Instructions

An arithmetic progression is a sequence of the form a1, a1+d, a1+2d..., a1+(n-1)d. Number d step progression.It is obvious that the general of an arbitrary n-th term of the arithmetic progression has the form: An = A1+(n-1)d. Then knowing one of the members progression, member progression and step progression, you can, that is, the number of the progress member. Obviously, it will be determined by the formula n = (An-A1+d)/d.

Let now the mth term be known progression and another member progression- nth, but n , as in the previous case, but it is known that n and m do not coincide. Step progression can be calculated using the formula: d = (An-Am)/(n-m). Then n = (An-Am+md)/d.

If the sum of several elements of an arithmetic equation is known progression, as well as its first and last, then the number of these elements can also be determined. The sum of the arithmetic progression will be equal to: S = ((A1+An)/2)n. Then n = 2S/(A1+An) - chdenov progression. Using the fact that An = A1+(n-1)d, this formula can be rewritten as: n = 2S/(2A1+(n-1)d). From this we can express n by solving a quadratic equation.

An arithmetic sequence is an ordered set of numbers, each member of which, except the first, differs from the previous one by the same amount. This constant value is called the difference of the progression or its step and can be calculated from the known terms of the arithmetic progression.

Instructions

If the values ​​of the first and second or any other pair of adjacent terms are known from the conditions of the problem, to calculate the difference (d) simply subtract the previous one from the subsequent term. The resulting value can be either a positive or a negative number - it depends on whether the progression is increasing. In general form, write the solution for an arbitrary pair (aᵢ and aᵢ₊₁) of neighboring terms of the progression as follows: d = aᵢ₊₁ - aᵢ.

For a pair of terms of such a progression, one of which is the first (a₁), and the other is any other arbitrarily chosen one, it is also possible to create a formula for finding the difference (d). However, in this case, the serial number (i) of an arbitrary selected member of the sequence must be known. To calculate the difference, add both numbers and divide the resulting result by the ordinal number of an arbitrary term reduced by one. In general, write this formula as follows: d = (a₁+ aᵢ)/(i-1).

If, in addition to an arbitrary member of an arithmetic progression with ordinal number i, another member with ordinal number u is known, change the formula from the previous step accordingly. In this case, the difference (d) of the progression will be the sum of these two terms divided by the difference of their ordinal numbers: d = (aᵢ+aᵥ)/(i-v).

The formula for calculating the difference (d) becomes somewhat more complicated if the problem conditions give the value of its first term (a₁) and the sum (Sᵢ) of a given number (i) of the first terms of the arithmetic sequence. To obtain the desired value, divide the sum by the number of terms that make it up, subtract the value of the first number in the sequence, and double the result. Divide the resulting value by the number of terms that make up the sum reduced by one. In general, write the formula for calculating the discriminant as follows: d = 2*(Sᵢ/i-a₁)/(i-1).

Sum of an arithmetic progression.

The sum of an arithmetic progression is a simple thing. Both in meaning and in formula. But there are all sorts of tasks on this topic. From basic to quite solid.

First, let's understand the meaning and formula of the amount. And then we'll decide. For your own pleasure.) The meaning of the amount is as simple as a moo. To find the sum of an arithmetic progression, you just need to carefully add all its terms. If these terms are few, you can add without any formulas. But if there is a lot, or a lot... addition is annoying.) In this case, the formula comes to the rescue.

The formula for the amount is simple:

Let's figure out what kind of letters are included in the formula. This will clear things up a lot.

S n - the sum of an arithmetic progression. Addition result everyone members, with first By last. It is important. They add up exactly All members in a row, without skipping or skipping. And, precisely, starting from first. In problems like finding the sum of the third and eighth terms, or the sum of the fifth to twentieth terms, direct application of the formula will disappoint.)

a 1 - first member of the progression. Everything is clear here, it's simple first row number.

a n- last member of the progression. The last number of the series. Not a very familiar name, but when applied to the amount, it’s very suitable. Then you will see for yourself.

n - number of the last member. It is important to understand that in the formula this number coincides with the number of added terms.

Let's define the concept last member a n. Tricky question: which member will be the last one if given endless arithmetic progression?)

To answer confidently, you need to understand the elementary meaning of arithmetic progression and... read the task carefully!)

In the task of finding the sum of an arithmetic progression, the last term always appears (directly or indirectly), which should be limited. Otherwise, a final, specific amount simply doesn't exist. For the solution, it does not matter whether the progression is given: finite or infinite. It doesn’t matter how it is given: a series of numbers, or a formula for the nth term.

The most important thing is to understand that the formula works from the first term of the progression to the term with number n. Actually, the full name of the formula looks like this: the sum of the first n terms of an arithmetic progression. The number of these very first members, i.e. n, is determined solely by the task. In a task, all this valuable information is often encrypted, yes... But never mind, in the examples below we reveal these secrets.)

Examples of tasks on the sum of an arithmetic progression.

First of all, useful information:

The main difficulty in tasks involving the sum of an arithmetic progression lies in the correct determination of the elements of the formula.

The task writers encrypt these very elements with boundless imagination.) The main thing here is not to be afraid. Understanding the essence of the elements, it is enough to simply decipher them. Let's look at a few examples in detail. Let's start with a task based on a real GIA.

1. The arithmetic progression is given by the condition: a n = 2n-3.5. Find the sum of its first 10 terms.

Good job. Easy.) To determine the amount using the formula, what do we need to know? First member a 1, last term a n, yes the number of the last member n.

Where can I get the last member's number? n? Yes, right there, on condition! It says: find the sum first 10 members. Well, what number will it be with? last, tenth member?) You won’t believe it, his number is tenth!) Therefore, instead of a n We will substitute into the formula a 10, and instead n- ten. I repeat, the number of the last member coincides with the number of members.

It remains to determine a 1 And a 10. This is easily calculated using the formula for the nth term, which is given in the problem statement. Don't know how to do this? Attend the previous lesson, without this there is no way.

a 1= 2 1 - 3.5 = -1.5

a 10=2·10 - 3.5 =16.5

S n = S 10.

We have found out the meaning of all elements of the formula for the sum of an arithmetic progression. All that remains is to substitute them and count:

That's it. Answer: 75.

Another task based on the GIA. A little more complicated:

2. Given an arithmetic progression (a n), the difference of which is 3.7; a 1 =2.3. Find the sum of its first 15 terms.

We immediately write the sum formula:

This formula allows us to find the value of any term by its number. We look for a simple substitution:

a 15 = 2.3 + (15-1) 3.7 = 54.1

It remains to substitute all the elements into the formula for the sum of an arithmetic progression and calculate the answer:

Answer: 423.

By the way, if in the sum formula instead of a n We simply substitute the formula for the nth term and get:

Let us present similar ones and obtain a new formula for the sum of terms of an arithmetic progression:

As you can see, the nth term is not required here a n. In some problems this formula helps a lot, yes... You can remember this formula. Or you can simply display it at the right time, like here. After all, you always need to remember the formula for the sum and the formula for the nth term.)

Now the task in the form of a short encryption):

3. Find the sum of all positive two-digit numbers that are multiples of three.

Wow! Neither your first member, nor your last, nor progression at all... How to live!?

You will have to think with your head and pull out all the elements of the sum of the arithmetic progression from the condition. We know what two-digit numbers are. They consist of two numbers.) What two-digit number will be first? 10, presumably.) A last thing double digit number? 99, of course! The three-digit ones will follow him...

Multiples of three... Hm... These are numbers that are divisible by three, here! Ten is not divisible by three, 11 is not divisible... 12... is divisible! So, something is emerging. You can already write down a series according to the conditions of the problem:

12, 15, 18, 21, ... 96, 99.

Will this series be an arithmetic progression? Certainly! Each term differs from the previous one by strictly three. If you add 2 or 4 to a term, say, the result, i.e. the new number is no longer divisible by 3. You can immediately determine the difference of the arithmetic progression: d = 3. It will come in handy!)

So, we can safely write down some progression parameters:

What will the number be? n last member? Anyone who thinks that 99 is fatally mistaken... The numbers always go in a row, but our members jump over three. They don't match.

There are two solutions here. One way is for the super hardworking. You can write down the progression, the entire series of numbers, and count the number of members with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If we apply the formula to our problem, we find that 99 is the thirtieth term of the progression. Those. n = 30.

Let's look at the formula for the sum of an arithmetic progression:

We look and rejoice.) We pulled out from the problem statement everything necessary to calculate the amount:

a 1= 12.

a 30= 99.

S n = S 30.

All that remains is elementary arithmetic. We substitute the numbers into the formula and calculate:

Answer: 1665

Another type of popular puzzle:

4. Given an arithmetic progression:

-21,5; -20; -18,5; -17; ...

Find the sum of terms from twentieth to thirty-four.

We look at the formula for the amount and... we get upset.) The formula, let me remind you, calculates the amount from the first member. And in the problem you need to calculate the sum since the twentieth... The formula won't work.

You can, of course, write out the entire progression in a series, and add terms from 20 to 34. But... it’s somehow stupid and takes a long time, right?)

There is a more elegant solution. Let's divide our series into two parts. The first part will be from the first term to the nineteenth. Second part - from twenty to thirty-four. It is clear that if we calculate the sum of the terms of the first part S 1-19, let's add it with the sum of the terms of the second part S 20-34, we get the sum of the progression from the first term to the thirty-fourth S 1-34. Like this:

S 1-19 + S 20-34 = S 1-34

From this we can see that find the sum S 20-34 can be done by simple subtraction

S 20-34 = S 1-34 - S 1-19

Both amounts on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Let's get started?

We extract the progression parameters from the problem statement:

d = 1.5.

a 1= -21,5.

To calculate the sums of the first 19 and first 34 terms, we will need the 19th and 34th terms. We calculate them using the formula for the nth term, as in problem 2:

a 19= -21.5 +(19-1) 1.5 = 5.5

a 34= -21.5 +(34-1) 1.5 = 28

There's nothing left. From the sum of 34 terms subtract the sum of 19 terms:

S 20-34 = S 1-34 - S 1-19 = 110.5 - (-152) = 262.5

Answer: 262.5

One important note! There is a very useful trick in solving this problem. Instead of direct calculation what you need (S 20-34), we counted something that would seem not to be needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the complete result. This kind of “feint with your ears” often saves you in wicked problems.)

In this lesson we looked at problems for which it is enough to understand the meaning of the sum of an arithmetic progression. Well, you need to know a couple of formulas.)

Practical advice:

When solving any problem involving the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.

Formula for the nth term:

These formulas will immediately tell you what to look for and in what direction to think in order to solve the problem. Helps.

And now the tasks for independent solution.

5. Find the sum of all two-digit numbers that are not divisible by three.

Cool?) The hint is hidden in the note to problem 4. Well, problem 3 will help.

6. The arithmetic progression is given by the condition: a 1 = -5.5; a n+1 = a n +0.5. Find the sum of its first 24 terms.

Unusual?) This is a recurrent formula. You can read about it in the previous lesson. Don’t ignore the link, such problems are often found in the State Academy of Sciences.

7. Vasya saved up money for the holiday. As much as 4550 rubles! And I decided to give my favorite person (myself) a few days of happiness). Live beautifully without denying yourself anything. Spend 500 rubles on the first day, and on each subsequent day spend 50 rubles more than the previous one! Until the money runs out. How many days of happiness did Vasya have?

Is it difficult?) The additional formula from task 2 will help.

Answers (in disarray): 7, 3240, 6.

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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.

First level

Arithmetic progression. Detailed theory with examples (2019)

Number sequence

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 a broader sense as an infinite numerical 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:

1. 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?
2. What is the sum of all odd numbers contained in.
3. 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:

1. 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.

2. 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.

3. 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

1. - a number sequence in which the difference between adjacent numbers is the same and equal. It can be increasing or decreasing.
2. Finding formula The th term of an arithmetic progression is written by the formula - , where is the number of numbers in the progression.
3. Property of members of an arithmetic progression- - where is the number of numbers in progression.
4. 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:

1. 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?
2. 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?
3. 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:

1. 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:
2. 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:

3. 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.

Where is the number of values.

The concept of a number sequence implies that each natural number corresponds to some real value. Such a series of numbers can be either arbitrary or have certain properties - a progression. In the latter case, each subsequent element (member) of the sequence can be calculated using the previous one.

An arithmetic progression is a sequence of numerical values ​​in which its neighboring members differ from each other by the same number (all elements of the series, starting from the 2nd, have a similar property). This number - the difference between the previous and subsequent terms - is constant and is called the progression difference.

Progression difference: definition

Consider a sequence consisting of j values ​​A = a(1), a(2), a(3), a(4) ... a(j), j belongs to the set of natural numbers N. An arithmetic progression, according to its definition, is a sequence , in which a(3) – a(2) = a(4) – a(3) = a(5) – a(4) = … = a(j) – a(j-1) = d. The value d is the desired difference of this progression.

d = a(j) – a(j-1).

Highlight:

• An increasing progression, in which case d > 0. Example: 4, 8, 12, 16, 20, ...
• Decreasing progression, then d< 0. Пример: 18, 13, 8, 3, -2, …

Difference progression and its arbitrary elements

If 2 arbitrary terms of the progression are known (i-th, k-th), then the difference for a given sequence can be determined based on the relationship:

a(i) = a(k) + (i – k)*d, which means d = (a(i) – a(k))/(i-k).

Difference of progression and its first term

This expression will help determine an unknown value only in cases where the number of the sequence element is known.

Progression difference and its sum

The sum of a progression is the sum of its terms. To calculate the total value of its first j elements, use the appropriate formula:

S(j) =((a(1) + a(j))/2)*j, but since a(j) = a(1) + d(j – 1), then S(j) = ((a(1) + a(1) + d(j – 1))/2)*j=(( 2a(1) + d(– 1))/2)*j.

What is the main essence of the formula?

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 the 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.

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