Someone treats the word "progression" with caution, as a very complex term from the sections of higher mathematics. Meanwhile, the simplest arithmetic progression- 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 nth 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 members
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
What is the essence of the formula?
This formula allows you to find any BY HIS NUMBER" n" .
Of course, you need to know the first term a 1 and progression difference d, well, without these parameters, you can’t write down a specific progression.
It is not enough to memorize (or cheat) this formula. It is necessary to assimilate its essence and apply the formula in various problems. Yes, and do not forget at the right time, yes ...) How not forget- I don't know. And here how to remember If needed, I'll give you a hint. For those who master the lesson to the end.)
So, let's deal with the formula of the n-th member of an arithmetic progression.
What is a formula in general - we imagine.) What is an arithmetic progression, a member number, a progression difference - is clearly stated in the previous lesson. Take a look if you haven't read it. Everything is simple there. It remains to figure out what nth member.
The progression in general can be written as a series of numbers:
a 1 , a 2 , a 3 , a 4 , a 5 , .....
a 1- denotes the first term of an arithmetic progression, a 3- third member a 4- fourth, and so on. If we are interested in the fifth term, let's say we are working with a 5, if one hundred and twentieth - from a 120.
How to define in general any member of an arithmetic progression, s any number? Very simple! Like this:
a n
That's what it is n-th member of an arithmetic progression. Under the letter n all the numbers of members are hidden at once: 1, 2, 3, 4, and so on.
And what does such a record give us? Just think, instead of a number, they wrote down a letter ...
This notation gives us a powerful tool for working with arithmetic progressions. Using the notation a n, we can quickly find any member any arithmetic progression. And a bunch of tasks to solve in progression. You will see further.
In the formula of the nth member of an arithmetic progression:
| a n = a 1 + (n-1)d |
a 1- the first member of the arithmetic progression;
n- member number.
The formula links the key parameters of any progression: a n ; a 1 ; d And n. Around these parameters, all the puzzles revolve in progression.
The nth term formula can also be used to write a specific progression. For example, in the problem it can be said that the progression is given by the condition:
a n = 5 + (n-1) 2.
Such a problem can even confuse ... There is no series, 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 can be even angrier!) If we take the same condition: a n = 5 + (n-1) 2, yes, open the brackets and give similar ones? We get a new formula:
an = 3 + 2n.
This Only not general, but for a specific progression. This is where the pitfall lies. Some people think that the first term is a three. Although in reality the first member is a five ... A little lower we will work with such a modified formula.
In tasks for progression, there is another notation - a n+1. This is, you guessed it, the "n plus the first" term of the progression. Its meaning is simple and harmless.) This is a member of the progression, the number of which is greater than the number n by one. For example, if in some problem we take for a n fifth term, then a n+1 will be the sixth member. Etc.
Most often the designation a n+1 occurs in recursive formulas. Do not be afraid of this terrible word!) This is just a way of expressing a term of an arithmetic progression through the previous one. Suppose we are given an arithmetic progression in this form, using the recurrent formula:
a n+1 = a n +3
a 2 = a 1 + 3 = 5+3 = 8
a 3 = a 2 + 3 = 8+3 = 11
The fourth - through the third, the fifth - through the fourth, and so on. And how to count immediately, say the twentieth term, a 20? But no way!) While the 19th term is not known, the 20th cannot be counted. This is the fundamental difference between the recursive formula and the formula of the nth term. Recursive works only through previous term, and the formula of the nth term - through first and allows straightaway find any member by its number. Not counting the whole series of numbers in order.
In an arithmetic progression, a recursive formula can easily be turned into a regular one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write the formula in the usual form, and work with it. In the GIA, such tasks are often found.
Application of the formula of the n-th member of an arithmetic progression.
First, let's look at the direct application of the formula. At the end of the previous lesson there was a problem:
Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.
This problem can be solved without any formulas, simply based on the meaning of the arithmetic progression. Add, yes add ... An hour or two.)
And according to the formula, the solution will take less than a minute. You can time it.) We decide.
The conditions provide all the data for using the formula: a 1 \u003d 3, d \u003d 1/6. It remains to be seen what n. No problem! We need to find a 121. Here we write:
Please pay attention! Instead of an index n a specific number appeared: 121. Which is quite logical.) We are interested in the member of the arithmetic progression number one hundred twenty one. This will be our n. It is this meaning n= 121 we will substitute further into the formula, in brackets. Substitute all the numbers in the formula and calculate:
a 121 = 3 + (121-1) 1/6 = 3+20 = 23
That's all there is to it. Just as quickly one could find the five hundred and tenth member, and the thousand and third, any. We put instead n the desired number in the index of the letter " a" and in brackets, and we consider.
Let me remind you the essence: this formula allows you to find any term of an arithmetic progression BY HIS NUMBER" n" .
Let's solve the problem smarter. Let's say we have the following problem:
Find the first term of the arithmetic progression (a n) if a 17 =-2; d=-0.5.
If you have any difficulties, I will suggest the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Hand write, right in your notebook:
| a n = a 1 + (n-1)d |
And now, looking at the letters of the formula, we understand what data we have and what is missing? Available d=-0.5, there is a seventeenth member ... Everything? If you think that's all, then you can't solve the problem, yes ...
We also have a number n! In the condition a 17 =-2 hidden two options. This is both the value of the seventeenth member (-2) and its number (17). Those. n=17. This "little thing" often slips past the head, and without it, (without the "little thing", not the head!) The problem cannot be solved. Although ... and without a head too.)
Now we can just stupidly substitute our data into the formula:
a 17 \u003d a 1 + (17-1) (-0.5)
Oh yes, a 17 we know it's -2. Okay, let's put it in:
-2 \u003d a 1 + (17-1) (-0.5)
That, in essence, is all. It remains to express the first term of the arithmetic progression from the formula, and calculate. You get the answer: a 1 = 6.
Such a technique - writing a formula and simply substituting known data - helps a lot in simple tasks. Well, you must, of course, be able to express a variable from a formula, but what to do!? Without this skill, mathematics can not be studied at all ...
Another popular problem:
Find the difference of the arithmetic progression (a n) if a 1 =2; a 15 =12.
What are we doing? You will be surprised, we write the formula!)
| a n = a 1 + (n-1)d |
Consider what we know: a 1 =2; a 15 =12; and (special highlight!) n=15. Feel free to substitute in the formula:
12=2 + (15-1)d
Let's do the arithmetic.)
12=2 + 14d
d=10/14 = 5/7
This is the correct answer.
So, tasks a n , a 1 And d decided. It remains to learn how to find the number:
The number 99 is a member of an arithmetic progression (a n), where a 1 =12; d=3. Find the number of this member.
We substitute the known quantities into the formula of the nth term:
a n = 12 + (n-1) 3
At first glance, there are two unknown quantities here: a n and n. But a n is some member of the progression with the number n... And this member of the progression we know! It's 99. We don't know his number. n, so this number also needs to be found. Substitute the progression term 99 into the formula:
99 = 12 + (n-1) 3
We express from the formula n, we think. We get the answer: n=30.
And now a problem on the same topic, but more creative):
Determine if the number 117 will be a member of an arithmetic progression (a n):
-3,6; -2,4; -1,2 ...
Let's write the formula again. What, there are no options? Hm... Why do we need eyes?) Do we 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 be determined from the series? It's easy if you know what the difference of an arithmetic progression is:
d = -2.4 - (-3.6) = 1.2
Yes, we did the simplest thing. It remains to deal with an unknown number n and an incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know that ... How to be!? Well, how to be, how to be... Turn on your creative abilities!)
We suppose that 117 is, after all, a member of our progression. With an unknown number n. And, just like in the previous problem, let's try to find this number. Those. we write the formula (yes-yes!)) and substitute our numbers:
117 = -3.6 + (n-1) 1.2
Again we express from the formulan, we count and get:
Oops! The number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion do we draw? Yes! Number 117 is not member of our progression. It is somewhere between the 101st and 102nd member. If the number turned out to be natural, i.e. positive integer, then the number would be a member of the progression with the found number. And in our case, the answer to the problem will be: No.
Task based on a real version of the GIA:
The arithmetic progression is given by the condition:
a n \u003d -4 + 6.8n
Find the first and tenth terms of the progression.
Here the progression is set in an unusual way. Some kind of formula ... It happens.) However, this formula (as I wrote above) - also the formula of the n-th member of an arithmetic progression! She also allows find any member of the progression by its number.
We are looking for the first member. The one who thinks. that the first term is minus four, is fatally mistaken!) Because the formula in the problem is modified. The first term of an arithmetic progression in it hidden. Nothing, we'll find it now.)
Just as in the previous tasks, we substitute n=1 into this formula:
a 1 \u003d -4 + 6.8 1 \u003d 2.8
Here! The first term is 2.8, not -4!
Similarly, we are looking for the tenth term:
a 10 \u003d -4 + 6.8 10 \u003d 64
That's all there is to it.
And now, for those who have read up to these lines, the promised bonus.)
Suppose, in a difficult combat situation of the GIA or the Unified State Exam, you forgot the useful formula of the n-th member of an arithmetic progression. Something comes to mind, but somehow uncertainly ... Whether n there, or n+1, or n-1... How to be!?
Calm! This formula is easy to derive. Not very strict, but definitely enough for confidence and the right decision!) For the conclusion, it is enough to remember the elementary meaning of the arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.
We draw a numerical axis and mark the first one on it. second, third, etc. members. And note the difference d between members. Like this:
We look at the picture and think: what is the second term equal to? Second one d:
a 2 =a 1 + 1 d
What is the third term? Third term equals first term plus two d.
a 3 =a 1 + 2 d
Do you get it? I don't put some words in bold for nothing. Okay, one more step.)
What is the fourth term? Fourth term equals first term plus three d.
a 4 =a 1 + 3 d
It's time to realize that the number of gaps, i.e. d, Always one less than the number of the member you are looking for n. That is, up to the number n, number of gaps will n-1. So, the formula will be (no options!):
| a n = a 1 + (n-1)d |
In general, visual pictures are very helpful in solving many problems in mathematics. Don't neglect the pictures. But if it's difficult to draw a picture, then ... only a formula!) In addition, the formula of the nth term allows you to connect the entire powerful arsenal of mathematics to the solution - equations, inequalities, systems, etc. You can't put a picture in an equation...
Tasks for independent decision.
For warm-up:
1. In arithmetic progression (a n) a 2 =3; a 5 \u003d 5.1. Find a 3 .
Hint: according to the picture, the problem is solved in 20 seconds ... According to the formula, it turns out more difficult. But for mastering the formula, it is more useful.) In Section 555, this problem is solved both by the picture and by the formula. Feel the difference!)
And this is no longer a warm-up.)
2. In arithmetic progression (a n) a 85 \u003d 19.1; a 236 =49, 3. Find a 3 .
What, reluctance to draw a picture?) Still! Better formula, yes...
3. Arithmetic progression is given by the condition:a 1 \u003d -5.5; a n+1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.
In this task, the progression is given in a recurrent way. But counting up to the one hundred and twenty-fifth term... Not everyone can do such a feat.) But the formula of the nth term is within the power of everyone!
4. Given an arithmetic progression (a n):
-148; -143,8; -139,6; -135,4, .....
Find the number of the smallest positive term of the progression.
5. According to the condition of task 4, find the sum of the smallest positive and largest negative members of the progression.
6. The product of the fifth and twelfth terms of an increasing arithmetic progression is -2.5, and the sum of the third and eleventh terms is zero. Find a 14 .
Not the easiest task, yes ...) Here the method "on the fingers" will not work. You have to write formulas and solve equations.
Answers (in disarray):
3,7; 3,5; 2,2; 37; 2,7; 56,5
Happened? It's nice!)
Not everything works out? Happens. By the way, in the last task there is one subtle point. Attentiveness when reading the problem will be required. And logic.
The solution to all these problems is discussed in detail in Section 555. And the fantasy element for the fourth, and the subtle moment for the sixth, and general approaches for solving any problems for the formula of the nth term - everything is painted. I recommend.
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.
Mathematics has its own beauty, as does painting and poetry.
Russian scientist, mechanic N.E. Zhukovsky
Very common tasks in the entrance tests in mathematics are tasks related to the concept of an arithmetic progression. To successfully solve such problems, it is necessary to know the properties of an arithmetic progression well and have certain skills in their application.
Let us first recall the main properties of an arithmetic progression and present the most important formulas, associated with this concept.
Definition. Numeric sequence, in which each subsequent term differs from the previous one by the same number, called an arithmetic progression. At the same time, the numberis called the progression difference.
For an arithmetic progression, the formulas are valid
, (1)
Where . Formula (1) is called the formula of the common term of an arithmetic progression, and formula (2) is the main property of an arithmetic progression: each member of the progression coincides with the arithmetic mean of its neighboring members and .
Note that it is precisely because of this property that the progression under consideration is called "arithmetic".
Formulas (1) and (2) above are summarized as follows:
(3)
To calculate the sum first members of an arithmetic progressionthe formula is usually used
(5) where and .
If we take into account the formula (1), then formula (5) implies
If we designate
Where . Since , then formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).
In particular , from formula (5) it follows, What
Among the little-known to most students is the property of an arithmetic progression, formulated by means of the following theorem.
Theorem. If , then
Proof. If , then
The theorem has been proven.
For example , using the theorem, it can be shown that
Let's move on to the consideration of typical examples of solving problems on the topic "Arithmetic progression".
Example 1 Let and . Find .
Solution. Applying formula (6), we obtain . Since and , then or .
Example 2 Let three times more, and when dividing by in the quotient, it turns out 2 and the remainder is 8. Determine and.
Solution. The system of equations follows from the condition of the example
Since , , and , then from the system of equations (10) we obtain
The solution of this system of equations are and .
Example 3 Find if and .
Solution. According to formula (5), we have or . However, using property (9), we obtain .
Since and , then from the equality the equation follows or .
Example 4 Find if .
Solution.By formula (5) we have
However, using the theorem, one can write
From here and from formula (11) we obtain .
Example 5. Given: . Find .
Solution. Since , then . However , therefore .
Example 6 Let , and . Find .
Solution. Using formula (9), we obtain . Therefore, if , then or .
Since and then here we have a system of equations
Solving which, we get and .
Natural root of the equation is .
Example 7 Find if and .
Solution. Since according to formula (3) we have that , then the system of equations follows from the condition of the problem
If we substitute the expressioninto the second equation of the system, then we get or .
The roots of the quadratic equation are And .
Let's consider two cases.
1. Let , then . Since and , then .
In this case, according to formula (6), we have
2. If , then , and
Answer: and.
Example 8 It is known that and Find .
Solution. Taking into account formula (5) and the condition of the example, we write and .
This implies the system of equations
If we multiply the first equation of the system by 2, and then add it to the second equation, we get
According to formula (9), we have. In this connection, from (12) it follows or .
Since and , then .
Answer: .
Example 9 Find if and .
Solution. Since , and by condition , then or .
From formula (5) it is known, What . Since , then .
Hence , here we have a system of linear equations
From here we get and . Taking into account formula (8), we write .
Example 10 Solve the equation.
Solution. It follows from the given equation that . Let's assume that , , and . In this case .
According to formula (1), we can write or .
Since , equation (13) has a unique suitable root .
Example 11. Find the maximum value provided that and .
Solution. Since , then the considered arithmetic progression is decreasing. In this regard, the expression takes on a maximum value when it is the number of the minimum positive member of the progression.
We use formula (1) and the fact, which and . Then we get that or .
Because , then or . However, in this inequalitylargest natural number, That's why .
If the values , and are substituted into formula (6), then we get .
Answer: .
Example 12. Find the sum of all two-digit natural numbers that, when divided by 6, have a remainder of 5.
Solution. Denote by the set of all two-valued natural numbers, i.e. . Next, we construct a subset consisting of those elements (numbers) of the set that, when divided by the number 6, give a remainder of 5.
Easy to install, What . Obviously , that the elements of the setform an arithmetic progression, in which and .
To determine the cardinality (number of elements) of the set, we assume that . Since and , then formula (1) implies or . Taking into account formula (5), we obtain .
The above examples of solving problems can by no means claim to be exhaustive. This article is written on the basis of an analysis of modern methods for solving typical problems on a given topic. For a deeper study of methods for solving problems related to arithmetic progression, it is advisable to refer to the list of recommended literature.
1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.
2. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. - 216 p.
3. Medynsky M.M. A complete course of elementary mathematics in tasks and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. - 208 p.
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Yes, yes: arithmetic progression is not a toy for you :) Well, friends, if you are reading this text, then the internal cap evidence tells me that you still do not know what an arithmetic progression is, but you really (no, like this: SOOOOO!) want to know. Therefore, I will not torment you with long introductions and will immediately get down to business.
To start, a couple of examples. Consider several sets of numbers:
- 1; 2; 3; 4; ...
- 15; 20; 25; 30; ...
- $\sqrt(2);\ 2\sqrt(2);\ 3\sqrt(2);...$
What do all these sets have in common? At first glance, nothing. But actually there is something. Namely: each next element differs from the previous one by the same number.
Judge for yourself. The first set is just consecutive numbers, each one more than the previous one. In the second case, the difference between adjacent numbers is already equal to five, but this difference is still constant. In the third case, there are roots in general. However, $2\sqrt(2)=\sqrt(2)+\sqrt(2)$, while $3\sqrt(2)=2\sqrt(2)+\sqrt(2)$, i.e. in which case each next element simply increases by $\sqrt(2)$ (and don't be scared that this number is irrational).
So: all such sequences are just called arithmetic progressions. Let's give a strict definition:
Definition. A sequence of numbers in which each next differs from the previous one by exactly the same amount is called an arithmetic progression. The very amount by which the numbers differ is called the progression difference and is most often denoted by the letter $d$.
Notation: $\left(((a)_(n)) \right)$ is the progression itself, $d$ is its difference.
And just a couple of important remarks. First, progression is considered only orderly sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. You can't rearrange or swap numbers.
Secondly, the sequence itself can be either finite or infinite. For example, the set (1; 2; 3) is obviously a finite arithmetic progression. But if you write something like (1; 2; 3; 4; ...) - this is already an infinite progression. The ellipsis after the four, as it were, hints that quite a lot of numbers go further. Infinitely many, for example. :)
I would also like to note that progressions are increasing and decreasing. We have already seen increasing ones - the same set (1; 2; 3; 4; ...). Here are examples of decreasing progressions:
- 49; 41; 33; 25; 17; ...
- 17,5; 12; 6,5; 1; −4,5; −10; ...
- $\sqrt(5);\ \sqrt(5)-1;\ \sqrt(5)-2;\ \sqrt(5)-3;...$
Okay, okay: the last example may seem overly complicated. But the rest, I think, you understand. Therefore, we introduce new definitions:
Definition. An arithmetic progression is called:
- increasing if each next element is greater than the previous one;
- decreasing, if, on the contrary, each subsequent element is less than the previous one.
In addition, there are so-called "stationary" sequences - they consist of the same repeating number. For example, (3; 3; 3; ...).
Only one question remains: how to distinguish an increasing progression from a decreasing one? Fortunately, everything here depends only on the sign of the number $d$, i.e. progression differences:
- If $d \gt 0$, then the progression is increasing;
- If $d \lt 0$, then the progression is obviously decreasing;
- Finally, there is the case $d=0$ — in this case the entire progression is reduced to a stationary sequence of identical numbers: (1; 1; 1; 1; ...), etc.
Let's try to calculate the difference $d$ for the three decreasing progressions above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract from the number on the right, the number on the left. It will look like this:
- 41−49=−8;
- 12−17,5=−5,5;
- $\sqrt(5)-1-\sqrt(5)=-1$.
As you can see, in all three cases the difference really turned out to be negative. And now that we have more or less figured out the definitions, it's time to figure out how progressions are described and what properties they have.
Members of the progression and the recurrent formula
Since the elements of our sequences cannot be interchanged, they can be numbered:
\[\left(((a)_(n)) \right)=\left\( ((a)_(1)),\ ((a)_(2)),((a)_(3 )),... \right\)\]
Individual elements of this set are called members of the progression. They are indicated in this way with the help of a number: the first member, the second member, and so on.
In addition, as we already know, neighboring members of the progression are related by the formula:
\[((a)_(n))-((a)_(n-1))=d\Rightarrow ((a)_(n))=((a)_(n-1))+d \]
In short, to find the $n$th term of the progression, you need to know the $n-1$th term and the difference $d$. Such a formula is called recurrent, because with its help you can find any number, only knowing the previous one (and in fact, all the previous ones). This is very inconvenient, so there is a more tricky formula that reduces any calculation to the first term and the difference:
\[((a)_(n))=((a)_(1))+\left(n-1 \right)d\]
You have probably come across this formula before. They like to give it in all sorts of reference books and reshebniks. And in any sensible textbook on mathematics, it is one of the first.
However, I suggest you practice a little.
Task number 1. Write down the first three terms of the arithmetic progression $\left(((a)_(n)) \right)$ if $((a)_(1))=8,d=-5$.
Solution. So, we know the first term $((a)_(1))=8$ and the progression difference $d=-5$. Let's use the formula just given and substitute $n=1$, $n=2$ and $n=3$:
\[\begin(align) & ((a)_(n))=((a)_(1))+\left(n-1 \right)d; \\ & ((a)_(1))=((a)_(1))+\left(1-1 \right)d=((a)_(1))=8; \\ & ((a)_(2))=((a)_(1))+\left(2-1 \right)d=((a)_(1))+d=8-5= 3; \\ & ((a)_(3))=((a)_(1))+\left(3-1 \right)d=((a)_(1))+2d=8-10= -2. \\ \end(align)\]
Answer: (8; 3; -2)
That's all! Note that our progression is decreasing.
Of course, $n=1$ could not have been substituted - we already know the first term. However, by substituting the unit, we made sure that even for the first term our formula works. In other cases, everything came down to banal arithmetic.
Task number 2. Write out the first three terms of an arithmetic progression if its seventh term is −40 and its seventeenth term is −50.
Solution. We write the condition of the problem in the usual terms:
\[((a)_(7))=-40;\quad ((a)_(17))=-50.\]
\[\left\( \begin(align) & ((a)_(7))=((a)_(1))+6d \\ & ((a)_(17))=((a) _(1))+16d \\ \end(align) \right.\]
\[\left\( \begin(align) & ((a)_(1))+6d=-40 \\ & ((a)_(1))+16d=-50 \\ \end(align) \right.\]
I put the sign of the system because these requirements must be met simultaneously. And now we note that if we subtract the first equation from the second equation (we have the right to do this, because we have a system), we get this:
\[\begin(align) & ((a)_(1))+16d-\left(((a)_(1))+6d \right)=-50-\left(-40 \right); \\ & ((a)_(1))+16d-((a)_(1))-6d=-50+40; \\ & 10d=-10; \\&d=-1. \\ \end(align)\]
Just like that, we found the progression difference! It remains to substitute the found number in any of the equations of the system. For example, in the first:
\[\begin(matrix) ((a)_(1))+6d=-40;\quad d=-1 \\ \Downarrow \\ ((a)_(1))-6=-40; \\ ((a)_(1))=-40+6=-34. \\ \end(matrix)\]
Now, knowing the first term and the difference, it remains to find the second and third terms:
\[\begin(align) & ((a)_(2))=((a)_(1))+d=-34-1=-35; \\ & ((a)_(3))=((a)_(1))+2d=-34-2=-36. \\ \end(align)\]
Ready! Problem solved.
Answer: (-34; -35; -36)
Notice a curious property of the progression that we discovered: if we take the $n$th and $m$th terms and subtract them from each other, we get the difference of the progression multiplied by the number $n-m$:
\[((a)_(n))-((a)_(m))=d\cdot \left(n-m \right)\]
A simple but very useful property that you should definitely know - with its help, you can significantly speed up the solution of many progression problems. Here is a prime example of this:
Task number 3. The fifth term of the arithmetic progression is 8.4, and its tenth term is 14.4. Find the fifteenth term of this progression.
Solution. Since $((a)_(5))=8.4$, $((a)_(10))=14.4$, and we need to find $((a)_(15))$, we note following:
\[\begin(align) & ((a)_(15))-((a)_(10))=5d; \\ & ((a)_(10))-((a)_(5))=5d. \\ \end(align)\]
But by condition $((a)_(10))-((a)_(5))=14.4-8.4=6$, so $5d=6$, whence we have:
\[\begin(align) & ((a)_(15))-14,4=6; \\ & ((a)_(15))=6+14.4=20.4. \\ \end(align)\]
Answer: 20.4
That's all! We did not need to compose any systems of equations and calculate the first term and the difference - everything was decided in just a couple of lines.
Now let's consider another type of problem - the search for negative and positive members of the progression. It is no secret that if the progression increases, while its first term is negative, then sooner or later positive terms will appear in it. And vice versa: the terms of a decreasing progression will sooner or later become negative.
At the same time, it is far from always possible to find this moment “on the forehead”, sequentially sorting through the elements. Often, problems are designed in such a way that without knowing the formulas, calculations would take several sheets - we would just fall asleep until we found the answer. Therefore, we will try to solve these problems in a faster way.
Task number 4. How many negative terms in an arithmetic progression -38.5; -35.8; …?
Solution. So, $((a)_(1))=-38.5$, $((a)_(2))=-35.8$, from which we immediately find the difference:
Note that the difference is positive, so the progression is increasing. The first term is negative, so indeed at some point we will stumble upon positive numbers. The only question is when this will happen.
Let's try to find out: how long (i.e., up to what natural number $n$) the negativity of the terms is preserved:
\[\begin(align) & ((a)_(n)) \lt 0\Rightarrow ((a)_(1))+\left(n-1 \right)d \lt 0; \\ & -38.5+\left(n-1 \right)\cdot 2.7 \lt 0;\quad \left| \cdot 10 \right. \\ & -385+27\cdot \left(n-1 \right) \lt 0; \\ & -385+27n-27 \lt 0; \\ & 27n \lt 412; \\ & n \lt 15\frac(7)(27)\Rightarrow ((n)_(\max ))=15. \\ \end(align)\]
The last line needs clarification. So we know that $n \lt 15\frac(7)(27)$. On the other hand, only integer values of the number will suit us (moreover: $n\in \mathbb(N)$), so the largest allowable number is precisely $n=15$, and in no case 16.
Task number 5. In arithmetic progression $(()_(5))=-150,(()_(6))=-147$. Find the number of the first positive term of this progression.
This would be exactly the same problem as the previous one, but we don't know $((a)_(1))$. But the neighboring terms are known: $((a)_(5))$ and $((a)_(6))$, so we can easily find the progression difference:
In addition, let's try to express the fifth term in terms of the first and the difference using the standard formula:
\[\begin(align) & ((a)_(n))=((a)_(1))+\left(n-1 \right)\cdot d; \\ & ((a)_(5))=((a)_(1))+4d; \\ & -150=((a)_(1))+4\cdot 3; \\ & ((a)_(1))=-150-12=-162. \\ \end(align)\]
Now we proceed by analogy with the previous problem. We find out at what point in our sequence positive numbers will appear:
\[\begin(align) & ((a)_(n))=-162+\left(n-1 \right)\cdot 3 \gt 0; \\ & -162+3n-3 \gt 0; \\ & 3n \gt 165; \\ & n \gt 55\Rightarrow ((n)_(\min ))=56. \\ \end(align)\]
The minimum integer solution of this inequality is the number 56.
Please note that in the last task everything was reduced to strict inequality, so the option $n=55$ will not suit us.
Now that we have learned how to solve simple problems, let's move on to more complex ones. But first, let's learn another very useful property of arithmetic progressions, which will save us a lot of time and unequal cells in the future. :)
Arithmetic mean and equal indents
Consider several consecutive terms of the increasing arithmetic progression $\left(((a)_(n)) \right)$. Let's try to mark them on a number line:
Arithmetic progression members on the number lineI specifically noted the arbitrary members $((a)_(n-3)),...,((a)_(n+3))$, and not any $((a)_(1)) ,\ ((a)_(2)),\ ((a)_(3))$ etc. Because the rule, which I will now tell you, works the same for any "segments".
And the rule is very simple. Let's remember the recursive formula and write it down for all marked members:
\[\begin(align) & ((a)_(n-2))=((a)_(n-3))+d; \\ & ((a)_(n-1))=((a)_(n-2))+d; \\ & ((a)_(n))=((a)_(n-1))+d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n+1))+d; \\ \end(align)\]
However, these equalities can be rewritten differently:
\[\begin(align) & ((a)_(n-1))=((a)_(n))-d; \\ & ((a)_(n-2))=((a)_(n))-2d; \\ & ((a)_(n-3))=((a)_(n))-3d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n))+2d; \\ & ((a)_(n+3))=((a)_(n))+3d; \\ \end(align)\]
Well, so what? But the fact that the terms $((a)_(n-1))$ and $((a)_(n+1))$ lie at the same distance from $((a)_(n)) $. And this distance is equal to $d$. The same can be said about the terms $((a)_(n-2))$ and $((a)_(n+2))$ - they are also removed from $((a)_(n))$ by the same distance equal to $2d$. You can continue indefinitely, but the picture illustrates the meaning well
The members of the progression lie at the same distance from the center What does this mean for us? This means that you can find $((a)_(n))$ if the neighboring numbers are known:
\[((a)_(n))=\frac(((a)_(n-1))+((a)_(n+1)))(2)\]
We have deduced a magnificent statement: each member of an arithmetic progression is equal to the arithmetic mean of the neighboring members! Moreover, we can deviate from our $((a)_(n))$ to the left and to the right not by one step, but by $k$ steps — and still the formula will be correct:
\[((a)_(n))=\frac(((a)_(n-k))+((a)_(n+k)))(2)\]
Those. we can easily find some $((a)_(150))$ if we know $((a)_(100))$ and $((a)_(200))$, because $(( a)_(150))=\frac(((a)_(100))+((a)_(200)))(2)$. At first glance, it may seem that this fact does not give us anything useful. However, in practice, many tasks are specially "sharpened" for the use of the arithmetic mean. Take a look:
Task number 6. Find all values of $x$ such that the numbers $-6((x)^(2))$, $x+1$ and $14+4((x)^(2))$ are consecutive members of an arithmetic progression (in specified order).
Solution. Since these numbers are members of a progression, the arithmetic mean condition is satisfied for them: the central element $x+1$ can be expressed in terms of neighboring elements:
\[\begin(align) & x+1=\frac(-6((x)^(2))+14+4((x)^(2)))(2); \\ & x+1=\frac(14-2((x)^(2)))(2); \\ & x+1=7-((x)^(2)); \\ & ((x)^(2))+x-6=0. \\ \end(align)\]
The result is a classic quadratic equation. Its roots: $x=2$ and $x=-3$ are the answers.
Answer: -3; 2.
Task number 7. Find the values of $$ such that the numbers $-1;4-3;(()^(2))+1$ form an arithmetic progression (in that order).
Solution. Again, we express the middle term in terms of the arithmetic mean of neighboring terms:
\[\begin(align) & 4x-3=\frac(x-1+((x)^(2))+1)(2); \\ & 4x-3=\frac(((x)^(2))+x)(2);\quad \left| \cdot 2\right.; \\ & 8x-6=((x)^(2))+x; \\ & ((x)^(2))-7x+6=0. \\ \end(align)\]
Another quadratic equation. And again two roots: $x=6$ and $x=1$.
Answer: 1; 6.
If in the process of solving a problem you get some brutal numbers, or you are not completely sure of the correctness of the answers found, then there is a wonderful trick that allows you to check: did we solve the problem correctly?
Let's say in problem 6 we got answers -3 and 2. How can we check that these answers are correct? Let's just plug them into the original condition and see what happens. Let me remind you that we have three numbers ($-6(()^(2))$, $+1$ and $14+4(()^(2))$), which should form an arithmetic progression. Substitute $x=-3$:
\[\begin(align) & x=-3\Rightarrow \\ & -6((x)^(2))=-54; \\ &x+1=-2; \\ & 14+4((x)^(2))=50. \end(align)\]
We got the numbers -54; −2; 50 that differ by 52 is undoubtedly an arithmetic progression. The same thing happens for $x=2$:
\[\begin(align) & x=2\Rightarrow \\ & -6((x)^(2))=-24; \\ &x+1=3; \\ & 14+4((x)^(2))=30. \end(align)\]
Again a progression, but with a difference of 27. Thus, the problem is solved correctly. Those who wish can check the second task on their own, but I’ll say right away: everything is correct there too.
In general, while solving the last problems, we stumbled upon another interesting fact that also needs to be remembered:
If three numbers are such that the second is the average of the first and last, then these numbers form an arithmetic progression.
In the future, understanding this statement will allow us to literally “construct” the necessary progressions based on the condition of the problem. But before we engage in such a "construction", we should pay attention to one more fact, which directly follows from what has already been considered.
Grouping and sum of elements
Let's go back to the number line again. We note there several members of the progression, between which, perhaps. worth a lot of other members:
6 elements marked on the number lineLet's try to express the "left tail" in terms of $((a)_(n))$ and $d$, and the "right tail" in terms of $((a)_(k))$ and $d$. It's very simple:
\[\begin(align) & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n))+2d; \\ & ((a)_(k-1))=((a)_(k))-d; \\ & ((a)_(k-2))=((a)_(k))-2d. \\ \end(align)\]
Now note that the following sums are equal:
\[\begin(align) & ((a)_(n))+((a)_(k))=S; \\ & ((a)_(n+1))+((a)_(k-1))=((a)_(n))+d+((a)_(k))-d= S; \\ & ((a)_(n+2))+((a)_(k-2))=((a)_(n))+2d+((a)_(k))-2d= S. \end(align)\]
Simply put, if we consider as a start two elements of the progression, which in total are equal to some number $S$, and then we start stepping from these elements in opposite directions (towards each other or vice versa to move away), then the sums of the elements that we will stumble upon will also be equal$S$. This can be best represented graphically:
Same indents give equal sums Understanding this fact will allow us to solve problems of a fundamentally higher level of complexity than those that we considered above. For example, these:
Task number 8. Determine the difference of an arithmetic progression in which the first term is 66, and the product of the second and twelfth terms is the smallest possible.
Solution. Let's write down everything we know:
\[\begin(align) & ((a)_(1))=66; \\&d=? \\ & ((a)_(2))\cdot ((a)_(12))=\min . \end(align)\]
So, we do not know the difference of the progression $d$. Actually, the whole solution will be built around the difference, since the product $((a)_(2))\cdot ((a)_(12))$ can be rewritten as follows:
\[\begin(align) & ((a)_(2))=((a)_(1))+d=66+d; \\ & ((a)_(12))=((a)_(1))+11d=66+11d; \\ & ((a)_(2))\cdot ((a)_(12))=\left(66+d \right)\cdot \left(66+11d \right)= \\ & =11 \cdot \left(d+66 \right)\cdot \left(d+6 \right). \end(align)\]
For those in the tank: I've taken the common factor 11 out of the second bracket. Thus, the desired product is a quadratic function with respect to the variable $d$. Therefore, consider the function $f\left(d \right)=11\left(d+66 \right)\left(d+6 \right)$ - its graph will be a parabola with branches up, because if we open the brackets, we get:
\[\begin(align) & f\left(d \right)=11\left(((d)^(2))+66d+6d+66\cdot 6 \right)= \\ & =11(( d)^(2))+11\cdot 72d+11\cdot 66\cdot 6 \end(align)\]
As you can see, the coefficient with the highest term is 11 - this is a positive number, so we are really dealing with a parabola with branches up:
graph of a quadratic function - parabola
Please note: this parabola takes its minimum value at its vertex with the abscissa $((d)_(0))$. Of course, we can calculate this abscissa according to the standard scheme (there is a formula $((d)_(0))=(-b)/(2a)\;$), but it would be much more reasonable to note that the desired vertex lies on the axis symmetry of the parabola, so the point $((d)_(0))$ is equidistant from the roots of the equation $f\left(d \right)=0$:
\[\begin(align) & f\left(d\right)=0; \\ & 11\cdot \left(d+66 \right)\cdot \left(d+6 \right)=0; \\ & ((d)_(1))=-66;\quad ((d)_(2))=-6. \\ \end(align)\]
That is why I was in no hurry to open the brackets: in the original form, the roots were very, very easy to find. Therefore, the abscissa is equal to the arithmetic mean of the numbers −66 and −6:
\[((d)_(0))=\frac(-66-6)(2)=-36\]
What gives us the discovered number? With it, the required product takes the smallest value (by the way, we did not calculate $((y)_(\min ))$ - this is not required of us). At the same time, this number is the difference of the initial progression, i.e. we found the answer. :)
Answer: -36
Task number 9. Insert three numbers between the numbers $-\frac(1)(2)$ and $-\frac(1)(6)$ so that together with the given numbers they form an arithmetic progression.
Solution. In fact, we need to make a sequence of five numbers, with the first and last number already known. Denote the missing numbers by the variables $x$, $y$ and $z$:
\[\left(((a)_(n)) \right)=\left\( -\frac(1)(2);x;y;z;-\frac(1)(6) \right\ )\]
Note that the number $y$ is the "middle" of our sequence - it is equidistant from the numbers $x$ and $z$, and from the numbers $-\frac(1)(2)$ and $-\frac(1)( 6)$. And if at the moment we cannot get $y$ from the numbers $x$ and $z$, then the situation is different with the ends of the progression. Remember the arithmetic mean:
Now, knowing $y$, we will find the remaining numbers. Note that $x$ lies between $-\frac(1)(2)$ and $y=-\frac(1)(3)$ just found. That's why
Arguing similarly, we find the remaining number:
Ready! We found all three numbers. Let's write them down in the answer in the order in which they should be inserted between the original numbers.
Answer: $-\frac(5)(12);\ -\frac(1)(3);\ -\frac(1)(4)$
Task number 10. Between the numbers 2 and 42, insert several numbers that, together with the given numbers, form an arithmetic progression, if it is known that the sum of the first, second, and last of the inserted numbers is 56.
Solution. An even more difficult task, which, however, is solved in the same way as the previous ones - through the arithmetic mean. The problem is that we don't know exactly how many numbers to insert. Therefore, for definiteness, we assume that after inserting there will be exactly $n$ numbers, and the first of them is 2, and the last is 42. In this case, the desired arithmetic progression can be represented as:
\[\left(((a)_(n)) \right)=\left\( 2;((a)_(2));((a)_(3));...;(( a)_(n-1));42 \right\)\]
\[((a)_(2))+((a)_(3))+((a)_(n-1))=56\]
Note, however, that the numbers $((a)_(2))$ and $((a)_(n-1))$ are obtained from the numbers 2 and 42 standing at the edges by one step towards each other, i.e. . to the center of the sequence. And this means that
\[((a)_(2))+((a)_(n-1))=2+42=44\]
But then the above expression can be rewritten like this:
\[\begin(align) & ((a)_(2))+((a)_(3))+((a)_(n-1))=56; \\ & \left(((a)_(2))+((a)_(n-1)) \right)+((a)_(3))=56; \\ & 44+((a)_(3))=56; \\ & ((a)_(3))=56-44=12. \\ \end(align)\]
Knowing $((a)_(3))$ and $((a)_(1))$, we can easily find the progression difference:
\[\begin(align) & ((a)_(3))-((a)_(1))=12-2=10; \\ & ((a)_(3))-((a)_(1))=\left(3-1 \right)\cdot d=2d; \\ & 2d=10\Rightarrow d=5. \\ \end(align)\]
It remains only to find the remaining members:
\[\begin(align) & ((a)_(1))=2; \\ & ((a)_(2))=2+5=7; \\ & ((a)_(3))=12; \\ & ((a)_(4))=2+3\cdot 5=17; \\ & ((a)_(5))=2+4\cdot 5=22; \\ & ((a)_(6))=2+5\cdot 5=27; \\ & ((a)_(7))=2+6\cdot 5=32; \\ & ((a)_(8))=2+7\cdot 5=37; \\ & ((a)_(9))=2+8\cdot 5=42; \\ \end(align)\]
Thus, already at the 9th step we will come to the left end of the sequence - the number 42. In total, only 7 numbers had to be inserted: 7; 12; 17; 22; 27; 32; 37.
Answer: 7; 12; 17; 22; 27; 32; 37
Text tasks with progressions
In conclusion, I would like to consider a couple of relatively simple problems. Well, as simple ones: for most students who study mathematics at school and have not read what is written above, these tasks may seem like a gesture. Nevertheless, it is precisely such tasks that come across in the OGE and the USE in mathematics, so I recommend that you familiarize yourself with them.
Task number 11. The team produced 62 parts in January, and in each subsequent month they produced 14 more parts than in the previous one. How many parts did the brigade produce in November?
Solution. Obviously, the number of parts, painted by month, will be an increasing arithmetic progression. And:
\[\begin(align) & ((a)_(1))=62;\quad d=14; \\ & ((a)_(n))=62+\left(n-1 \right)\cdot 14. \\ \end(align)\]
November is the 11th month of the year, so we need to find $((a)_(11))$:
\[((a)_(11))=62+10\cdot 14=202\]
Therefore, 202 parts will be manufactured in November.
Task number 12. The bookbinding workshop bound 216 books in January, and each month it bound 4 more books than the previous month. How many books did the workshop bind in December?
Solution. All the same:
$\begin(align) & ((a)_(1))=216;\quad d=4; \\ & ((a)_(n))=216+\left(n-1 \right)\cdot 4. \\ \end(align)$
December is the last, 12th month of the year, so we are looking for $((a)_(12))$:
\[((a)_(12))=216+11\cdot 4=260\]
This is the answer - 260 books will be bound in December.
Well, if you have read this far, I hasten to congratulate you: you have successfully completed the “young fighter course” in arithmetic progressions. We can safely move on to the next lesson, where we will study the progression sum formula, as well as important and very useful consequences from it.
