Arithmetic progression formulas how to find e. Independent work in pairs

Arithmetic and geometric progressions

Theoretical information

Theoretical information

	Arithmetic progression	Geometric progression
Definition	Arithmetic progression a n a sequence is called, each member of which, starting from the second, is equal to the previous member, added with the same number d (d- progression difference)	geometric progression b n a sequence of non-zero numbers is called, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression)
Recurrent formula	For any natural n a n + 1 = a n + d	For any natural n b n + 1 = b n ∙ q, b n ≠ 0
nth term formula	a n = a 1 + d (n - 1)	b n \u003d b 1 ∙ q n - 1, b n ≠ 0
characteristic property
Sum of the first n terms

Examples of tasks with comments

Exercise 1

In arithmetic progression ( a n) a 1 = -6, a 2

According to the formula of the nth term:

a 22 = a 1+ d (22 - 1) = a 1+ 21d

By condition:

a 1= -6, so a 22= -6 + 21d.

It is necessary to find the difference of progressions:

d= a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = - 48.

Answer : a 22 = -48.

Task 2

Find the fifth term of the geometric progression: -3; 6;....

1st way (using n-term formula)

According to the formula of the n-th member of a geometric progression:

b 5 \u003d b 1 ∙ q 5 - 1 = b 1 ∙ q 4.

As b 1 = -3,

2nd way (using recursive formula)

Since the denominator of the progression is -2 (q = -2), then:

b 3 = 6 ∙ (-2) = -12;

b 4 = -12 ∙ (-2) = 24;

b 5 = 24 ∙ (-2) = -48.

Answer : b 5 = -48.

Task 3

In arithmetic progression ( a n) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.

For an arithmetic progression, the characteristic property has the form .

Therefore:

.

Substitute the data in the formula:

Answer: 95.

Task 4

In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.

To find the sum of the first n terms of an arithmetic progression, two formulas are used:

.

Which of them is more convenient to apply in this case?

By condition, the formula of the nth member of the original progression is known ( a n) a n= 3n - 4. Can be found immediately and a 1, and a 16 without finding d . Therefore, we use the first formula.

Answer: 368.

Task 5

In arithmetic progression a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.

According to the formula of the nth term:

a 22 = a 1 + d (22 – 1) = a 1+ 21d.

By condition, if a 1= -6, then a 22= -6 + 21d. It is necessary to find the difference of progressions:

d= a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = -48.

Answer : a 22 = -48.

Task 6

Several consecutive terms of a geometric progression are recorded:

Find the term of the progression, denoted by the letter x .

When solving, we use the formula for the nth term b n \u003d b 1 ∙ q n - 1 for geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of these terms of the progression and divide by the previous one. In our example, you can take and divide by. We get that q \u003d 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.

Substituting the found values ​​into the formula, we get:

.

Answer : .

Task 7

From the arithmetic progressions given by the formula of the nth term, choose the one for which the condition is satisfied a 27 > 9:

Since the specified condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:

.

Answer: 4.

Task 8

In arithmetic progression a 1= 3, d = -1.5. Specify the largest value of n for which the inequality holds a n > -6.

Before we start to decide arithmetic progression problems, consider what a number sequence is, since an arithmetic progression is a special case of a number sequence.

A numerical sequence is a numerical set, each element of which has its own serial number. The elements of this set are called members of the sequence. The ordinal number of a sequence element is indicated by an index:

The first element of the sequence;

The fifth element of the sequence;

- "nth" element of the sequence, i.e. the element "standing in the queue" at number n.

There is a dependency between the value of a sequence element and its ordinal number. Therefore, we can consider a sequence as a function whose argument is the ordinal number of an element of the sequence. In other words, one can say that the sequence is a function of the natural argument:

The sequence can be specified in three ways:

1 . The sequence can be specified using a table. In this case, we simply set the value of each member of the sequence.

For example, Someone decided to do personal time management, and to begin with, to calculate how much time he spends on VKontakte during the week. By writing the time in a table, he will get a sequence consisting of seven elements:

The first line of the table contains the number of the day of the week, the second - the time in minutes. We see that, that is, on Monday Someone spent 125 minutes on VKontakte, that is, on Thursday - 248 minutes, and, that is, on Friday, only 15.

2 . The sequence can be specified using the nth member formula.

In this case, the dependence of the value of a sequence element on its number is expressed directly as a formula.

For example, if , then

To find the value of a sequence element with a given number, we substitute the element number into the formula for the nth member.

We do the same if we need to find the value of a function if the value of the argument is known. We substitute the value of the argument instead in the equation of the function:

If, for example, , then

Once again, I note that in a sequence, in contrast to an arbitrary numeric function, only a natural number can be an argument.

3 . The sequence can be specified using a formula that expresses the dependence of the value of the member of the sequence with number n on the value of the previous members. In this case, it is not enough for us to know only the number of a sequence member in order to find its value. We need to specify the first member or first few members of the sequence.

For example, consider the sequence ,

We can find the values ​​of the members of a sequence in sequence, starting from the third:

That is, each time to find the value of the nth member of the sequence, we return to the previous two. This way of sequencing is called recurrent, from the Latin word recurro- come back.

Now we can define an arithmetic progression. An arithmetic progression is a simple special case of a numerical sequence.

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

The number is called the difference of an arithmetic progression. The difference of an arithmetic progression can be positive, negative, or zero.

If title="(!LANG:d>0">, то каждый член арифметической прогрессии больше предыдущего, и прогрессия является !} increasing.

For example, 2; 5; eight; eleven;...

If , then each term of the arithmetic progression is less than the previous one, and the progression is waning.

For example, 2; -one; -4; -7;...

If , then all members of the progression are equal to the same number, and the progression is stationary.

For example, 2;2;2;2;...

The main property of an arithmetic progression:

Let's look at the picture.

We see that

, and at the same time

Adding these two equalities, we get:

.

Divide both sides of the equation by 2:

So, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of two neighboring ones:

Moreover, since

, and at the same time

, then

, and hence

Each member of the arithmetic progression starting with title="(!LANG:k>l">, равен среднему арифметическому двух равноотстоящих. !}

th member formula.

We see that for the members of the arithmetic progression, the following relations hold:

and finally

We got formula of the nth term.

IMPORTANT! Any member of an arithmetic progression can be expressed in terms of and . Knowing the first term and the difference of an arithmetic progression, you can find any of its members.

The sum of n members of an arithmetic progression.

In an arbitrary arithmetic progression, the sums of terms equally spaced from the extreme ones are equal to each other:

Consider an arithmetic progression with n members. Let the sum of n members of this progression be equal to .

Arrange the terms of the progression first in ascending order of numbers, and then in descending order:

Let's pair it up:

The sum in each parenthesis is , the number of pairs is n.

We get:

So, the sum of n members of an arithmetic progression can be found using the formulas:

Consider solving arithmetic progression problems.

1 . The sequence is given by the formula of the nth member: . Prove that this sequence is an arithmetic progression.

Let us prove that the difference between two adjacent members of the sequence is equal to the same number.

We have obtained that the difference of two adjacent members of the sequence does not depend on their number and is a constant. Therefore, by definition, this sequence is an arithmetic progression.

2 . Given an arithmetic progression -31; -27;...

a) Find the 31 terms of the progression.

b) Determine if the number 41 is included in this progression.

a) We see that ;

Let's write down the formula for the nth term for our progression.

In general

In our case , That's why

Attention!
There are additional
material in Special Section 555.
For those who strongly "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 is often difficult and incomprehensible. Letter indices, the nth term of the progression, the difference of the progression - all this is somehow confusing, yes ... Let's figure out the meaning of the arithmetic progression and everything will work out right away.)

The concept of arithmetic progression.

Arithmetic progression is a very simple and clear concept. Doubt? In vain.) See for yourself.

I'll write an unfinished series of numbers:

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

Can you extend this line? What numbers will go next, after the five? Everyone ... uh ..., in short, everyone will figure out that the numbers 6, 7, 8, 9, etc. will go further.

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

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

You can catch the pattern, extend the series, and name seventh row number?

If you figured out that this number is 20 - I congratulate you! You not only felt key points of an arithmetic progression, but also successfully used them in business! If you don't understand, 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, building graphs and all that ... And then 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 with series of numbers and expressions. Get used to it.)

Second key point.

In an arithmetic progression, any number differs 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 times greater than the previous one. Actually, it is this moment that gives us the opportunity to catch the pattern and calculate the subsequent numbers.

Third key point.

This moment is not striking, yes ... But very, very important. There he is: each progression number is in its place. There is the first number, there is the seventh, there is the forty-fifth, and so on. If you confuse them haphazardly, the pattern will disappear. The arithmetic progression will also disappear. It's just a series of numbers.

That's the whole point.

Of course, new terms and notation appear in the new topic. They need to know. Otherwise, you won't understand the task. For example, you have to decide something like:

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

Does it inspire?) Letters, some indexes... And the task, by the way, couldn't be easier. You just need to understand the meaning of the terms and notation. 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 value is called . Let's deal with this concept in more detail.

Arithmetic progression difference.

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

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

To calculate, let's say second numbers of the row, it is necessary 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 of 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 adding positive number, +5 to the previous one.

The difference can 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 every number is obtained too adding to the previous, but already negative number, -5.

By the way, when working with a progression, it is very useful to immediately determine its nature - whether it is increasing or decreasing. It helps a lot to find your bearings in the decision, to detect 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 of the series previous number. Subtract. By the way, the result of subtraction is called "difference".)

Let's define, for example, d for an increasing arithmetic progression:

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

We take any number of the row that we want, for example, 11. Subtract from it the previous number those. eight:

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

You can just take any number of progressions, 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 can not take only the very first number. Just because the very first number no previous.)

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

Let's define d for a decreasing arithmetic progression:

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

I remind you that, regardless of the signs, to determine d needed from any number take away the previous one. We choose any number of progression, 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.

Other terms and designations.

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

Each member of the progression has his 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 member, five is the second, eleven is the fourth, well, you understand ...) Please clearly understand - the numbers themselves can be absolutely any, whole, fractional, negative, whatever, but numbering- strictly in order!

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

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

a 1 is the first number a 3- third, etc. Nothing tricky. You can write this series briefly like this: (a n).

There are progressions finite and infinite.

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

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

You can write a final progression through a series like this, all members 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 a 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 already solve tasks. The tasks are simple, purely for understanding the meaning of the arithmetic progression.

Examples of tasks for arithmetic progression.

Let's take a closer look at the task above:

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

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

For clarity, I will write down a series according to the condition of the problem. The first six members, where the second member is five:

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

a 3 = a 2 + d

We substitute in the expression a 2 = 5 and d=-2.5. Don't forget the minus!

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

The third term is less than the second. Everything is logical. If the number is greater than the previous one negative value, so the number itself will be less than the previous one. Progression is decreasing. Okay, let's take it into account.) We consider the fourth member 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, the terms from the third to the sixth have been calculated. This resulted in a 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.) Hence, 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 all there is to it. Task response:

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

In passing, I note that we solved this task recurrent way. This terrible word means, only, the search for a member of the progression by the previous (adjacent) number. Other ways to work with progression will be discussed later.

One important conclusion can be drawn from this simple task.

Remember:

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

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

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

For example, consider some popular tasks on this topic.

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

Everything is simple here. Everything is already given. You need to remember how the members of an arithmetic progression are calculated, count, and write down. It is advisable not to skip the words in the task condition: "final" and " n=5". In order not to count until you are completely blue in the face.) There are only 5 (five) members in this progression:

a 2 \u003d a 1 + d \u003d 3.6 + 0.4 \u003d 4

a 3 \u003d a 2 + d \u003d 4 + 0.4 \u003d 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 if the number 7 will be a member of an arithmetic progression (a n) if a 1 \u003d 4.1; d = 1.2.

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

How-how ... Yes, write down the progression in the form of a series and see if there will be a seven or not! We believe:

a 2 \u003d a 1 + d \u003d 4.1 + 1.2 \u003d 5.3

a 3 \u003d a 2 + d \u003d 5.3 + 1.2 \u003d 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 seen that we are just seven slipped through between 6.5 and 7.7! The seven did not get into our series of numbers, and, therefore, the seven will not be a member of the given progression.

Answer: no.

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

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

...; fifteen; X; nine; 6; ...

Here is a series 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 see and see what we can discover from this line? What are the parameters of the three main ones?

Member numbers? There is not a single number here.

But there are three numbers and - attention! - word "consecutive" 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. So we can calculate the difference of an arithmetic progression! We subtract from the six previous number, i.e. nine:

There are empty spaces left. What number will be the previous one for x? Fifteen. So x can be easily found by simple addition. To 15 add the difference of an arithmetic progression:

That's all. Answer: x=12

We solve the following problems ourselves. Note: these puzzles are not for formulas. Purely for understanding the meaning of an arithmetic progression.) We just write down a series of numbers-letters, look and think.

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

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

7. It is known that in an arithmetic progression a 2 = 4; a 5 \u003d 15.1. Find a 3 .

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

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

Find the term of the progression, denoted by the letter x.

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

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

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

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

Didn't everything work out? No problem. In Special Section 555, all these puzzles are broken down piece by piece.) And, of course, a simple practical technique is described that immediately highlights the solution of such tasks clearly, clearly, as in the palm of your hand!

By the way, in the puzzle about the train there are two problems on which people often stumble. One - purely by progression, and the second - common to any tasks 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 examined 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 decided.

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

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

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

And what, we will add 1/6 many, many times?! Is it possible to kill yourself!?

You can.) If you do not know a simple formula by which you can solve such tasks in a minute. This formula will be in the next lesson. And that 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. Learning - with interest!)

you can get acquainted with functions and derivatives.

The 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 elementary to quite solid.

First, let's deal with the meaning and formula of the sum. And then we'll decide. For your own pleasure.) The meaning of the sum is as simple as lowing. To find the sum of an arithmetic progression, you just need to carefully add all its members. 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 saves.

The sum formula is simple:

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

S n is the sum of an arithmetic progression. Addition result all members, with first on last. It is important. Add up exactly all members in a row, without gaps and jumps. And, exactly, starting from first. In problems like finding the sum of the third and eighth terms, or the sum of terms five through twentieth, direct application of the formula will be disappointing.)

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 row. Not a very familiar name, but, when applied to the amount, it is very suitable. Then you will see for yourself.

n is the 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. Filling question: what kind of member will last, if given endless arithmetic progression?

For a confident answer, you need to understand the elementary meaning of an arithmetic progression and ... read the assignment 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 finite, specific amount just doesn't exist. For the solution, it does not matter what kind of progression is given: finite or infinite. It doesn't matter how it is given: by a series of numbers, or by the formula of the nth member.

The most important thing is to understand that the formula works from the first term of the progression to the term with the 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 the task, all this valuable information is often encrypted, yes ... But nothing, in the examples below we will reveal these secrets.)

Examples of tasks for the sum of an arithmetic progression.

First of all, useful information:

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

The authors of the assignments 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 just to decipher them. Let's take a 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 the first 10 terms.

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

Where to get the last member number n? Yes, in the same place, in the condition! It says find the sum first 10 members. Well, what number will it be 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, but instead n- ten. Again, the number of the last member is the same as the number of members.

It remains to be determined a 1 and a 10. This is easily calculated by the formula of the nth term, which is given in the problem statement. Don't know how to do it? Visit the previous lesson, without this - nothing.

a 1= 2 1 - 3.5 = -1.5

a 10\u003d 2 10 - 3.5 \u003d 16.5

S n = S 10.

We found out the meaning of all elements of the formula for the sum of an arithmetic progression. It remains to substitute them, and count:

That's all there is to 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 \u003d 2.3. Find the sum of the first 15 terms.

We immediately write the sum formula:

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

a 15 \u003d 2.3 + (15-1) 3.7 \u003d 54.1

It remains to substitute all the elements in 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 just substitute the formula of the nth term, we get:

We give similar ones, we get a new formula for the sum of members of an arithmetic progression:

As you can see, the nth term is not required here. a n. In some tasks, this formula helps out a lot, yes ... You can remember this formula. And you can simply withdraw it at the right time, as here. After all, the formula for the sum and the formula for the nth term must be remembered in every way.)

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.

How! No first member, no last, no progression at all... How to live!?

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

Multiples of three... Hm... These are numbers that are evenly 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 a series according to the condition of the problem:

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

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

So, we can safely write down some progression parameters:

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

There are two solutions here. One way is for the super hardworking. You can paint the progression, the whole series of numbers, and count the number of terms with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If the formula is applied to our problem, we get that 99 is the thirtieth member of the progression. Those. n = 30.

We look at the formula for the sum of an arithmetic progression:

We look and rejoice.) We pulled out everything necessary for calculating the amount from the condition of the problem:

a 1= 12.

a 30= 99.

S n = S 30.

What remains is elementary arithmetic. Substitute the numbers in the formula and calculate:

Answer: 1665

Another type of popular puzzles:

4. An arithmetic progression is given:

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

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

We look at the sum formula and ... we are upset.) The formula, let me remind you, calculates the sum 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, paint the entire progression in a row, and put the members from 20 to 34. But ... somehow it turns out stupidly and for a long time, right?)

There is a more elegant solution. Let's break our series into two parts. The first part will from the first term to the nineteenth. Second part - 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 to the sum of the members 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

This shows that to find the sum S 20-34 can be done by simple subtraction

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

Both sums on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Are we getting started?

We extract the progression parameters from the task condition:

d = 1.5.

a 1= -21,5.

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

a 19\u003d -21.5 + (19-1) 1.5 \u003d 5.5

a 34\u003d -21.5 + (34-1) 1.5 \u003d 28

There is nothing left. Subtract the sum of 19 terms from the sum of 34 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 feature in solving this problem. Instead of direct calculation what you need (S 20-34), we counted what, it would seem, is not needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the full result. Such a "feint with the ears" often saves in evil puzzles.)

In this lesson, we examined 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 for the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.

Formula of the nth term:

These formulas will immediately tell you what to look for, in which 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. Arithmetic progression is given by the condition: a 1 =-5.5; a n+1 = a n +0.5. Find the sum of the first 24 terms.

Unusual?) This is a recurrent formula. You can read about it in the previous lesson. Do not ignore the link, such puzzles are often found in the GIA.

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

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

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

If you like this site...

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

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

you can get acquainted with functions and derivatives.

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

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

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

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

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

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

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

For example,

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

a n= 2n- 1,

and the sequence of alternating 1 and -1 - formula

b n = (-1)n +1 . ◄

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

For example,

if a 1 = 1 , a a n +1 = a n + 5

a 1 = 1,

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

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

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

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

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

a 1 = 1,

a 2 = 1,

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

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

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

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

a 7 = a 5 + a 6 = 5 + 8 = 13. ◄

Sequences can be final and endless .

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

For example,

sequence of two-digit natural numbers:

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

final.

Prime number sequence:

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

endless. ◄

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

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

For example,

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

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

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

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

Arithmetic progression

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

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

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

a n +1 = a n + d,

where d - some number.

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

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

Number d called the difference of an arithmetic progression.

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

For example,

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

a 1 =3,

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

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

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

a 5 = a 4 + d= 15 + 4 = 19. ◄

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

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

For example,

find the thirtieth term of an arithmetic progression

1, 4, 7, 10, . . .

a 1 =1, d = 3,

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

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

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

a n +1 = a 1 + nd,

then obviously

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

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

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

For example,

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

Let's use the statement above. We have:

a n = 2n- 7,

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

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

Hence,

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

◄

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

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

For example,

for a 5 can be written

a 5 = a 1 + 4d,

a 5 = a 2 + 3d,

a 5 = a 3 + 2d,

a 5 = a 4 + d. ◄

a n = a n-k + kd,

a n = a n+k - kd,

then obviously

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

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

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

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

m + n = k + l.

For example,

in arithmetic progression

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

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

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

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

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

a 5 + a 9 = 13 + 25 = 38. ◄

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

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

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

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

then the previous formula retains its structure:

For example,

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

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

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

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

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

An arithmetic progression is a monotonic sequence. Wherein:

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

Geometric progression

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

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

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

b n +1 = b n · q,

where q ≠ 0 - some number.

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

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

Number q called denominator of a geometric progression.

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

For example,

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

b 1 = 1,

b 2 = b 1 · q = 1 · (-3) = -3,

b 3 = b 2 · q= -3 · (-3) = 9,

b 4 = b 3 · q= 9 · (-3) = -27,

b 5 = b 4 · q= -27 · (-3) = 81. ◄

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

b n = b 1 · q n -1 .

For example,

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

b 1 = 1, q = 2,

b 7 = b 1 · q 6 = 1 2 6 = 64. ◄

bn-1 = b 1 · q n -2 ,

b n = b 1 · q n -1 ,

b n +1 = b 1 · q n,

then obviously

b n 2 = b n -1 · b n +1 ,

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

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

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

For example,

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

b n= -3 2 n,

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

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

Hence,

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

which proves the required assertion. ◄

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

b n = b k · q n - k.

For example,

for b 5 can be written

b 5 = b 1 · q 4 ,

b 5 = b 2 · q 3,

b 5 = b 3 · q2,

b 5 = b 4 · q. ◄

b n = b k · q n - k,

b n = b n - k · q k,

then obviously

b n 2 = b n - k· b n + k

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

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

b m· b n= b k· b l,

m+ n= k+ l.

For example,

exponentially

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

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

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

4) b 2 · b 7 = b 4 · b 5 , as

b 2 · b 7 = 2 · 64 = 128,

b 4 · b 5 = 8 · 16 = 128. ◄

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

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

And when q = 1 - according to the formula

S n= n.b. 1

Note that if we need to sum the terms

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

then the formula is used:

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

For example,

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

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

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

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

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

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

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

b 1 > 0 and q> 1;

b 1 < 0 and 0 < q< 1;

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

b 1 > 0 and 0 < q< 1;

b 1 < 0 and q> 1.

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

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

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

For example,

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

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

Infinitely decreasing geometric progression

Infinitely decreasing geometric progression is called an infinite geometric progression whose denominator modulus is less than 1 , i.e

|q| < 1 .

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

1 < q< 0 .

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

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

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

S= b 1 + b 2 + b 3 + . . . =	b 1	.
	1 - q

For example,

10 + 1 + 0,1 + 0,01 + . . . = 10 / (1 - 0,1) = 11 1 / 9 ,

10 - 1 + 0,1 - 0,01 + . . . = 10 / (1 + 0,1) = 9 1 / 11 . ◄

Relationship between arithmetic and geometric progressions

Arithmetic and geometric progressions are closely related. Let's consider just two examples.

a 1 , a 2 , a 3 , . . . d , then

b a 1 , b a 2 , b a 3 , . . . b d .

For example,

1, 3, 5, . . . — arithmetic progression with difference 2 and

7 1 , 7 3 , 7 5 , . . . is a geometric progression with a denominator 7 2 . ◄

b 1 , b 2 , b 3 , . . . is a geometric progression with a denominator q , then

log a b 1, log a b 2, log a b 3, . . . — arithmetic progression with difference log aq .

For example,

2, 12, 72, . . . is a geometric progression with a denominator 6 and

lg 2, lg 12, lg 72, . . . — arithmetic progression with difference lg 6 . ◄