Algebraic progression examples with solution. Arithmetic progression. Detailed theory with examples (2019)

The topic "arithmetic progression" is studied in the general algebra course in schools in the 9th grade. This topic is important for further in-depth study of the mathematics of number series. In this article we will get acquainted with the arithmetic progression, its difference, as well as typical problems that schoolchildren may encounter.

The concept of algebraic progression

A number progression is a sequence of numbers in which each subsequent element can be obtained from the previous one if some mathematical law is applied. There are two simple types of progression: geometric and arithmetic, which is also called algebraic. Let's look at it in more detail.

Let's imagine some rational number, denote it by the symbol a 1, where the index indicates its serial number in the series under consideration. Let's add some other number to a 1 and call it d. Then the second element of the series can be reflected as follows: a 2 = a 1 + d. Now add d again, we get: a 3 = a 2 +d. Continuing this mathematical operation, you can get a whole series of numbers, which will be called an arithmetic progression.

As can be understood from the above, to find the nth element of this sequence, you need to use the formula: a n = a 1 + (n-1)*d. Indeed, substituting n=1 into the expression, we get a 1 = a 1, if n = 2, then the formula follows: a 2 = a 1 + 1*d, and so on.

For example, if the difference of the arithmetic progression is 5, and a 1 = 1, then this means that the number series of the type under consideration has the form: 1, 6, 11, 16, 21, ... As you can see, each of its members is 5 more than the previous one .

Arithmetic progression difference formulas

From the above definition of the series of numbers under consideration, it follows that to define it you need to know two numbers: a 1 and d. The latter is called the difference of this progression. It uniquely determines the behavior of the entire series. Indeed, if d is positive, then the number series will constantly increase; on the contrary, if d is negative, the numbers in the series will increase only in absolute value, while their absolute value will decrease with increasing number n.

What is the difference of the arithmetic progression? Let's consider two basic formulas that are used to calculate this value:

1. d = a n+1 -a n, this formula follows directly from the definition of the series of numbers under consideration.
2. d = (-a 1 +a n)/(n-1), this expression is obtained if we express d from the formula given in the previous paragraph of the article. Note that this expression becomes undefined (0/0) if n=1. This is due to the fact that knowledge of at least 2 elements of the series is necessary to determine its difference.

These two basic formulas are used to solve any problems involving finding the difference of a progression. However, there is another formula that you also need to know about.

Sum of first elements

The formula with which you can determine the sum of any number of terms of an algebraic progression, according to historical evidence, was first obtained by the “prince” of mathematics in the 18th century, Carl Gauss. A German scientist, while still a boy in the primary classes of a village school, noticed that in order to add natural numbers in the series from 1 to 100, it is necessary to first sum the first element and the last (the resulting value will be equal to the sum of the penultimate and second, penultimate and third elements , and so on), and then this number should be multiplied by the number of these amounts, that is, by 50.

The formula, which reflects the stated result in a particular example, can be generalized to an arbitrary case. It will look like: S n = n/2*(a n +a 1). Note that to find the indicated value, knowledge of the difference d is not required if two terms of the progression are known (a n and a 1).

Example No. 1. Determine the difference, knowing two terms of the series a1 and an

We will show you how to apply the formulas mentioned above in the article. Let's give a simple example: the difference of the arithmetic progression is unknown, it is necessary to determine what it will be equal to if a 13 = -5.6 and a 1 = -12.1.

Since we know the values ​​of two elements of a number sequence, and one of them is the first number, we can use formula No. 2 to determine the difference d. We have: d =(-1*(-12.1)+(-5.6))/12 = 0.54167. In the expression, we used the value n=13, since the term with this particular ordinal number is known.

The resulting difference indicates that the progression is increasing, despite the fact that the elements given in the task conditions have a negative value. It can be seen that a 13 >a 1, although |a 13 |<|a 1 |.

Example No. 2. Positive terms of the progression in example No. 1

Let's use the result obtained in the previous example to solve a new problem. It is formulated as follows: from what serial number will the elements of the progression in example No. 1 begin to take on positive values?

As has been shown, the progression in which a 1 = -12.1 and d = 0.54167 is increasing, therefore, from a certain number the numbers will begin to take only positive values. To determine this number n, it is necessary to solve a simple inequality, which is written mathematically as follows: a n >0 or, using the appropriate formula, we rewrite the inequality: a 1 + (n-1)*d>0. It is necessary to find the unknown n, let's express it: n>-1*a 1 /d + 1. Now it remains to substitute the known values ​​of the difference and the first term of the sequence. We get: n>-1*(-12.1) /0.54167 + 1= 23.338 or n>23.338. Since n can only take integer values, it follows from the resulting inequality that any terms in the series that have a number greater than 23 will be positive.

Let's check the answer we received by using the above formula to calculate the 23rd and 24th elements of this arithmetic progression. We have: a 23 = -12.1 + 22*0.54167 = -0.18326 (negative number); a 24 = -12.1 + 23*0.54167 =0.3584 (positive value). Thus, the result obtained is correct: starting from n=24, all members of the number series will be greater than zero.

Example No. 3. How many logs will fit?

Let us present one interesting problem: during logging, it was decided to stack the sawn logs on top of each other as shown in the figure below. How many logs can be stacked in this way, knowing that a total of 10 rows will fit?

One interesting thing can be noticed in this method of folding logs: each subsequent row will contain one log less than the previous one, that is, an algebraic progression takes place, the difference of which is d = 1. Assuming that the number of logs in each row is a member of this progression, and also taking into account that a 1 = 1 (only one log will fit at the very top), we find the number a 10. We have: a 10 = 1 + 1*(10-1) = 10. That is, in the 10th row, which lies on the ground, there will be 10 logs.

The total sum of this “pyramidal” structure can be obtained by using the Gauss formula. We get: S 10 = 10/2*(10+1) = 55 logs.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

An arithmetic progression is a series of numbers in which each number is greater (or less) than the previous one by the same amount.

This topic often seems complex and incomprehensible. The indices of the letters, the nth term of the progression, the difference of the progression - all this is somehow confusing, yes... Let's figure out the meaning of the arithmetic progression and everything will get better right away.)

The concept of arithmetic progression.

Arithmetic progression is a very simple and clear concept. Do you have any doubts? In vain.) See for yourself.

I'll write an unfinished series of numbers:

1, 2, 3, 4, 5, ...

Can you extend this series? What numbers will come next, after the five? Everyone... uh..., in short, everyone will realize that the numbers 6, 7, 8, 9, etc. will come next.

Let's complicate the task. I give you an unfinished series of numbers:

2, 5, 8, 11, 14, ...

You will be able to catch the pattern, extend the series, and name seventh row number?

If you realized that this number is 20, congratulations! Not only did you feel key points of arithmetic progression, but also successfully used them in business! If you haven’t figured it out, read on.

Now let’s translate the key points from sensations into mathematics.)

First key point.

Arithmetic progression deals with series of numbers. This is confusing at first. We are used to solving equations, drawing graphs and all that... But here we extend the series, find the number of the series...

It's OK. It’s just that progressions are the first acquaintance with a new branch of mathematics. The section is called "Series" and works specifically with series of numbers and expressions. Get used to it.)

Second key point.

In an arithmetic progression, any number is different from the previous one by the same amount.

In the first example, this difference is one. Whatever number you take, it is one more than the previous one. In the second - three. Any number is three more than the previous one. Actually, it is this moment that gives us the opportunity to grasp the pattern and calculate subsequent numbers.

Third key point.

This moment is not striking, yes... But it is very, very important. Here he is: Each progression number is in its place. There is the first number, there is the seventh, there is the forty-fifth, etc. If you mix them up at random, the pattern will disappear. Arithmetic progression will also disappear. What's left is just a series of numbers.

That's the whole point.

Of course, new terms and designations appear in a new topic. You need to know them. Otherwise you won’t understand the task. For example, you will have to decide something like:

Write down the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.

Inspiring?) Letters, some indexes... And the task, by the way, couldn’t be simpler. You just need to understand the meaning of the terms and designations. Now we will master this matter and return to the task.

Terms and designations.

Arithmetic progression is a series of numbers in which each number is different from the previous one by the same amount.

This quantity is called . Let's look at this concept in more detail.

Arithmetic progression difference.

Arithmetic progression difference is the amount by which any progression number more previous one.

One important point. Please pay attention to the word "more". Mathematically, this means that each progression number is by adding difference of arithmetic progression to the previous number.

To calculate, let's say second numbers of the series, you need to first number add this very difference of an arithmetic progression. For calculation fifth- the difference is necessary add To fourth, well, etc.

Arithmetic progression difference May be positive, then each number in the series will turn out to be real more than the previous one. This progression is called increasing. For example:

8; 13; 18; 23; 28; .....

Here each number is obtained by adding positive number, +5 to the previous one.

The difference may be negative, then each number in the series will be less than the previous one. This progression is called (you won’t believe it!) decreasing.

For example:

8; 3; -2; -7; -12; .....

Here each number is also obtained by adding to the previous one, but already a negative number, -5.

By the way, when working with progression, it is very useful to immediately determine its nature - whether it is increasing or decreasing. This helps a lot to navigate the decision, spot your mistakes and correct them before it’s too late.

Arithmetic progression difference usually denoted by the letter d.

How to find d? Very simple. It is necessary to subtract from any number in the series previous number. Subtract. By the way, the result of subtraction is called "difference".)

Let us define, for example, d for increasing arithmetic progression:

2, 5, 8, 11, 14, ...

We take any number in the series that we want, for example, 11. We subtract from it previous number those. 8:

This is the correct answer. For this arithmetic progression, the difference is three.

You can take it any progression number, because for a specific progression d-always the same. At least somewhere at the beginning of the row, at least in the middle, at least anywhere. You cannot take only the very first number. Simply because the very first number no previous one.)

By the way, knowing that d=3, finding the seventh number of this progression is very simple. Let's add 3 to the fifth number - we get the sixth, it will be 17. Let's add three to the sixth number, we get the seventh number - twenty.

Let's define d for descending arithmetic progression:

8; 3; -2; -7; -12; .....

I remind you that, regardless of the signs, to determine d need from any number take away the previous one. Choose any progression number, for example -7. His previous number is -2. Then:

d = -7 - (-2) = -7 + 2 = -5

The difference of an arithmetic progression can be any number: integer, fractional, irrational, any number.

Other terms and designations.

Each number in the series is called member of an arithmetic progression.

Each member of the progression has its own number. The numbers are strictly in order, without any tricks. First, second, third, fourth, etc. For example, in the progression 2, 5, 8, 11, 14, ... two is the first term, five is the second, eleven is the fourth, well, you understand...) Please clearly understand - the numbers themselves can be absolutely anything, whole, fractional, negative, whatever, but numbering of numbers- strictly in order!

How to write a progression in general form? No problem! Each number in a series is written as a letter. To denote an arithmetic progression, the letter is usually used a. The member number is indicated by an index at the bottom right. We write terms separated by commas (or semicolons), like this:

a 1, a 2, a 3, a 4, a 5, .....

a 1- this is the first number, a 3- third, etc. Nothing fancy. This series can be written briefly like this: (a n).

Progressions happen finite and infinite.

Ultimate the progression has a limited number of members. Five, thirty-eight, whatever. But it's a finite number.

Infinite progression - has an infinite number of members, as you might guess.)

You can write the final progression through a series like this, all terms and a dot at the end:

a 1, a 2, a 3, a 4, a 5.

Or like this, if there are many members:

a 1, a 2, ... a 14, a 15.

In the short entry you will have to additionally indicate the number of members. For example (for twenty members), like this:

(a n), n = 20

An infinite progression can be recognized by the ellipsis at the end of the row, as in the examples in this lesson.

Now you can solve the tasks. The tasks are simple, purely for understanding the meaning of an arithmetic progression.

Examples of tasks on arithmetic progression.

Let's look at the task given above in detail:

1. Write out the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.

We translate the task into understandable language. An infinite arithmetic progression is given. The second number of this progression is known: a 2 = 5. The progression difference is known: d = -2.5. We need to find the first, third, fourth, fifth and sixth terms of this progression.

For clarity, I will write down a series according to the conditions of the problem. The first six terms, where the second term is five:

a 1, 5, a 3, a 4, a 5, a 6,....

a 3 = a 2 + d

Substitute into expression a 2 = 5 And d = -2.5. Don't forget about the minus!

a 3=5+(-2,5)=5 - 2,5 = 2,5

The third term turned out to be smaller than the second. Everything is logical. If the number is greater than the previous one negative value, which means the number itself will be less than the previous one. Progression is decreasing. Okay, let's take it into account.) We count the fourth term of our series:

a 4 = a 3 + d

a 4=2,5+(-2,5)=2,5 - 2,5 = 0

a 5 = a 4 + d

a 5=0+(-2,5)= - 2,5

a 6 = a 5 + d

a 6=-2,5+(-2,5)=-2,5 - 2,5 = -5

So, terms from the third to the sixth were calculated. The result is the following series:

a 1, 5, 2.5, 0, -2.5, -5, ....

It remains to find the first term a 1 according to the well-known second. This is a step in the other direction, to the left.) So, the difference of the arithmetic progression d should not be added to a 2, A take away:

a 1 = a 2 - d

a 1=5-(-2,5)=5 + 2,5=7,5

That's it. Assignment answer:

7,5, 5, 2,5, 0, -2,5, -5, ...

In passing, I would like to note that we solved this task recurrent way. This terrible word means only the search for a member of the progression according to the previous (adjacent) number. We'll look at other ways to work with progression below.

One important conclusion can be drawn from this simple task.

Remember:

If we know at least one term and the difference of an arithmetic progression, we can find any term of this progression.

Do you remember? This simple conclusion allows you to solve most of the problems of the school course on this topic. All tasks revolve around three main parameters: member of an arithmetic progression, difference of a progression, number of a member of the progression. All.

Of course, all previous algebra is not canceled.) Inequalities, equations, and other things are attached to progression. But according to the progression itself- everything revolves around three parameters.

As an example, let's look at some popular tasks on this topic.

2. Write the finite arithmetic progression as a series if n=5, d = 0.4, and a 1 = 3.6.

Everything is simple here. Everything has already been given. You need to remember how the members of an arithmetic progression are counted, count them, and write them down. It is advisable not to miss the words in the task conditions: “final” and “ n=5". So as not to count until you are completely blue in the face.) There are only 5 (five) members in this progression:

a 2 = a 1 + d = 3.6 + 0.4 = 4

a 3 = a 2 + d = 4 + 0.4 = 4.4

a 4 = a 3 + d = 4.4 + 0.4 = 4.8

a 5 = a 4 + d = 4.8 + 0.4 = 5.2

It remains to write down the answer:

3,6; 4; 4,4; 4,8; 5,2.

Another task:

3. Determine whether the number 7 will be a member of the arithmetic progression (a n), if a 1 = 4.1; d = 1.2.

Hmm... Who knows? How to determine something?

How-how... Write down the progression in the form of a series and see whether there will be a seven there or not! We count:

a 2 = a 1 + d = 4.1 + 1.2 = 5.3

a 3 = a 2 + d = 5.3 + 1.2 = 6.5

a 4 = a 3 + d = 6.5 + 1.2 = 7.7

4,1; 5,3; 6,5; 7,7; ...

Now it is clearly visible that we are just seven slipped through between 6.5 and 7.7! Seven did not fall into our series of numbers, and, therefore, seven will not be a member of the given progression.

Answer: no.

And here is a problem based on a real version of the GIA:

4. Several consecutive terms of the arithmetic progression are written out:

...; 15; X; 9; 6; ...

Here is a series written without end and beginning. No member numbers, no difference d. It's OK. To solve the problem, it is enough to understand the meaning of an arithmetic progression. Let's look and see what's possible to know from this series? What are the three main parameters?

Member numbers? There is not a single number here.

But there are three numbers and - attention! - word "consistent" in condition. This means that the numbers are strictly in order, without gaps. Are there two in this row? neighboring known numbers? Yes, I have! These are 9 and 6. Therefore, we can calculate the difference of the arithmetic progression! Subtract from six previous number, i.e. nine:

There are mere trifles left. What number will be the previous one for X? Fifteen. This means that X can be easily found by simple addition. Add the difference of the arithmetic progression to 15:

That's all. Answer: x=12

We solve the following problems ourselves. Note: these problems are not based on formulas. Purely to understand the meaning of an arithmetic progression.) We just write down a series of numbers and letters, look and figure it out.

5. Find the first positive term of the arithmetic progression if a 5 = -3; d = 1.1.

6. It is known that the number 5.5 is a member of the arithmetic progression (a n), where a 1 = 1.6; d = 1.3. Determine the number n of this member.

7. It is known that in arithmetic progression a 2 = 4; a 5 = 15.1. Find a 3 .

8. Several consecutive terms of the arithmetic progression are written out:

...; 15.6; X; 3.4; ...

Find the term of the progression indicated by the letter x.

9. The train began moving from the station, uniformly increasing speed by 30 meters per minute. What will be the speed of the train in five minutes? Give your answer in km/hour.

10. It is known that in arithmetic progression a 2 = 5; a 6 = -5. Find a 1.

Answers (in disarray): 7.7; 7.5; 9.5; 9; 0.3; 4.

Everything worked out? Amazing! You can master arithmetic progression at a higher level in the following lessons.

Didn't everything work out? No problem. In Special Section 555, all these problems are sorted out piece by piece.) And, of course, a simple practical technique is described that immediately highlights the solution to such tasks clearly, clearly, at a glance!

By the way, in the train puzzle there are two problems that people often stumble over. One is purely in terms of progression, and the second is general for any problems in mathematics, and physics too. This is a translation of dimensions from one to another. It shows how these problems should be solved.

In this lesson we looked at the elementary meaning of an arithmetic progression and its main parameters. This is enough to solve almost all problems on this topic. Add d to the numbers, write a series, everything will be solved.

The finger solution works well for very short pieces of a row, as in the examples in this lesson. If the series is longer, the calculations become more complicated. For example, if in problem 9 in the question you replace "five minutes" on "thirty-five minutes" the problem will become significantly worse.)

And there are also tasks that are simple in essence, but absurd in terms of calculations, for example:

An arithmetic progression (a n) is given. Find a 121 if a 1 =3 and d=1/6.

So what, are we going to add 1/6 many, many times?! You can kill yourself!?

You can.) If you don’t know a simple formula by which you can solve such tasks in a minute. This formula will be in the next lesson. And this problem is solved there. In a minute.)

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. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

For example, the sequence \(2\); \(5\); \(8\); \(eleven\); \(14\)... is an arithmetic progression, because each subsequent 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 \(16\); \(10\); \(4\); \(-2\); \(-8\)... the progression difference \(d\) is equal to minus six.

And in this case, each next element will be smaller than the previous one. These progressions are called decreasing.

Arithmetic progression notation

Progression is indicated by a small Latin letter.

Numbers that form a progression are called members(or elements).

They are denoted by the same letter as an arithmetic progression, but with a numerical index equal to the number of the element 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 arithmetic progression problems

In principle, the information presented above is already enough to solve almost any arithmetic progression problem (including those offered at the OGE).

Example (OGE). The arithmetic progression is specified 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:

Answer: \(-3\)

Example (OGE). Given several consecutive elements of an arithmetic progression: \(…5; x; 10; 12.5...\) Find the value of the element designated by the letter \(x\).
Solution:

Answer: \(7,5\).

Example (OGE). The arithmetic progression is defined 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:

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 formulas for arithmetic progression

As you can see, many problems on arithmetic progression can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each subsequent 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 deciding “head-on” is very inconvenient. 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 \(b_(386)\). Should we add four \(385\) times? Or imagine that in the penultimate example you need to find the sum of the first seventy-three elements. You'll be tired of counting...

Therefore, in such cases they do not solve things “head-on”, 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 of \(n\) first terms.

Formula of the \(n\)th term: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first term of the progression;
\(n\) – number of the required element;
\(a_n\) – term of the progression with number \(n\).

This formula allows us to quickly find even the three-hundredth or the millionth element, knowing only the first and the difference of the progression.

Example. The arithmetic progression is specified by the conditions: \(b_1=-159\); \(d=8.2\). Find \(b_(246)\).
Solution:

Answer: \(b_(246)=1850\).

Formula for the sum of the first n terms: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where

\(a_n\) – the last summed term;

Example (OGE). The arithmetic progression is specified by the conditions \(a_n=3.4n-0.6\). Find the sum of the first \(25\) terms of this progression.
Solution:

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:

Formula for the sum of the first n terms: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where

\(S_n\) – the required sum of \(n\) first elements;
\(a_1\) – the first summed term;
\(d\) – progression difference;
\(n\) – number of elements in total.

Example. Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15.5\); \(14\)…
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 not only need to 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:

Answer: \(S_(65)=-630.5\).

Example (OGE). The arithmetic progression is specified by the conditions: \(a_1=-33\); \(a_(n+1)=a_n+4\). Find the sum from the \(26\)th to the \(42\) element inclusive.
Solution:

Answer: \(S=1683\).

For arithmetic progression, there are several more formulas that we did not consider in this article due to their low practical usefulness. However, you can easily find them.

Many people have heard about arithmetic progression, but not everyone has a good idea of ​​what it is. In this article we will give the corresponding definition, and also consider the question of how to find the difference of an arithmetic progression, and give a number of examples.

Mathematical definition

So, if we are talking about an arithmetic or algebraic progression (these concepts define the same thing), then this means that there is a certain number series that satisfies the following law: every two adjacent numbers in the series differ by the same value. Mathematically it is written like this:

Here n means the number of element a n in the sequence, and the number d is the difference of the progression (its name follows from the presented formula).

What does knowing the difference d mean? About how “far” neighboring numbers are from each other. However, knowledge of d is a necessary but not sufficient condition for determining (restoring) the entire progression. You need to know one more number, which can be absolutely any element of the series in question, for example, a 4, a10, but, as a rule, they use the first number, that is, a 1.

Formulas for determining progression elements

In general, the information above is already enough to move on to solving specific problems. Nevertheless, before the arithmetic progression is given, and it will be necessary to find its difference, we will present a couple of useful formulas, thereby facilitating the subsequent process of solving problems.

It is easy to show that any element of the sequence with number n can be found as follows:

a n = a 1 + (n - 1) * d

Indeed, anyone can check this formula by simple search: if you substitute n = 1, you get the first element, if you substitute n = 2, then the expression gives the sum of the first number and the difference, and so on.

The conditions of many problems are composed in such a way that, given a known pair of numbers, the numbers of which are also given in the sequence, it is necessary to reconstruct the entire number series (find the difference and the first element). Now we will solve this problem in general form.

So, let two elements with numbers n and m be given. Using the formula obtained above, you can create a system of two equations:

a n = a 1 + (n - 1) * d;

a m = a 1 + (m - 1) * d

To find unknown quantities, we will use a well-known simple technique for solving such a system: subtract the left and right sides in pairs, the equality will remain valid. We have:

a n = a 1 + (n - 1) * d;

a n - a m = (n - 1) * d - (m - 1) * d = d * (n - m)

Thus, we have excluded one unknown (a 1). Now we can write the final expression for determining d:

d = (a n - a m) / (n - m), where n > m

We received a very simple formula: to calculate the difference d in accordance with the conditions of the problem, you only need to take the ratio of the differences between the elements themselves and their serial numbers. One important point should be paid attention to: the differences are taken between the “senior” and “junior” members, that is, n > m (“senior” means standing further from the beginning of the sequence, its absolute value can be either greater or less more "junior" element).

The expression for the difference d progression should be substituted into any of the equations at the beginning of solving the problem to obtain the value of the first term.

In our age of computer technology development, many schoolchildren try to find solutions for their assignments on the Internet, so questions of this type often arise: find the difference of an arithmetic progression online. For such a request, the search engine will return a number of web pages, by going to which you will need to enter the data known from the condition (this can be either two terms of the progression or the sum of a certain number of them) and instantly receive an answer. However, this approach to solving the problem is unproductive in terms of the student’s development and understanding of the essence of the task assigned to him.

Solution without using formulas

Let's solve the first problem without using any of the given formulas. Let the elements of the series be given: a6 = 3, a9 = 18. Find the difference of the arithmetic progression.

Known elements stand close to each other in a row. How many times must the difference d be added to the smallest to get the largest? Three times (the first time adding d, we get the 7th element, the second time - the eighth, finally, the third time - the ninth). What number must be added to three three times to get 18? This is the number five. Really:

Thus, the unknown difference d = 5.

Of course, the solution could have been carried out using the appropriate formula, but this was not done intentionally. A detailed explanation of the solution to the problem should become a clear and clear example of what an arithmetic progression is.

A task similar to the previous one

Now let's solve a similar problem, but change the input data. So, you should find if a3 = 2, a9 = 19.

Of course, you can again resort to the “head-on” solution method. But since the elements of the series are given, which are relatively far from each other, this method will not be entirely convenient. But using the resulting formula will quickly lead us to the answer:

d = (a 9 - a 3) / (9 - 3) = (19 - 2) / (6) = 17 / 6 ≈ 2.83

Here we have rounded the final number. The extent to which this rounding led to an error can be judged by checking the result obtained:

a 9 = a 3 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 = 18.98

This result differs by only 0.1% from the value given in the condition. Therefore, the rounding used to the nearest hundredths can be considered a successful choice.

Problems involving applying the formula for the an term

Let's consider a classic example of a problem to determine the unknown d: find the difference of an arithmetic progression if a1 = 12, a5 = 40.

When two numbers of an unknown algebraic sequence are given, and one of them is the element a 1, then you do not need to think long, but should immediately apply the formula for the a n term. In this case we have:

a 5 = a 1 + d * (5 - 1) => d = (a 5 - a 1) / 4 = (40 - 12) / 4 = 7

We received the exact number when dividing, so there is no point in checking the accuracy of the calculated result, as was done in the previous paragraph.

Let's solve another similar problem: we need to find the difference of an arithmetic progression if a1 = 16, a8 = 37.

We use an approach similar to the previous one and get:

a 8 = a 1 + d * (8 - 1) => d = (a 8 - a 1) / 7 = (37 - 16) / 7 = 3

What else should you know about arithmetic progression?

In addition to problems of finding an unknown difference or individual elements, it is often necessary to solve problems of the sum of the first terms of a sequence. Consideration of these problems is beyond the scope of the article, however, for completeness of information, we present a general formula for the sum of n numbers in a series:

∑ n i = 1 (a i) = n * (a 1 + a n) / 2

Well, friends, if you are reading this text, then the internal cap-evidence tells me that you do not yet 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 get straight to the point.

First, a couple of examples. Let's look at 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 simply consecutive numbers, each next being one more than the previous one. In the second case, the difference between adjacent numbers is already five, but this difference is still constant. In the third case, there are roots altogether. However, $2\sqrt(2)=\sqrt(2)+\sqrt(2)$, and $3\sqrt(2)=2\sqrt(2)+\sqrt(2)$, i.e. and in this case, each next element simply increases by $\sqrt(2)$ (and don’t be afraid that this number is irrational).

So: all such sequences are called arithmetic progressions. Let's give a strict definition:

Definition. A sequence of numbers in which each next one 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 notes. Firstly, progression is only considered ordered sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. Numbers cannot be rearranged or swapped.

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 in the spirit (1; 2; 3; 4; ...) - this is already an infinite progression. The ellipsis after the four seems to hint that there are quite a few more numbers to come. Infinitely many, for example. :)

I would also like to note that progressions can be increasing or 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:

1. increasing if each next element is greater than the previous one;
2. 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:

1. If $d \gt 0$, then the progression increases;
2. If $d \lt 0$, then the progression is obviously decreasing;
3. 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 given above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract the number on the left from the number on the right. It will look like this:

• 41−49=−8;
• 12−17,5=−5,5;
• $\sqrt(5)-1-\sqrt(5)=-1$.

As we can see, in all three cases the difference actually 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.

Progression terms and recurrence formula

Since the elements of our sequences cannot be swapped, they can be numbered:

\[\left(((a)_(n)) \right)=\left\( ((a)_(1)),\ ((a)_(2)),((a)_(3 )),... \right\)\]

The individual elements of this set are called members of a progression. They are indicated by a number: first member, second member, etc.

In addition, as we already know, neighboring terms 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 a progression, you need to know the $n-1$th term and the difference $d$. This formula is called recurrent, because with its help you can find any number only by knowing the previous one (and in fact, all the previous ones). This is very inconvenient, so there is a more cunning formula that reduces any calculations to the first term and the difference:

\[((a)_(n))=((a)_(1))+\left(n-1 \right)d\]

You've probably already come across this formula. They like to give it in all sorts of reference books and problem books. And in any sensible mathematics textbook it is one of the first.

However, I suggest you practice a little.

Task No. 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 difference of the progression $d=-5$. 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! Please note: our progression is decreasing.

Of course, $n=1$ could not be substituted - the first term is already known to us. However, by substituting unity, we were convinced that even for the first term our formula works. In other cases, everything came down to banal arithmetic.

Task No. 2. Write down the first three terms of an arithmetic progression if its seventh term is equal to −40 and its seventeenth term is equal to −50.

Solution. Let us write the problem condition in familiar 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 system sign because these requirements must be met simultaneously. Now let’s note that if we subtract the first from the second equation (we have the right to do this, since we have a system), we get this:

\[\begin(align) & ((a)_(1))+16d-\left(((a)_(1))+6d \right)=-50-\left(-40 \right); \\ & ((a)_(1))+16d-((a)_(1))-6d=-50+40; \\&10d=-10; \\&d=-1. \\ \end(align)\]

That's how easy it is to find the progression difference! All that remains is to substitute the found number into 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! The problem is solved.

Answer: (−34; −35; −36)

Notice the interesting property of 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 $n-m$ number:

\[((a)_(n))-((a)_(m))=d\cdot \left(n-m \right)\]

A simple but very useful property that you definitely need to know - with its help you can significantly speed up the solution of many progression problems. Here is a clear example of this:

Task No. 3. The fifth term of an 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 $((a)_(10))-((a)_(5))=14.4-8.4=6$, therefore $5d=6$, from which 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 didn’t need to create any systems of equations and calculate the first term and the difference - everything was solved in just a couple of lines.

Now let's look at another type of problem - searching for negative and positive terms of a progression. It is no secret that if a progression increases, and 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 not always possible to find this moment “head-on” by sequentially going through the elements. Often, problems are written in such a way that without knowing the formulas, the calculations would take several sheets of paper—we would simply fall asleep while we found the answer. Therefore, let's try to solve these problems in a faster way.

Task No. 4. How many negative terms are there in the arithmetic progression −38.5; −35.8; ...?

Solution. So, $((a)_(1))=-38.5$, $((a)_(2))=-35.8$, from where we immediately find the difference:

Note that the difference is positive, so the progression increases. 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 remains:

\[\begin(align) & ((a)_(n)) \lt 0\Rightarrow ((a)_(1))+\left(n-1 \right)d \lt 0; \\ & -38.5+\left(n-1 \right)\cdot 2.7 \lt 0;\quad \left| \cdot 10 \right. \\ & -385+27\cdot \left(n-1 \right) \lt 0; \\ & -385+27n-27 \lt 0; \\ & 27n \lt 412; \\ & n \lt 15\frac(7)(27)\Rightarrow ((n)_(\max ))=15. \\ \end(align)\]

The last line requires some explanation. So we know that $n \lt 15\frac(7)(27)$. On the other hand, we are satisfied with only integer values ​​of the number (moreover: $n\in \mathbb(N)$), so the largest permissible number is precisely $n=15$, and in no case 16.

Task No. 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 do not know $((a)_(1))$. But the neighboring terms are known: $((a)_(5))$ and $((a)_(6))$, so we can easily find the difference of the progression:

In addition, let's try to express the fifth term through 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 task. Let's 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 to this inequality is the number 56.

Please note: in the last task everything came down 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 study 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 indentations

Let's consider several consecutive terms of the increasing arithmetic progression $\left(((a)_(n)) \right)$. Let's try to mark them on the number line:

I specifically marked arbitrary terms $((a)_(n-3)),...,((a)_(n+3))$, and not some $((a)_(1)) ,\ ((a)_(2)),\ ((a)_(3))$, etc. Because the rule that I’ll tell you about now works the same for any “segments”.

And the rule is very simple. Let's remember the recurrent formula and write it for all marked terms:

\[\begin(align) & ((a)_(n-2))=((a)_(n-3))+d; \\ & ((a)_(n-1))=((a)_(n-2))+d; \\ & ((a)_(n))=((a)_(n-1))+d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n+1))+d; \\ \end(align)\]

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? And 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 $((a)_(n))$ at the same distance equal to $2d$. We can continue ad infinitum, but the meaning is well illustrated by the picture

What does this mean for us? This means that $((a)_(n))$ can be found if the neighboring numbers are known:

\[((a)_(n))=\frac(((a)_(n-1))+((a)_(n+1)))(2)\]

We have derived an excellent statement: every term of an arithmetic progression is equal to the arithmetic mean of its neighboring terms! Moreover: we can step back from our $((a)_(n))$ to the left and to the right not by one step, but by $k$ steps - and the formula will still 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 problems are specially tailored to use the arithmetic mean. Take a look:

Task No. 6. Find all values ​​of $x$ for which the numbers $-6((x)^(2))$, $x+1$ and $14+4((x)^(2))$ are consecutive terms of an arithmetic progression (in in the order indicated).

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 No. 7. Find the values ​​of $$ for which the numbers $-1;4-3;(()^(2))+1$ form an arithmetic progression (in that order).

Solution. Let us again express the middle term through the arithmetic mean of neighboring terms:

\[\begin(align) & 4x-3=\frac(x-1+((x)^(2))+1)(2); \\ & 4x-3=\frac(((x)^(2))+x)(2);\quad \left| \cdot 2 \right.; \\ & 8x-6=((x)^(2))+x; \\ & ((x)^(2))-7x+6=0. \\ \end(align)\]

Quadratic equation again. And again there are two roots: $x=6$ and $x=1$.

Answer: 1; 6.

If in the process of solving a problem you come up with some brutal numbers, or you are not entirely sure of the correctness of the answers found, then there is a wonderful technique that allows you to check: have we solved the problem correctly?

Let's say in problem No. 6 we received 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 must form an arithmetic progression. Let's 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 was solved correctly. Those who wish can check the second problem on their own, but I’ll say right away: everything is correct there too.

In general, while solving the last problems, we came across another interesting fact that also needs to be remembered:

If three numbers are such that the second is the arithmetic mean 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 conditions of the problem. But before we engage in such “construction”, we should pay attention to one more fact, which directly follows from what has already been discussed.

Grouping and summing elements

Let's return to the number axis again. Let us note there several members of the progression, between which, perhaps. is worth a lot of other members:

Let's try to express the “left tail” through $((a)_(n))$ and $d$, and the “right tail” through $((a)_(k))$ 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 amounts 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 begin to step from these elements in opposite directions (toward 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 most clearly represented graphically:

Understanding this fact will allow us to solve problems of a fundamentally higher level of complexity than those we considered above. For example, these:

Task No. 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 progression difference $d$. Actually, the entire 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 took the total multiplier of 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 expand 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 of the highest term is 11 - this is a positive number, so we are really dealing with a parabola with upward branches:

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 using the standard scheme (there is the 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, therefore 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 particular hurry to open the brackets: in their 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 does the discovered number give us? With it, the required product takes on the smallest value (by the way, we never calculated $((y)_(\min ))$ - this is not required of us). At the same time, this number is the difference of the original progression, i.e. we found the answer. :)

Answer: −36

Task No. 9. Between the numbers $-\frac(1)(2)$ and $-\frac(1)(6)$ insert three numbers so that together with these numbers they form an arithmetic progression.

Solution. Essentially, we need to make a sequence of five numbers, with the first and last number already known. Let's 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 we currently cannot obtain $y$ from the numbers $x$ and $z$, then the situation is different with the ends of the progression. Let's remember the arithmetic mean:

Now, knowing $y$, we will find the remaining numbers. Note that $x$ lies between the numbers $-\frac(1)(2)$ and the $y=-\frac(1)(3)$ we just found. That's why

Using similar reasoning, we find the remaining number:

Ready! We found all three numbers. Let's write them 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 No. 10. Between the numbers 2 and 42, insert several numbers that, together with these numbers, form an arithmetic progression, if you know that the sum of the first, second and last of the inserted numbers is 56.

Solution. An even more complex problem, which, however, is solved according to the same scheme as the previous ones - through the arithmetic mean. The problem is that we don’t know exactly how many numbers need to be inserted. Therefore, let us assume for definiteness that after inserting everything there will be exactly $n$ numbers, and the first of them is 2, and the last is 42. In this case, the required arithmetic progression can be represented in the form:

\[\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 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 expression written above can be rewritten as follows:

\[\begin(align) & ((a)_(2))+((a)_(3))+((a)_(n-1))=56; \\ & \left(((a)_(2))+((a)_(n-1)) \right)+((a)_(3))=56; \\ & 44+((a)_(3))=56; \\ & ((a)_(3))=56-44=12. \\ \end(align)\]

Knowing $((a)_(3))$ and $((a)_(1))$, we can easily find the difference of the progression:

\[\begin(align) & ((a)_(3))-((a)_(1))=12-2=10; \\ & ((a)_(3))-((a)_(1))=\left(3-1 \right)\cdot d=2d; \\ & 2d=10\Rightarrow d=5. \\ \end(align)\]

All that remains is to find the remaining terms:

\[\begin(align) & ((a)_(1))=2; \\ & ((a)_(2))=2+5=7; \\ & ((a)_(3))=12; \\ & ((a)_(4))=2+3\cdot 5=17; \\ & ((a)_(5))=2+4\cdot 5=22; \\ & ((a)_(6))=2+5\cdot 5=27; \\ & ((a)_(7))=2+6\cdot 5=32; \\ & ((a)_(8))=2+7\cdot 5=37; \\ & ((a)_(9))=2+8\cdot 5=42; \\ \end(align)\]

Thus, already at the 9th step we will arrive at 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

Word problems with progressions

In conclusion, I would like to consider a couple of relatively simple problems. Well, as simple as that: for most students who study mathematics at school and have not read what is written above, these problems may seem tough. Nevertheless, these are the types of problems that appear in the OGE and the Unified State Exam in mathematics, so I recommend that you familiarize yourself with them.

Task No. 11. The team produced 62 parts in January, and in each subsequent month they produced 14 more parts than in the previous month. How many parts did the team produce in November?

Solution. Obviously, the number of parts listed by month will represent an increasing arithmetic progression. Moreover:

\[\begin(align) & ((a)_(1))=62;\quad d=14; \\ & ((a)_(n))=62+\left(n-1 \right)\cdot 14. \\ \end(align)\]

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 produced in November.

Task No. 12. The bookbinding workshop bound 216 books in January, and in each subsequent month it bound 4 more books than in 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’s course” in arithmetic progressions. You can safely move on to the next lesson, where we will study the formula for the sum of progression, as well as important and very useful consequences from it.