How to find an arithmetic progression formula. Arithmetic progression: what is it? Each group presents the solution to the problem on the board

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 - we imagine.) What is arithmetic progression, member number, progression difference - clearly stated in the previous lesson. 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 view can be written as a series of numbers:

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

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

Formula binds key parameters 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 give 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 it fundamental difference recurrent formula from 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 in the usual form, and work with her. 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 direct application formulas. At the end of the previous lesson there was a problem:

An arithmetic progression (a n) is given. Find a 121 if a 1 =3 and d=1/6.

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

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

The conditions provide all the data for using the formula: a 1 =3, d=1/6. It remains to figure out what is equal n. No problem! We need to find a 121. So we write:

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

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

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

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

Let's solve the problem in a more cunning way. Let us come across the following problem:

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

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

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

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

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

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

a 17 = a 1 + (17-1)·(-0.5)

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

-2 = a 1 + (17-1)·(-0.5)

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

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

Another popular puzzle:

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

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

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

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

12=2 + (15-1)d

We do the arithmetic.)

12=2 + 14d

d=10/14 = 5/7

This is the correct answer.

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

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

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

a n = 12 + (n-1) 3

At first glance, there are two unknown quantities here: a n and n. But a n- this is some member of the progression with a number n...And we know this member of the progression! It's 99. We don't know its number. n, So this number is what you need to find. We substitute the term of the progression 99 into the formula:

99 = 12 + (n-1) 3

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

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

Determine whether the number 117 is a member of the arithmetic progression (a n):

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

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

d = -2.4 - (-3.6) = 1.2

So, we did the simplest thing. It remains to deal with the unknown number n and the incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know... What to do!? Well, 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.

Task based real option 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 V 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, State Examination or Unified State Examination, you forgot useful formula 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. Not very strictly, but for confidence and the right decision definitely enough!) To make a conclusion, it is 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 is 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 can do such a feat.) But the formula for 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 to solve any problems involving the formula of the nth term - everything is written out. 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.

For example, the sequence \(2\); \(5\); \(8\); \(eleven\); \(14\)... is an arithmetic progression, because each subsequent element differs from the previous one by three (can be obtained from the previous one by adding three):

In this progression, the difference \(d\) is positive (equal to \(3\)), and therefore each next term is greater than the previous one. Such progressions are called increasing.

However, \(d\) can also be negative number. For example, in arithmetic progression \(16\); \(10\); \(4\); \(-2\); \(-8\)... the progression difference \(d\) is equal to minus six.

And in this case, each next element will be smaller than the previous one. These progressions are called decreasing.

Arithmetic progression notation

Progression is indicated by a small Latin letter.

Numbers that form a progression are called members(or elements).

They are denoted by the same letter as an arithmetic progression, but with a numerical index equal to the number of the element in order.

For example, the arithmetic progression \(a_n = \left\( 2; 5; 8; 11; 14…\right\)\) consists of the elements \(a_1=2\); \(a_2=5\); \(a_3=8\) and so on.

In other words, for the progression \(a_n = \left\(2; 5; 8; 11; 14…\right\)\)

Solving arithmetic progression problems

In principle, the information presented above is already enough to solve almost any arithmetic progression problem (including those offered at the OGE).

Example (OGE). The arithmetic progression is specified by the conditions \(b_1=7; d=4\). Find \(b_5\).
Solution:

Answer: \(b_5=23\)

Example (OGE). The first three terms of an arithmetic progression are given: \(62; 49; 36…\) Find the value of the first negative term of this progression..
Solution:

	We are given the first elements of the sequence and know that it is an arithmetic progression. That is, each element differs from its neighbor by the same number. Let's find out which one by subtracting the previous one from the next element: \(d=49-62=-13\).
	Now we can restore our progression to the (first negative) element we need.
	Ready. You can write an answer.

Answer: \(-3\)

Example (OGE). Given several consecutive elements of an arithmetic progression: \(…5; x; 10; 12.5...\) Find the value of the element designated by the letter \(x\).
Solution:

	To find \(x\), we need to know how much the next element differs from the previous one, in other words, the progression difference. Let's find it from two known neighboring elements: \(d=12.5-10=2.5\).
	And now we can easily find what we are looking for: \(x=5+2.5=7.5\).
	Ready. You can write an answer.

Answer: \(7,5\).

Example (OGE). Arithmetic progression is given the following conditions: \(a_1=-11\); \(a_(n+1)=a_n+5\) Find the sum of the first six terms of this progression.
Solution:

We need to find the sum of the first six terms of the progression. But we do not know their meanings; we are given only the first element. Therefore, we first calculate the values one by one, using what is given to us:

\(n=1\); \(a_(1+1)=a_1+5=-11+5=-6\)
\(n=2\); \(a_(2+1)=a_2+5=-6+5=-1\)
\(n=3\); \(a_(3+1)=a_3+5=-1+5=4\)
And having calculated the six elements we need, we find their sum.

\(S_6=a_1+a_2+a_3+a_4+a_5+a_6=\)
\(=(-11)+(-6)+(-1)+4+9+14=9\)

The required amount has been found.

Answer: \(S_6=9\).

Example (OGE). In arithmetic progression \(a_(12)=23\); \(a_(16)=51\). Find the difference of this progression.
Solution:

Answer: \(d=7\).

Important formulas for arithmetic progression

As you can see, many problems on arithmetic progression can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each subsequent element in this chain is obtained by adding the same number to the previous one (the difference of the progression).

However, sometimes there are situations when deciding “head-on” is very inconvenient. For example, imagine that in the very first example we need to find not the fifth element \(b_5\), but the three hundred and eighty-sixth \(b_(386)\). Should we add four \(385\) times? Or imagine that in the penultimate example you need to find the sum of the first seventy-three elements. You'll be tired of counting...

Therefore, in such cases they do not solve things “head-on”, but use special formulas derived for arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum of \(n\) first terms.

Formula of the \(n\)th term: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first term of the progression;
\(n\) – number of the required element;
\(a_n\) – term of the progression with number \(n\).

This formula allows us to quickly find even the three-hundredth or the millionth element, knowing only the first and the difference of the progression.

Example. The arithmetic progression is specified by the conditions: \(b_1=-159\); \(d=8.2\). Find \(b_(246)\).
Solution:

Answer: \(b_(246)=1850\).

Formula for the sum of the first n terms: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where

\(a_n\) – the last summed term;

Example (OGE). The arithmetic progression is specified by the conditions \(a_n=3.4n-0.6\). Find the sum of the first \(25\) terms of this progression.
Solution:

\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\)		To calculate the sum of the first twenty-five terms, we need to know the value of the first and twenty-fifth terms. Our progression is given by the formula of the nth term depending on its number (for more details, see). Let's calculate the first element by substituting one for \(n\).
\(n=1;\) \(a_1=3.4·1-0.6=2.8\)		Now let's find the twenty-fifth term by substituting twenty-five instead of \(n\).
\(n=25;\) \(a_(25)=3.4·25-0.6=84.4\)		Well, now we can easily calculate the required amount.
\(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25=\) \(=\) \(\frac(2.8+84.4)(2)\) \(\cdot 25 =\)\(1090\)		The answer is ready.

Answer: \(S_(25)=1090\).

For the sum \(n\) of the first terms, you can get another formula: you just need to \(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25\ ) instead of \(a_n\) substitute the formula for it \(a_n=a_1+(n-1)d\). We get:

Formula for the sum of the first n terms: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where

\(S_n\) – the required sum of \(n\) first elements;
\(a_1\) – the first summed term;
\(d\) – progression difference;
\(n\) – number of elements in the sum.

Example. Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15.5\); \(14\)…
Solution:

Answer: \(S_(33)=-231\).

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;

a 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: the numerical value of 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 number sequence 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 is 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

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

Number sequence
For example, for our sequence:

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

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

In our case:

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

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

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

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

a)
b)
c)
d)

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

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

1. Method

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

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

2. Method

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

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

In other words:

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

Did you calculate? Compare your notes with the answer:

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

Arithmetic progression equation.

Arithmetic progressions can be increasing or decreasing.

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

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

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

Since then:

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

Let's compare the results:

Arithmetic progression property

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

Let, ah, then:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Method 1.

Method 2.

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

Training

Tasks:

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

Answers:

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

Let's sum it up

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

, where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

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

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

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

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

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

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

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

sets the sequence:

And the formula is the following sequence:

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

nth term formula

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

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

Well, is it clear now what the formula is?

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

Much more convenient now, right? We check:

Decide for yourself:

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

Solution:

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

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

So, the formula:

Then the hundredth term is equal to:

What is the sum of all natural numbers from to?

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

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

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

Solution:

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

Formula of the th term for this progression:

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

Very easy: .

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

Answer: .

Now decide for yourself:

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

Answers:

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

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

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

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

For example:

Formula for finding the nth term of an arithmetic progression

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

Property of members of an arithmetic progression

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

Sum of terms of an arithmetic progression

There are two ways to find the amount:

Where is the number of values.

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\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\)		The task is very similar to the previous one. We begin to solve the same thing: first we find \(d\).
\(d=a_2-a_1=-19-(-19.3)=0.3\)		Now I would like to substitute \(d\) into the formula for the sum... and here a small nuance emerges - we do not know \(n\). In other words, we don’t know how many terms will need to be added. How to find out? Let's think. We will stop adding elements when we reach the first positive element. That is, you need to find out the number of this element. How? Let's write down the formula for calculating any element of an arithmetic progression: \(a_n=a_1+(n-1)d\) for our case.
\(a_n=a_1+(n-1)d\)
\(a_n=-19.3+(n-1)·0.3\)		We need \(a_n\) to become greater than zero. Let's find out at what \(n\) this will happen.
\(-19.3+(n-1)·0.3>0\)
\((n-1)·0.3>19.3\) \(\|:0.3\)		We divide both sides of the inequality by \(0.3\).
\(n-1>\)\(\frac(19.3)(0.3)\)		We transfer minus one, not forgetting to change the signs
\(n>\)\(\frac(19.3)(0.3)\) \(+1\)		Let's calculate...
\(n>65,333…\)		...and it turns out that the first positive element will have the number \(66\). Accordingly, the last negative one has \(n=65\). Just in case, let's check this.
\(n=65;\) \(a_(65)=-19.3+(65-1)·0.3=-0.1\) \(n=66;\) \(a_(66)=-19.3+(66-1)·0.3=0.2\)		So we need to add the first \(65\) elements.
\(S_(65)=\) \(\frac(2 \cdot (-19.3)+(65-1)0.3)(2)\)\(\cdot 65\) \(S_(65)=\)\((-38.6+19.2)(2)\)\(\cdot 65=-630.5\)		The answer is ready.

\(a_1=-33;\) \(a_(n+1)=a_n+4\)	In this problem you also need to find the sum of elements, but starting not from the first, but from the \(26\)th. For such a case we do not have a formula. How to decide? It’s easy - to get the sum from the \(26\)th to the \(42\)th, you must first find the sum from the \(1\)th to the \(42\)th, and then subtract from it the sum from first to \(25\)th (see picture).
	For our progression \(a_1=-33\), and the difference \(d=4\) (after all, we add the four to the previous element to find the next one). Knowing this, we find the sum of the first \(42\)-y elements.
\(S_(42)=\) \(\frac(2 \cdot (-33)+(42-1)4)(2)\)\(\cdot 42=\) \(=\)\(\frac(-66+164)(2)\) \(\cdot 42=2058\)	Now the sum of the first \(25\) elements.
\(S_(25)=\) \(\frac(2 \cdot (-33)+(25-1)4)(2)\)\(\cdot 25=\) \(=\)\(\frac(-66+96)(2)\) \(\cdot 25=375\)	And finally, we calculate the answer.
\(S=S_(42)-S_(25)=2058-375=1683\)

How to find an arithmetic progression formula. Arithmetic progression: what is it? Each group presents the solution to the problem on the board

Application of the formula for the nth term of an arithmetic progression.

If you like this site...

Arithmetic progression notation

Progression is indicated by a small Latin letter.

Solving arithmetic progression problems

Important formulas for arithmetic progression

Formula of the \(n\)th term: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first term of the progression;
\(n\) – number of the required element;
\(a_n\) – term of the progression with number \(n\).

Formula for the sum of the first n terms: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where

Formula for the sum of the first n terms: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where

More complex arithmetic progression problems

Mathematical number sequence

Definition

Specified member value

An example of calculating the value of a given term

Sum of a given number of terms

Calculation example

Example of practical application of arithmetic progression

Another type of number sequence is geometric

Arithmetic progression property

Training

Let's sum it up

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

nth term formula

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

Formula for finding the nth term of an arithmetic progression

Property of members of an arithmetic progression

Sum of terms of an arithmetic progression

THE REMAINING 2/3 ARTICLES ARE AVAILABLE ONLY TO YOUCLEVER STUDENTS!

Latest

You need to know this

How to find an arithmetic progression formula. Arithmetic progression: what is it? Each group presents the solution to the problem on the board

Application of the formula for the nth term of an arithmetic progression.

If you like this site...

Arithmetic progression notation

Progression is indicated by a small Latin letter.

Solving arithmetic progression problems

Important formulas for arithmetic progression

Formula of the \(n\)th term: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first term of the progression; \(n\) – number of the required element; \(a_n\) – term of the progression with number \(n\).

Formula for the sum of the first n terms: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where

Formula for the sum of the first n terms: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where

More complex arithmetic progression problems

Mathematical number sequence

Definition

Specified member value

An example of calculating the value of a given term

Sum of a given number of terms

Calculation example

Example of practical application of arithmetic progression

Another type of number sequence is geometric

Arithmetic progression property

Training

Let's sum it up

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

nth term formula

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

Formula for finding the nth term of an arithmetic progression

Property of members of an arithmetic progression

Sum of terms of an arithmetic progression

THE REMAINING 2/3 ARTICLES ARE AVAILABLE ONLY TO YOUCLEVER STUDENTS!

Latest

You need to know this

Formula of the \(n\)th term: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first term of the progression;
\(n\) – number of the required element;
\(a_n\) – term of the progression with number \(n\).