums of the first 19 and the first 34 terms, we will need the 19th and 34th terms. We count them according to the formula of the nth term, as in problem 2:
a 19\u003d -21.5 + (19-1) 1.5 \u003d 5.5
a 34\u003d -21.5 + (34-1) 1.5 \u003d 28
There is nothing left. Subtract the sum of 19 terms from the sum of 34 terms:
S 20-34 = S 1-34 - S 1-19 = 110.5 - (-152) = 262.5
Answer: 262.5
One important note! There is a very useful feature in solving this problem. Instead of direct calculation what you need (S 20-34), we counted what, it would seem, is not needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the full result. Such a "feint with the ears" often saves in evil puzzles.)
In this lesson, we examined problems for which it is enough to understand the meaning of the sum of an arithmetic progression. Well, you need to know a couple of formulas.)
When solving any problem for the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.
Formula of the nth member:
These formulas will immediately tell you what to look for, in which direction to think in order to solve the problem. Helps.
And now the tasks for independent solution.
5. Find the sum of all two-digit numbers that are not divisible by three.
Cool?) The hint is hidden in the note to problem 4. Well, problem 3 will help.
6. Arithmetic progression is given by the condition: a 1 =-5.5; a n+1 = a n +0.5. Find the sum of the first 24 terms.
Unusual?) This is a recurrent formula. You can read about it in the previous lesson. Do not ignore the link, such puzzles are often found in the GIA.
7. Vasya saved up money for the Holiday. As much as 4550 rubles! And I decided to give the most beloved person (myself) a few days of happiness). Live beautifully without denying yourself anything. Spend 500 rubles on the first day, and spend 50 rubles more on each subsequent day than on the previous one! Until the money runs out. How many days of happiness did Vasya have?
Is it difficult?) An additional formula from task 2 will help.
Answers (in disarray): 7, 3240, 6.
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 general education school(Grade 9) one of 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.
Arithmetic or is such a set of ordered rational numbers, each member of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of an ordered series of numbers and the difference, you can restore the entire arithmetic progression.
Let's take an example. The next sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the considered type of progression, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).
Important formulas
We now give the basic formulas that will be needed to solve problems using an arithmetic progression. Let a n denote the nth member of the sequence, where n is an integer. Let us denote the difference Latin letter d. Then the following expressions are true:
- To determine the value of the nth term, the formula is suitable: a n \u003d (n-1) * d + a 1.
- To determine the sum of the first n terms: S n = (a n + a 1)*n/2.
To understand any examples of an arithmetic progression with a solution in grade 9, it is enough to remember these two formulas, since any problems of the type under consideration are built on their use. Also, do not forget that the progression difference is determined by the formula: d = a n - a n-1 .
Example #1: Finding an Unknown Member
We give a simple example of an arithmetic progression and the formulas that must be used to solve.
Let the sequence 10, 8, 6, 4, ... be given, it is necessary to find five terms in it.
It already follows from the conditions of the problem that the first 4 terms are known. The fifth can be defined in two ways:
- Let's calculate the difference first. We have: d = 8 - 10 = -2. Similarly, one could take any two other terms standing next to each other. For example, d = 4 - 6 = -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 = 4 + (-2) = 2.
- The second method also requires knowledge of the difference of the progression in question, so you first need to determine it, as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for the n number of the sequence. We have: a n \u003d (n - 1) * d + a 1 \u003d (n - 1) * (-2) + 10 \u003d 12 - 2 * n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.
As you can see, both solutions lead to the same result. Note that in this example the difference d of the progression is negative. Such sequences are called decreasing because each successive term is less than the previous one.
Example #2: progression difference
Now let's complicate the task a bit, give an example of how to find the difference of an arithmetic progression.
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 \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and split common task into separate subtasks (in this case first find the terms a n and a m).
If there are doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. How to find an arithmetic progression, found out. Once you figure it out, it's not that hard.
IV Yakovlev | Materials on mathematics | MathUs.ru
Arithmetic progression
An arithmetic progression is a special kind of sequence. Therefore, before defining the arithmetic (and then geometric) progression, we need to briefly discuss important concept number sequence.
Subsequence
Imagine a device on the screen of which some numbers are displayed one after another. Let's say 2; 7; 13; 1; 6; 0; 3; : : : Such a set of numbers is just an example of a sequence.
Definition. A numerical sequence is a set of numbers in which each number can be assigned a unique number (that is, put in correspondence with a single natural number)1. The number with number n is called nth member sequences.
So, in the above example, the first number has the number 2, which is the first member of the sequence, which can be denoted by a1 ; the number five has the number 6 which is the fifth member of the sequence, which can be denoted a5 . In general, the nth member of a sequence is denoted by an (or bn , cn , etc.).
A very convenient situation is when the nth member of the sequence can be specified by some formula. For example, the formula an = 2n 3 specifies the sequence: 1; 1; 3; 5; 7; : : : The formula an = (1)n defines the sequence: 1; 1; 1; 1; : : :
Not every set of numbers is a sequence. So, a segment is not a sequence; it contains ¾too many¿ numbers to be renumbered. The set R of all real numbers is also not a sequence. These facts are proved in the course of mathematical analysis.
Arithmetic progression: basic definitions
Now we are ready to define an arithmetic progression.
Definition. An arithmetic progression is a sequence in which each term (starting from the second) is equal to the sum of the previous term and some fixed number (called the difference of the arithmetic progression).
For example, sequence 2; 5; 8; eleven; : : : is an arithmetic progression with first term 2 and difference 3. Sequence 7; 2; 3; 8; : : : is an arithmetic progression with first term 7 and difference 5. Sequence 3; 3; 3; : : : is an arithmetic progression with zero difference.
Equivalent definition: A sequence an is called an arithmetic progression if the difference an+1 an is a constant value (not dependent on n).
An arithmetic progression is said to be increasing if its difference is positive, and decreasing if its difference is negative.
1 And here is a more concise definition: a sequence is a function defined on the set of natural numbers. For example, the sequence of real numbers is the function f: N! R.
By default, sequences are considered infinite, that is, containing an infinite number of numbers. But no one bothers to consider finite sequences as well; in fact, any finite set of numbers can be called a finite sequence. For example, the final sequence 1; 2; 3; 4; 5 consists of five numbers.
Formula of the nth member of an arithmetic progression
It is easy to understand that an arithmetic progression is completely determined by two numbers: the first term and the difference. Therefore, the question arises: how, knowing the first term and the difference, find an arbitrary term of an arithmetic progression?
It is not difficult to obtain the desired formula for the nth term of an arithmetic progression. Let an
arithmetic progression with difference d. We have: | |
an+1 = an + d (n = 1; 2; : ::): | |
In particular, we write: | |
a2 = a1 + d; | |
a3 = a2 + d = (a1 + d) + d = a1 + 2d; | |
a4 = a3 + d = (a1 + 2d) + d = a1 + 3d; | |
and now it becomes clear that the formula for an is: | |
an = a1 + (n 1)d: |
Task 1. In arithmetic progression 2; 5; 8; eleven; : : : find the formula of the nth term and calculate the hundredth term.
Solution. According to formula (1) we have:
an = 2 + 3(n 1) = 3n 1:
a100 = 3 100 1 = 299:
Property and sign of arithmetic progression
property of an arithmetic progression. In arithmetic progression an for any
In other words, each member of the arithmetic progression (starting from the second) is the arithmetic mean of the neighboring members.
Proof. We have: | ||||
a n 1+ a n+1 | (an d) + (an + d) | |||
which is what was required.
More generally, the arithmetic progression an satisfies the equality
a n = a n k+ a n+k
for any n > 2 and any natural k< n. Попробуйте самостоятельно доказать эту формулу тем же самым приёмом, что и формулу (2 ).
It turns out that formula (2) is not only a necessary but also a sufficient condition for a sequence to be an arithmetic progression.
Sign of an arithmetic progression. If equality (2) holds for all n > 2, then the sequence an is an arithmetic progression.
Proof. Let's rewrite the formula (2) as follows:
a na n 1= a n+1a n:
This shows that the difference an+1 an does not depend on n, and this just means that the sequence an is an arithmetic progression.
The property and sign of an arithmetic progression can be formulated as one statement; for convenience, we will do this for three numbers (this is the situation that often occurs in problems).
Characterization of an arithmetic progression. Three numbers a, b, c form an arithmetic progression if and only if 2b = a + c.
Problem 2. (Moscow State University, Faculty of Economics, 2007) Three numbers 8x, 3 x2 and 4 in the specified order form a decreasing arithmetic progression. Find x and write the difference of this progression.
Solution. By the property of an arithmetic progression, we have:
2(3 x2 ) = 8x 4 , 2x2 + 8x 10 = 0 , x2 + 4x 5 = 0 , x = 1; x=5:
If x = 1, then a decreasing progression of 8, 2, 4 is obtained with a difference of 6. If x = 5, then an increasing progression of 40, 22, 4 is obtained; this case does not work.
Answer: x = 1, the difference is 6.
The sum of the first n terms of an arithmetic progression
The legend says that once the teacher told the children to find the sum of numbers from 1 to 100 and sat down to read the newspaper quietly. However, within a few minutes, one boy said that he had solved the problem. It was 9-year-old Carl Friedrich Gauss, later one of the greatest mathematicians in history.
Little Gauss's idea was this. Let
S = 1 + 2 + 3 + : : : + 98 + 99 + 100:
Let's write this sum in reverse order:
S = 100 + 99 + 98 + : : : + 3 + 2 + 1;
and add these two formulas:
2S = (1 + 100) + (2 + 99) + (3 + 98) + : : : + (98 + 3) + (99 + 2) + (100 + 1):
Each term in brackets is equal to 101, and there are 100 such terms in total. Therefore
2S = 101 100 = 10100;
We use this idea to derive the sum formula
S = a1 + a2 + : : : + an + a n n: (3)
A useful modification of formula (3) is obtained by substituting the formula for the nth term an = a1 + (n 1)d into it:
2a1 + (n 1)d | |||||
Task 3. Find the sum of all positive three-digit numbers divisible by 13.
Solution. Three-digit numbers that are multiples of 13 form an arithmetic progression with the first term 104 and the difference 13; The nth term of this progression is:
an = 104 + 13(n 1) = 91 + 13n:
Let's find out how many members our progression contains. To do this, we solve the inequality:
an 6999; 91 + 13n 6999;
n 6 908 13 = 6911 13; n 6 69:
So there are 69 members in our progression. According to the formula (4) we find the required amount:
S = 2 104 + 68 13 69 = 37674: 2
