How many positive numbers are there in an arithmetic progression? Formula for the nth term of an arithmetic progression

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 bring it into general form and we 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, asked in class next task: “Count the sum of all natural numbers from to (according to other sources up 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 closer look at the highlighted numbers and try to perform various mathematical operations with them.

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

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

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

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

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

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

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

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

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

Method 1.

Method 2.

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

Training

Tasks:

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

Answers:

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

Let's sum it up

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

, where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

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

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

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

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

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

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

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

sets the sequence:

And the formula is the following sequence:

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

Formula nth term

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 double digit numbers, multiples.

Solution:

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

Formula of the th term for this progression:

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

Very easy: .

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

Answer: .

Now decide for yourself:

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

Answers:

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

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

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

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

For example:

Formula for finding the nth term of an arithmetic progression

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

Property of members of an arithmetic progression

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

Sum of terms of an arithmetic progression

There are two ways to find the amount:

Where is the number of values.

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Arithmetic progression

An arithmetic progression is a special type of sequence. Therefore, before defining an arithmetic (and then geometric) progression, we need to briefly discuss important concept number sequence.

Subsequence

Imagine a device on the screen of which certain numbers are displayed one after another. Let's say 2; 7; 13; 1; 6; 0; 3; : : : This set of numbers is precisely an example of a sequence.

Definition. A number sequence is a set of numbers in which each number can be assigned a unique number (that is, associated with a single natural number)1. The number with number n is called nth term sequences.

So, in the example above, the first number is 2, this is the first member of the sequence, which can be denoted by a1; number five has the number 6 is the fifth term of the sequence, which can be denoted by a5. At all, nth term sequences are denoted by an (or bn, cn, etc.).

A very convenient situation is when the nth term of the sequence can be specified by some formula. For example, the formula an = 2n 3 specifies the sequence: 1; 1; 3; 5; 7; : : : The formula an = (1)n specifies the sequence: 1; 1; 1; 1; : : :

Not every set of numbers is a sequence. Thus, a segment is not a sequence; it contains “too many” numbers to be renumbered. The set R of all real numbers is also not a sequence. These facts are proven in the course of mathematical analysis.

Arithmetic progression: basic definitions

Now we are ready to define an arithmetic progression.

Definition. An arithmetic progression is a sequence in which each term (starting from the second) is equal to the sum of the previous term and some fixed number (called the difference of the arithmetic progression).

For example, sequence 2; 5; 8; eleven; : : : is an arithmetic progression with first term 2 and difference 3. Sequence 7; 2; 3; 8; : : : is an arithmetic progression with first term 7 and difference 5. Sequence 3; 3; 3; : : : is an arithmetic progression with a difference equal to zero.

Equivalent definition: the sequence an is called an arithmetic progression if the difference an+1 an is a constant value (independent of n).

An arithmetic progression is called increasing if its difference is positive, and decreasing if its difference is negative.

1 But here is a more concise definition: a sequence is a function defined on the set of natural numbers. For example, a sequence of real numbers is a function f: N ! R.

By default, sequences are considered infinite, that is, containing an infinite number of numbers. But no one bothers us to consider finite sequences; in fact, any finite set of numbers can be called a finite sequence. For example, the ending sequence is 1; 2; 3; 4; 5 is made up of five numbers.

Formula for the nth term of an arithmetic progression

It is easy to understand that an arithmetic progression is completely determined by two numbers: the first term and the difference. Therefore, the question arises: how, knowing the first term and the difference, find an arbitrary term of an arithmetic progression?

It is not difficult to obtain the required formula for the nth term of an arithmetic progression. Let an

arithmetic progression with difference d. We have:
an+1 = an + d (n = 1; 2; : : :):
In particular, we write:
a2 = a1 + d;
a3 = a2 + d = (a1 + d) + d = a1 + 2d;
a4 = a3 + d = (a1 + 2d) + d = a1 + 3d;
and now it becomes clear that the formula for an is:
an = a1 + (n 1)d:

Problem 1. In arithmetic progression 2; 5; 8; eleven; : : : find the formula for the nth term and calculate the hundredth term.

Solution. According to formula (1) we have:

an = 2 + 3(n 1) = 3n 1:

a100 = 3 100 1 = 299:

Property and sign of arithmetic progression

Property of arithmetic progression. In arithmetic progression an for any

In other words, each member of an arithmetic progression (starting from the second) is the arithmetic mean of its neighboring members.

	Proof. We have:
	a n 1+ a n+1		(an d) + (an + d)

which is what was required.

More generally, the arithmetic progression an satisfies the equality

a n = a n k+ a n+k

for any n > 2 and any natural k< n. Попробуйте самостоятельно доказать эту формулу тем же самым приёмом, что и формулу (2 ).

It turns out that formula (2) serves not only as a necessary but also as a sufficient condition for the sequence to be an arithmetic progression.

Arithmetic progression sign. If equality (2) holds for all n > 2, then the sequence an is an arithmetic progression.

Proof. Let's rewrite formula (2) as follows:

a na n 1= a n+1a n:

From this we can see that the difference an+1 an does not depend on n, and this precisely means that the sequence an is an arithmetic progression.

The property and sign of an arithmetic progression can be formulated in the form of one statement; For convenience, we will do this for three numbers(this is the situation that often occurs in problems).

Characterization of an arithmetic progression. Three numbers a, b, c form an arithmetic progression if and only if 2b = a + c.

Problem 2. (MSU, Faculty of Economics, 2007) Three numbers 8x, 3 x2 and 4 in the indicated order form a decreasing arithmetic progression. Find x and indicate the difference of this progression.

Solution. By the property of arithmetic progression we have:

2(3 x2 ) = 8x 4 , 2x2 + 8x 10 = 0 , x2 + 4x 5 = 0 , x = 1; x = 5:

If x = 1, then we get a decreasing progression of 8, 2, 4 with a difference of 6. If x = 5, then we get an increasing progression of 40, 22, 4; this case is not suitable.

Answer: x = 1, the difference is 6.

Sum of the first n terms of an arithmetic progression

Legend has it that one day the teacher told the children to find the sum of the numbers from 1 to 100 and sat down quietly to read the newspaper. However, within a few minutes, one boy said that he had solved the problem. This was 9-year-old Carl Friedrich Gauss, later one of the greatest mathematicians in history.

Little Gauss's idea was as follows. Let

S = 1 + 2 + 3 + : : : + 98 + 99 + 100:

Let's write this amount in reverse order:

S = 100 + 99 + 98 + : : : + 3 + 2 + 1;

and add these two formulas:

2S = (1 + 100) + (2 + 99) + (3 + 98) + : : : + (98 + 3) + (99 + 2) + (100 + 1):

Each term in brackets is equal to 101, and there are 100 such terms in total. Therefore

2S = 101 100 = 10100;

We use this idea to derive the sum formula

S = a1 + a2 + : : : + an + a n n: (3)

A useful modification of formula (3) is obtained if we substitute the formula of the nth term an = a1 + (n 1)d into it:

		2a1 + (n 1)d

Problem 3. Find the sum of all positive three-digit numbers divisible by 13.

Solution. Three-digit numbers that are multiples of 13 form an arithmetic progression with the first term being 104 and the difference being 13; The nth term of this progression has the form:

an = 104 + 13(n 1) = 91 + 13n:

Let's find out how many terms our progression contains. To do this, let's solve the inequality:

an 6 999; 91 + 13n 6 999;

n 6 908 13 = 6911 13; n 6 69:

So, there are 69 members in our progression. Using formula (4) we find the required amount:

S = 2 104 + 68 13 69 = 37674: 2

Some people treat the word “progression” with caution, as a very complex term from the branches of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi meter (where they still exist). And understand the essence (and in mathematics there is nothing more important than “getting the essence”) arithmetic sequence It's not that difficult once you understand a few basic concepts.

Mathematical number sequence

A numerical sequence is usually called a series of numbers, each of which has its own number.

a 1 is the first member of the sequence;

and 2 is the second term of the sequence;

and 7 is the seventh member of the sequence;

and n is the nth member of the sequence;

However, not any arbitrary set of numbers and numbers interests us. We will focus our attention on a numerical sequence in which the value of the nth term is related to its ordinal number by a relationship that can be clearly formulated mathematically. In other words: numerical value The nth number is some function of n.

a is the value of a member of a numerical sequence;

n is its serial number;

f(n) is a function, where the ordinal number in the numerical sequence n is the argument.

Definition

An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth term of an arithmetic sequence is as follows:

a n - the value of the current member of the arithmetic progression;

a n+1 - formula of the next number;

d - difference (certain number).

It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one and such an arithmetic progression will be increasing.

In the graph below it is easy to see why the number sequence is called “increasing”.

In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.

Specified member value

Sometimes it is necessary to determine the value of any arbitrary term a n of an arithmetic progression. This can be done by sequentially calculating the values of all members of the arithmetic progression, starting from the first to the desired one. However, this path is not always acceptable if, for example, it is necessary to find the value of the five-thousandth or eight-millionth term. Traditional calculations will take a lot of time. However, a specific arithmetic progression can be studied using certain formulas. There is also a formula for the nth term: the value of any term of an arithmetic progression can be determined as the sum of the first term of the progression with the difference of the progression, multiplied by the number of the desired term, reduced by one.

The formula is universal for increasing and decreasing progression.

An example of calculating the value of a given term

Let us solve the following problem of finding the value of the nth term of an arithmetic progression.

Condition: there is an arithmetic progression with parameters:

The first term of the sequence is 3;

The difference in the number series is 1.2.

Task: you need to find the value of 214 terms

Solution: to determine the value of a given term, we use the formula:

a(n) = a1 + d(n-1)

Substituting the data from the problem statement into the expression, we have:

a(214) = a1 + d(n-1)

a(214) = 3 + 1.2 (214-1) = 258.6

Answer: The 214th term of the sequence is equal to 258.6.

The advantages of this method of calculation are obvious - the entire solution takes no more than 2 lines.

Sum of a given number of terms

Very often, in a given arithmetic series, it is necessary to determine the sum of the values of some of its segments. To do this, there is also no need to calculate the values of each term and then add them up. This method is applicable if the number of terms whose sum needs to be found is small. In other cases, it is more convenient to use the following formula.

The sum of the terms of an arithmetic progression from 1 to n is equal to the sum of the first and nth terms, multiplied by the number of the term n and divided by two. If in the formula the value of the nth term is replaced by the expression from the previous paragraph of the article, we get:

Calculation example

For example, let’s solve a problem with the following conditions:

The first term of the sequence is zero;

The difference is 0.5.

The problem requires determining the sum of the terms of the series from 56 to 101.

Solution. Let's use the formula for determining the amount of progression:

s(n) = (2∙a1 + d∙(n-1))∙n/2

First, we determine the sum of the values of 101 terms of the progression by substituting the given conditions of our problem into the formula:

s 101 = (2∙0 + 0.5∙(101-1))∙101/2 = 2,525

Obviously, in order to find out the sum of the terms of the progression from the 56th to the 101st, it is necessary to subtract S 55 from S 101.

s 55 = (2∙0 + 0.5∙(55-1))∙55/2 = 742.5

Thus, the sum of the arithmetic progression for this example is:

s 101 - s 55 = 2,525 - 742.5 = 1,782.5

Example of practical application of arithmetic progression

At the end of the article, let's return to the example of an arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider this example.

Boarding a taxi (which includes 3 km of travel) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles/km. Travel distance 30 km. Calculate the cost of the trip.

1. Let’s discard the first 3 km, the price of which is included in the cost of landing.

30 - 3 = 27 km.

2. Further calculation is nothing more than parsing an arithmetic number series.

Member number - the number of kilometers traveled (minus the first three).

The value of the member is the sum.

The first term in this problem will be equal to a 1 = 50 rubles.

Progression difference d = 22 r.

the number we are interested in is the value of the (27+1)th term of the arithmetic progression - the meter reading at the end of the 27th kilometer is 27.999... = 28 km.

a 28 = 50 + 22 ∙ (28 - 1) = 644

Calendar data calculations for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of the celestial body to the star. In addition, various number series are successfully used in statistics and other applied areas of mathematics.

Another type of number sequence is geometric

Geometric progression is characterized by greater rates of change compared to arithmetic progression. It is no coincidence that in politics, sociology, and medicine, in order to show the high speed of spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops in geometric progression.

The Nth term of the geometric number series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first term is 1, the denominator is correspondingly equal to 2, then:

n=1: 1 ∙ 2 = 2

n=2: 2 ∙ 2 = 4

n=3: 4 ∙ 2 = 8

n=4: 8 ∙ 2 = 16

n=5: 16 ∙ 2 = 32,

b n - the value of the current term of the geometric progression;

b n+1 - formula of the next term of the geometric progression;

q is the denominator of the geometric progression (a constant number).

If the graph of an arithmetic progression is a straight line, then a geometric progression paints a slightly different picture:

As in the case of arithmetic, geometric progression has a formula for the value of an arbitrary term. Any nth term of a geometric progression is equal to the product of the first term and the denominator of the progression to the power of n reduced by one:

Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Let's find the 5th term of the progression

b 5 = b 1 ∙ q (5-1) = 3 ∙ 1.5 4 = 15.1875

The sum of a given number of terms is also calculated using a special formula. The sum of the first n terms of a geometric progression is equal to the difference between the product of the nth term of the progression and its denominator and the first term of the progression, divided by the denominator reduced by one:

If b n is replaced using the formula discussed above, the value of the sum of the first n terms of the number series under consideration will take the form:

Example. The geometric progression starts with the first term equal to 1. The denominator is set to 3. Let's find the sum of the first eight terms.

s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280

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 next to 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 determine from a 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, how to be, how to be... Turn on your creative abilities!)

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:

The arithmetic progression is given by the condition:

a n = -4 + 6.8n

Find the first and tenth terms of the progression.

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

We are looking for the first member. The one who thinks. that the first term is minus four is fatally mistaken!) Because the formula in the problem is modified. The first term of the arithmetic progression in it hidden. It’s okay, we’ll find it now.)

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

a 1 = -4 + 6.8 1 = 2.8

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

We look for the tenth term in the same way:

a 10 = -4 + 6.8 10 = 64

That's it.

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

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

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

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

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

a 2 =a 1 + 1 d

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

a 3 =a 1 + 2 d

Do you get it? It’s not for nothing that I highlight some words in bold. Okay, one more step).

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

a 4 =a 1 + 3 d

It's time to realize that the number of gaps, i.e. d, Always one less than the number of the member you are looking for n. That is, to the number n, number of spaces will n-1. Therefore, the formula will be (without variations!):

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

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

Tasks for independent solution.

To warm up:

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

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

And this is no longer a warm-up.)

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

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

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

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

4. Given an arithmetic progression (a n):

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

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

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

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

Not the easiest task, yes...) The “fingertip” method won’t work here. You will have to write formulas and solve equations.

Answers (in disarray):

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

Happened? It's nice!)

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

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

If you like this site...

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

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

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.