Arithmetic progression find the sum of the first five. Formula for the nth term of an arithmetic progression. Formula for finding the nth term of an arithmetic progression

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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. Letter indices nth term progressions, progression differences - all this is somehow confusing, yes... Let's figure out the meaning 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, in new topic new terms and designations appear. 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 problems 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 term.

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 for more high 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 tutorial. If the series is longer, the calculations become more complicated. For example, if in problem 9 in the question we 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.

What the main point formulas?

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

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

Memorizing (or cribing) this formula is not enough. You need to understand its essence and apply the formula in various problems. And also not to forget at the right moment, yes...) How not forget- I don't know. And here how to remember If necessary, I will definitely advise you. For those who complete the lesson to the end.)

So, let's look at the formula for the nth term of an arithmetic progression.

What is a formula in general? By the way, take a look if you haven’t read it. Everything is simple there. It remains to figure out what it is nth term.

Progression in general can be written as a series of numbers:

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

a 1- denotes the first term of an arithmetic progression, a 3- third member, a 4- the 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 - s a 120.

How can we define it in general terms? any term of an arithmetic progression, with any number? Very simple! Like this:

a n

That's what it is nth term of an arithmetic progression. The letter n hides all the member numbers 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 progression. Using the notation a n, we can quickly find any member any arithmetic progression. And solve a bunch of other progression problems. You'll see for yourself further.

In the formula for the nth term of an arithmetic progression:

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

a 1- the first term of an arithmetic progression;

n- member number.

The formula connects the key parameters of any progression: a n ; a 1 ; d And n. All progression problems revolve around these parameters.

The nth term formula can also be used to write a specific progression. For example, the problem may say that the progression is specified by the condition:

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

Such a problem can be a dead end... There is neither a series nor a difference... But, comparing the condition with the formula, it is easy to understand that in this progression a 1 =5, and d=2.

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

a n = 3 + 2n.

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

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

Most often the designation a n+1 found in recurrence formulas. Don't be afraid of this scary word!) This is just a way of expressing a member of an arithmetic progression through the previous one. Let's say we are given an arithmetic progression in this form, using a 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. How can we immediately count, say, the twentieth term? a 20? But there’s no way!) Until we find out the 19th term, we can’t count the 20th. This is the fundamental difference between the recurrent formula and the formula of the nth term. Recurrent works only through previous term, and the formula of the nth term is through first and allows straightaway find any member by its number. Without calculating the entire series of numbers in order.

In an arithmetic progression, it is easy to turn a recurrent formula 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 its usual form, and work with it. Such tasks are often encountered in the State Academy of Sciences.

Application of the formula for the nth term 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:

An arithmetic progression (a n) is given. 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 an arithmetic progression. Add and add... An hour or two.)

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

The conditions provide all the data for using the formula: a 1 =3, d=1/6. It remains to figure out what is equal n. No problem! We need to find a 121. So 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 ours n. This is the meaning n= 121 we will substitute further into the formula, in brackets. We substitute all the numbers into the formula and calculate:

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

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

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

Let's solve the problem in a more cunning way. Let us come across 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 tell you the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Write down with your hands, 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... Is that it? If you think that’s it, then you won’t solve the problem, yes...

We still have a number n! In condition a 17 =-2 hidden two parameters. This is both the value of the seventeenth term (-2) and its number (17). Those. n=17. This “trifle” often slips past the head, and without it, (without the “trifle”, not the head!) the problem cannot be solved. Although... and without a head too.)

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

a 17 = a 1 + (17-1)·(-0.5)

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

-2 = a 1 + (17-1)·(-0.5)

That's basically all. It remains to express the first term of the arithmetic progression from the formula and calculate it. The answer will be: a 1 = 6.

This technique - writing down a formula and simply substituting known data - is a great help in simple tasks. Well, of course, you must be able to express a variable from a formula, but what to do!? Without this skill, mathematics may not be studied at all...

Another popular puzzle:

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 are writing the formula!)

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

Let's consider what we know: a 1 =2; a 15 =12; and (I’ll especially highlight!) n=15. Feel free to substitute this into the formula:

12=2 + (15-1)d

We do the arithmetic.)

12=2 + 14d

d=10/14 = 5/7

This is the correct answer.

So, the tasks for a n, a 1 And d decided. All that remains is to learn how to find the number:

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

We substitute the quantities known to us 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- this is some member of the progression with a number n...And we know this member of the progression! It's 99. We don't know its number. n, So this number is what you need to find. We substitute the term of the progression 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 whether the number 117 is a member of the arithmetic progression (a n):

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

Let's write the formula again. What, there are no parameters? Hm... Why are we given eyes?) Do we see the first term of the progression? We see. This is -3.6. You can safely write: a 1 = -3.6. Difference d Can you tell from the series? It’s easy if you know what the difference of an arithmetic progression is:

d = -2.4 - (-3.6) = 1.2

So, we did the simplest thing. It remains to deal with the unknown number n and the 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... What to do!? Well, what to do, what to do... 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 can we draw? Yes! Number 117 is not member of our progression. It is somewhere between the one hundred and first and one hundred and second terms. If the number turned out natural, i.e. is a positive integer, then the number would be a member of the progression with the number found. And in our case, the answer to the problem will be: No.

A task based on a real version of the GIA:

An arithmetic progression is given by the condition:

a n = -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 for the nth term 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 the arithmetic progression in it hidden. It’s okay, we’ll find it now.)

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

a 1 = -4 + 6.8 1 = 2.8

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

We look for the tenth term in the same way:

a 10 = -4 + 6.8 10 = 64

That's it.

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

Suppose, in a difficult combat situation of the State Examination or Unified State Examination, you have forgotten the useful formula for the nth term of an arithmetic progression. I remember something, but somehow uncertainly... Or n there, or n+1, or n-1... How to be!?

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

Draw a number line and mark the first one on it. second, third, etc. members. And we note the difference d between members. Like this:

We look at the picture and think: what does the second term equal? 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? It’s not for nothing that I highlight some words in bold. 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, to the number n, number of spaces will n-1. Therefore, the formula will be (without variations!):

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 insert a picture into the equation...

Tasks for independent solution.

To warm up:

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

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

And this is no longer a warm-up.)

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

What, you don’t want to draw a picture?) Of course! Better according to the formula, yes...

3. The arithmetic progression is given by the condition:a 1 = -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 specified in a recurrent manner. But counting to the one hundred and twenty-fifth term... Not everyone is capable of 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 conditions of task 4, find the sum of the smallest positive and largest negative terms of the progression.

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

Not the easiest task, yes...) The “fingertip” method won’t work here. You will 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, there is one subtle point in the last task. Care will be required when reading the problem. And logic.

The solution to all these problems is discussed in detail in Section 555. And the element of fantasy for the fourth, and the subtle point for the sixth, and general approaches for solving any problems involving the formula of the nth term - everything is described. 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. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

So, let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many of them as you like (in our case, there are them). No matter how many numbers we write, we can always say which one is first, which one is second, and so on until the last, that is, we can number them. This is an example of a number sequence:

Number sequence
For example, for our sequence:

The assigned number is specific to only one number in the sequence. In other words, there are no three second numbers in the sequence. The second number (like the th number) is always the same.
The number with number is called the th term of the sequence.

We usually call the entire sequence by some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member: .

In our case:

Let's say we have a number sequence in which the difference between adjacent numbers is the same and equal.
For example:

etc.
This number sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius back in the 6th century and was understood in more in a broad sense, like an infinite number sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which was studied by the ancient Greeks.

This is a number sequence, each member of which is equal to the previous one added to the same number. This number is called the difference of an arithmetic progression and is designated.

Try to determine which number sequences are an arithmetic progression and which are not:

a)
b)
c)
d)

Got it? Let's compare our answers:
Is arithmetic progression - b, c.
Is not arithmetic progression - a, d.

Let's return to the given progression () and try to find the value of its th term. Exists two way to find it.

1. Method

We can add the progression number to the previous value until we reach the th term of the progression. It’s good that we don’t have much to summarize - only three values:

So, the th term of the described arithmetic progression is equal to.

2. Method

What if we needed to find the value of the th term of the progression? The summation would take us more than one hour, and it is not a fact that we would not make mistakes when adding numbers.
Of course, mathematicians have come up with a way in which it is not necessary to add the difference of an arithmetic progression to the previous value. Take a closer look at the drawn picture... Surely you have already noticed a certain pattern, namely:

For example, let’s see what the value of the th term of this arithmetic progression consists of:

In other words:

Try to find the value of a member of a given arithmetic progression yourself in this way.

Did you calculate? Compare your notes with the answer:

Please note that you got exactly the same number as in the previous method, when we sequentially added the terms of the arithmetic progression to the previous value.
Let’s try to “depersonalize” this formula - let’s put it in general form and get:

Arithmetic progression equation.

Arithmetic progressions can be increasing or decreasing.

Increasing- progressions in which each subsequent value of the terms is greater than the previous one.
For example:

Descending- progressions in which each subsequent value of the terms is less than the previous one.
For example:

The derived formula is used in the calculation of terms in both increasing and decreasing terms of an arithmetic progression.
Let's check this in practice.
We are given an arithmetic progression consisting of the following numbers: Let's check what the th number of this arithmetic progression will be if we use our formula to calculate it:

Since then:

Thus, we are convinced that the formula operates in both decreasing and increasing arithmetic progression.
Try to find the th and th terms of this arithmetic progression yourself.

Let's compare the results:

Arithmetic progression property

Let's complicate the problem - we will derive the property of arithmetic progression.
Let's say we are given the following condition:
- arithmetic progression, find the value.
Easy, you say and start counting according to the formula you already know:

Let, ah, then:

Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but what if we are given numbers in the condition? Agree, there is a possibility of making a mistake in the calculations.
Now think about whether it is possible to solve this problem in one step using any formula? Of course yes, and that’s what we’ll try to bring out now.

Let us denote the required term of the arithmetic progression as, the formula for finding it is known to us - this is the same formula we derived at the beginning:
, Then:

the previous term of the progression is:
the next term of the progression is:

Let's sum up the previous and subsequent terms of the progression:

It turns out that the sum of the previous and subsequent terms of the progression is the double value of the progression term located between them. In other words, to find the value of a progression term with known previous and successive values, you need to add them and divide by.

That's right, we got the same number. Let's secure the material. Calculate the value for the progression yourself, it’s not at all difficult.

Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, was easily deduced by one of the greatest mathematicians of all time, the “king of mathematicians” - Karl Gauss...

When Carl Gauss was 9 years old, a teacher, busy checking the work of students in other classes, assigned the following task in class: “Calculate the sum of all natural numbers from to (according to other sources to) inclusive.” Imagine the teacher’s surprise when one of his students (this was Karl Gauss) a minute later gave the correct answer to the task, while most of the daredevil’s classmates, after long calculations, received the wrong result...

Young Carl Gauss noticed a certain pattern that you can easily notice too.
Let's say we have an arithmetic progression consisting of -th terms: We need to find the sum of these terms of the arithmetic progression. Of course, we can manually sum all the values, but what if the task requires finding the sum of its terms, as Gauss was looking for?

Let us depict the progression given to us. Take a close look at the highlighted numbers and try to perform various mathematical operations with them.

Have you tried it? What did you notice? Right! Their sums are equal

Now tell me, how many such pairs are there in total in the progression given to us? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two terms of an arithmetic progression is equal, and similar pairs are equal, we obtain that the total sum is equal to:
.
Thus, the formula for the sum of the first terms of any arithmetic progression will be:

In some problems we do not know the th term, but we know the difference of the progression. Try to substitute the formula of the th term into the sum formula.
What did you get?

Well done! Now let's return to the problem that was asked to Carl Gauss: calculate for yourself what the sum of numbers starting from the th is equal to and the sum of the numbers starting from the th.

How much did you get?
Gauss found that the sum of the terms is equal, and the sum of the terms. Is that what you decided?

In fact, the formula for the sum of the terms of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people made full use of the properties of the arithmetic progression.
For example, imagine Ancient Egypt and the largest construction project of that time - the construction of a pyramid... The picture shows one side of it.

Where is the progression here, you say? Look carefully and find a pattern in the number of sand blocks in each row of the pyramid wall.

Why not an arithmetic progression? Calculate how many blocks are needed to build one wall if block bricks are placed at the base. I hope you won’t count while moving your finger across the monitor, you remember the last formula and everything we said about arithmetic progression?

In this case, the progression looks like this: .
Arithmetic progression difference.
The number of terms of an arithmetic progression.
Let's substitute our data into the last formulas (calculate the number of blocks in 2 ways).

Method 1.

Method 2.

And now you can calculate on the monitor: compare the obtained values with the number of blocks that are in our pyramid. Got it? Well done, you have mastered the sum of the nth terms of an arithmetic progression.
Of course, you can’t build a pyramid from blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:

Training

Tasks:

Masha is getting in shape for summer. Every day she increases the number of squats by. How many times will Masha do squats in a week if she did squats at the first training session?
What is the sum of all odd numbers contained in.
When storing logs, loggers stack them in such a way that each top layer contains one log less than the previous one. How many logs are in one masonry, if the foundation of the masonry is logs?

Answers:

Let us define the parameters of the arithmetic progression. In this case
(weeks = days).
Answer: In two weeks, Masha should do squats once a day.
First odd number, last number.
Arithmetic progression difference.
The number of odd numbers in is half, however, let’s check this fact using the formula for finding the th term of an arithmetic progression:
Numbers do contain odd numbers.
Let's substitute the available data into the formula:
Answer: The sum of all odd numbers contained in is equal.
Let's remember the problem about pyramids. For our case, a , since each top layer is reduced by one log, then in total there are a bunch of layers, that is.
Let's substitute the data into the formula:
Answer: There are logs in the masonry.

Let's sum it up

- a number sequence in which the difference between adjacent numbers is the same and equal. It can be increasing or decreasing.
Finding formula The th term of an arithmetic progression is written by the formula - , where is the number of numbers in the progression.
Property of members of an arithmetic progression- - where is the number of numbers in progression.
The sum of the terms of an arithmetic progression can be found in two ways:

, where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

Let's sit down and start writing some numbers. For example:

You can write any numbers, and there can be as many of them as you like. But we can always say which one is first, which one is second, and so on, that is, we can number them. This is an example of a number sequence.

Number sequence is a set of numbers, each of which can be assigned a unique number.

In other words, each number can be associated with a certain natural number, and a unique one. And we will not assign this number to any other number from this set.

The number with number is called the th member of the sequence.

We usually call the entire sequence by some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member: .

It is very convenient if the th term of the sequence can be specified by some formula. For example, the formula

sets the sequence:

And the formula is the following sequence:

For example, an arithmetic progression is a sequence (the first term here is equal, and the difference is). Or (, difference).

nth term formula

We call a formula recurrent in which, in order to find out the th term, you need to know the previous or several previous ones:

To find, for example, the th term of the progression using this formula, we will have to calculate the previous nine. For example, let it. Then:

Well, is it clear now what the formula is?

In each line we add to, multiplied by some number. Which one? Very simple: this is the number of the current member minus:

Much more convenient now, right? We check:

Decide for yourself:

In an arithmetic progression, find the formula for the nth term and find the hundredth term.

Solution:

The first term is equal. What is the difference? Here's what:

(This is why it is called difference because it is equal to the difference of successive terms of the progression).

So, the formula:

Then the hundredth term is equal to:

What is the sum of all natural numbers from to?

According to legend, the great mathematician Carl Gauss, as a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last numbers is equal, the sum of the second and penultimate is the same, the sum of the third and 3rd from the end is the same, and so on. How many such pairs are there in total? That's right, exactly half the number of all numbers, that is. So,

The general formula for the sum of the first terms of any arithmetic progression will be:

Example:
Find the sum of all two-digit multiples.

Solution:

The first such number is this. Each subsequent number is obtained by adding to the previous number. Thus, the numbers we are interested in form an arithmetic progression with the first term and the difference.

Formula of the th term for this progression:

How many terms are there in the progression if they all have to be two-digit?

Very easy: .

The last term of the progression will be equal. Then the sum:

Answer: .

Now decide for yourself:

Every day the athlete runs more meters than the previous day. How many total kilometers will he run in a week if he ran km m on the first day?
A cyclist travels more kilometers every day than the previous day. On the first day he traveled km. How many days does he need to travel to cover a kilometer? How many kilometers will he travel during the last day of his journey?
The price of a refrigerator in a store decreases by the same amount every year. Determine how much the price of a refrigerator decreased each year if, put up for sale for rubles, six years later it was sold for rubles.

Answers:

The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
.
Answer:
Here it is given: , must be found.
Obviously, you need to use the same sum formula as in the previous problem:
.
Substitute the values:
The root obviously doesn't fit, so the answer is.
Let's calculate the path traveled over the last day using the formula of the th term:
(km).
Answer:
Given: . Find: .
It couldn't be simpler:
(rub).
Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

This is a number sequence in which the difference between adjacent numbers is the same and equal.

Arithmetic progression can be increasing () and decreasing ().

For example:

Formula for finding the nth term of an arithmetic progression

is written by the formula, where is the number of numbers in progression.

Property of members of an arithmetic progression

It allows you to easily find a term of a progression if its neighboring terms are known - where is the number of numbers in the progression.

Sum of terms of an arithmetic progression

There are two ways to find the amount:

Where is the number of values.

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Mathematics has its own beauty, just like painting and poetry.

Russian scientist, mechanic N.E. Zhukovsky

Very common tasks in entrance examinations in mathematics are problems related to the concept of arithmetic progression. To successfully solve such problems, you must have a good knowledge of the properties of arithmetic progression and have certain skills in their application.

Let us first recall the basic properties of an arithmetic progression and present the most important formulas, related to this concept.

Definition. Number sequence, in which each subsequent term differs from the previous one by the same number, called an arithmetic progression. In this case the numbercalled the progression difference.

For an arithmetic progression, the following formulas are valid:

, (1)

Where . Formula (1) is called the formula of the general term of an arithmetic progression, and formula (2) represents the main property of an arithmetic progression: each term of the progression coincides with the arithmetic mean of its neighboring terms and .

Note that it is precisely because of this property that the progression under consideration is called “arithmetic”.

The above formulas (1) and (2) are generalized as follows:

(3)

To calculate the amount first terms of an arithmetic progressionthe formula is usually used

(5) where and .

If we take into account the formula (1), then from formula (5) it follows

If we denote , then

Where . Since , formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).

In particular , from formula (5) it follows, What

Little known to most students is the property of arithmetic progression, formulated through the following theorem.

Theorem. If , then

Proof. If , then

The theorem has been proven.

For example , using the theorem, it can be shown that

Let's move on to consider typical examples of solving problems on the topic “Arithmetic progression”.

Example 1. Let it be. Find .

Solution. Applying formula (6), we obtain . Since and , then or .

Example 2. Let it be three times greater, and when divided by the quotient, the result is 2 and the remainder is 8. Determine and .

Solution. From the conditions of the example, the system of equations follows

Since , , and , then from the system of equations (10) we obtain

The solution to this system of equations is and .

Example 3. Find if and .

Solution. According to formula (5) we have or . However, using property (9), we obtain .

Since and , then from the equality the equation follows or .

Example 4. Find if .

Solution.According to formula (5) we have

However, using the theorem, we can write

From here and from formula (11) we obtain .

Example 5. Given: . Find .

Solution. Since, then. However, therefore.

Example 6. Let , and . Find .

Solution. Using formula (9), we obtain . Therefore, if , then or .

Since and then here we have a system of equations

Solving which, we get and .

Natural root of the equation is .

Example 7. Find if and .

Solution. Since according to formula (3) we have that , then the system of equations follows from the problem conditions

If we substitute the expressioninto the second equation of the system, then we get or .

The roots of a quadratic equation are And .

Let's consider two cases.

1. Let , then . Since and , then .

In this case, according to formula (6), we have

2. If , then , and

Answer: and.

Example 8. It is known that and. Find .

Solution. Taking into account formula (5) and the condition of the example, we write and .

This implies the system of equations

If we multiply the first equation of the system by 2 and then add it to the second equation, we get

According to formula (9) we have. In this regard, it follows from (12) or .

Since and , then .

Answer: .

Example 9. Find if and .

Solution. Since , and by condition , then or .

From formula (5) it is known, What . Since, then.

Hence , here we have a system of linear equations

From here we get and . Taking into account formula (8), we write .

Example 10. Solve the equation.

Solution. From the given equation it follows that . Let us assume that , , and . In this case .

According to formula (1), we can write or .

Since , then equation (13) has the only suitable root .

Example 11. Find the maximum value provided that and .

Solution. Since , then the arithmetic progression under consideration is decreasing. In this regard, the expression takes on its maximum value when it is the number of the minimum positive term of the progression.

Let us use formula (1) and the fact, that and . Then we get that or .

Since , then or . However, in this inequalitylargest natural number, That's why .

If the values of , and are substituted into formula (6), we get .

Answer: .

Example 12. Determine the sum of all two-digit natural numbers that, when divided by the number 6, leave a remainder of 5.

Solution. Let us denote by the set of all two-digit natural numbers, i.e. . Next, we will construct a subset consisting of those elements (numbers) of the set that, when divided by the number 6, give a remainder of 5.

Easy to install, What . Obviously , that the elements of the setform an arithmetic progression, in which and .

To establish the cardinality (number of elements) of the set, we assume that . Since and , it follows from formula (1) or . Taking into account formula (5), we obtain .

The above examples of problem solving can by no means claim to be exhaustive. This article is written based on the analysis modern methods solving typical problems on a given topic. For a more in-depth study of methods for solving problems related to arithmetic progression, it is advisable to refer to the list of recommended literature.

1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace and Education, 2013. – 608 p.

2. Suprun V.P. Mathematics for high school students: additional sections school curriculum. – M.: Lenand / URSS, 2014. – 216 p.

3. Medynsky M.M. A complete course of elementary mathematics in problems and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. – 208 p.

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Lesson type: learning new material.

Lesson objectives:

expanding and deepening students’ understanding of problems solved using arithmetic progression; organizing students' search activities when deriving the formula for the sum of the first n terms of an arithmetic progression;
developing the ability to independently acquire new knowledge and use already acquired knowledge to achieve a given task;
developing the desire and need to generalize the facts obtained, developing independence.

Tasks:

summarize and systematize existing knowledge on the topic “Arithmetic progression”;
derive formulas for calculating the sum of the first n terms of an arithmetic progression;
teach how to apply the obtained formulas when solving various problems;
draw students' attention to the procedure for finding the value of a numerical expression.

Equipment:

cards with tasks for working in groups and pairs;
evaluation paper;
presentation“Arithmetic progression.”

I. Updating of basic knowledge.

1. Independent work in pairs.

1st option:

Define arithmetic progression. Write down a recurrence formula that defines an arithmetic progression. Please provide an example of an arithmetic progression and indicate its difference.

2nd option:

Write down the formula for the nth term of an arithmetic progression. Find the 100th term of the arithmetic progression ( a n}: 2, 5, 8 …
At this time, two students on the back of the board are preparing answers to the same questions.
Students evaluate their partner's work by checking them on the board. (Sheets with answers are handed in.)

2. Game moment.

Exercise 1.

Teacher. I thought of some arithmetic progression. Ask me only two questions so that after the answers you can quickly name the 7th term of this progression. (1, 3, 5, 7, 9, 11, 13, 15…)

Questions from students.

What is the sixth term of the progression and what is the difference?
What is the eighth term of the progression and what is the difference?

If there are no more questions, then the teacher can stimulate them - a “ban” on d (difference), that is, it is not allowed to ask what the difference is equal to. You can ask questions: what is the 6th term of the progression equal to and what is the 8th term of the progression equal to?

Task 2.

There are 20 numbers written on the board: 1, 4, 7 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58.

The teacher stands with his back to the board. Students call out the number, and the teacher instantly calls out the number itself. Explain how I can do this?

The teacher remembers the formula for the nth term a n = 3n – 2 and, substituting the specified values n, finds the corresponding values a n.

II. Setting a learning task.

I propose to solve an ancient problem dating back to the 2nd millennium BC, found in Egyptian papyri.

Task:“Let it be said to you: divide 10 measures of barley among 10 people, the difference between each person and his neighbor is 1/8 of the measure.”

How is this problem related to the topic arithmetic progression? (Each next person receives 1/8 of the measure more, which means the difference is d=1/8, 10 people, which means n=10.)
What do you think the number 10 measures means? (Sum of all terms of the progression.)
What else do you need to know to make it easy and simple to divide the barley according to the conditions of the problem? (First term of progression.)

Lesson Objective– obtaining the dependence of the sum of the terms of the progression on their number, the first term and the difference, and checking whether the problem was solved correctly in ancient times.

Before we deduce the formula, let's look at how the ancient Egyptians solved the problem.

And they solved it as follows:

1) 10 measures: 10 = 1 measure – average share;
2) 1 measure ∙ = 2 measures – doubled average share.
Doubled average share is the sum of the shares of the 5th and 6th person.
3) 2 measures – 1/8 measures = 1 7/8 measures – double the share of the fifth person.
4) 1 7/8: 2 = 5/16 – fraction of a fifth; and so on, you can find the share of each previous and subsequent person.

We get the sequence:

III. Solving the problem.

1. Work in groups

Group I: Find the sum of 20 consecutive natural numbers: S 20 =(20+1)∙10 =210.

In general

II group: Find the sum of natural numbers from 1 to 100 (The Legend of Little Gauss).

S 100 = (1+100)∙50 = 5050

Conclusion:

III group: Find the sum of natural numbers from 1 to 21.

Solution: 1+21=2+20=3+19=4+18…

Conclusion:

IV group: Find the sum of natural numbers from 1 to 101.

Conclusion:

This method of solving the problems considered is called the “Gauss Method”.

2. Each group presents the solution to the problem on the board.

3. Generalization of the proposed solutions for an arbitrary arithmetic progression:

a 1, a 2, a 3,…, a n-2, a n-1, a n.
S n =a 1 + a 2 + a 3 + a 4 +…+ a n-3 + a n-2 + a n-1 + a n.

Let's find this sum using similar reasoning:

4. Have we solved the problem?(Yes.)

IV. Primary understanding and application of the obtained formulas when solving problems.

1. Checking the solution to an ancient problem using the formula.

2. Application of the formula in solving various problems.

3. Exercises to develop the ability to apply formulas when solving problems.

A) No. 613

Given: ( a n) – arithmetic progression;

(a n): 1, 2, 3, …, 1500

Find: S 1500

Solution: , a 1 = 1, and 1500 = 1500,

B) Given: ( a n) – arithmetic progression;
(a n): 1, 2, 3, …
S n = 210

Find: n
Solution:

V. Independent work with mutual verification.

Denis started working as a courier. In the first month his salary was 200 rubles, in each subsequent month it increased by 30 rubles. How much did he earn in total in a year?

Given: ( a n) – arithmetic progression;
a 1 = 200, d=30, n=12
Find: S 12
Solution:

Answer: Denis received 4380 rubles for the year.

VI. Homework instruction.

Section 4.3 – learn the derivation of the formula.
№№ 585, 623 .
Create a problem that can be solved using the formula for the sum of the first n terms of an arithmetic progression.

VII. Summing up the lesson.

1. Score sheet

2. Continue the sentences