First level

Arithmetic progression. Detailed theory with examples (2019)

Numeric sequence

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

Numeric sequence
For example, for our sequence:

The assigned number is specific to only one sequence number. 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 the number is called the -th member of the sequence.

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

In our case:

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

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

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

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

a)
b)
c)
d)

Got it? Compare our answers:
Is an arithmetic progression - b, c.
Is not arithmetic progression - a, d.

Back to given progression() and try to find the value of its th member. Exist two way to find it.

1. Method

We can add to the previous value of the progression number 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 member of the described arithmetic progression is equal to.

2. Way

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

For example, let's see what makes up the value of the -th member of this arithmetic progression:


In other words:

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

Calculated? Compare your entries with the answer:

Pay attention that you got exactly the same number as in the previous method, when we successively added the members of an arithmetic progression to the previous value.
Let's try to "depersonalize" this formula- bring her to general form and get:

Arithmetic progression equation.

Arithmetic progressions are either 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 it out in practice.
We are given an arithmetic progression consisting of following numbers: Let's check what the -th number of this arithmetic progression will turn out if we use our formula when calculating it:

Since then:

Thus, we were convinced that the formula works both in decreasing and in increasing arithmetic progression.
Try to find the -th and -th members of this arithmetic progression on your own.

Let's compare the results:

Arithmetic progression property

Let's complicate the task - we derive the property of an arithmetic progression.
Suppose we are given the following condition:
- arithmetic progression, find the value.
It's easy, you say, and start counting according to the formula you already know:

Let, a, 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 mistakes in the calculations.
Now think, is it possible to solve this problem in one step using any formula? Of course, yes, and we will try to bring it out now.

Let's denote the desired term of the arithmetic progression as, we know the formula for finding it - this is the same formula that we derived at the beginning:
, then:

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

Let's sum the previous and next members of the progression:

It turns out that the sum of the previous and subsequent members of the progression is twice the value of the member of the progression located between them. In other words, to find the value of the progression term with known previous and consecutive values, you need to add them up and divide by.

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

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

When Carl Gauss was 9 years old, the teacher, busy checking the work of students in other classes, asked the following task at the lesson: “Calculate the sum of all natural numbers from to (according to other sources up to) inclusive. What was the surprise of the teacher when one of his students (it was Karl Gauss) after a minute gave the correct answer to the task, while most of the classmates of the daredevil after long calculations received the wrong result ...

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

Let's depict the progression given to us. Look closely at the highlighted numbers and try to perform various mathematical operations with them.


Tried? What did you notice? Correctly! Their sums are equal

Now answer, how many such pairs will there be in the progression given to us? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two members of an arithmetic progression is equal, and similar equal pairs, we get that total amount 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 progression difference. Try to substitute in the sum formula, the formula of the th member.
What did you get?

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

How much did you get?
Gauss turned out that the sum of the terms is equal, and the sum of the terms. Is that how you decided?

In fact, the formula for the sum of members of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time witty people used the properties of an arithmetic progression with might and main.
For example, imagine Ancient Egypt and the largest construction site of that time - the construction of a pyramid ... The figure 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? Count how many blocks are needed to build one wall if block bricks are placed in the base. I hope you will not count by moving your finger across the monitor, do you remember the last formula and everything we said about arithmetic progression?

AT this case the progression looks like this:
Arithmetic progression difference.
The number of members of an arithmetic progression.
Let's substitute our data into the last formulas (we count the number of blocks in 2 ways).

Method 1.

Method 2.

And now you can also calculate on the monitor: compare the obtained values ​​​​with the number of blocks that are in our pyramid. Did it agree? Well done, you have mastered the sum of the th terms of an arithmetic progression.
Of course, you can’t build a pyramid from the 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:

Workout

Tasks:

1. Masha is getting in shape for the summer. Every day she increases the number of squats by. How many times will Masha squat in weeks if she did squats at the first workout.
2. What is the sum of all odd numbers contained in.
3. When storing logs, lumberjacks stack them in such a way that each top layer contains one less log than the previous one. How many logs are in one masonry, if the base of the masonry is logs.

Answers:

1. Let us define the parameters of the arithmetic progression. In this case
   (weeks = days).

   Answer: In two weeks, Masha should squat once a day.

2. First odd number, last number.
   Arithmetic progression difference.
   The number of odd numbers in - half, however, check this fact using the formula for finding the -th member of an arithmetic progression:

   The numbers do contain odd numbers.
   We substitute the available data into the formula:

   Answer: The sum of all odd numbers contained in is equal to.

3. Recall the problem about the pyramids. For our case, a , since each top layer is reduced by one log, there are only a bunch of layers, that is.
   Substitute the data in the formula:

   Answer: There are logs in the masonry.

Summing up

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

ARITHMETIC PROGRESSION. MIDDLE LEVEL

Numeric sequence

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

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

Numeric 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 only 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 whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .

It is very convenient if the -th member of the sequence can be given 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). Or (, difference).

nth term formula

We call recurrent a formula 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 such a formula, we have to calculate the previous nine. For example, let. Then:

Well, now it's clear what the formula is?

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

Much more comfortable now, right? We check:

Decide for yourself:

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

Decision:

The first term is equal. And what is the difference? And here's what:

(after all, it is called the difference because it is equal to the difference of successive members of the progression).

So the formula is:

Then the hundredth term is:

What is the sum of all natural numbers from to?

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

Decision:

The first such number is this. Each of the following is obtained by adding to previous number. Thus, the numbers of interest to us form an arithmetic progression with the first term and the difference.

The formula for the th term for this progression is:

How many terms are in the progression if they must all be two digits?

Very easy: .

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

Answer: .

Now decide for yourself:

1. Every day the athlete runs 1m more than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
2. A cyclist rides more miles each day than the previous one. On the first day he traveled km. How many days does he have to drive to cover a kilometer? How many kilometers will he travel on the last day of the journey?
3. The price of a refrigerator in the store is reduced by the same amount every year. Determine how much the price of a refrigerator decreased every year if, put up for sale for rubles, six years later it was sold for rubles.

Answers:

1. The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
   .
   Answer:
2. Here it is given:, it is necessary to find.
   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.
   Let's calculate the distance traveled over the last day using the formula of the -th term:
   (km).
   Answer:

3. Given: . To find: .
   It doesn't get easier:
   (rub).
   Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN

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

Arithmetic progression is increasing () and decreasing ().

For example:

The formula for finding the n-th member of an arithmetic progression

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

Property of members of an arithmetic progression

It makes it easy to find a member of the progression if its neighboring members are known - where is the number of numbers in the progression.

The sum of the members of an arithmetic progression

There are two ways to find the sum:

Where is the number of values.

Where is the number of values.

Online calculator.
Arithmetic progression solution.
Given: a n , d, n
Find: a 1

This math program finds \(a_1\) of an arithmetic progression based on user-specified numbers \(a_n, d \) and \(n \).
The numbers \(a_n\) and \(d \) can be specified not only as integers, but also as fractions. Moreover, fractional number can be entered as a decimal (\(2.5 \)) and as common fraction(\(-5\frac(2)(7) \)).

The program not only gives the answer to the problem, but also displays the process of finding a solution.

This online calculator can be useful for high school students general education schools in preparation for control work and exams, when testing knowledge before the exam, parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as soon as possible? homework math or algebra? In this case, you can also use our programs with a detailed solution.

Thus, you can carry out your own training and/or training their younger brothers or sisters, while the level of education in the field of tasks being solved increases.

If you are not familiar with the rules for entering numbers, we recommend that you familiarize yourself with them.

Rules for entering numbers

The numbers \(a_n\) and \(d \) can be specified not only as integers, but also as fractions.
The number \(n\) can only be a positive integer.

Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a dot or a comma.
For example, you can enter decimals so 2.5 or so 2.5

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative.

When you enter numeric fraction The numerator is separated from the denominator by a division sign: /
Input:
Result: \(-\frac(2)(3) \)

whole part separated from the fraction by an ampersand: &
Input:
Result: \(-1\frac(2)(3) \)

Enter numbers a n , d, n

Find a 1

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A bit of theory.

Numeric sequence

Numbering is often used in everyday practice. various items to indicate their order. For example, the houses on each street are numbered. In the library, reader's subscriptions are numbered and then arranged in the order of the assigned numbers in special file cabinets.

In a savings bank, by the number of the depositor's personal account, you can easily find this account and see what kind of deposit it has. Let there be a deposit of a1 rubles on account No. 1, a deposit of a2 rubles on account No. 2, etc. It turns out numerical sequence
a 1 , a 2 , a 3 , ..., a N
where N is the number of all accounts. Here, each natural number n from 1 to N is assigned a number a n .

Mathematics also studies infinite number sequences:
a 1 , a 2 , a 3 , ..., a n , ... .
The number a 1 is called the first member of the sequence, number a 2 - the second member of the sequence, number a 3 - the third member of the sequence etc.
The number a n is called nth (nth) member of the sequence, and the natural number n is its number.

For example, in the sequence of squares of natural numbers 1, 4, 9, 16, 25, ..., n 2 , (n + 1) 2 , ... and 1 = 1 is the first member of the sequence; and n = n 2 is nth member sequences; a n+1 = (n + 1) 2 is the (n + 1)th (en plus the first) member of the sequence. Often a sequence can be specified by the formula of its nth member. For example, the formula \(a_n=\frac(1)(n), \; n \in \mathbb(N) \) gives the sequence \(1, \; \frac(1)(2) , \; \frac( 1)(3) , \; \frac(1)(4) , \dots,\frac(1)(n) , \dots \)

Arithmetic progression

The length of a year is approximately 365 days. A more accurate value is \(365\frac(1)(4) \) days, so every four years an error of one day accumulates.

To account for this error, a day is added to every fourth year, and the elongated year is called a leap year.

For example, in the third millennium leap years the years are 2004, 2008, 2012, 2016, ... .

In this sequence, each member, starting from the second, is equal to the previous one, added with the same number 4. Such sequences are called arithmetic progressions.

Definition.
The numerical sequence a 1 , a 2 , a 3 , ..., a n , ... is called arithmetic progression, if for all natural n the equality
\(a_(n+1) = a_n+d, \)
where d is some number.

It follows from this formula that a n+1 - a n = d. The number d is called the difference arithmetic progression.

By definition of an arithmetic progression, we have:
\(a_(n+1)=a_n+d, \quad a_(n-1)=a_n-d, \)
where
\(a_n= \frac(a_(n-1) +a_(n+1))(2) \), where \(n>1 \)

Thus, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the two members adjacent to it. This explains the name "arithmetic" progression.

Note that if a 1 and d are given, then the remaining terms of the arithmetic progression can be calculated using the recursive formula a n+1 = a n + d. In this way, it is not difficult to calculate the first few terms of the progression, however, for example, for a 100, a lot of calculations will already be required. Usually, the nth term formula is used for this. According to the definition of an arithmetic progression
\(a_2=a_1+d, \)
\(a_3=a_2+d=a_1+2d, \)
\(a_4=a_3+d=a_1+3d\)
etc.
Generally,
\(a_n=a_1+(n-1)d, \)
as nth member arithmetic progression is obtained from the first term by adding (n-1) times the number d.
This formula is called formula of the nth member of an arithmetic progression.

The sum of the first n terms of an arithmetic progression

Let's find the sum of all natural numbers from 1 to 100.
We write this sum in two ways:
S = l + 2 + 3 + ... + 99 + 100,
S = 100 + 99 + 98 + ... + 2 + 1.
We add these equalities term by term:
2S = 101 + 101 + 101 + ... + 101 + 101.
There are 100 terms in this sum.
Therefore, 2S = 101 * 100, whence S = 101 * 50 = 5050.

Consider now an arbitrary arithmetic progression
a 1 , a 2 , a 3 , ..., a n , ...
Let S n be the sum of the first n terms of this progression:
S n \u003d a 1, a 2, a 3, ..., a n
Then the sum of the first n terms of an arithmetic progression is
\(S_n = n \cdot \frac(a_1+a_n)(2) \)

Since \(a_n=a_1+(n-1)d \), then replacing a n in this formula, we get another formula for finding the sums of the first n terms of an arithmetic progression:
\(S_n = n \cdot \frac(2a_1+(n-1)d)(2) \)

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For example, the sequence \(2\); \(5\); \(eight\); \(eleven\); \(14\)… is an arithmetic progression, because each next 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 \(16\); \(ten\); \(4\); \(-2\); \(-8\)… the progression difference \(d\) is equal to minus six.

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

Arithmetic progression notation

Progression is denoted by a small Latin letter.

The numbers that form a progression are called it members(or elements).

They are denoted by the same letter as the arithmetic progression, but with a numerical index equal to the element number 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 problems on an arithmetic progression

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

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

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..
Decision:

	We are given the first elements of the sequence and know that it is an arithmetic progression. That is, each element differs from the neighboring one by the same number. 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 desired (first negative) element.
	Ready. You can write an answer.

Answer: \(-3\)

Example (OGE). Several successive elements of an arithmetic progression are given: \(...5; x; 10; 12.5...\) Find the value of the element denoted by the letter \(x\).
Decision:

	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 find what we are looking for without any problems: \(x=5+2.5=7.5\).
	Ready. You can write an answer.

Answer: \(7,5\).

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

	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 ​​​​in turn, using the 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 requested 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.
Decision:

Answer: \(d=7\).

Important Arithmetic Progression Formulas

As you can see, many arithmetic progression problems can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each next 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 it is very inconvenient to solve "on the forehead". 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 \(b_(386)\). What is it, we \ (385 \) times to add four? Or imagine that in the penultimate example, you need to find the sum of the first seventy-three elements. Counting is confusing...

Therefore, in such cases, they do not solve “on the forehead”, 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 \(n\) of the first terms.

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

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

Example. The arithmetic progression is given by the conditions: \(b_1=-159\); \(d=8,2\). Find \(b_(246)\).
Decision:

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

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

\(a_n\) is the last summed term;

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

\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\)		To calculate the sum of the first twenty-five elements, we need to know the value of the first and twenty-fifth term. Our progression is given by the formula of the nth term depending on its number (see details). Let's compute the first element by replacing \(n\) with one.
\(n=1;\) \(a_1=3.4 1-0.6=2.8\)		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 calculate the required amount without any problems.
\(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25=\) \(=\) \(\frac(2,8+84,4)(2)\) \(\cdot 25 =\)\(1090\)		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:

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

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

Example. Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15,5\); \(fourteen\)…
Decision:

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

More complex arithmetic progression problems

Now you have all the necessary information for solving almost any problem on an arithmetic progression. Let's finish the topic by considering problems in which you need to not only apply formulas, but also think a little (in mathematics, this can be useful ☺)

Example (OGE). Find the sum of all negative terms of the progression: \(-19.3\); \(-nineteen\); \(-18.7\)…
Decision:

\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\)		The task is very similar to the previous one. We start solving the same way: first we find \(d\).
\(d=a_2-a_1=-19-(-19.3)=0.3\)		Now we would substitute \(d\) into the formula for the sum ... and here a small nuance pops up - we don't know \(n\). In other words, we do not know how many terms will need to be added. How to find out? Let's think. We will stop adding elements when we get to 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 Above zero. Let's find out for 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 signs
\(n>\)\(\frac(19,3)(0,3)\) \(+1\)		Computing...
\(n>65,333…\)		…and it turns out that the first positive element will have the number \(66\). Accordingly, the last negative has \(n=65\). Just in case, let's check it out.
\(n=65;\) \(a_(65)=-19.3+(65-1) 0.3=-0.1\) \(n=66;\) \(a_(66)=-19.3+(66-1) 0.3=0.2\)		Thus, 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.

Answer: \(S_(65)=-630.5\).

Example (OGE). The arithmetic progression is given by the conditions: \(a_1=-33\); \(a_(n+1)=a_n+4\). Find the sum from \(26\)th to \(42\) element inclusive.
Decision:

\(a_1=-33;\) \(a_(n+1)=a_n+4\)	In this problem, you also need to find the sum of elements, but starting not from the first, but from the \(26\)th. We don't have a formula for this. How to decide? Easy - to get the sum from \(26\)th to \(42\)th, you must first find the sum from \(1\)th to \(42\)th, and then subtract from it the sum from the first to \ (25 \) th (see picture).
	For our progression \(a_1=-33\), and the difference \(d=4\) (after all, we add four to the previous element to find the next one). Knowing this, we find the sum of the first \(42\)-uh 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 \(25\)-th 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.
\(S=S_(42)-S_(25)=2058-375=1683\)

Answer: \(S=1683\).

For an arithmetic progression, there are several more formulas that we have not considered in this article due to their low practical usefulness. However, you can easily find them.

Matrix A -1 is called the inverse matrix with respect to matrix A, if A * A -1 \u003d E, where E - identity matrix nth order. The inverse matrix can only exist for square matrices.

Service assignment. With this service, online mode one can find algebraic complements, the transposed matrix A T , the union matrix, and the inverse matrix. The solution is carried out directly on the site (online) and is free. The calculation results are presented in a report in Word format and in Excel format (that is, it is possible to check the solution). see design example.

Instruction. To obtain a solution, you must specify the dimension of the matrix. Next, in the new dialog box, fill in the matrix A .

Matrix dimension 2 3 4 5 6 7 8 9 10

See also Inverse Matrix by the Jordan-Gauss Method

Algorithm for finding the inverse matrix

1. Finding the transposed matrix A T .
2. Definition of algebraic additions. Replace each element of the matrix with its algebraic complement.
3. Drafting inverse matrix from algebraic additions: each element of the resulting matrix is ​​divided by the determinant of the original matrix. The resulting matrix is ​​the inverse of the original matrix.

Next inverse matrix algorithm similar to the previous one, except for some steps: first calculate algebraic additions, and then the union matrix C is determined.

1. Determine if the matrix is ​​square. If not, then there is no inverse matrix for it.
2. Calculation of the determinant of the matrix A . If it is not equal to zero, we continue the solution, otherwise, the inverse matrix does not exist.
3. Definition of algebraic additions.
4. Filling in the union (mutual, adjoint) matrix C .
5. Compilation of the inverse matrix from algebraic additions: each element of the adjoint matrix C is divided by the determinant of the original matrix. The resulting matrix is ​​the inverse of the original matrix.
6. Make a check: multiply the original and the resulting matrices. The result should be an identity matrix.

Example #1. We write the matrix in the form:

Algebraic additions.

A 1.1 = (-1) 1+1

-1 -2 5 4

∆ 1,1 = (-1 4-5 (-2)) = 6

A 1,2 = (-1) 1+2

2 -2 -2 4

∆ 1,2 = -(2 4-(-2 (-2))) = -4

A 1.3 = (-1) 1+3

2 -1 -2 5

∆ 1,3 = (2 5-(-2 (-1))) = 8

A 2.1 = (-1) 2+1

2 3 5 4

∆ 2,1 = -(2 4-5 3) = 7

A 2.2 = (-1) 2+2

-1 3 -2 4

∆ 2,2 = (-1 4-(-2 3)) = 2

A 2.3 = (-1) 2+3

-1 2 -2 5

∆ 2,3 = -(-1 5-(-2 2)) = 1

A 3.1 = (-1) 3+1

2 3 -1 -2

∆ 3,1 = (2 (-2)-(-1 3)) = -1

A 3.2 = (-1) 3+2

-1 3 2 -2

∆ 3,2 = -(-1 (-2)-2 3) = 4

A 3.3 = (-1) 3+3

-1 2 2 -1

∆ 3,3 = (-1 (-1)-2 2) = -3
Then inverse matrix can be written as:

A -1 = 1 / 10

6 -4 8 7 2 1 -1 4 -3

A -1 =

0,6 -0,4 0,8 0,7 0,2 0,1 -0,1 0,4 -0,3

Another algorithm for finding the inverse matrix

We present another scheme for finding the inverse matrix.

1. We find the determinant of this square matrix A.
2. We find algebraic additions to all elements of the matrix A .
3. We write the algebraic complements of the elements of the rows into the columns (transposition).
4. We divide each element of the resulting matrix by the determinant of the matrix A .

As you can see, the transposition operation can be applied both at the beginning, over the original matrix, and at the end, over the resulting algebraic additions.

A special case: The inverse, with respect to the identity matrix E , is the identity matrix E .

When studying algebra in general education school(Grade 9) one of important topics is the study number sequences, which include progressions - geometric and arithmetic. In this article, we will consider an arithmetic progression and examples with solutions.

What is an arithmetic progression?

To understand this, it is necessary to give a definition of the progression under consideration, as well as to give the basic formulas that will be further used in solving problems.

Arithmetic or is such a set of ordered rational numbers, each member of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of an ordered series of numbers and the difference, you can restore the entire arithmetic progression.

Let's take an example. The next sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the type of progression under consideration, since the difference for it is not constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).

Important Formulas

We now give the basic formulas that will be needed to solve problems using an arithmetic progression. Let a n denote the nth member of the sequence, where n is an integer. Let us denote the difference Latin letter d. Then the following expressions:

1. To determine the value of the nth term, the formula is suitable: a n \u003d (n-1) * d + a 1.
2. To determine the sum of the first n terms: S n = (a n + a 1)*n/2.

To understand any examples of an arithmetic progression with a solution in grade 9, it is enough to remember these two formulas, since any problems of the type in question are built on their use. Also, do not forget that the progression difference is determined by the formula: d = a n - a n-1 .

Example #1: Finding an Unknown Member

We give a simple example of an arithmetic progression and the formulas that must be used to solve.

Let the sequence 10, 8, 6, 4, ... be given, it is necessary to find five terms in it.

It already follows from the conditions of the problem that the first 4 terms are known. The fifth can be defined in two ways:

1. Let's calculate the difference first. We have: d = 8 - 10 = -2. Similarly, one could take any two other terms standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d \u003d a n - a n-1, then d \u003d a 5 - a 4, from where we get: a 5 \u003d a 4 + d. Substitute known values: a 5 = 4 + (-2) = 2.
2. The second method also requires knowledge of the difference of the progression in question, so you first need to determine it, as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for the n number of the sequence. We have: a n \u003d (n - 1) * d + a 1 \u003d (n - 1) * (-2) + 10 \u003d 12 - 2 * n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.

As you can see, both solutions lead to the same result. Note that in this example the difference d of the progression is negative value. Such sequences are called decreasing because each successive term is less than the previous one.

Example #2: progression difference

Now let's complicate the task a bit, give an example of how to find the difference of an arithmetic progression.

It is known that in some algebraic progression the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.

Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . We substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 \u003d 6 + 6 * d. From this expression, you can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, the first part of the problem was answered.

To restore a sequence up to 7 terms, one should use the definition algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2=8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16 and 7 = 18.

Example #3: making a progression

Let's make it harder stronger condition tasks. Now you need to answer the question of how to find an arithmetic progression. The following example can be given: two numbers are given, for example, 4 and 5. It is necessary to make an algebraic progression so that three more terms are placed between these.

Before starting to solve this problem, it is necessary to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 \u003d -4 and a 5 \u003d 5. Having established this, we proceed to a task that is similar to the previous one. Again, for the nth term, we use the formula, we get: a 5 \u003d a 1 + 4 * d. From: d \u003d (a 5 - a 1) / 4 \u003d (5 - (-4)) / 4 \u003d 2.25. Here we received not an integer value of the difference, but it is rational number, so the formulas for the algebraic progression remain the same.

Now let's add the found difference to a 1 and restore the missing members of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 \u003d 2.75 + 2.25 \u003d 5, which coincided with the condition of the problem.

Example #4: The first member of the progression

We continue to give examples of an arithmetic progression with a solution. In all previous problems, the first number of the algebraic progression was known. Now consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find from what number this sequence begins.

The formulas that have been used so far assume knowledge of a 1 and d. Nothing is known about these numbers in the condition of the problem. Nevertheless, let's write out the expressions for each term about which we have information: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We got two equations in which there are 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.

The specified system is easiest to solve if you express a 1 in each equation, and then compare the resulting expressions. First equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 \u003d a 43 - 42 * d \u003d 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d \u003d 37 - 42 * d, whence the difference d \u003d (37 - 50) / (42 - 14) \u003d - 0.464 (only 3 decimal places are given).

Knowing d, you can use any of the 2 expressions above for a 1 . For example, first: a 1 \u003d 50 - 14 * d \u003d 50 - 14 * (- 0.464) \u003d 56.496.

If there are doubts about the result, you can check it, for example, determine the 43rd member of the progression, which is specified in the condition. We get: a 43 \u003d a 1 + 42 * d \u003d 56.496 + 42 * (- 0.464) \u003d 37.008. A small error is due to the fact that rounding to thousandths was used in the calculations.

Example #5: Sum

Now let's look at some examples with solutions for the sum of an arithmetic progression.

Let it be given numerical progression the following kind: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?

Thanks to the development computer technology you can solve this problem, that is, sequentially add all the numbers, which Calculating machine will do as soon as the person presses the Enter key. However, the problem can be solved mentally if you pay attention that the presented series of numbers is an algebraic progression, and its difference is 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.

It is curious to note that this problem is called "Gaussian" because in early XVIII of the century, the famous German, still at the age of only 10 years old, was able to solve it in his mind in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add pairs of numbers located at the edges of the sequence, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer, it is enough to multiply 50 by 101.

Example #6: sum of terms from n to m

Another a typical example the sum of an arithmetic progression is as follows: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its members from 8 to 14 will be.

The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them up sequentially. Since there are few terms, this method is not laborious enough. Nevertheless, it is proposed to solve this problem by the second method, which is more universal.

The idea is to get a formula for the sum of an algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:

1. S m \u003d m * (a m + a 1) / 2.
2. S n \u003d n * (a n + a 1) / 2.

Since n > m, it is obvious that the 2 sum includes the first one. The last conclusion means that if we take the difference between these sums, and add the term a m to it (in the case of taking the difference, it is subtracted from the sum S n), then we get the necessary answer to the problem. We have: S mn \u003d S n - S m + a m \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m \u003d a 1 * (n - m) / 2 + a n * n / 2 + a m * (1- m / 2). It is necessary to substitute formulas for a n and a m into this expression. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d * (3 * m - m 2 - 2) / 2.

The resulting formula is somewhat cumbersome, however, the sum S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.

As can be seen from the above solutions, all problems are based on the knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before you start solving any of these problems, it is recommended that you carefully read the condition, clearly understand what you want to find, and only then proceed with the solution.

Another tip is to strive for simplicity, that is, if you can answer the question without using complex mathematical calculations, then you need to do just that, since in this case the probability of making a mistake is less. For example, in the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and split common task into separate subproblems (in this case, first find the terms a n and a m).

If there are doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. How to find an arithmetic progression, found out. Once you figure it out, it's not that hard.