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.
Arithmetic or is such a set of ordered rational numbers, each member of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of an ordered series of numbers and the difference, you can restore the entire arithmetic progression.
Let's take an example. The next sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the considered type of progression, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).
Important formulas
We now give the basic formulas that will be needed to solve problems using an arithmetic progression. Let a n denote the nth member of the sequence, where n is an integer. The difference is denoted by the Latin letter d. Then the following expressions are true:
- To determine the value of the nth term, the formula is suitable: a n \u003d (n-1) * d + a 1.
- To determine the sum of the first n terms: S n = (a n + a 1)*n/2.
To understand any examples of an arithmetic progression with a solution in grade 9, it is enough to remember these two formulas, since any problems of the type in question are built on their use. Also, do not forget that the progression difference is determined by the formula: d = a n - a n-1 .
Example #1: Finding an Unknown Member
We give a simple example of an arithmetic progression and the formulas that must be used to solve.
Let the sequence 10, 8, 6, 4, ... be given, it is necessary to find five terms in it.
It already follows from the conditions of the problem that the first 4 terms are known. The fifth can be defined in two ways:
- Let's calculate the difference first. We have: d = 8 - 10 = -2. Similarly, one could take any two other terms standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d \u003d a n - a n-1, then d \u003d a 5 - a 4, from where we get: a 5 \u003d a 4 + d. We substitute the known values: a 5 = 4 + (-2) = 2.
- The second method also requires knowledge of the difference of the progression in question, so you first need to determine it, as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for the n number of the sequence. We have: a n \u003d (n - 1) * d + a 1 \u003d (n - 1) * (-2) + 10 \u003d 12 - 2 * n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.
As you can see, both solutions lead to the same result. Note that in this example the difference d of the progression is negative. Such sequences are called decreasing because each successive term is less than the previous one.
Example #2: progression difference
Now let's complicate the task a bit, give an example of how to find the difference of an arithmetic progression.
It is known that in some algebraic progression the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.
Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . We substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 \u003d 6 + 6 * d. From this expression, you can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, the first part of the problem was answered.
To restore the sequence to the 7th member, you should use the definition of an algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d, and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2=8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16 and 7 = 18.
Example #3: making a progression
Let us complicate the condition of the problem even more. Now you need to answer the question of how to find an arithmetic progression. We can give the following example: two numbers are given, for example, 4 and 5. It is necessary to make an algebraic progression so that three more terms fit between these.
Before starting to solve this problem, it is necessary to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 \u003d -4 and a 5 \u003d 5. Having established this, we proceed to a task that is similar to the previous one. Again, for the nth term, we use the formula, we get: a 5 \u003d a 1 + 4 * d. From: d \u003d (a 5 - a 1) / 4 \u003d (5 - (-4)) / 4 \u003d 2.25. Here, the difference is not an integer value, but it is a rational number, so the formulas for the algebraic progression remain the same.
Now let's add the found difference to a 1 and restore the missing members of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 \u003d 2.75 + 2.25 \u003d 5, which coincided with the condition of the problem.
Example #4: The first member of the progression
We continue to give examples of an arithmetic progression with a solution. In all previous problems, the first number of the algebraic progression was known. Now consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find from what number this sequence begins.
The formulas that have been used so far assume knowledge of a 1 and d. Nothing is known about these numbers in the condition of the problem. Nevertheless, let's write out the expressions for each term about which we have information: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We got two equations in which there are 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.
The specified system is easiest to solve if you express a 1 in each equation, and then compare the resulting expressions. First equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 \u003d a 43 - 42 * d \u003d 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d \u003d 37 - 42 * d, whence the difference d \u003d (37 - 50) / (42 - 14) \u003d - 0.464 (only 3 decimal places are given).
Knowing d, you can use any of the 2 expressions above for a 1 . For example, first: a 1 \u003d 50 - 14 * d \u003d 50 - 14 * (- 0.464) \u003d 56.496.
If there are doubts about the result, you can check it, for example, determine the 43rd member of the progression, which is specified in the condition. We get: a 43 \u003d a 1 + 42 * d \u003d 56.496 + 42 * (- 0.464) \u003d 37.008. A small error is due to the fact that rounding to thousandths was used in the calculations.
Example #5: Sum
Now let's look at some examples with solutions for the sum of an arithmetic progression.
Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?
Thanks to the development of computer technology, this problem can be solved, that is, sequentially add up all the numbers, which the computer will do as soon as a person presses the Enter key. However, the problem can be solved mentally if you pay attention that the presented series of numbers is an algebraic progression, and its difference is 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.
It is curious to note that this problem is called "Gaussian", since at the beginning of the 18th century the famous German, still at the age of only 10 years old, was able to solve it in his mind in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add pairs of numbers located at the edges of the sequence, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer, it is enough to multiply 50 by 101.
Example #6: sum of terms from n to m
Another typical example of the sum of an arithmetic progression is the following: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its terms from 8 to 14 will be.
The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them up sequentially. Since there are few terms, this method is not laborious enough. Nevertheless, it is proposed to solve this problem by the second method, which is more universal.
The idea is to get a formula for the sum of an algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:
- S m \u003d m * (a m + a 1) / 2.
- S n \u003d n * (a n + a 1) / 2.
Since n > m, it is obvious that the 2 sum includes the first one. The last conclusion means that if we take the difference between these sums, and add the term a m to it (in the case of taking the difference, it is subtracted from the sum S n), then we get the necessary answer to the problem. We have: S mn \u003d S n - S m + a m \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m \u003d a 1 * (n - m) / 2 + a n * n / 2 + a m * (1- m / 2). It is necessary to substitute formulas for a n and a m into this expression. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d * (3 * m - m 2 - 2) / 2.
The resulting formula is somewhat cumbersome, however, the sum S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.
As can be seen from the above solutions, all problems are based on the knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before you start solving any of these problems, it is recommended that you carefully read the condition, clearly understand what you want to find, and only then proceed with the solution.
Another tip is to strive for simplicity, that is, if you can answer the question without using complex mathematical calculations, then you need to do just that, since in this case the probability of making a mistake is less. For example, in the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn = 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.
First level
Arithmetic progression. Detailed theory with examples (2019)
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. Way
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:
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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? Correctly! 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 proved 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 values obtained 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:
Workout
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 a recurrent formula such 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 each 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 member:
(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.
For example, the sequence \(2\); \(5\); \(eight\); \(eleven\); \(14\)… is an arithmetic progression, because each next element differs from the previous one by three (can be obtained from the previous one by adding three):
In this progression, the difference \(d\) is positive (equal to \(3\)), and therefore each next term is greater than the previous one. Such progressions are called increasing.
However, \(d\) can also be a negative number. For example, in arithmetic progression \(16\); \(ten\); \(four\); \(-2\); \(-8\)… the progression difference \(d\) is equal to minus six.
And in this case, each next element will be less than the previous one. These progressions are called decreasing.
Arithmetic progression notation
Progression is denoted by a small Latin letter.
The numbers that form a progression are called it members(or elements).
They are denoted by the same letter as the arithmetic progression, but with a numerical index equal to the element number in order.
For example, the arithmetic progression \(a_n = \left\( 2; 5; 8; 11; 14…\right\)\) consists of the elements \(a_1=2\); \(a_2=5\); \(a_3=8\) and so on.
In other words, for the progression \(a_n = \left\(2; 5; 8; 11; 14…\right\)\)
Solving problems on an arithmetic progression
In principle, the above information is already enough to solve almost any problem on an arithmetic progression (including those offered at the OGE).
Example (OGE).
The arithmetic progression is given by the conditions \(b_1=7; d=4\). Find \(b_5\).
Solution:
Answer: \(b_5=23\)
Example (OGE).
The first three terms of an arithmetic progression are given: \(62; 49; 36…\) Find the value of the first negative term of this progression..
Solution:
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We are given the first elements of the sequence and know that it is an arithmetic progression. That is, each element differs from the neighboring one by the same number. Find out which one by subtracting the previous one from the next element: \(d=49-62=-13\). |
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Now we can restore our progression to the desired (first negative) element. |
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Ready. You can write an answer. |
Answer: \(-3\)
Example (OGE).
Several successive elements of an arithmetic progression are given: \(...5; x; 10; 12.5...\) Find the value of the element denoted by the letter \(x\).
Solution:
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To find \(x\), we need to know how much the next element differs from the previous one, in other words, the progression difference. Let's find it from two known neighboring elements: \(d=12.5-10=2.5\). |
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And now we find what we are looking for without any problems: \(x=5+2.5=7.5\). |
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Ready. You can write an answer. |
Answer: \(7,5\).
Example (OGE).
The arithmetic progression is given by the following conditions: \(a_1=-11\); \(a_(n+1)=a_n+5\) Find the sum of the first six terms of this progression.
Solution:
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We need to find the sum of the first six terms of the progression. But we do not know their meanings, we are given only the first element. Therefore, we first calculate the values in turn, using the given to us: \(n=1\); \(a_(1+1)=a_1+5=-11+5=-6\) |
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\(S_6=a_1+a_2+a_3+a_4+a_5+a_6=\) |
The requested amount has been found. |
Answer: \(S_6=9\).
Example (OGE).
In arithmetic progression \(a_(12)=23\); \(a_(16)=51\). Find the difference of this progression.
Solution:
Answer: \(d=7\).
Important Arithmetic Progression Formulas
As you can see, many arithmetic progression problems can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each next element in this chain is obtained by adding the same number to the previous one (the difference of the progression).
However, sometimes there are situations when it is very inconvenient to solve "on the forehead". For example, imagine that in the very first example, we need to find not the fifth element \(b_5\), but the three hundred and eighty-sixth \(b_(386)\). What is it, we \ (385 \) times to add four? Or imagine that in the penultimate example, you need to find the sum of the first seventy-three elements. Counting is confusing...
Therefore, in such cases, they do not solve “on the forehead”, but use special formulas derived for arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum \(n\) of the first terms.
Formula for the \(n\)th member: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first member of the progression;
\(n\) – number of the required element;
\(a_n\) is a member of the progression with the number \(n\).
This formula allows us to quickly find at least the three hundredth, even the millionth element, knowing only the first and the progression difference.
Example.
The arithmetic progression is given by the conditions: \(b_1=-159\); \(d=8,2\). Find \(b_(246)\).
Solution:
Answer: \(b_(246)=1850\).
The formula for the sum of the first n terms is: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where
\(a_n\) is the last summed term;
Example (OGE).
The arithmetic progression is given by the conditions \(a_n=3.4n-0.6\). Find the sum of the first \(25\) terms of this progression.
Solution:
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\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\) |
To calculate the sum of the first twenty-five elements, we need to know the value of the first and twenty-fifth term. |
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\(n=1;\) \(a_1=3.4 1-0.6=2.8\) |
Now let's find the twenty-fifth term by substituting twenty-five instead of \(n\). |
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\(n=25;\) \(a_(25)=3.4 25-0.6=84.4\) |
Well, now we calculate the required amount without any problems. |
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\(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25=\) |
The answer is ready. |
Answer: \(S_(25)=1090\).
For the sum \(n\) of the first terms, you can get another formula: you just need to \(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25\ ) instead of \(a_n\) substitute the formula for it \(a_n=a_1+(n-1)d\). We get:
The formula for the sum of the first n terms is: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where
\(S_n\) – the required sum \(n\) of the first elements;
\(a_1\) is the first term to be summed;
\(d\) – progression difference;
\(n\) - the number of elements in the sum.
Example.
Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15,5\); \(fourteen\)…
Solution:
Answer: \(S_(33)=-231\).
More complex arithmetic progression problems
Now you have all the information you need to solve almost any arithmetic progression problem. Let's finish the topic by considering problems in which you need to not only apply formulas, but also think a little (in mathematics, this can be useful ☺)
Example (OGE).
Find the sum of all negative terms of the progression: \(-19.3\); \(-19\); \(-18.7\)…
Solution:
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\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\) |
The task is very similar to the previous one. We start solving the same way: first we find \(d\). |
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\(d=a_2-a_1=-19-(-19.3)=0.3\) |
Now we would substitute \(d\) into the formula for the sum ... and here a small nuance pops up - we don't know \(n\). In other words, we do not know how many terms will need to be added. How to find out? Let's think. We will stop adding elements when we get to the first positive element. That is, you need to find out the number of this element. How? Let's write down the formula for calculating any element of an arithmetic progression: \(a_n=a_1+(n-1)d\) for our case. |
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\(a_n=a_1+(n-1)d\) |
||
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\(a_n=-19.3+(n-1) 0.3\) |
We need \(a_n\) to be greater than zero. Let's find out for what \(n\) this will happen. |
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\(-19.3+(n-1) 0.3>0\) |
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\((n-1) 0.3>19.3\) \(|:0.3\) |
We divide both sides of the inequality by \(0,3\). |
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\(n-1>\)\(\frac(19,3)(0,3)\) |
We transfer minus one, not forgetting to change signs |
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\(n>\)\(\frac(19,3)(0,3)\) \(+1\) |
Computing... |
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\(n>65,333…\) |
…and it turns out that the first positive element will have the number \(66\). Accordingly, the last negative has \(n=65\). Just in case, let's check it out. |
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\(n=65;\) \(a_(65)=-19.3+(65-1) 0.3=-0.1\) |
Thus, we need to add the first \(65\) elements. |
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\(S_(65)=\) \(\frac(2 \cdot (-19,3)+(65-1)0,3)(2)\)\(\cdot 65\) |
The answer is ready. |
Answer: \(S_(65)=-630.5\).
Example (OGE).
The arithmetic progression is given by the conditions: \(a_1=-33\); \(a_(n+1)=a_n+4\). Find the sum from \(26\)th to \(42\) element inclusive.
Solution:
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\(a_1=-33;\) \(a_(n+1)=a_n+4\) |
In this problem, you also need to find the sum of elements, but starting not from the first, but from the \(26\)th. We don't have a formula for this. How to decide? |
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For our progression \(a_1=-33\), and the difference \(d=4\) (after all, we add four to the previous element to find the next one). Knowing this, we find the sum of the first \(42\)-uh elements. |
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\(S_(42)=\) \(\frac(2 \cdot (-33)+(42-1)4)(2)\)\(\cdot 42=\) |
Now the sum of the first \(25\)-th elements. |
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\(S_(25)=\) \(\frac(2 \cdot (-33)+(25-1)4)(2)\)\(\cdot 25=\) |
And finally, we calculate the answer. |
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\(S=S_(42)-S_(25)=2058-375=1683\) |
Answer: \(S=1683\).
For an arithmetic progression, there are several more formulas that we have not considered in this article due to their low practical usefulness. However, you can easily find them.
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)
Pay attention to 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, then 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 from the numbers $x$ and $z$ we are in this moment we cannot get $y$, 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.
