How to find the sum of an arithmetic progression. The sum of the first n-terms of an arithmetic progression

Someone treats the word "progression" with caution, as a very complex term from the sections of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi counter (where they still remain). And to understand the essence (and in mathematics there is nothing more important than “to understand the essence”) of an arithmetic sequence is not so difficult, having analyzed a few elementary concepts.

Mathematical number sequence

It is customary to call a numerical sequence a series of numbers, each of which has its own number.

and 1 is the first member of the sequence;

and 2 is the second member of the sequence;

and 7 is the seventh member of the sequence;

and n is the nth member of the sequence;

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

a - value of a member of the numerical sequence;

n is its serial number;

f(n) is a function where the ordinal in the numeric 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 member of an arithmetic sequence is as follows:

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

a n+1 - the formula of the next number;

d - difference (a certain number).

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

In the graph below, it is easy to see why the number sequence is called "increasing".

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

The value of the specified member

Sometimes it is necessary to determine the value of some arbitrary term a n of an arithmetic progression. You can do this by calculating successively the values ​​of all members of the arithmetic progression, from the first to the desired one. However, this way is not always acceptable if, for example, it is necessary to find the value of the five thousandth or eight millionth term. The traditional calculation will take a long time. However, a specific arithmetic progression can be investigated using certain formulas. There is also a formula for the nth term: the value of any member of an arithmetic progression can be determined as the sum of the first member of the progression with the difference of the progression, multiplied by the number of the desired member, minus one.

The formula is universal for increasing and decreasing progression.

An example of calculating the value of a given member

Let's solve the following problem of finding the value of the n-th member of an arithmetic progression.

Condition: there is an arithmetic progression with parameters:

The first member of the sequence is 3;

The difference in the number series is 1.2.

Task: it is necessary to find the value of 214 terms

Solution: to determine the value of a given member, 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 member of the sequence is equal to 258.6.

The advantages of this calculation method 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 required to determine the sum of the values ​​of some of its segments. It also doesn't need to calculate the values ​​of each term and then sum them up. This method is applicable if the number of terms whose sum must be found is small. In other cases, it is more convenient to use the following formula.

The sum of the members of an arithmetic progression from 1 to n is equal to the sum of the first and nth members, multiplied by the member number n and divided by two. If in the formula the value of the n-th member 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.

In the problem, it is required to determine the sum of the terms of the series from 56 to 101.

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

s(n) = (2∙a1 + d∙(n-1))∙n/2

First, we determine the sum of the values ​​of 101 members 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

So the sum of the arithmetic progression for this example is:

s 101 - s 55 \u003d 2,525 - 742.5 \u003d 1,782.5

Example of practical application of arithmetic progression

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

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

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

30 - 3 = 27 km.

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

The member number is 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 p.

the number of interest to us - the value of the (27 + 1)th member of the arithmetic progression - the meter reading at the end of the 27th kilometer - 27.999 ... = 28 km.

a 28 \u003d 50 + 22 ∙ (28 - 1) \u003d 644

Calculations of calendar data 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 luminary. In addition, various numerical series are successfully used in statistics and other applied branches of mathematics.

Another kind of number sequence is geometric

A geometric progression is characterized by a large, compared with an arithmetic, rate of change. It is no coincidence that in politics, sociology, medicine, often, in order to show the high speed of the spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops exponentially.

The N-th member 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 member is 1, the denominator is 2, respectively, 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 member of the geometric progression;

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

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

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

As in the case of arithmetic, a geometric progression has a formula for the value of an arbitrary member. Any n-th 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. Find the 5th term of the progression

b 5 \u003d b 1 ∙ q (5-1) \u003d 3 ∙ 1.5 4 \u003d 15.1875

The sum of a given number of members is also calculated using a special formula. The sum of the first n members of a geometric progression is equal to the difference between the product of the nth member of the progression and its denominator and the first member 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 members of the considered number series will take the form:

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

s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280

When studying algebra in a secondary school (grade 9), one of the important topics is the study of numerical 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.

An arithmetic or algebraic progression is such a set of ordered rational numbers, each member of which differs from the previous one by some constant amount. 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 considered type of progression, since the difference for it is not a 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. The difference is denoted by the Latin letter d. Then the following expressions are true:

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 under consideration 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. We substitute the 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. 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 little, give an example of how

It is known that in some 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 the sequence to the 7th member, 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 and 7 = 18.

Example #3: making a progression

Let us complicate the condition of the problem even more. Now you need to answer the question of how to find an arithmetic progression. We can give the following example: two numbers are given, for example, 4 and 5. It is necessary to make an algebraic progression so that three more terms fit 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, the difference is not an integer value, but it is a 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 a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?

Thanks to the development of computer technology, this problem can be solved, that is, sequentially add up all the numbers, which the computer will do as soon as a 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", since at the beginning of the 18th 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 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.

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 = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and break the general task into separate subtasks (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.

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 a negative number. For example, in arithmetic progression \(16\); \(ten\); \(four\); \(-2\); \(-8\)… 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 \(b_5\).
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:

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 \(x\).
Solution:

Answer: \(7,5\).

Example (OGE). The arithmetic progression is given 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:

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 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 \(b_(246)\).
Solution:

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.
Solution:

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: \(17\); \(15,5\); \(fourteen\)…
Solution:

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

More complex arithmetic progression problems

Now you have all the information you need to solve almost any arithmetic progression problem. 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: \(-19.3\); \(-19\); \(-18.7\)…
Solution:

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.
Solution:

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.

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 broader sense as an endless numerical 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 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 member. Exists 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 - we bring it into a general form and get:

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 the following numbers:

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, in order to find the value of a progression member with known previous and successive values, it is necessary to add them 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 from other classes, asked the following task at the lesson: "Calculate the sum of all natural numbers from up 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 terms of an arithmetic progression is equal, and similar equal pairs, we get 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 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?

In 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. AVERAGE 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.

Solution:

The first member 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 legend, the 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 multiples.

Solution:

The first such number is this. Each next is obtained by adding a number to the previous one. 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 member:
   (km).
   Answer:

3. Given: . 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.

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Lesson type: learning new material.

Lesson Objectives:

• expansion and deepening of students' ideas about tasks solved using arithmetic progression; organization of search activity of students when deriving the formula for the sum of the first n members of an arithmetic progression;
• development of skills to independently acquire new knowledge, use already acquired knowledge to achieve the task;
• development of the desire and need to generalize the facts obtained, the development of independence.

Tasks:

• generalize and systematize the existing knowledge on the topic “Arithmetic progression”;
• derive formulas for calculating the sum of the first n members of an arithmetic progression;
• teach how to apply the obtained formulas in solving various problems;
• draw students' attention to the procedure for finding the value of a numerical expression.

Equipment:

• cards with tasks for work in groups and pairs;
• evaluation paper;
• presentation"Arithmetic progression".

I. Actualization of basic knowledge.

1. Independent work in pairs.

1st option:

Define an arithmetic progression. Write down a recursive formula that defines an arithmetic progression. Give an example of an arithmetic progression and indicate its difference.

2nd option:

Write down the formula for the nth term of an arithmetic progression. Find the 100th term of an arithmetic progression ( a n}: 2, 5, 8 …
At this time, two students on the back of the board are preparing answers to the same questions.
Students evaluate the partner's work by comparing it with the board. (Leaflets with answers are handed over).

2. Game moment.

Exercise 1.

Teacher. I conceived some arithmetic progression. Ask me only two questions so that after the answers you can quickly name the 7th member of this progression. (1, 3, 5, 7, 9, 11, 13, 15…)

Questions from students.

1. What is the sixth term of the progression and what is the difference?
2. What is the eighth term of the progression and what is the difference?

If there are no more questions, then the teacher can stimulate them - a “ban” on d (difference), that is, it is not allowed to ask what the difference is. You can ask questions: what is the 6th term of the progression and what is the 8th term of the progression?

Task 2.

There are 20 numbers written on the board: 1, 4, 7 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58.

The teacher stands with his back to the blackboard. The students say the number of the number, and the teacher immediately calls the number itself. Explain how I can do it?

The teacher remembers the formula of the nth term a n \u003d 3n - 2 and, substituting the given values ​​of n, finds the corresponding values a n .

II. Statement of the educational task.

I propose to solve an old problem dating back to the 2nd millennium BC, found in Egyptian papyri.

A task:“Let it be said to you: divide 10 measures of barley between 10 people, the difference between each person and his neighbor is 1/8 of the measure.”

• How does this problem relate to the topic of arithmetic progression? (Each next person gets 1/8 of the measure more, so the difference is d=1/8, 10 people, so n=10.)
• What do you think the number 10 means? (The sum of all members of the progression.)
• What else do you need to know to make it easy and simple to divide barley according to the condition of the problem? (The first term of the progression.)

Lesson objective- obtaining the dependence of the sum of the terms of the progression on their number, the first term and the difference, and checking whether the problem was solved correctly in ancient times.

Before deriving the formula, let's see how the ancient Egyptians solved the problem.

And they solved it like this:

1) 10 measures: 10 = 1 measure - average share;
2) 1 measure ∙ = 2 measures - doubled average share.
doubled average the share is the sum of the shares of the 5th and 6th person.
3) 2 measures - 1/8 measure = 1 7/8 measures - twice the share of the fifth person.
4) 1 7/8: 2 = 5/16 - the share of the fifth; and so on, you can find the share of each previous and subsequent person.

We get the sequence:

III. The solution of the task.

1. Work in groups

1st group: Find the sum of 20 consecutive natural numbers: S 20 \u003d (20 + 1) ∙ 10 \u003d 210.

In general

II group: Find the sum of natural numbers from 1 to 100 (Legend of Little Gauss).

S 100 \u003d (1 + 100) ∙ 50 \u003d 5050

Conclusion:

III group: Find the sum of natural numbers from 1 to 21.

Solution: 1+21=2+20=3+19=4+18…

Conclusion:

IV group: Find the sum of natural numbers from 1 to 101.

Conclusion:

This method of solving the considered problems is called the “Gauss method”.

2. Each group presents the solution to the problem on the board.

3. Generalization of the proposed solutions for an arbitrary arithmetic progression:

a 1 , a 2 , a 3 ,…, a n-2 , a n-1 , a n .
S n \u003d a 1 + a 2 + a 3 + a 4 + ... + a n-3 + a n-2 + a n-1 + a n.

We find this sum by arguing similarly:

4. Have we solved the task?(Yes.)

IV. Primary comprehension and application of the obtained formulas in solving problems.

1. Checking the solution of an old problem by the formula.

2. Application of the formula in solving various problems.

3. Exercises for the formation of the ability to apply the formula in solving problems.

A) No. 613

Given :( and n) - arithmetic progression;

(a n): 1, 2, 3, ..., 1500

Find: S 1500

Solution: , and 1 = 1, and 1500 = 1500,

B) Given: ( and n) - arithmetic progression;
(and n): 1, 2, 3, ...
S n = 210

Find: n
Solution:

V. Independent work with mutual verification.

Denis went to work as a courier. In the first month, his salary was 200 rubles, in each subsequent month it increased by 30 rubles. How much did he earn in a year?

Given :( and n) - arithmetic progression;
a 1 = 200, d=30, n=12
Find: S 12
Solution:

Answer: Denis received 4380 rubles for the year.

VI. Homework instruction.

1. p. 4.3 - learn the derivation of the formula.
2. №№ 585, 623 .
3. Compose a problem that would be solved using the formula for the sum of the first n terms of an arithmetic progression.

VII. Summing up the lesson.

1. Score sheet

2. Continue the sentences

• Today in class I learned...
• Learned Formulas...
• I think that …

3. Can you find the sum of numbers from 1 to 500? What method will you use to solve this problem?

Bibliography.

1. Algebra, 9th grade. Textbook for educational institutions. Ed. G.V. Dorofeeva. Moscow: Enlightenment, 2009.