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, V 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). The arithmetic progression is defined by 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 total.

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

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

More complex arithmetic progression problems

Now you have everything necessary information for solving almost any arithmetic progression problem. Let’s finish the topic by considering problems in which you not only need to 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: \(-19.3\); \(-19\); \(-18.7\)…
Solution:

\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\)		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 we 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.

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

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

\(a_1=-33;\) \(a_(n+1)=a_n+4\)	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, it is the four that we add 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.
\(S=S_(42)-S_(25)=2058-375=1683\)

Answer: \(S=1683\).

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

Or arithmetic is a type of ordered numerical sequence, the properties of which are studied in school course algebra. This article discusses in detail the question of how to find the sum of an arithmetic progression.

What kind of progression is this?

Before moving on to the question (how to find the sum of an arithmetic progression), it is worth understanding what we are talking about.

Any sequence of real numbers that is obtained by adding (subtracting) some value from each previous number is called an algebraic (arithmetic) progression. This definition, when translated into mathematical language, takes the form:

Here i - serial number element of the series a i. Thus, knowing just one starting number, you can easily restore the entire series. The parameter d in the formula is called the progression difference.

It can be easily shown that for the series of numbers under consideration the following equality holds:

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

That is, to find the value of the nth element in order, you should add the difference d to the first element a 1 n-1 times.

What is the sum of an arithmetic progression: formula

Before giving the formula for the indicated amount, it is worth considering a simple special case. The progression is given natural numbers from 1 to 10, you need to find their sum. Since there are few terms in the progression (10), it is possible to solve the problem head-on, that is, sum all the elements in order.

S 10 = 1+2+3+4+5+6+7+8+9+10 = 55.

One thing worth considering interesting thing: since each term differs from the next one by the same value d = 1, then the pairwise summation of the first with the tenth, the second with the ninth, and so on will give the same result. Really:

11 = 1+10 = 2+9 = 3+8 = 4+7 = 5+6.

As you can see, there are only 5 of these sums, that is, exactly two times less than the number of elements of the series. Then multiplying the number of sums (5) by the result of each sum (11), you will arrive at the result obtained in the first example.

If we generalize these arguments, we can write the following expression:

S n = n * (a 1 + a n) / 2.

This expression shows that it is not at all necessary to sum all the elements in a row; it is enough to know the value of the first a 1 and the last a n , as well as total number n terms.

It is believed that Gauss was the first to think of this equality when he was looking for a solution to a given problem. school teacher task: sum the first 100 integers.

Sum of elements from m to n: formula

The formula given in the previous paragraph answers the question of how to find the sum of an arithmetic progression (the first elements), but often in problems it is necessary to sum a series of numbers in the middle of the progression. How to do it?

The easiest way to answer this question is by considering the following example: let it be necessary to find the sum of terms from the mth to the nth. To solve the problem, you should present the given segment from m to n of the progression in the form of a new number series. In such m-th representation the term a m will be the first, and a n will be numbered n-(m-1). In this case, applying the standard formula for the sum, the following expression will be obtained:

S m n = (n - m + 1) * (a m + a n) / 2.

Example of using formulas

Knowing how to find the sum of an arithmetic progression, it is worth considering a simple example of using the above formulas.

Below is a numerical sequence, you should find the sum of its terms, starting from the 5th and ending with the 12th:

The given numbers indicate that the difference d is equal to 3. Using the expression for the nth element, you can find the values ​​of the 5th and 12th terms of the progression. It turns out:

a 5 = a 1 + d * 4 = -4 + 3 * 4 = 8;

a 12 = a 1 + d * 11 = -4 + 3 * 11 = 29.

Knowing the values ​​of the numbers at the ends of the given algebraic progression, and also knowing what numbers in the row they occupy, you can use the formula for the amount obtained in the previous paragraph. It will turn out:

S 5 12 = (12 - 5 + 1) * (8 + 29) / 2 = 148.

It is worth noting that this value could be obtained differently: first find the sum of the first 12 elements using the standard formula, then calculate the sum of the first 4 elements using the same formula, then subtract the second from the first sum.

First level

Arithmetic progression. Detailed theory with examples (2019)

Number sequence

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

Number sequence
For example, for our sequence:

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

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

In our case:

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

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

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

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

a)
b)
c)
d)

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

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

1. Method

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

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

2. Method

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

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


In other words:

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

Did you calculate? Compare your notes with the answer:

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

Arithmetic progression equation.

Arithmetic progressions can be increasing or decreasing.

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

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

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

Since then:

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

Let's compare the results:

Arithmetic progression property

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

Let, ah, then:

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

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

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

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

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

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

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

When Carl Gauss was 9 years old, a teacher, busy checking the work of students in other classes, asked 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 on your own what the sum of the numbers starting from the th is equal to and the sum of the numbers starting from the th.

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

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

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


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

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

Method 1.

Method 2.

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

Training

Tasks:

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

Answers:

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

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

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

   Numbers do contain odd numbers.
   Let's substitute the available data into the formula:

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

3. Let's remember the problem about pyramids. For our case, a , since each top layer is reduced by one log, then in total there are a bunch of layers, that is.
   Let's substitute the data into the formula:

   Answer: There are logs in the masonry.

Let's sum it up

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

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

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

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

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

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

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

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

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

sets the sequence:

And the formula is the following sequence:

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

nth term formula

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

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

Well, is it clear now what the formula is?

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

Much more convenient now, right? We check:

Decide for yourself:

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

Solution:

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

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

So, the formula:

Then the hundredth term is equal to:

What is the sum of all natural numbers from to?

According to legend, the great mathematician Carl Gauss, as a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last date is equal, the sum of the second and the 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 in total? That's right, exactly half the number of all numbers, that is. So,

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

Example:
Find the sum of all double digit numbers, multiples.

Solution:

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

Formula of the th term for this progression:

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

Very easy: .

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

Answer: .

Now decide for yourself:

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

Answers:

1. The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
   .
   Answer:
2. Here it is given: , must be found.
   Obviously, you need to use the same sum formula as in the previous problem:
   .
   Substitute the values:

   The root obviously doesn't fit, so the answer is.
   Let's calculate the path traveled over the last day using the formula of the th term:
   (km).
   Answer:

3. Given: . Find: .
   It couldn't be simpler:
   (rub).
   Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

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

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

For example:

Formula for finding the nth term of an arithmetic progression

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

Property of members of an arithmetic progression

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

Sum of terms of an arithmetic progression

There are two ways to find the amount:

Where is the number of values.

Where is the number of values.

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

What is an arithmetic progression?

To understand this, it is necessary to define the progression in question, as well as provide the basic formulas that will be used later in solving problems.

Arithmetic or is 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 give an example. The following 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 a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).

Important Formulas

Let us now present the basic formulas that will be needed to solve problems using arithmetic progression. Let us denote by the symbol a n nth term sequences where n is an integer. We denote the difference Latin letter d. Then the following expressions are valid:

1. To determine the value of the nth term, the following formula is suitable: a n = (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 arithmetic progression with solutions in 9th grade, it is enough to remember these two formulas, since any problems of the type under consideration are based on their use. You should also remember that the progression difference is determined by the formula: d = a n - a n-1.

Example #1: finding an unknown member

Let's give a simple example of an arithmetic progression and the formulas that need to be used to solve it.

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

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

1. Let's first calculate the difference. We have: d = 8 - 10 = -2. Similarly, you could take any two other members standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d = a n - a n-1, then d = a 5 - a 4, from which we get: a 5 = a 4 + d. Let's 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 = (n - 1) * d + a 1 = (n - 1) * (-2) + 10 = 12 - 2*n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.

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

Example #2: progression difference

Now let's complicate the task a little, 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 . Let's substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 = 6 + 6 * d. From this expression you can easily calculate the difference: d = (18 - 6) /6 = 2. Thus, we have answered the first part of the problem.

To restore the sequence to the 7th term, you should use the definition of an 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, a 7 = 18.

Example No. 3: drawing up a progression

Let's complicate the problem even more. Now we 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 create an algebraic progression so that three more terms are placed between these.

Before you start solving this problem, you need 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 = -4 and a 5 = 5. Having established this, we move on to the problem, which is similar to the previous one. Again, for the nth term we use the formula, we get: a 5 = a 1 + 4 * d. From: d = (a 5 - a 1)/4 = (5 - (-4)) / 4 = 2.25. What we got here is 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 terms 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 = 2.75 + 2.25 = 5, which coincided with the conditions of the problem.

Example No. 4: first term of progression

Let's continue to give examples of arithmetic progression with solutions. In all previous problems, the first number of the algebraic progression was known. Now let's 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 which number this sequence begins with.

The formulas used so far assume knowledge of a 1 and d. In the problem statement, nothing is known about these numbers. Nevertheless, we will write down expressions for each term about which information is available: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We received 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 easiest way to solve this system is to 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 = a 43 - 42 * d = 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d = 37 - 42 * d, whence the difference d = (37 - 50) / (42 - 14) = - 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 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.

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

Example No. 5: amount

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

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

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

It is interesting to note that this problem is called “Gaussian” because at the beginning of the 18th century the famous German, still only 10 years old, was able to solve it in his head 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 the numbers at the ends of the sequence in pairs, 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 No. 6: sum of terms from n to m

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

The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them sequentially. Since there are few terms, this method is not quite labor-intensive. Nevertheless, it is proposed to solve this problem using a second method, which is more universal.

The idea is to obtain a formula for the sum of the 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 = m * (a m + a 1) / 2.
2. S n = n * (a n + a 1) / 2.

Since n > m, it is obvious that the 2nd sum includes the first. 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), we will obtain the necessary answer to the problem. We have: S mn = S n - S m + a m =n * (a 1 + a n) / 2 - m *(a 1 + a m)/2 + a m = 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 knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before starting to solve any of these problems, it is recommended that you carefully read the condition, clearly understand what you need to find, and only then proceed with the solution.

Another tip is to strive for simplicity, that is, if you can answer a question without using complex mathematical calculations, then you need to do just that, since in this case the likelihood 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 = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and divide the overall problem into separate subtasks (in this case, first find the terms a n and a m).

If you have doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. We found out how to find an arithmetic progression. If you figure it out, it's not that difficult.