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Make an arithmetic progression of the difference. Arithmetic progression - number sequence

Online calculator.
Arithmetic progression solution.
Given: a n , d, n
Find: a 1

This math program finds $a_1$ of an arithmetic progression based on user-specified numbers $a_n, d $ and $n $.
The numbers $a_n$ and $d $ can be specified not only as integers, but also as fractions. Moreover, a fractional number can be entered in the form of a decimal fraction (\ (2.5 \)) and in the form common fraction($-5\frac(2)(7) $).

The program not only gives the answer to the problem, but also displays the process of finding a solution.

This online calculator can be useful for high school students general education schools in preparation for control work and exams, when testing knowledge before the exam, parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as soon as possible? homework math or algebra? In this case, you can also use our programs with a detailed solution.

Thus, you can carry out your own training and/or training of their younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.

If you are not familiar with the rules for entering numbers, we recommend that you familiarize yourself with them.

Rules for entering numbers

The numbers $a_n$ and $d $ can be specified not only as integers, but also as fractions.
The number $n$ can only be a positive integer.

Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a dot or a comma.
For example, you can enter decimals like 2.5 or like 2.5

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative.

When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
Input:
Result: $-\frac(2)(3) $

The integer part is separated from the fraction by an ampersand: &
Input:
Result: $-1\frac(2)(3) $

Enter numbers a n , d, n Find a 1

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A bit of theory.

Numeric sequence

Numbering is often used in everyday practice. various items to indicate their order. For example, the houses on each street are numbered. In the library, reader's subscriptions are numbered and then arranged in the order of the assigned numbers in special file cabinets.

In a savings bank, by the number of the depositor's personal account, you can easily find this account and see what kind of deposit it has. Let there be a deposit of a1 rubles on account No. 1, a deposit of a2 rubles on account No. 2, etc. It turns out numerical sequence
a 1 , a 2 , a 3 , ..., a N
where N is the number of all accounts. Here, each natural number n from 1 to N is assigned a number a n .

Mathematics also studies infinite number sequences:
a 1 , a 2 , a 3 , ..., a n , ... .
The number a 1 is called the first member of the sequence, number a 2 - the second member of the sequence, number a 3 - the third member of the sequence etc.
The number a n is called nth (nth) member of the sequence, and the natural number n is its number.

For example, in the sequence of squares of natural numbers 1, 4, 9, 16, 25, ..., n 2 , (n + 1) 2 , ... and 1 = 1 is the first member of the sequence; and n = n 2 is nth member sequences; a n+1 = (n + 1) 2 is the (n + 1)th (en plus the first) member of the sequence. Often a sequence can be specified by the formula of its nth term. For example, the formula $a_n=\frac(1)(n), \; n \in \mathbb(N) $ gives the sequence $1, \; \frac(1)(2) , \; \frac( 1)(3) , \; \frac(1)(4) , \dots,\frac(1)(n) , \dots $

Arithmetic progression

The length of a year is approximately 365 days. More exact value equals $365\frac(1)(4) $ days, so every four years an error of one day accumulates.

To account for this error, a day is added to every fourth year, and the elongated year is called a leap year.

For example, in the third millennium, leap years are 2004, 2008, 2012, 2016, ... .

In this sequence, each member, starting from the second, is equal to the previous one, added with the same number 4. Such sequences are called arithmetic progressions.

Definition.
The numerical sequence a 1 , a 2 , a 3 , ..., a n , ... is called arithmetic progression, if for all natural n the equality
$a_(n+1) = a_n+d, $
where d is some number.

It follows from this formula that a n+1 - a n = d. The number d is called the difference arithmetic progression.

By definition of an arithmetic progression, we have:
$a_(n+1)=a_n+d, \quad a_(n-1)=a_n-d, $
where
$a_n= \frac(a_(n-1) +a_(n+1))(2) $, where $n>1 $

Thus, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the two members adjacent to it. This explains the name "arithmetic" progression.

Note that if a 1 and d are given, then the remaining terms of the arithmetic progression can be calculated using the recursive formula a n+1 = a n + d. In this way, it is not difficult to calculate the first few terms of the progression, however, for example, for a 100, a lot of calculations will already be required. Usually, the nth term formula is used for this. According to the definition of an arithmetic progression
$a_2=a_1+d, $
$a_3=a_2+d=a_1+2d, $
$a_4=a_3+d=a_1+3d$
etc.
At all,
$a_n=a_1+(n-1)d, $
because nth member arithmetic progression is obtained from the first term by adding (n-1) times the number d.
This formula is called formula of the nth member of an arithmetic progression.

The sum of the first n terms of an arithmetic progression

Let's find the sum of all natural numbers from 1 to 100.
We write this sum in two ways:
S = l + 2 + 3 + ... + 99 + 100,
S = 100 + 99 + 98 + ... + 2 + 1.
We add these equalities term by term:
2S = 101 + 101 + 101 + ... + 101 + 101.
There are 100 terms in this sum.
Therefore, 2S = 101 * 100, whence S = 101 * 50 = 5050.

Consider now an arbitrary arithmetic progression
a 1 , a 2 , a 3 , ..., a n , ...
Let S n be the sum of the first n terms of this progression:
S n \u003d a 1, a 2, a 3, ..., a n
Then the sum of the first n terms of an arithmetic progression is
$S_n = n \cdot \frac(a_1+a_n)(2) $

Since $a_n=a_1+(n-1)d $, then replacing a n in this formula, we get another formula for finding the sums of the first n terms of an arithmetic progression:
$S_n = n \cdot \frac(2a_1+(n-1)d)(2) $

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Or arithmetic - this is a type of ordered numerical sequence, the properties of which are studied in a school algebra course. This article discusses in detail the question of how to find the sum of an arithmetic progression.

What is this progression?

Before proceeding to the consideration of the question (how to find the sum of an arithmetic progression), it is worth understanding what will be discussed.

Any sequence of real numbers that is obtained by adding (subtracting) some value from each previous number is called an algebraic (arithmetic) progression. This definition, translated into the language of mathematics, takes the form:

Here i is the ordinal number of the element of the series a i . Thus, knowing only one initial number, you can easily restore the entire series. The parameter d in the formula is called the progression difference.

It can be easily shown that the following equality holds for the series of numbers under consideration:

a n \u003d a 1 + d * (n - 1).

That is, to find the value of the n-th element in order, add the difference d to the first element a 1 n-1 times.

What is the sum of an arithmetic progression: formula

Before giving the formula for the indicated amount, it is worth considering a simple special case. Given a progression of natural numbers from 1 to 10, you need to find their sum. Since there are few terms in the progression (10), it is possible to solve the problem head-on, that is, sum all the elements in order.

S 10 \u003d 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 \u003d 55.

It is worth considering one interesting thing: since each term differs from the next one by the same value d \u003d 1, then the pairwise summation of the first with the tenth, the second with the ninth, and so on will give the same result. Really:

11 = 1+10 = 2+9 = 3+8 = 4+7 = 5+6.

As you can see, there are only 5 of these sums, that is, exactly two times less than the number of elements in the series. Then multiplying the number of sums (5) by the result of each sum (11), you will come to the result obtained in the first example.

If we generalize these arguments, we can write the following expression:

S n \u003d n * (a 1 + a n) / 2.

This expression shows that it is not at all necessary to sum all the elements in a row, it is enough to know the value of the first a 1 and the last a n , and also total number terms n.

It is believed that Gauss first thought of this equality when he was looking for a solution to the problem set by his school teacher: to sum the first 100 integers.

Sum of elements from m to n: formula

The formula given in the previous paragraph answers the question of how to find the sum of an arithmetic progression (of the first elements), but often in tasks it is necessary to sum a series of numbers in the middle of the progression. How to do it?

The easiest way to answer this question is by considering the following example: let it be necessary to find the sum of terms from the mth to the nth. To solve the problem, a given segment from m to n of the progression should be represented as a new number series. In such representation m-th term a m will be the first, and a n will be numbered n-(m-1). In this case, applying the standard formula for the sum, the following expression will be obtained:

S m n \u003d (n - m + 1) * (a m + a n) / 2.

Example of using formulas

Knowing how to find the sum of an arithmetic progression, it is worth considering a simple example of using the above formulas.

Below is a numerical sequence, you should find the sum of its members, starting from the 5th and ending with the 12th:

The given numbers indicate that the difference d is equal to 3. Using the expression for the nth element, you can find the values of the 5th and 12th members of the progression. It turns out:

a 5 \u003d a 1 + d * 4 \u003d -4 + 3 * 4 \u003d 8;
a 12 \u003d a 1 + d * 11 \u003d -4 + 3 * 11 \u003d 29.

Knowing the values \u200b\u200bof the numbers at the ends of the algebraic progression under consideration, and also knowing which numbers in the series they occupy, you can use the formula for the sum obtained in the previous paragraph. Get:

S 5 12 \u003d (12 - 5 + 1) * (8 + 29) / 2 \u003d 148.

It is worth noting that this value could be obtained differently: first, find the sum of the first 12 elements using the standard formula, then calculate the sum of the first 4 elements using the same formula, and then subtract the second from the first sum.

What main point formulas?

This formula allows you to find any BY HIS NUMBER" n" .

Of course, you need to know the first term a 1 and progression difference d, well, without these parameters, you can’t write down a specific progression.

It is not enough to memorize (or cheat) this formula. It is necessary to assimilate its essence and apply the formula in various problems. Yes, and do not forget at the right time, yes ...) How not forget- I don't know. And here how to remember If needed, I'll give you a hint. For those who master the lesson to the end.)

So, let's deal with the formula of the n-th member of an arithmetic progression.

What is a formula in general - we imagine.) What is an arithmetic progression, a member number, a progression difference - is clearly stated in the previous lesson. Take a look if you haven't read it. Everything is simple there. It remains to figure out what nth member.

progression in general view can be written as a series of numbers:

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

a 1- denotes the first term of an arithmetic progression, a 3- third member a 4- fourth, and so on. If we are interested in the fifth term, let's say we are working with a 5, if one hundred and twentieth - from a 120.

How to define in general any member of an arithmetic progression, s any number? Very simple! Like this:

a n

That's what it is n-th member of an arithmetic progression. Under the letter n all the numbers of members are hidden at once: 1, 2, 3, 4, and so on.

And what does such a record give us? Just think, instead of a number, they wrote down a letter ...

This notation gives us a powerful tool for working with arithmetic progressions. Using the notation a n, we can quickly find any member any arithmetic progression. And a bunch of tasks to solve in progression. You will see further.

In the formula of the nth member of an arithmetic progression:

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

a 1- the first member of the arithmetic progression;

n- member number.

The formula links the key parameters of any progression: a n ; a 1 ; d And n. Around these parameters, all the puzzles revolve in progression.

The nth term formula can also be used to write a specific progression. For example, in the problem it can be said that the progression is given by the condition:

a n = 5 + (n-1) 2.

Such a problem can even confuse ... There is no series, no difference ... But, comparing the condition with the formula, it is easy to figure out that in this progression a 1 \u003d 5, and d \u003d 2.

And it can be even angrier!) If we take the same condition: a n = 5 + (n-1) 2, yes, open the brackets and give similar ones? We get a new formula:

an = 3 + 2n.

This Only not general, but for a specific progression. This is where the pitfall lies. Some people think that the first term is a three. Although in reality the first member is a five ... A little lower we will work with such a modified formula.

In tasks for progression, there is another notation - a n+1. This is, you guessed it, the "n plus the first" term of the progression. Its meaning is simple and harmless.) This is a member of the progression, the number of which is greater than the number n by one. For example, if in some problem we take for a n fifth term, then a n+1 will be the sixth member. Etc.

Most often the designation a n+1 occurs in recursive formulas. Do not be afraid of this terrible word!) This is just a way of expressing a term of an arithmetic progression through the previous one. Suppose we are given an arithmetic progression in this form, using the recurrent formula:

a n+1 = a n +3

a 2 = a 1 + 3 = 5+3 = 8

a 3 = a 2 + 3 = 8+3 = 11

The fourth - through the third, the fifth - through the fourth, and so on. And how to count immediately, say the twentieth term, a 20? But no way!) While the 19th term is not known, the 20th cannot be counted. This is the fundamental difference between the recursive formula and the formula of the nth term. Recursive works only through previous term, and the formula of the nth term - through first and allows straightaway find any member by its number. Not counting the whole series of numbers in order.

In an arithmetic progression, a recursive formula can easily be turned into a regular one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write the formula in the usual form, and work with it. In the GIA, such tasks are often found.

Application of the formula of the n-th member of an arithmetic progression.

First, let's look at the direct application of the formula. At the end of the previous lesson there was a problem:

Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.

This problem can be solved without any formulas, simply based on the meaning of the arithmetic progression. Add, yes add ... An hour or two.)

And according to the formula, the solution will take less than a minute. You can time it.) We decide.

The conditions provide all the data for using the formula: a 1 \u003d 3, d \u003d 1/6. It remains to be seen what n. No problem! We need to find a 121. Here we write:

Please pay attention! Instead of an index n a specific number appeared: 121. Which is quite logical.) We are interested in the member of the arithmetic progression number one hundred twenty one. This will be our n. It is this meaning n= 121 we will substitute further into the formula, in brackets. Substitute all the numbers in the formula and calculate:

a 121 = 3 + (121-1) 1/6 = 3+20 = 23

That's all there is to it. Just as quickly one could find the five hundred and tenth member, and the thousand and third, any. We put instead n the desired number in the index of the letter " a" and in brackets, and we consider.

Let me remind you the essence: this formula allows you to find any term of an arithmetic progression BY HIS NUMBER" n" .

Let's solve the problem smarter. Let's say we have the following problem:

Find the first term of the arithmetic progression (a n) if a 17 =-2; d=-0.5.

If you have any difficulties, I will suggest the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Hand write, right in your notebook:

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

And now, looking at the letters of the formula, we understand what data we have and what is missing? Available d=-0.5, there is a seventeenth member ... Everything? If you think that's all, then you can't solve the problem, yes ...

We also have a number n! In the condition a 17 =-2 hidden two options. This is both the value of the seventeenth member (-2) and its number (17). Those. n=17. This "little thing" often slips past the head, and without it, (without the "little thing", not the head!) The problem cannot be solved. Although ... and without a head too.)

Now we can just stupidly substitute our data into the formula:

a 17 \u003d a 1 + (17-1) (-0.5)

Oh yes, a 17 we know it's -2. Okay, let's put it in:

-2 \u003d a 1 + (17-1) (-0.5)

That, in essence, is all. It remains to express the first term of the arithmetic progression from the formula, and calculate. You get the answer: a 1 = 6.

Such a technique - writing a formula and simply substituting known data - helps a lot in simple tasks. Well, you must, of course, be able to express a variable from a formula, but what to do!? Without this skill, mathematics can not be studied at all ...

Another popular problem:

Find the difference of the arithmetic progression (a n) if a 1 =2; a 15 =12.

What are we doing? You will be surprised, we write the formula!)

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

Consider what we know: a 1 =2; a 15 =12; and (special highlight!) n=15. Feel free to substitute in the formula:

12=2 + (15-1)d

Let's do the arithmetic.)

12=2 + 14d

d=10/14 = 5/7

This is the correct answer.

So, tasks a n , a 1 And d decided. It remains to learn how to find the number:

The number 99 is a member of an arithmetic progression (a n), where a 1 =12; d=3. Find the number of this member.

We substitute the known quantities into the formula of the nth term:

a n = 12 + (n-1) 3

At first glance, there are two unknown quantities here: a n and n. But a n is some member of the progression with the number n... And this member of the progression we know! It's 99. We don't know his number. n, so this number also needs to be found. Substitute the progression term 99 into the formula:

99 = 12 + (n-1) 3

We express from the formula n, we think. We get the answer: n=30.

And now a problem on the same topic, but more creative):

Determine if the number 117 will be a member of an arithmetic progression (a n):

-3,6; -2,4; -1,2 ...

Let's write the formula again. What, there are no options? Hm... Why do we need eyes?) Do we see the first member of the progression? We see. This is -3.6. You can safely write: a 1 \u003d -3.6. Difference d can be determined from the series? It's easy if you know what the difference of an arithmetic progression is:

d = -2.4 - (-3.6) = 1.2

Yes, we did the simplest thing. It remains to deal with an unknown number n and an incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know that ... How to be!? Well, how to be, how to be... Turn on Creative skills!)

We suppose that 117 is, after all, a member of our progression. With an unknown number n. And, just like in the previous problem, let's try to find this number. Those. we write the formula (yes-yes!)) and substitute our numbers:

117 = -3.6 + (n-1) 1.2

Again we express from the formulan, we count and get:

Oops! The number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion do we draw? Yes! Number 117 is not member of our progression. It is somewhere between the 101st and 102nd member. If the number turned out to be natural, i.e. positive integer, then the number would be a member of the progression with the found number. And in our case, the answer to the problem will be: No.

Task based on a real version of the GIA:

Arithmetic progression given by the condition:

a n \u003d -4 + 6.8n

Find the first and tenth terms of the progression.

Here the progression is set in an unusual way. Some kind of formula ... It happens.) However, this formula (as I wrote above) - also the formula of the n-th member of an arithmetic progression! She also allows find any member of the progression by its number.

We are looking for the first member. The one who thinks. that the first term is minus four, is fatally mistaken!) Because the formula in the problem is modified. The first term of an arithmetic progression in it hidden. Nothing, we'll find it now.)

Just as in the previous tasks, we substitute n=1 into this formula:

a 1 \u003d -4 + 6.8 1 \u003d 2.8

Here! The first term is 2.8, not -4!

Similarly, we are looking for the tenth term:

a 10 \u003d -4 + 6.8 10 \u003d 64

That's all there is to it.

And now, for those who have read up to these lines, the promised bonus.)

Suppose, in a difficult combat situation of the GIA or the Unified State Exam, you forgot the useful formula of the n-th member of an arithmetic progression. Something comes to mind, but somehow uncertainly ... Whether n there, or n+1, or n-1... How to be!?

Calm! This formula is easy to derive. Not very strict, but definitely enough for confidence and the right decision!) For the conclusion, it is enough to remember the elementary meaning of the arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.

We draw a numerical axis and mark the first one on it. second, third, etc. members. And note the difference d between members. Like this:

We look at the picture and think: what is the second term equal to? Second one d:

a 2 =a 1 + 1 d

What is the third term? Third term equals first term plus two d.

a 3 =a 1 + 2 d

Do you get it? I don't put some words in bold for nothing. Okay, one more step.)

What is the fourth term? Fourth term equals first term plus three d.

a 4 =a 1 + 3 d

It's time to realize that the number of gaps, i.e. d, Always one less than the number of the member you are looking for n. That is, up to the number n, number of gaps will n-1. So, the formula will be (no options!):

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

In general, visual pictures are very helpful in solving many problems in mathematics. Don't neglect the pictures. But if it's difficult to draw a picture, then ... only a formula!) In addition, the formula of the nth term allows you to connect the entire powerful arsenal of mathematics to the solution - equations, inequalities, systems, etc. You can't put a picture in an equation...

Tasks for independent decision.

For warm-up:

1. In arithmetic progression (a n) a 2 =3; a 5 \u003d 5.1. Find a 3 .

Hint: according to the picture, the problem is solved in 20 seconds ... According to the formula, it turns out more difficult. But for mastering the formula, it is more useful.) In Section 555, this problem is solved both by the picture and by the formula. Feel the difference!)

And this is no longer a warm-up.)

2. In arithmetic progression (a n) a 85 \u003d 19.1; a 236 =49, 3. Find a 3 .

What, reluctance to draw a picture?) Still! It's better in the formula, yes ...

3. Arithmetic progression is given by the condition:a 1 \u003d -5.5; a n+1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.

In this task, the progression is given in a recurrent way. But counting up to the one hundred and twenty-fifth term... Not everyone can do such a feat.) But the formula of the nth term is within the power of everyone!

4. Given an arithmetic progression (a n):

-148; -143,8; -139,6; -135,4, .....

Find the number of the smallest positive term of the progression.

5. According to the condition of task 4, find the sum of the smallest positive and largest negative members of the progression.

6. The product of the fifth and twelfth terms of an increasing arithmetic progression is -2.5, and the sum of the third and eleventh terms is zero. Find a 14 .

Not the easiest task, yes ...) Here the method "on the fingers" will not work. You have to write formulas and solve equations.

Answers (in disarray):

3,7; 3,5; 2,2; 37; 2,7; 56,5

Happened? It's nice!)

Not everything works out? Happens. By the way, in last assignment there is one subtle point. Attentiveness when reading the problem will be required. And logic.

The solution to all these problems is discussed in detail in Section 555. And the fantasy element for the fourth, and the subtle moment for the sixth, and general approaches for solving any problems for the formula of the nth term - everything is painted. I recommend.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

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 in the spirit (1; 2; 3; 4; ...) - this is already 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; ...
$\sqrt(5);\ \sqrt(5)-1;\ \sqrt(5)-2;\ \sqrt(5)-3;...$

OK OK: last example may seem overly complicated. But the rest, I think, you understand. Therefore, we introduce new definitions:

Definition. An arithmetic progression is called:

increasing if each next element is greater than the previous one;

decreasing, if, on the contrary, each subsequent element is less than the previous one.

In addition, there are so-called "stationary" sequences - they consist of the same repeating number. For example, (3; 3; 3; ...).

Only one question remains: how to distinguish an increasing progression from a decreasing one? Fortunately, everything here depends only on the sign of the number $d$, i.e. progression differences:

If $d \gt 0$, then the progression is increasing;
If $d \lt 0$, then the progression is obviously decreasing;
Finally, there is the case $d=0$ — in this case the entire progression is reduced to a stationary sequence of identical numbers: (1; 1; 1; 1; ...), etc.

Let's try to calculate the difference $d$ for the three decreasing progressions above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract from the number on the right, the number on the left. It will look like this:

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

As you can see, in all three cases the difference really turned out to be negative. And now that we have more or less figured out the definitions, it's time to figure out how progressions are described and what properties they have.

Members of the progression and the recurrent formula

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

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

Individual elements of this set are called members of the progression. They are indicated in this way with the help of a number: the first member, the second member, and so on.

In addition, as we already know, neighboring members of the progression are related by the formula:

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

In short, to find the $n$th term of the progression, you need to know the $n-1$th term and the difference $d$. Such a formula is called recurrent, because with its help you can find any number, only knowing the previous one (and in fact, all the previous ones). This is very inconvenient, so there is a more tricky formula that reduces any calculation to the first term and the difference:

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

You have probably come across this formula before. They like to give it in all sorts of reference books and reshebniks. And in any sensible textbook on mathematics, it is one of the first.

However, I suggest you practice a little.

Task number 1. Write down the first three terms of the arithmetic progression $\left(((a)_(n)) \right)$ if $((a)_(1))=8,d=-5$.

Solution. So, we know the first term $((a)_(1))=8$ and the progression difference $d=-5$. Let's use the formula just given and substitute $n=1$, $n=2$ and $n=3$:

\[\begin(align) & ((a)_(n))=((a)_(1))+\left(n-1 \right)d; \\ & ((a)_(1))=((a)_(1))+\left(1-1 \right)d=((a)_(1))=8; \\ & ((a)_(2))=((a)_(1))+\left(2-1 \right)d=((a)_(1))+d=8-5= 3; \\ & ((a)_(3))=((a)_(1))+\left(3-1 \right)d=((a)_(1))+2d=8-10= -2. \\ \end(align)\]

Answer: (8; 3; -2)

That's all! Note that our progression is decreasing.

Of course, $n=1$ could not have been substituted - we already know the first term. However, by substituting the unit, we made sure that even for the first term our formula works. In other cases, everything came down to banal arithmetic.

Task number 2. Write out the first three terms of an arithmetic progression if its seventh term is −40 and its seventeenth term is −50.

Solution. We write the condition of the problem in the usual terms:

\[((a)_(7))=-40;\quad ((a)_(17))=-50.\]

\[\left\( \begin(align) & ((a)_(7))=((a)_(1))+6d \\ & ((a)_(17))=((a) _(1))+16d \\ \end(align) \right.\]

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

I put the sign of the system because these requirements must be met simultaneously. And now we note that if we subtract the first equation from the second equation (we have the right to do this, because we have a system), we get this:

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

Just like that, we found the progression difference! It remains to substitute the found number in any of the equations of the system. For example, in the first:

\[\begin(matrix) ((a)_(1))+6d=-40;\quad d=-1 \\ \Downarrow \\ ((a)_(1))-6=-40; \\ ((a)_(1))=-40+6=-34. \\ \end(matrix)\]

Now, knowing the first term and the difference, it remains to find the second and third terms:

\[\begin(align) & ((a)_(2))=((a)_(1))+d=-34-1=-35; \\ & ((a)_(3))=((a)_(1))+2d=-34-2=-36. \\ \end(align)\]

Ready! Problem solved.

Answer: (-34; -35; -36)

Notice a curious property of the progression that we discovered: if we take the $n$th and $m$th terms and subtract them from each other, we get the difference of the progression multiplied by the number $n-m$:

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

Simple but very useful property, which you definitely need to know - with its help you can significantly speed up the solution of many problems in progressions. 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 line

I 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:

\[\begin(align) & x+1=\frac(-6((x)^(2))+14+4((x)^(2)))(2); \\ & x+1=\frac(14-2((x)^(2)))(2); \\ & x+1=7-((x)^(2)); \\ & ((x)^(2))+x-6=0. \\ \end(align)\]

It turned out 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 tasks, we stumbled upon another interesting fact, which 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 line

Let'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 fundamentally more high level complexity than those discussed 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 at the highest term is 11 - this is 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 smallest value(By the way, we did not calculate $((y)_(\min ))$ - we are not required to do this). 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. You can safely go to next lesson, where we will study the progression sum formula, as well as important and very useful consequences from it.

If every natural number n match a real number a n , then they say that given number sequence :

a 1 , a 2 , a 3 , . . . , a n , . . . .

So, a numerical sequence is a function of a natural argument.

Number a 1 called the first member of the sequence , number a 2 — the second member of the sequence , number a 3 — third and so on. Number a n called nth member of the sequence , and the natural number n — his number .

From two neighboring members a n And a n +1 member sequences a n +1 called subsequent (towards a n ), A a n — previous (towards a n +1 ).

To specify a sequence, you must specify a method that allows you to find a sequence member with any number.

Often the sequence is given with nth term formulas , that is, a formula that allows you to determine a sequence member by its number.

For example,

the sequence of positive odd numbers can be given by the formula

a n= 2n- 1,

and the sequence of alternating 1 And -1 - formula

b n = (-1)n +1 . ◄

The sequence can be determined recurrent formula, that is, a formula that expresses any member of the sequence, starting with some, through the previous (one or more) members.

For example,

If a 1 = 1 , A a n +1 = a n + 5

a 1 = 1,

a 2 = a 1 + 5 = 1 + 5 = 6,

a 3 = a 2 + 5 = 6 + 5 = 11,

a 4 = a 3 + 5 = 11 + 5 = 16,

a 5 = a 4 + 5 = 16 + 5 = 21.

If a 1= 1, a 2 = 1, a n +2 = a n + a n +1 , then the first seven members of the numerical sequence are set as follows:

a 1 = 1,

a 2 = 1,

a 3 = a 1 + a 2 = 1 + 1 = 2,

a 4 = a 2 + a 3 = 1 + 2 = 3,

a 5 = a 3 + a 4 = 2 + 3 = 5,

a 6 = a 4 + a 5 = 3 + 5 = 8,

a 7 = a 5 + a 6 = 5 + 8 = 13. ◄

Sequences can be final And endless .

The sequence is called ultimate if it has a finite number of members. The sequence is called endless if it has infinitely many members.

For example,

sequence of two-digit natural numbers:

10, 11, 12, 13, . . . , 98, 99

final.

Prime number sequence:

2, 3, 5, 7, 11, 13, . . .

endless. ◄

The sequence is called increasing , if each of its members, starting from the second, is greater than the previous one.

The sequence is called waning , if each of its members, starting from the second, is less than the previous one.

For example,

2, 4, 6, 8, . . . , 2n, . . . is an ascending sequence;

1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . is a descending sequence. ◄

A sequence whose elements do not decrease with increasing number, or, conversely, do not increase, is called monotonous sequence .

Monotonic sequences, in particular, are increasing sequences and decreasing sequences.

Arithmetic progression

Arithmetic progression a sequence is called, each member of which, starting from the second, is equal to the previous one, to which the same number is added.

a 1 , a 2 , a 3 , . . . , a n, . . .

is an arithmetic progression if for any natural number n condition is met:

a n +1 = a n + d,

Where d - some number.

Thus, the difference between the next and the previous members of a given arithmetic progression is always constant:

a 2 - a 1 = a 3 - a 2 = . . . = a n +1 - a n = d.

Number d called the difference of an arithmetic progression.

To set an arithmetic progression, it is enough to specify its first term and difference.

For example,

If a 1 = 3, d = 4 , then the first five terms of the sequence are found as follows:

a 1 =3,

a 2 = a 1 + d = 3 + 4 = 7,

a 3 = a 2 + d= 7 + 4 = 11,

a 4 = a 3 + d= 11 + 4 = 15,

a 5 = a 4 + d= 15 + 4 = 19. ◄

For an arithmetic progression with the first term a 1 and difference d her n

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

For example,

find the thirtieth term of an arithmetic progression

1, 4, 7, 10, . . .

a 1 =1, d = 3,

a 30 = a 1 + (30 - 1)d= 1 + 29· 3 = 88. ◄

a n-1 = a 1 + (n- 2)d,

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

a n +1 = a 1 + nd,

then obviously

a n=	a n-1 + a n+1
	2

each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the previous and subsequent members.

numbers a, b and c are consecutive members of some arithmetic progression if and only if one of them is equal to the arithmetic mean of the other two.

For example,

a n = 2n- 7 , is an arithmetic progression.

Let's use the statement above. We have:

a n = 2n- 7,

a n-1 = 2(n- 1) - 7 = 2n- 9,

a n+1 = 2(n+ 1) - 7 = 2n- 5.

Hence,

a n+1 + a n-1	=	2n- 5 + 2n- 9	= 2n- 7 = a n,
2		2

◄

Note that n -th member of an arithmetic progression can be found not only through a 1 , but also any previous a k

a n = a k + (n- k)d.

For example,

For a 5 can be written

a 5 = a 1 + 4d,

a 5 = a 2 + 3d,

a 5 = a 3 + 2d,

a 5 = a 4 + d. ◄

a n = a n-k + kd,

a n = a n+k - kd,

then obviously

a n=	a n-k +a n+k
	2

any member of an arithmetic progression, starting from the second, is equal to half the sum of the members of this arithmetic progression equally spaced from it.

In addition, for any arithmetic progression, the equality is true:

a m + a n = a k + a l,

m + n = k + l.

For example,

in arithmetic progression

1) a 10 = 28 = (25 + 31)/2 = (a 9 + a 11 )/2;

2) 28 = a 10 = a 3 + 7d= 7 + 7 3 = 7 + 21 = 28;

3) a 10= 28 = (19 + 37)/2 = (a 7 + a 13)/2;

4) a 2 + a 12 = a 5 + a 9, because

a 2 + a 12= 4 + 34 = 38,

a 5 + a 9 = 13 + 25 = 38. ◄

S n= a 1 + a 2 + a 3 + . . .+ a n,

first n members of an arithmetic progression is equal to the product of half the sum of the extreme terms by the number of terms:

From this, in particular, it follows that if it is necessary to sum the terms

a k, a k +1 , . . . , a n,

then the previous formula retains its structure:

For example,

in arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, . . .

S 10 = 1 + 4 + . . . + 28 = (1 + 28) · 10/2 = 145;

10 + 13 + 16 + 19 + 22 + 25 + 28 = S 10 - S 3 = (10 + 28 ) · (10 - 4 + 1)/2 = 133. ◄

If an arithmetic progression is given, then the quantities a 1 , a n, d, n AndS n linked by two formulas:

Therefore, if the values of three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.

An arithmetic progression is a monotonic sequence. Wherein:

If d > 0 , then it is increasing;
If d < 0 , then it is decreasing;
If d = 0 , then the sequence will be stationary.

Geometric progression

geometric progression a sequence is called, each term of which, starting from the second, is equal to the previous one, multiplied by the same number.

b 1 , b 2 , b 3 , . . . , b n, . . .

is a geometric progression if for any natural number n condition is met:

b n +1 = b n · q,

Where q ≠ 0 - some number.

Thus, the ratio of the next term of this geometric progression to the previous one is a constant number:

b 2 / b 1 = b 3 / b 2 = . . . = b n +1 / b n = q.

Number q called denominator of a geometric progression.

To set a geometric progression, it is enough to specify its first term and denominator.

For example,

If b 1 = 1, q = -3 , then the first five terms of the sequence are found as follows:

b 1 = 1,

b 2 = b 1 · q = 1 · (-3) = -3,

b 3 = b 2 · q= -3 · (-3) = 9,

b 4 = b 3 · q= 9 · (-3) = -27,

b 5 = b 4 · q= -27 · (-3) = 81. ◄

b 1 and denominator q her n -th term can be found by the formula:

b n = b 1 · q n -1 .

For example,

find the seventh term of a geometric progression 1, 2, 4, . . .

b 1 = 1, q = 2,

b 7 = b 1 · q 6 = 1 2 6 = 64. ◄

bn-1 = b 1 · q n -2 ,

b n = b 1 · q n -1 ,

b n +1 = b 1 · q n,

then obviously

b n 2 = b n -1 · b n +1 ,

each member of the geometric progression, starting from the second, is equal to the geometric mean (proportional) of the previous and subsequent members.

Since the converse is also true, the following assertion holds:

numbers a, b and c are consecutive members of some geometric progression if and only if the square of one of them is equal to the product of the other two, that is, one of the numbers is the geometric mean of the other two.

For example,

let us prove that the sequence given by the formula b n= -3 2 n , is a geometric progression. Let's use the statement above. We have:

b n= -3 2 n,

b n -1 = -3 2 n -1 ,

b n +1 = -3 2 n +1 .

Hence,

b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) (-3 2 n +1 ) = b n -1 · b n +1 ,

which proves the required assertion. ◄

Note that n th term of a geometric progression can be found not only through b 1 , but also any previous term b k , for which it suffices to use the formula

b n = b k · q n - k.

For example,

For b 5 can be written

b 5 = b 1 · q 4 ,

b 5 = b 2 · q 3,

b 5 = b 3 · q2,

b 5 = b 4 · q. ◄

b n = b k · q n - k,

b n = b n - k · q k,

then obviously

b n 2 = b n - k· b n + k

the square of any member of a geometric progression, starting from the second, is equal to the product of the members of this progression equidistant from it.

In addition, for any geometric progression, the equality is true:

b m· b n= b k· b l,

m+ n= k+ l.

For example,

exponentially

1) b 6 2 = 32 2 = 1024 = 16 · 64 = b 5 · b 7 ;

2) 1024 = b 11 = b 6 · q 5 = 32 · 2 5 = 1024;

3) b 6 2 = 32 2 = 1024 = 8 · 128 = b 4 · b 8 ;

4) b 2 · b 7 = b 4 · b 5 , because

b 2 · b 7 = 2 · 64 = 128,

b 4 · b 5 = 8 · 16 = 128. ◄

S n= b 1 + b 2 + b 3 + . . . + b n

first n members of a geometric progression with a denominator q ≠ 0 calculated by the formula:

And when q = 1 - according to the formula

S n= n.b. 1

Note that if we need to sum the terms

b k, b k +1 , . . . , b n,

then the formula is used:

S n- S k -1 = b k + b k +1 + . . . + b n = b k ·	1 - q n - k +1	.
	1 - q

For example,

exponentially 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, . . .

S 10 = 1 + 2 + . . . + 512 = 1 · (1 - 2 10) / (1 - 2) = 1023;

64 + 128 + 256 + 512 = S 10 - S 6 = 64 · (1 - 2 10-7+1) / (1 - 2) = 960. ◄

If given geometric progression, then the quantities b 1 , b n, q, n And S n linked by two formulas:

Therefore, if the values of any three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.

For a geometric progression with the first term b 1 and denominator q the following take place monotonicity properties :

the progression is increasing if one of the following conditions is met:

b 1 > 0 And q> 1;

b 1 < 0 And 0 < q< 1;

A progression is decreasing if one of the following conditions is met:

b 1 > 0 And 0 < q< 1;

b 1 < 0 And q> 1.

If q< 0 , then the geometric progression is sign-alternating: its odd-numbered terms have the same sign as its first term, and even-numbered terms have the opposite sign. It is clear that an alternating geometric progression is not monotonic.

Product of the first n terms of a geometric progression can be calculated by the formula:

P n= b 1 · b 2 · b 3 · . . . · b n = (b 1 · b n) n / 2 .

For example,

1 · 2 · 4 · 8 · 16 · 32 · 64 · 128 = (1 · 128) 8/2 = 128 4 = 268 435 456;

3 · 6 · 12 · 24 · 48 = (3 · 48) 5/2 = (144 1/2) 5 = 12 5 = 248 832.◄

Infinitely decreasing geometric progression

Infinitely decreasing geometric progression is called an infinite geometric progression whose denominator modulus is less than 1 , that is

|q| < 1 .

Note that an infinitely decreasing geometric progression may not be a decreasing sequence. This fits the case

1 < q< 0 .

With such a denominator, the sequence is sign-alternating. For example,

1, - 1 / 2 , 1 / 4 , - 1 / 8 , . . . .

The sum of an infinitely decreasing geometric progression name the number to which the sum of the first n terms of the progression with an unlimited increase in the number n . This number is always finite and is expressed by the formula