For example, the sequence \(2\); \(5\); \(8\); \(eleven\); \(14\)… 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 \(16\); \(10\); \(4\); \(-2\); \(-8\)… 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 \(b_5\).
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 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 \(x\).
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 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). 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:

	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.
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:

\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\)		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: \(17\); \(15,5\); \(14\)…
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:

\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\)		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 be greater than 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.
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. 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.

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.

What is the essence of the formula?

This formula allows you to find any BY HIS NUMBER" n" .

Of course, you need to know the first term a 1 and progression difference d, well, without these parameters, you can’t write down a specific progression.

It is not enough to memorize (or cheat) this formula. It is necessary to assimilate its essence and apply the formula in various problems. Yes, and do not forget at the right time, yes ...) How not forget- I don't know. And here how to remember If needed, I'll give you a hint. For those who master the lesson to the end.)

So, let's deal with the formula of the n-th member of an arithmetic progression.

What is a formula in general - we imagine.) What is an arithmetic progression, a member number, a progression difference - is clearly stated in the previous lesson. Take a look if you haven't read it. Everything is simple there. It remains to figure out what nth member.

The progression in general can be written as a series of numbers:

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

a 1- denotes the first term of an arithmetic progression, a 3- third member a 4- fourth, and so on. If we are interested in the fifth term, let's say we are working with a 5, if one hundred and twentieth - from a 120.

How to define in general any member of an arithmetic progression, s any number? Very simple! Like this:

a n

That's what it is n-th member of an arithmetic progression. Under the letter n all the numbers of members are hidden at once: 1, 2, 3, 4, and so on.

And what does such a record give us? Just think, instead of a number, they wrote down a letter ...

This notation gives us a powerful tool for working with arithmetic progressions. Using the notation a n, we can quickly find any member any arithmetic progression. And a bunch of tasks to solve in progression. You will see further.

In the formula of the nth member of an arithmetic progression:

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

a 1- the first member of the arithmetic progression;

n- member number.

The formula links the key parameters of any progression: a n ; a 1 ; d And n. Around these parameters, all the puzzles revolve in progression.

The nth term formula can also be used to write a specific progression. For example, in the problem it can be said that the progression is given by the condition:

a n = 5 + (n-1) 2.

Such a problem can even confuse ... There is no series, no difference ... But, comparing the condition with the formula, it is easy to figure out that in this progression a 1 \u003d 5, and d \u003d 2.

And it can be even angrier!) If we take the same condition: a n = 5 + (n-1) 2, yes, open the brackets and give similar ones? We get a new formula:

an = 3 + 2n.

This Only not general, but for a specific progression. This is where the pitfall lies. Some people think that the first term is a three. Although in reality the first member is a five ... A little lower we will work with such a modified formula.

In tasks for progression, there is another notation - a n+1. This is, you guessed it, the "n plus the first" term of the progression. Its meaning is simple and harmless.) This is a member of the progression, the number of which is greater than the number n by one. For example, if in some problem we take for a n fifth term, then a n+1 will be the sixth member. Etc.

Most often the designation a n+1 occurs in recursive formulas. Do not be afraid of this terrible word!) This is just a way of expressing a term of an arithmetic progression through the previous one. Suppose we are given an arithmetic progression in this form, using the recurrent formula:

a n+1 = a n +3

a 2 = a 1 + 3 = 5+3 = 8

a 3 = a 2 + 3 = 8+3 = 11

The fourth - through the third, the fifth - through the fourth, and so on. And how to count immediately, say the twentieth term, a 20? But no way!) While the 19th term is not known, the 20th cannot be counted. This is the fundamental difference between the recursive formula and the formula of the nth term. Recursive works only through previous term, and the formula of the nth term - through first and allows straightaway find any member by its number. Not counting the whole series of numbers in order.

In an arithmetic progression, a recursive formula can easily be turned into a regular one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write the formula in the usual form, and work with it. In the GIA, such tasks are often found.

Application of the formula of the n-th member of an arithmetic progression.

First, let's look at the direct application of the formula. At the end of the previous lesson there was a problem:

Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.

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

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

The conditions provide all the data for using the formula: a 1 \u003d 3, d \u003d 1/6. It remains to be seen what n. No problem! We need to find a 121. Here we write:

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

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

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

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

Let's solve the problem smarter. Let's say we have the following problem:

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

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

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

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

We also have a number n! In the condition a 17 =-2 hidden two options. This is both the value of the seventeenth member (-2) and its number (17). Those. n=17. This "little thing" often slips past the head, and without it, (without the "little thing", not the head!) The problem cannot be solved. Although ... and without a head too.)

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

a 17 \u003d a 1 + (17-1) (-0.5)

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

-2 \u003d a 1 + (17-1) (-0.5)

That, in essence, is all. It remains to express the first term of the arithmetic progression from the formula, and calculate. You get the answer: a 1 = 6.

Such a technique - writing a formula and simply substituting known data - helps a lot in simple tasks. Well, you must, of course, be able to express a variable from a formula, but what to do!? Without this skill, mathematics can not be studied at all ...

Another popular problem:

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

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

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

Consider what we know: a 1 =2; a 15 =12; and (special highlight!) n=15. Feel free to substitute in the formula:

12=2 + (15-1)d

Let's do the arithmetic.)

12=2 + 14d

d=10/14 = 5/7

This is the correct answer.

So, tasks a n , a 1 And d decided. It remains to learn how to find the number:

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

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

a n = 12 + (n-1) 3

At first glance, there are two unknown quantities here: a n and n. But a n is some member of the progression with the number n... And this member of the progression we know! It's 99. We don't know his number. n, so this number also needs to be found. Substitute the progression term 99 into the formula:

99 = 12 + (n-1) 3

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

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

Determine if the number 117 will be a member of an arithmetic progression (a n):

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

Let's write the formula again. What, there are no options? Hm... Why do we need eyes?) Do we see the first member of the progression? We see. This is -3.6. You can safely write: a 1 \u003d -3.6. Difference d can be determined from the series? It's easy if you know what the difference of an arithmetic progression is:

d = -2.4 - (-3.6) = 1.2

Yes, we did the simplest thing. It remains to deal with an unknown number n and an incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know that ... How to be!? Well, how to be, how to be... Turn on your creative abilities!)

We suppose that 117 is, after all, a member of our progression. With an unknown number n. And, just like in the previous problem, let's try to find this number. Those. we write the formula (yes-yes!)) and substitute our numbers:

117 = -3.6 + (n-1) 1.2

Again we express from the formulan, we count and get:

Oops! The number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion do we draw? Yes! Number 117 is not member of our progression. It is somewhere between the 101st and 102nd member. If the number turned out to be natural, i.e. positive integer, then the number would be a member of the progression with the found number. And in our case, the answer to the problem will be: No.

Task based on a real version of the GIA:

The arithmetic progression is given by the condition:

a n \u003d -4 + 6.8n

Find the first and tenth terms of the progression.

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

We are looking for the first member. The one who thinks. that the first term is minus four, is fatally mistaken!) Because the formula in the problem is modified. The first term of an arithmetic progression in it hidden. Nothing, we'll find it now.)

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

a 1 \u003d -4 + 6.8 1 \u003d 2.8

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

Similarly, we are looking for the tenth term:

a 10 \u003d -4 + 6.8 10 \u003d 64

That's all there is to it.

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

Suppose, in a difficult combat situation of the GIA or the Unified State Exam, you forgot the useful formula of the n-th member of an arithmetic progression. Something comes to mind, but somehow uncertainly ... Whether n there, or n+1, or n-1... How to be!?

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

We draw a numerical axis and mark the first one on it. second, third, etc. members. And note the difference d between members. Like this:

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

a 2 =a 1 + 1 d

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

a 3 =a 1 + 2 d

Do you get it? I don't put some words in bold for nothing. Okay, one more step.)

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

a 4 =a 1 + 3 d

It's time to realize that the number of gaps, i.e. d, Always one less than the number of the member you are looking for n. That is, up to the number n, number of gaps will n-1. So, the formula will be (no options!):

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

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

Tasks for independent decision.

For warm-up:

1. In arithmetic progression (a n) a 2 =3; a 5 \u003d 5.1. Find a 3 .

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

And this is no longer a warm-up.)

2. In arithmetic progression (a n) a 85 \u003d 19.1; a 236 =49, 3. Find a 3 .

What, reluctance to draw a picture?) Still! Better formula, yes...

3. Arithmetic progression is given by the condition:a 1 \u003d -5.5; a n+1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.

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

4. Given an arithmetic progression (a n):

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

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

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

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

Not the easiest task, yes ...) Here the method "on the fingers" will not work. You have to write formulas and solve equations.

Answers (in disarray):

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

Happened? It's nice!)

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

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

If you like this site...

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

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

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.

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 believe 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.