When studying algebra in a secondary school (grade 9), one of the important topics is the study of numerical sequences, which include progressions - geometric and arithmetic. In this article, we will consider an arithmetic progression and examples with solutions.
What is an arithmetic progression?
To understand this, it is necessary to give a definition of the progression under consideration, as well as to give the basic formulas that will be further used in solving problems.
It is known that in some algebraic progression the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.
Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . We substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 \u003d 6 + 6 * d. From this expression, you can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, the first part of the problem was answered.
To restore the sequence to the 7th member, you should use the definition of an algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d, and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2=8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16 and 7 = 18.
Example #3: making a progression
Let us complicate the condition of the problem even more. Now you need to answer the question of how to find an arithmetic progression. We can give the following example: two numbers are given, for example, 4 and 5. It is necessary to make an algebraic progression so that three more terms fit between these.
Before starting to solve this problem, it is necessary to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 \u003d -4 and a 5 \u003d 5. Having established this, we proceed to a task that is similar to the previous one. Again, for the nth term, we use the formula, we get: a 5 \u003d a 1 + 4 * d. From: d \u003d (a 5 - a 1) / 4 \u003d (5 - (-4)) / 4 \u003d 2.25. Here the difference is not an integer value, but it is a rational number, so the formulas for the algebraic progression remain the same.
Now let's add the found difference to a 1 and restore the missing members of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 \u003d 2.75 + 2.25 \u003d 5, which coincided with the condition of the problem.
Example #4: The first member of the progression
We continue to give examples of an arithmetic progression with a solution. In all previous problems, the first number of the algebraic progression was known. Now consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find from what number this sequence begins.
The formulas that have been used so far assume knowledge of a 1 and d. Nothing is known about these numbers in the condition of the problem. Nevertheless, let's write out the expressions for each term about which we have information: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We got two equations in which there are 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.
The specified system is easiest to solve if you express a 1 in each equation, and then compare the resulting expressions. First equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 \u003d a 43 - 42 * d \u003d 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d \u003d 37 - 42 * d, whence the difference d \u003d (37 - 50) / (42 - 14) \u003d - 0.464 (only 3 decimal places are given).
Knowing d, you can use any of the 2 expressions above for a 1 . For example, first: a 1 \u003d 50 - 14 * d \u003d 50 - 14 * (- 0.464) \u003d 56.496.
If there are doubts about the result, you can check it, for example, determine the 43rd member of the progression, which is specified in the condition. We get: a 43 \u003d a 1 + 42 * d \u003d 56.496 + 42 * (- 0.464) \u003d 37.008. A small error is due to the fact that rounding to thousandths was used in the calculations.
Example #5: Sum
Now let's look at some examples with solutions for the sum of an arithmetic progression.
Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?
Thanks to the development of computer technology, this problem can be solved, that is, sequentially add up all the numbers, which the computer will do as soon as a person presses the Enter key. However, the problem can be solved mentally if you pay attention that the presented series of numbers is an algebraic progression, and its difference is 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.
It is curious to note that this problem is called "Gaussian", since at the beginning of the 18th century the famous German, still at the age of only 10 years old, was able to solve it in his mind in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add pairs of numbers located at the edges of the sequence, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer, it is enough to multiply 50 by 101.
Example #6: sum of terms from n to m
Another typical example of the sum of an arithmetic progression is the following: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its terms from 8 to 14 will be.
The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them up sequentially. Since there are few terms, this method is not laborious enough. Nevertheless, it is proposed to solve this problem by the second method, which is more universal.
The idea is to get a formula for the sum of an algebraic progression between terms m and n, where n > m are integers. Let us write out two expressions for the sum for both cases:
- S m \u003d m * (a m + a 1) / 2.
- S n \u003d n * (a n + a 1) / 2.
Since n > m, it is obvious that the 2 sum includes the first one. The last conclusion means that if we take the difference between these sums, and add the term a m to it (in the case of taking the difference, it is subtracted from the sum S n), then we get the necessary answer to the problem. We have: S mn \u003d S n - S m + a m \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m \u003d a 1 * (n - m) / 2 + a n * n / 2 + a m * (1- m / 2). It is necessary to substitute formulas for a n and a m into this expression. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d * (3 * m - m 2 - 2) / 2.
The resulting formula is somewhat cumbersome, however, the sum S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.
As can be seen from the above solutions, all problems are based on the knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before you start solving any of these problems, it is recommended that you carefully read the condition, clearly understand what you want to find, and only then proceed with the solution.
Another tip is to strive for simplicity, that is, if you can answer the question without using complex mathematical calculations, then you need to do just that, since in this case the probability of making a mistake is less. For example, in the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and break the general task into separate subtasks (in this case, first find the terms a n and a m).
If there are doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. How to find an arithmetic progression, found out. Once you figure it out, it's not that hard.
Someone treats the word "progression" with caution, as a very complex term from the sections of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi counter (where they still remain). And to understand the essence (and in mathematics there is nothing more important than “to understand the essence”) of an arithmetic sequence is not so difficult, having analyzed a few elementary concepts.
Mathematical number sequence
It is customary to call a numerical sequence a series of numbers, each of which has its own number.
and 1 is the first member of the sequence;
and 2 is the second member of the sequence;
and 7 is the seventh member of the sequence;
and n is the nth member of the sequence;
However, not any arbitrary set of figures and numbers interests us. We will focus our attention on a numerical sequence in which the value of the n-th member is related to its ordinal number by a dependence that can be clearly formulated mathematically. In other words: the numerical value of the n-th number is some function of n.
a - value of a member of the numerical sequence;
n is its serial number;
f(n) is a function where the ordinal in the numeric sequence n is the argument.
Definition
An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth member of an arithmetic sequence is as follows:
a n - the value of the current member of the arithmetic progression;
a n+1 - the formula of the next number;
d - difference (a certain number).
It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one, and such an arithmetic progression will be increasing.
In the graph below, it is easy to see why the number sequence is called "increasing".
In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.
The value of the specified member
Sometimes it is necessary to determine the value of some arbitrary term a n of an arithmetic progression. You can do this by calculating successively the values of all members of the arithmetic progression, from the first to the desired one. However, this way is not always acceptable if, for example, it is necessary to find the value of the five thousandth or eight millionth term. The traditional calculation will take a long time. However, a specific arithmetic progression can be investigated using certain formulas. There is also a formula for the nth term: the value of any member of an arithmetic progression can be determined as the sum of the first member of the progression with the difference of the progression, multiplied by the number of the desired member, minus one.
The formula is universal for increasing and decreasing progression.
An example of calculating the value of a given member
Let's solve the following problem of finding the value of the n-th member of an arithmetic progression.
Condition: there is an arithmetic progression with parameters:
The first member of the sequence is 3;
The difference in the number series is 1.2.
Task: it is necessary to find the value of 214 terms
Solution: to determine the value of a given member, we use the formula:
a(n) = a1 + d(n-1)
Substituting the data from the problem statement into the expression, we have:
a(214) = a1 + d(n-1)
a(214) = 3 + 1.2 (214-1) = 258.6
Answer: The 214th member of the sequence is equal to 258.6.
The advantages of this calculation method are obvious - the entire solution takes no more than 2 lines.
Sum of a given number of terms
Very often, in a given arithmetic series, it is required to determine the sum of the values of some of its segments. It also doesn't need to calculate the values of each term and then sum them up. This method is applicable if the number of terms whose sum must be found is small. In other cases, it is more convenient to use the following formula.
The sum of the members of an arithmetic progression from 1 to n is equal to the sum of the first and nth members, multiplied by the member number n and divided by two. If in the formula the value of the n-th member is replaced by the expression from the previous paragraph of the article, we get:
Calculation example
For example, let's solve a problem with the following conditions:
The first term of the sequence is zero;
The difference is 0.5.
In the problem, it is required to determine the sum of the terms of the series from 56 to 101.
Solution. Let's use the formula for determining the sum of the progression:
s(n) = (2∙a1 + d∙(n-1))∙n/2
First, we determine the sum of the values of 101 members of the progression by substituting the given conditions of our problem into the formula:
s 101 = (2∙0 + 0.5∙(101-1))∙101/2 = 2 525
Obviously, in order to find out the sum of the terms of the progression from the 56th to the 101st, it is necessary to subtract S 55 from S 101.
s 55 = (2∙0 + 0.5∙(55-1))∙55/2 = 742.5
So the sum of the arithmetic progression for this example is:
s 101 - s 55 \u003d 2,525 - 742.5 \u003d 1,782.5
Example of practical application of arithmetic progression
At the end of the article, let's return to the example of the arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider such an example.
Getting into a taxi (which includes 3 km) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles / km. Travel distance 30 km. Calculate the cost of the trip.
1. Let's discard the first 3 km, the price of which is included in the landing cost.
30 - 3 = 27 km.
2. Further calculation is nothing more than parsing an arithmetic number series.
The member number is the number of kilometers traveled (minus the first three).
The value of the member is the sum.
The first term in this problem will be equal to a 1 = 50 rubles.
Progression difference d = 22 p.
the number of interest to us - the value of the (27 + 1)th member of the arithmetic progression - the meter reading at the end of the 27th kilometer - 27.999 ... = 28 km.
a 28 \u003d 50 + 22 ∙ (28 - 1) \u003d 644
Calculations of calendar data for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of the celestial body to the luminary. In addition, various numerical series are successfully used in statistics and other applied branches of mathematics.
Another kind of number sequence is geometric
A geometric progression is characterized by a large, compared with an arithmetic, rate of change. It is no coincidence that in politics, sociology, medicine, often, in order to show the high speed of the spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops exponentially.
The N-th member of the geometric number series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first member is 1, the denominator is 2, respectively, then:
n=1: 1 ∙ 2 = 2
n=2: 2 ∙ 2 = 4
n=3: 4 ∙ 2 = 8
n=4: 8 ∙ 2 = 16
n=5: 16 ∙ 2 = 32,
b n - the value of the current member of the geometric progression;
b n+1 - the formula of the next member of the geometric progression;
q is the denominator of a geometric progression (constant number).
If the graph of an arithmetic progression is a straight line, then the geometric one draws a slightly different picture:
As in the case of arithmetic, a geometric progression has a formula for the value of an arbitrary member. Any n-th term of a geometric progression is equal to the product of the first term and the denominator of the progression to the power of n reduced by one:
Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Find the 5th term of the progression
b 5 \u003d b 1 ∙ q (5-1) \u003d 3 ∙ 1.5 4 \u003d 15.1875
The sum of a given number of members is also calculated using a special formula. The sum of the first n members of a geometric progression is equal to the difference between the product of the nth member of the progression and its denominator and the first member of the progression, divided by the denominator reduced by one:
If b n is replaced using the formula discussed above, the value of the sum of the first n members of the considered number series will take the form:
Example. The geometric progression starts with the first term equal to 1. The denominator is set equal to 3. Let's find the sum of the first eight terms.
s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280
Many have heard of an arithmetic progression, but not everyone is well aware of what it is. In this article, we will give the corresponding definition, and also consider the question of how to find the difference of an arithmetic progression, and give a number of examples.
Mathematical definition
So, if we are talking about an arithmetic or algebraic progression (these concepts define the same thing), then this means that there is some number series that satisfies the following law: every two adjacent numbers in the series differ by the same value. Mathematically, this is written like this:
Here n means the number of the element a n in the sequence, and the number d is the difference of the progression (its name follows from the presented formula).
What does knowing the difference d mean? About how far apart adjacent numbers are. However, knowledge of d is a necessary but not sufficient condition for determining (restoring) the entire progression. You need to know one more number, which can be absolutely any element of the series under consideration, for example, a 4, a10, but, as a rule, the first number is used, that is, a 1.
Formulas for determining the elements of the progression
In general, the information above is already enough to move on to solving specific problems. Nevertheless, before an arithmetic progression is given, and it will be necessary to find its difference, we present a couple of useful formulas, thereby facilitating the subsequent process of solving problems.
It is easy to show that any element of the sequence with number n can be found as follows:
a n \u003d a 1 + (n - 1) * d
Indeed, everyone can check this formula by simple enumeration: if we substitute n = 1, then we get the first element, if we substitute n = 2, then the expression gives the sum of the first number and the difference, and so on.
The conditions of many problems are compiled in such a way that for a known pair of numbers, the numbers of which are also given in the sequence, it is necessary to restore the entire number series (find the difference and the first element). Now we will solve this problem in a general way.
So, let's say we are given two elements with numbers n and m. Using the formula obtained above, we can compose a system of two equations:
a n \u003d a 1 + (n - 1) * d;
a m = a 1 + (m - 1) * d
To find unknown quantities, we use a well-known simple method for solving such a system: we subtract the left and right parts in pairs, while the equality remains valid. We have:
a n \u003d a 1 + (n - 1) * d;
a n - a m = (n - 1) * d - (m - 1) * d = d * (n - m)
Thus, we have eliminated one unknown (a 1). Now we can write the final expression for determining d:
d = (a n - a m) / (n - m), where n > m
We have obtained a very simple formula: in order to calculate the difference d in accordance with the conditions of the problem, it is only necessary to take the ratio of the differences between the elements themselves and their serial numbers. Attention should be paid to one important point: the differences are taken between the "senior" and "junior" members, that is, n\u003e m ("senior" - meaning standing farther from the beginning of the sequence, its absolute value can be either more or less more "younger" element).
The expression for the difference d of the progression should be substituted into any of the equations at the beginning of the solution of the problem in order to obtain the value of the first term.
In our age of computer technology development, many schoolchildren try to find solutions for their tasks on the Internet, so questions of this type often arise: find the difference of an arithmetic progression online. Upon such a request, the search engine will display a number of web pages, by going to which, you will need to enter the data known from the condition (it can be either two members of the progression, or the sum of some of them) and instantly get an answer. Nevertheless, such an approach to solving the problem is unproductive in terms of the development of the student and understanding the essence of the task assigned to him.
Solution without using formulas
Let's solve the first problem, while we will not use any of the above formulas. Let the elements of the series be given: a6 = 3, a9 = 18. Find the difference of the arithmetic progression.
Known elements are close to each other in a row. How many times must the difference d be added to the smallest one to get the largest one? Three times (the first time adding d, we get the 7th element, the second time - the eighth, finally, the third time - the ninth). What number must be added to three three times to get 18? This is the number five. Really:
Thus, the unknown difference is d = 5.
Of course, the solution could be done using the appropriate formula, but this was not done intentionally. A detailed explanation of the solution to the problem should become a clear and vivid example of what an arithmetic progression is.
A task similar to the previous one
Now let's solve a similar problem, but change the input data. So, you should find if a3 = 2, a9 = 19.
Of course, you can resort again to the method of solving "on the forehead". But since the elements of the series are given, which are relatively far apart, such a method becomes not very convenient. But using the resulting formula will quickly lead us to the answer:
d \u003d (a 9 - a 3) / (9 - 3) \u003d (19 - 2) / (6) \u003d 17 / 6 ≈ 2.83
Here we have rounded the final number. How much this rounding led to an error can be judged by checking the result:
a 9 \u003d a 3 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 \u003d 18.98
This result differs by only 0.1% from the value given in the condition. Therefore, rounding to hundredths used can be considered a good choice.
Tasks for applying the formula for an member
Let's consider a classic example of the problem of determining the unknown d: find the difference of the arithmetic progression if a1 = 12, a5 = 40.
When two numbers of an unknown algebraic sequence are given, and one of them is the element a 1 , then you do not need to think long, but you should immediately apply the formula for the a n member. In this case we have:
a 5 = a 1 + d * (5 - 1) => d = (a 5 - a 1) / 4 = (40 - 12) / 4 = 7
We got the exact number when dividing, so there is no point in checking the accuracy of the calculated result, as was done in the previous paragraph.
Let's solve another similar problem: we should find the difference of the arithmetic progression if a1 = 16, a8 = 37.
We use a similar approach to the previous one and get:
a 8 = a 1 + d * (8 - 1) => d = (a 8 - a 1) / 7 = (37 - 16) / 7 = 3
What else you should know about arithmetic progression
In addition to problems of finding an unknown difference or individual elements, it is often necessary to solve problems of the sum of the first terms of a sequence. The consideration of these problems is beyond the scope of the topic of the article, however, for completeness of information, we present a general formula for the sum of n numbers of the series:
∑ n i = 1 (a i) = n * (a 1 + a n) / 2
Numeric sequence
So let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many as you like (in our case, them). No matter how many numbers we write, we can always say which of them is the first, which is the second, and so on to the last, that is, we can number them. This is an example of a number sequence:
Numeric sequence
For example, for our sequence:
The assigned number is specific to only one sequence number. In other words, there are no three second numbers in the sequence. The second number (like the -th number) is always the same.
The number with the number is called the -th member of the sequence.
We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .
In our case: 
Let's say we have a numerical sequence in which the difference between adjacent numbers is the same and equal.
For example: 
etc.
Such a numerical sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius as early as the 6th century and was understood in a broader sense as an endless numerical sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which the ancient Greeks were engaged in.
This is a numerical sequence, each member of which is equal to the previous one, added with the same number. This number is called the difference of an arithmetic progression and is denoted.
Try to determine which number sequences are an arithmetic progression and which are not:
a)
b)
c)
d)
Got it? Compare our answers:
Is arithmetic progression - b, c.
Is not arithmetic progression - a, d.
Let's return to the given progression () and try to find the value of its th member. Exists two way to find it.
1. Method
We can add to the previous value of the progression number until we reach the th term of the progression. It’s good that we don’t have much to summarize - only three values:
So, the -th member of the described arithmetic progression is equal to.
2. Method
What if we needed to find the value of the th term of the progression? The summation would have taken us more than one hour, and it is not a fact that we would not have made mistakes when adding the numbers.
Of course, mathematicians have come up with a way in which you do not need to add the difference of an arithmetic progression to the previous value. Look closely at the drawn picture ... Surely you have already noticed a certain pattern, namely:
For example, let's see what makes up the value of the -th member of this arithmetic progression: 
In other words:
Try to independently find in this way the value of a member of this arithmetic progression.
Calculated? Compare your entries with the answer:
Pay attention that you got exactly the same number as in the previous method, when we successively added the members of an arithmetic progression to the previous value.
Let's try to "depersonalize" this formula - we bring it into a general form and get:
|
Arithmetic progression equation. |
Arithmetic progressions are either increasing or decreasing.
Increasing- progressions in which each subsequent value of the terms is greater than the previous one.
For example:
Descending- progressions in which each subsequent value of the terms is less than the previous one.
For example:
The derived formula is used in the calculation of terms in both increasing and decreasing terms of an arithmetic progression.
Let's check it out in practice.
We are given an arithmetic progression consisting of the following numbers:
Since then:
Thus, we were convinced that the formula works both in decreasing and in increasing arithmetic progression.
Try to find the -th and -th members of this arithmetic progression on your own.
Let's compare the results:
Arithmetic progression property
Let's complicate the task - we derive the property of an arithmetic progression.
Suppose we are given the following condition:
- arithmetic progression, find the value.
It's easy, you say, and start counting according to the formula you already know:
Let, a, then:
Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but what if we are given numbers in the condition? Agree, there is a possibility of making mistakes in the calculations.
Now think, is it possible to solve this problem in one step using any formula? Of course, yes, and we will try to bring it out now.
Let's denote the desired term of the arithmetic progression as, we know the formula for finding it - this is the same formula that we derived at the beginning:
, Then:
- the previous member of the progression is:
- the next term of the progression is:
Let's sum the previous and next members of the progression:
It turns out that the sum of the previous and subsequent members of the progression is twice the value of the member of the progression located between them. In other words, in order to find the value of a progression member with known previous and successive values, it is necessary to add them and divide by.
That's right, we got the same number. Let's fix the material. Calculate the value for the progression yourself, because it is not difficult at all.
Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, one of the greatest mathematicians of all time, the "king of mathematicians" - Karl Gauss, easily deduced for himself ...
When Carl Gauss was 9 years old, the teacher, busy checking the work of students from other classes, asked the following task at the lesson: "Calculate the sum of all natural numbers from up to (according to other sources up to) inclusive." What was the surprise of the teacher when one of his students (it was Karl Gauss) after a minute gave the correct answer to the task, while most of the classmates of the daredevil after long calculations received the wrong result ...
Young Carl Gauss noticed a pattern that you can easily notice.
Let's say we have an arithmetic progression consisting of -ti members: We need to find the sum of the given members of the arithmetic progression. Of course, we can manually sum all the values, but what if we need to find the sum of its terms in the task, as Gauss was looking for?
Let's depict the progression given to us. Look closely at the highlighted numbers and try to perform various mathematical operations with them. 
Tried? What did you notice? Right! Their sums are equal
Now answer, how many such pairs will there be in the progression given to us? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two terms of an arithmetic progression is equal, and similar equal pairs, we get that the total sum is equal to:
.
Thus, the formula for the sum of the first terms of any arithmetic progression will be:
In some problems, we do not know the th term, but we know the progression difference. Try to substitute in the sum formula, the formula of the th member.
What did you get?
Well done! Now let's return to the problem that was given to Carl Gauss: calculate for yourself what the sum of numbers starting from the -th is, and the sum of the numbers starting from the -th.
How much did you get?
Gauss turned out that the sum of the terms is equal, and the sum of the terms. Is that how you decided?
In fact, the formula for the sum of members of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people used the properties of an arithmetic progression with might and main.
For example, imagine Ancient Egypt and the largest construction site of that time - the construction of a pyramid ... The figure shows one side of it. 
Where is the progression here you say? Look carefully and find a pattern in the number of sand blocks in each row of the pyramid wall. 
Why not an arithmetic progression? Count how many blocks are needed to build one wall if block bricks are placed in the base. I hope you will not count by moving your finger across the monitor, do you remember the last formula and everything we said about arithmetic progression?
In this case, the progression looks like this:
Arithmetic progression difference.
The number of members of an arithmetic progression.
Let's substitute our data into the last formulas (we count the number of blocks in 2 ways).
Method 1.
Method 2.
And now you can also calculate on the monitor: compare the obtained values with the number of blocks that are in our pyramid. Did it agree? Well done, you have mastered the sum of the th terms of an arithmetic progression.
Of course, you can’t build a pyramid from the blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:
Training
Tasks:
- Masha is getting in shape for the summer. Every day she increases the number of squats by. How many times will Masha squat in weeks if she did squats at the first workout.
- What is the sum of all odd numbers contained in.
- When storing logs, lumberjacks stack them in such a way that each top layer contains one less log than the previous one. How many logs are in one masonry, if the base of the masonry is logs.
Answers:
- Let us define the parameters of the arithmetic progression. In this case
(weeks = days).Answer: In two weeks, Masha should squat once a day.
- First odd number, last number.
Arithmetic progression difference.
The number of odd numbers in - half, however, check this fact using the formula for finding the -th member of an arithmetic progression:The numbers do contain odd numbers.
We substitute the available data into the formula:Answer: The sum of all odd numbers contained in is equal to.
- Recall the problem about the pyramids. For our case, a , since each top layer is reduced by one log, there are only a bunch of layers, that is.
Substitute the data in the formula:Answer: There are logs in the masonry.
Summing up
- - a numerical sequence in which the difference between adjacent numbers is the same and equal. It is increasing and decreasing.
- Finding formula th member of an arithmetic progression is written by the formula - , where is the number of numbers in the progression.
- Property of members of an arithmetic progression- - where - the number of numbers in the progression.
- The sum of the members of an arithmetic progression can be found in two ways:
, where is the number of values.
ARITHMETIC PROGRESSION. AVERAGE LEVEL
Numeric sequence
Let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many as you like. But you can always tell which of them is the first, which is the second, and so on, that is, we can number them. This is an example of a number sequence.
Numeric sequence is a set of numbers, each of which can be assigned a unique number.
In other words, each number can be associated with a certain natural number, and only one. And we will not assign this number to any other number from this set.
The number with the number is called the -th member of the sequence.
We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .
It is very convenient if the -th member of the sequence can be given by some formula. For example, the formula
sets the sequence:
And the formula is the following sequence:
For example, an arithmetic progression is a sequence (the first term here is equal, and the difference). Or (, difference).
nth term formula
We call recurrent a formula in which, in order to find out the -th term, you need to know the previous or several previous ones:
To find, for example, the th term of the progression using such a formula, we have to calculate the previous nine. For example, let. Then:
Well, now it's clear what the formula is?
In each line, we add to, multiplied by some number. For what? Very simple: this is the number of the current member minus:
Much more comfortable now, right? We check:
Decide for yourself:
In an arithmetic progression, find the formula for the nth term and find the hundredth term.
Solution:
The first member is equal. And what is the difference? And here's what:
(after all, it is called the difference because it is equal to the difference of successive members of the progression).
So the formula is:
Then the hundredth term is:
What is the sum of all natural numbers from to?
According to legend, the great mathematician Carl Gauss, being a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last number is equal, the sum of the second and penultimate is the same, the sum of the third and the 3rd from the end is the same, and so on. How many such pairs are there? That's right, exactly half the number of all numbers, that is. So,
The general formula for the sum of the first terms of any arithmetic progression will be:
Example:
Find the sum of all two-digit multiples.
Solution:
The first such number is this. Each next is obtained by adding a number to the previous one. Thus, the numbers of interest to us form an arithmetic progression with the first term and the difference.
The formula for the th term for this progression is:
How many terms are in the progression if they must all be two digits?
Very easy: .
The last term of the progression will be equal. Then the sum:
Answer: .
Now decide for yourself:
- Every day the athlete runs 1m more than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
- A cyclist rides more miles each day than the previous one. On the first day he traveled km. How many days does he have to drive to cover a kilometer? How many kilometers will he travel on the last day of the journey?
- The price of a refrigerator in the store is reduced by the same amount every year. Determine how much the price of a refrigerator decreased every year if, put up for sale for rubles, six years later it was sold for rubles.
Answers:
- The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
.
Answer: - Here it is given:, it is necessary to find.
Obviously, you need to use the same sum formula as in the previous problem:
.
Substitute the values:The root obviously doesn't fit, so the answer.
Let's calculate the distance traveled over the last day using the formula of the -th term:
(km).
Answer: - Given: . Find: .
It doesn't get easier:
(rub).
Answer:
ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN
This is a numerical sequence in which the difference between adjacent numbers is the same and equal.
Arithmetic progression is increasing () and decreasing ().
For example:
The formula for finding the n-th member of an arithmetic progression
is written as a formula, where is the number of numbers in the progression.
Property of members of an arithmetic progression
It makes it easy to find a member of the progression if its neighboring members are known - where is the number of numbers in the progression.
The sum of the members of an arithmetic progression
There are two ways to find the sum:
Where is the number of values.
Where is the number of values.
Well, the topic is over. If you are reading these lines, then you are very cool.
Because only 5% of people are able to master something on their own. And if you have read to the end, then you are in the 5%!
Now the most important thing.
You've figured out the theory on this topic. And, I repeat, it's ... it's just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough ...
For what?
For the successful passing of the exam, for admission to the institute on the budget and, MOST IMPORTANTLY, for life.
I will not convince you of anything, I will just say one thing ...
People who have received a good education earn much more than those who have not received it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because much more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the exam and be ultimately ... happier?
FILL YOUR HAND, SOLVING PROBLEMS ON THIS TOPIC.
On the exam, you will not be asked theory.
You will need solve problems on time.
And, if you haven’t solved them (LOTS!), you will definitely make a stupid mistake somewhere or simply won’t make it in time.
It's like in sports - you need to repeat many times to win for sure.
Find a collection anywhere you want necessarily with solutions, detailed analysis and decide, decide, decide!
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“Understood” and “I know how to solve” are completely different skills. You need both.
Find problems and solve!
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. The last number of the row. Not a very familiar name, but, when applied to the amount, it is very suitable. Then you will see for yourself.
n is the number of the last member. It is important to understand that in the formula this number coincides with the number of added members.
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, useful information:
The main difficulty in tasks for the sum of an arithmetic progression is the correct determination of the elements of the formula.
The authors of the assignments encrypt these very elements with boundless imagination.) The main thing here is not to be afraid. Understanding the essence of the elements, it is enough just to decipher them. Let's take a look at a few examples in detail. Let's start with a task based on a real GIA.
1. The arithmetic progression is given by the condition: a n = 2n-3.5. Find the sum of the first 10 terms.
Good job. Easy.) To determine the amount according to the formula, what do we need to know? First member a 1, last term a n, yes the number of the last term n.
Where to get the last member number n? Yes, in the same place, in the condition! It says find the sum first 10 members. Well, what number will it be last, tenth member?) You won’t believe it, his number is tenth!) Therefore, instead of a n we will substitute into the formula a 10, but instead n- ten. Again, the number of the last member is the same as the number of members.
It remains to be determined a 1 And a 10. This is easily calculated by the formula of the nth term, which is given in the problem statement. Don't know how to do it? Visit the previous lesson, without this - nothing.
a 1= 2 1 - 3.5 = -1.5
a 10\u003d 2 10 - 3.5 \u003d 16.5
S n = S 10.
We found out the meaning of all elements of the formula for the sum of an arithmetic progression. It remains to substitute them, and count:
That's all there is to it. Answer: 75.
Another task based on the GIA. A little more complicated:
2. Given an arithmetic progression (a n), the difference of which is 3.7; a 1 \u003d 2.3. Find the sum of the first 15 terms.
We immediately write the sum formula:
This formula allows us to find the value of any member by its number. We are looking for a simple substitution:
a 15 \u003d 2.3 + (15-1) 3.7 \u003d 54.1
It remains to substitute all the elements in the formula for the sum of an arithmetic progression and calculate the answer:
Answer: 423.
By the way, if in the sum formula instead of a n just substitute the formula of the nth term, we get:
We give similar ones, we get a new formula for the sum of members of an arithmetic progression:
 |
As you can see, the nth term is not required here. a n. In some tasks, this formula helps out a lot, yes ... You can remember this formula. And you can simply withdraw it at the right time, as here. After all, the formula for the sum and the formula for the nth term must be remembered in every way.)
Now the task in the form of a short encryption):
3. Find the sum of all positive two-digit numbers that are multiples of three.
How! No first member, no last, no progression at all... How to live!?
You will have to think with your head and pull out from the condition all the elements of the sum of an arithmetic progression. What are two-digit numbers - we know. They consist of two numbers.) What two-digit number will first? 10, presumably.) last thing two digit number? 99, of course! The three-digit ones will follow him ...
Multiples of three... Hm... These are numbers that are evenly divisible by three, here! Ten is not divisible by three, 11 is not divisible... 12... is divisible! So, something is emerging. You can already write a series according to the condition of the problem:
12, 15, 18, 21, ... 96, 99.
Will this series be an arithmetic progression? 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.)
Practical advice:
When solving any problem for the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.
Formula of the nth term:
These formulas will immediately tell you what to look for, in which direction to think in order to solve the problem. Helps.
And now the tasks for independent solution.
5. Find the sum of all two-digit numbers that are not divisible by three.
Cool?) The hint is hidden in the note to problem 4. Well, problem 3 will help.
6. Arithmetic progression is given by the condition: a 1 =-5.5; a n+1 = a n +0.5. Find the sum of the first 24 terms.
Unusual?) This is a recurrent formula. You can read about it in the previous lesson. Do not ignore the link, such puzzles are often found in the GIA.
7. Vasya saved up money for the Holiday. As much as 4550 rubles! And I decided to give the most beloved person (myself) a few days of happiness). Live beautifully without denying yourself anything. Spend 500 rubles on the first day, and spend 50 rubles more on each subsequent day than on the previous one! Until the money runs out. How many days of happiness did Vasya have?
Is it difficult?) An additional formula from task 2 will help.
Answers (in disarray): 7, 3240, 6.
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