How to find n number in arithmetic progression. Arithmetic progression

The general term of the sequence has the form $u_n=n^2$. Substituting $n=1$, we get:

$$ u_1=1^2=1. $$

This is the first member of the sequence. Substituting $n=2$ into $u_n=n^2$, we get the second term of the sequence:

$$ u_2=2^2=4. $$

If we substitute $n=3$, then we get the third term of the sequence:

$$ u_3=3^2=9. $$

Similarly, we find the fourth, fifth, sixth and other members of the sequence. This is how we get the corresponding numbers:

$$1;\; 4;\; nine;\; sixteen;\; 25;\; 36;\; 49;\; 64; \;81; \ldots $$

It is also worth bearing in mind the members of the sequence $u_n=n^3$. Here are some of its first members:

\begin(equation)1;\; eight;\; 27;\; 64;\; 125;\; 216;\; 343;\; 512;\;729; \ldots \end(equation)

In addition, to form a common member of a series, the sequence $u_n=n!$ is often used, the first few members of which are as follows:

\begin(equation)1;\; 2;\; 6;\; 24;\; 120;\; 720;\; 5040; \ldots \end(equation)

Recording "n!" (read "en factorial") denotes the product of all natural numbers from 1 to n, i.e.

$$ n!=1\cdot2\cdot 3\cdot \ldots\cdot n. $$

By definition, it is assumed that $0!=1!=1$. For example, let's find 5!:

$$ 5!=1\cdot 2\cdot 3\cdot 4\cdot 5=120. $$

Arithmetic and geometric progressions are also often used. If the first member arithmetic progression is equal to $a_1$, and the difference is equal to $d$, then the common term of the arithmetic progression is written using the following formula:

\begin(equation)a_n=a_1+d\cdot (n-1) \end(equation)

What is an arithmetic progression? show/hide

An arithmetic progression is a sequence of numbers in which the difference between the next and the previous members is unchanged. This constant difference is called progression difference

$$ 3;\; ten;\; 17;\; 24;\; 31;\; 38;\; 45;\; 52; \ldots $$

Please note that no matter what pair of neighboring elements we take, the difference between the next and previous members will always be constant and equal to 7:

\begin(aligned) & 10-3=7;\\ & 17-10=7;\\ & 31-24=7; \ldots \end(aligned)

This number, i.e. 7, and there is a progression difference. Usually it is denoted by the letter $d$, i.e. $d=7$. The first element of the progression is $a_1=3$. We write the common term of this progression using the formula . Substituting $a_1=3$ and $d=7$ into it, we will have:

$$ a_n=3+7\cdot (n-1)=3+7n-7=7n-4. $$

For clarity, let's find the first few members of the arithmetic progression using the formula $a_n=7n-4$:

\begin(aligned) & a_1=7\cdot 1-4=3;\\ & a_2=7\cdot 2-4=10;\\ & a_3=7\cdot 3-4=17;\\ & a_4= 7\cdot 4-4=24;\\ & a_5=7\cdot 5-4=31. \end(aligned)

By substituting any value of the number $n$ into the formula $a_n=7n-4$, you can get any member of the arithmetic progression.

It is also worth noting the geometric progression. If the first term of the progression is equal to $b_1$, and the denominator is equal to $q$, then the common term of the geometric progression is given by the following formula:

\begin(equation)b_n=b_1\cdot q^(n-1) \end(equation)

What geometric progression? show/hide

A geometric progression is a sequence of numbers in which the ratio between the next and the previous members is constant. This constant relationship is called denominator of progression. For example, consider the following sequence:

$$6;\; eighteen;\; 54;\; 162;\; 486;\; 1458;\; 4374; \ldots $$

Please note that no matter what pair of neighboring elements we take, the ratio of the next to the previous will always be constant and equal to 3:

\begin(aligned) & \frac(18)(6)=3;\\ & \frac(54)(18)=3;\\ & \frac(1458)(486)=3;\\ & \ldots \end(aligned)

This number, i.e. 3, and there is the denominator of the progression. Usually it is denoted by the letter $q$, i.e. $q=3$. The first element of the progression is $b_1=6$. We write the common term of this progression using the formula . Substituting $b_1=6$ and $q=3$ into it, we will have:

$$ b_n=6\cdot 3^(n-1). $$

For clarity, we find the first few terms of the geometric progression using the formula $b_n=6\cdot 3^(n-1)$:

\begin(aligned) & b_1=6\cdot 3^0=6;\\ & b_2=6\cdot 3^1=18;\\ & b_3=6\cdot 3^2=54;\\ & b_4= 6\cdot 3^3=162;\\ & b_5=6\cdot 3^4=486. \end(aligned)

Substituting any value of the number $n$ into the formula $b_n=6\cdot 3^(n-1)$, one can obtain any member of the geometric progression.

In all the examples below, the members of the series will be denoted by the letters $u_1$ (the first member of the series), $u_2$ (the second member of the series), and so on. The notation $u_n$ will denote the common member of the series.

Example #1

Find the common term of the series $\frac(1)(7)+\frac(2)(9)+\frac(3)(11)+\frac(4)(13)+\ldots$.

The essence of such tasks is to notice the pattern that is inherent in the first members of the series. And on the basis of this pattern, draw a conclusion about the form of the common term. What does the phrase "find a common term" mean? It means that it is necessary to find such an expression, substituting $n=1$ into which we get the first term of the series, i.e. $\frac(1)(7)$; substituting $n=2$ we get the second term of the series, i.e. $\frac(2)(9)$; substituting $n=3$ we get the third term of the series, i.e. $\frac(3)(11)$ and so on. We know the first four terms of the series:

$$ u_1=\frac(1)(7);\; u_2=\frac(2)(9);\; u_3=\frac(3)(11);\; u_4=\frac(4)(13). $$

Let's move gradually. All the members of the series known to us are fractions, therefore it is reasonable to assume that the common member of the series is also represented by a fraction:

$$ u_n=\frac(?)(?) $$

Our task is to find out what is hidden under the question marks in the numerator and denominator. Let's look at the numerator first. The numerators of the members of the series known to us are the numbers 1, 2, 3 and 4. Note that the number of each member of the series is equal to the numerator. The first term has a 1 in the numerator, the second has a two, the third has a three, and the fourth has a four.

It is logical to assume that the nth term will have $n$ in the numerator:

$$ u_n=\frac(n)(?) $$

By the way, we can come to this conclusion in another way, more formal. What is the sequence 1, 2, 3, 4? Note that each subsequent term of this sequence is 1 more than the previous one. We are dealing with four members of an arithmetic progression, the first member of which is $a_1=1$, and the difference is $d=1$. Using the formula , we obtain the expression for the common term of the progression:

$$ a_n=1+1\cdot (n-1)=1+n-1=n. $$

So, guessing or formal calculation is a matter of taste. The main thing is that we have written down the numerator of the common term of the series. Let's move on to the denominator.

In the denominators we have the sequence 7, 9, 11, 13. These are four members of an arithmetic progression, the first member of which is equal to $b_1=7$, and the difference is $d=2$. We find the common term of the progression using the formula:

$$ b_n=7+2\cdot (n-1)=7+2n-2=2n+5. $$

The resulting expression, i.e. $2n+5$, and will be the denominator of the common term of the series. So:

$$ u_n=\frac(n)(2n+5). $$

The common term of the series is received. Let's check if the formula $u_n=\frac(n)(2n+5)$ we found is suitable for calculating the already known terms of the series. Let's find the terms $u_1$, $u_2$, $u_3$ and $u_4$ using the formula $u_n=\frac(n)(2n+5)$. The results, of course, must coincide with the first four terms of the series given to us by condition.

$$ u_1=\frac(1)(2\cdot 1+5)=\frac(1)(7);\; u_2=\frac(2)(2\cdot 2+5)=\frac(2)(9);\; u_3=\frac(3)(2\cdot 3+5)=\frac(3)(11);\; u_4=\frac(4)(2\cdot 4+5)=\frac(4)(13). $$

That's right, the results are the same. The series specified in the condition can now be written in the following form: $\sum\limits_(n=1)^(\infty)\frac(n)(2n+5)$. The general term of the series has the form $u_n=\frac(n)(2n+5)$.

$$ \frac(1)(7)+\frac(2)(9)+\frac(3)(11)+\frac(4)(13)+0+0+0+0+0+0+ 0+\ldots $$

Doesn't such a series have a right to exist? Even as it has. And for this series, we can write that

$$ u_1=\frac(1)(7);\; u_2=\frac(2)(9);\; u_3=\frac(3)(11);\; u_4=\frac(4)(13); \; u_n=0\; (n≥ 5). $$

You can write another extension. For example, this:

$$ \frac(1)(7)+\frac(2)(9)+\frac(3)(11)+\frac(4)(13)+\frac(1)(5)+\frac( 1)(6)+\frac(1)(7)+\frac(1)(8)+\frac(1)(9)+\frac(1)(10)+\ldots $$

And such a continuation does not contradict anything. At the same time, it can be written that

$$ u_1=\frac(1)(7);\; u_2=\frac(2)(9);\; u_3=\frac(3)(11);\; u_4=\frac(4)(13); \; u_n=\frac(1)(n)\; (n≥ 5). $$

If the first two options seemed too formal to you, then I will offer the third. Let's write the common term like this:

$$ u_n=\frac(n)(n^4-10n^3+35n^2-48n+29). $$

We calculate the first four terms of the series using the proposed general term formula:

\begin(aligned) & u_1=\frac(1)(1^4-10\cdot 1^3+35\cdot 1^2-48\cdot 1+29)=\frac(1)(7);\ \ & u_2=\frac(2)(2^4-10\cdot 2^3+35\cdot 2^2-48\cdot 2+29)=\frac(2)(9);\\ & u_3= \frac(3)(3^4-10\cdot 3^3+35\cdot 3^2-48\cdot 3+29)=\frac(3)(11);\\ & u_4=\frac(4 )(4^4-10\cdot 4^3+35\cdot 4^2-48\cdot 4+29)=\frac(4)(13). \end(aligned)

As you can see, the proposed general term formula is quite correct. And there are infinitely many such variations, their number is not limited by anything. AT standard examples, of course, a standard set of some well-known sequences (progressions, degrees, factorials, etc.) is used. However, in such problems there is always uncertainty, and it is advisable to keep this in mind.

In all subsequent examples, this ambiguity will not be stipulated. We will decide in standard ways, which are accepted in most problem books.

Answer: common term of the series: $u_n=\frac(n)(2n+5)$.

Example #2

Write the common term of the series $\frac(1)(1\cdot 5)+\frac(1)(3\cdot 8)+\frac(1)(5\cdot 11)+\frac(1)(7\cdot 14)+\frac(1)(9\cdot 17)+\ldots$.

We know the first five members of the series:

$$ u_1=\frac(1)(1\cdot 5);\; u_2=\frac(1)(3\cdot 8); \; u_3=\frac(1)(5\cdot 11); \; u_4=\frac(1)(7\cdot 14); \; u_5=\frac(1)(9\cdot 17). $$

All the members of the series known to us are fractions, which means that we will look for the common term of the series in the form of a fraction:

$$ u_n=\frac(?)(?). $$

Let's take a look at the numerator. All numerators contain units, therefore, there will be one in the numerator of the common term of the series, i.e.

$$ u_n=\frac(1)(?). $$

Now let's look at the denominator. The denominators of the first terms of the series known to us are the products of numbers: $1\cdot 5$, $3\cdot 8$, $5\cdot 11$, $7\cdot 14$, $9\cdot 17$. The first of these numbers are: 1, 3, 5, 7, 9. This sequence has the first term $a_1=1$, and each subsequent one is obtained from the previous one by adding the number $d=2$. In other words, these are the first five members of an arithmetic progression, the common term of which can be written using the formula:

$$ a_n=1+2\cdot (n-1)=1+2n-2=2n-1. $$

In the products $1\cdot 5$, $3\cdot 8$, $5\cdot 11$, $7\cdot 14$, $9\cdot 17$ the second numbers are: 5, 8, 11, 14, 17. These are the elements of an arithmetic progression, whose first term is $b_1=5$ and whose denominator is $d=3$. We write the common term of this progression using the same formula:

$$ b_n=5+3\cdot (n-1)=5+3n-3=3n+2. $$

Let's bring the results together. The product in the denominator of the common term of the series is $(2n-1)(3n+2)$. And the general term of the series itself has the following form:

$$ u_n=\frac(1)((2n-1)(3n+2)). $$

To check the result obtained, we find by the formula $u_n=\frac(1)((2n-1)(3n+2))$ the first four terms of the series that we know:

\begin(aligned) & u_1=\frac(1)((2\cdot 1-1)(3\cdot 1+2))=\frac(1)(1\cdot 5);\\ & u_2=\ frac(1)((2\cdot 2-1)(3\cdot 2+2))=\frac(1)(3\cdot 8);\\ & u_3=\frac(1)((2\cdot 3-1)(3\cdot 3+2))=\frac(1)(5\cdot 11);\\ & u_4=\frac(1)((2\cdot 4-1)(3\cdot 4 +2))=\frac(1)(7\cdot 14);\\ & u_5=\frac(1)((2\cdot 5-1)(3\cdot 5+2))=\frac(1 )(9\cdot 17). \end(aligned)

So, the formula $u_n=\frac(1)((2n-1)(3n+2))$ allows us to accurately calculate the terms of the series known from the condition. If desired, the given series can be written as follows:

$$ \sum\limits_(n=1)^(\infty)\frac(1)((2n-1)(3n+2))=\frac(1)(1\cdot 5)+\frac(1 )(3\cdot 8)+\frac(1)(5\cdot 11)+\frac(1)(7\cdot 14)+\frac(1)(9\cdot 17)+\ldots $$

Answer: common term of the series: $u_n=\frac(1)((2n-1)(3n+2))$.

We will continue this topic in the second and third parts.

Many have heard of an arithmetic progression, but not everyone is well aware of what it is. In this article, we will give an appropriate definition, and also consider the question of how to find the difference of an arithmetic progression, and give a number of examples.

Mathematical definition

So if we are talking about an arithmetic or algebraic progression (these concepts define the same thing), then this means that there is some number series, which satisfies the following law: every two adjacent numbers in the series differ by the same value. Mathematically, this is written like this:

Here n means the number of the element a n in the sequence, and the number d is the difference of the progression (its name follows from the presented formula).

What does knowing the difference d mean? About how far apart adjacent numbers are. However, knowledge of d is necessary, but not sufficient condition to determine (restore) the entire progression. You need to know one more number, which can be absolutely any element of the series under consideration, for example, a 4, a10, but, as a rule, the first number is used, that is, a 1.

Formulas for determining the elements of the progression

In general, the information above is already enough to move on to solving specific problems. Nevertheless, before an arithmetic progression is given, and it will be necessary to find its difference, we present a couple of useful formulas, thereby facilitating the subsequent process of solving problems.

It is easy to show that any element of the sequence with number n can be found as follows:

a n \u003d a 1 + (n - 1) * d

Indeed, everyone can check this formula with a simple enumeration: if you substitute n = 1, then you get the first element, if you substitute n = 2, then the expression gives the sum of the first number and the difference, and so on.

The conditions of many problems are compiled in such a way that for a known pair of numbers, the numbers of which are also given in the sequence, it is necessary to restore the entire number series (find the difference and the first element). We will now solve this problem in general view.

So, let's say we are given two elements with numbers n and m. Using the formula obtained above, we can compose a system of two equations:

a n \u003d a 1 + (n - 1) * d;
a m = a 1 + (m - 1) * d

To find unknown quantities, we use the known simple trick solutions of such a system: we subtract pairwise the left and right parts, while the equality remains valid. We have:

a n \u003d a 1 + (n - 1) * d;
a n - a m = (n - 1) * d - (m - 1) * d = d * (n - m)

Thus, we have eliminated one unknown (a 1). Now we can write the final expression for determining d:

d = (a n - a m) / (n - m), where n > m

We have received very a simple formula: to calculate the difference d in accordance with the conditions of the problem, it is only necessary to take the ratio of the differences of the elements themselves and their serial numbers. Should focus on one important point attention: the differences are taken between the "senior" and "junior" members, that is, n > m ("senior" - meaning standing further from the beginning of the sequence, its absolute value can be either greater or less than the "younger" element).

The expression for the difference d of the progression should be substituted into any of the equations at the beginning of the solution of the problem in order to obtain the value of the first term.

In our age of development computer technology many schoolchildren try to find solutions for their tasks on the Internet, so questions of this type often arise: find the difference of an arithmetic progression online. Upon such a request, the search engine will display a number of web pages, by going to which, you will need to enter the data known from the condition (it can be either two members of the progression or the sum of some of them) and instantly get an answer. Nevertheless, such an approach to solving the problem is unproductive in terms of the development of the student and understanding the essence of the task assigned to him.

Solution without using formulas

Let's solve the first problem, while we will not use any of the above formulas. Let the elements of the series be given: a6 = 3, a9 = 18. Find the difference of the arithmetic progression.

Known elements are close to each other in a row. How many times must the difference d be added to the smallest one to get the largest one? Three times (the first time adding d, we get the 7th element, the second time - the eighth, finally, the third time - the ninth). What number must be added to three three times to get 18? This is the number five. Really:

Thus, the unknown difference is d = 5.

Of course, the solution could be done using the appropriate formula, but this was not done intentionally. Detailed explanation problem solving should be clear and a prime example What is an arithmetic progression.

A task similar to the previous one

Now let's solve a similar problem, but change the input data. So, you should find if a3 = 2, a9 = 19.

Of course, you can resort again to the method of solving "on the forehead". But since the elements of the series are given, which are relatively far apart, such a method becomes not very convenient. But using the resulting formula will quickly lead us to the answer:

d \u003d (a 9 - a 3) / (9 - 3) \u003d (19 - 2) / (6) \u003d 17 / 6 ≈ 2.83

Here we have rounded finite number. How much this rounding led to an error can be judged by checking the result:

a 9 \u003d a 3 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 \u003d 18.98

This result differs by only 0.1% from the value given in the condition. Therefore, rounding to hundredths used can be considered a good choice.

Tasks for applying the formula for an member

Consider classic example tasks to determine the unknown d: find the difference of the arithmetic progression if a1 = 12, a5 = 40.

When two numbers of an unknown algebraic sequence are given, and one of them is the element a 1 , then you do not need to think long, but you should immediately apply the formula for the a n member. AT this case we have:

a 5 = a 1 + d * (5 - 1) => d = (a 5 - a 1) / 4 = (40 - 12) / 4 = 7

We got the exact number when dividing, so it makes no sense to check the accuracy of the calculated result, as was done in the previous paragraph.

Let's solve another similar problem: we should find the difference of the arithmetic progression if a1 = 16, a8 = 37.

We use a similar approach to the previous one and get:

a 8 = a 1 + d * (8 - 1) => d = (a 8 - a 1) / 7 = (37 - 16) / 7 = 3

What else you should know about arithmetic progression

In addition to problems of finding an unknown difference or individual elements, it is often necessary to solve problems of the sum of the first terms of a sequence. Consideration of these problems is beyond the scope of the topic of the article; nevertheless, for completeness of information, we present general formula for the sum of n numbers of the series:

∑ n i = 1 (a i) = n * (a 1 + a n) / 2

Goals:

Introduce the concept of arithmetic progression.
Consider the main types of tasks for applying the formula of the n-th member of an arithmetic progression.
Use elements of developmental learning in the lesson.
Develop analytical thinking students.

During the classes

Teacher. In the previous lesson, we introduced the concept of infinite number sequence, as a function defined on the set of natural numbers and found out that sequences are infinite and finite, increasing and decreasing, and also learned about how to set them. List them.

students.

Analytical (using a formula).
Verbal (setting a sequence with a description).
Recurrent (when any member of the sequence, starting from some, is expressed through the previous members).

Exercise 1. Indicate, if possible, the 7th term of each sequence.

(a n): 6; ten; fourteen; eighteen; 22; 26;…
(bn): 49; 25; 81; 4; 121; 64...
(cn): 22; 17; 12; 7; 2; -3…
(xn): -3.8; -2.6; -1.4; -0.2; one; 2.2…
(yn): -12; 7; eight; fourteen; -23; 41…

Teacher. Why is it impossible to answer the question for sequences b n and y n?

students. There is no definite pattern in these sequences, although (b n) consists of squares of natural numbers, but they are taken in an arbitrary order, and (y n) is an arbitrary series of numbers, so any number can take the seventh place.

Teacher. For sequences (and n); (cn); (x n) all of you were able to find the 7th term correctly.

Task 2. Come up with your similar example such a sequence. List the first 4 terms. Exchange notebooks with a desk mate and determine the 5th member of this sequence.

Teacher. How common property have similar sequences?

student. Each subsequent term differs from the previous one by the same number.

Teacher. Sequences of this type are called arithmetic progressions. They will be the subject of our study today. Formulate the topic of the lesson.

(The student can easily formulate the first part of the topic. The teacher can formulate the second part himself)

Teacher. Formulate the objectives of the lesson based on this topic.

(It is important that students articulate as fully and accurately as possible learning goals, then they accept them and strive to achieve)

Students.

Define an arithmetic progression.
Derive the formula for the nth term of an arithmetic progression.
Learn to solve problems on the topic (consider Various types tasks).

Then it is useful to project on the screen the goals set by the teacher for the students, so that they are convinced that they have common goals.

Teacher. A bit of history. The term "progression" comes from the Latin progression, which means "moving forward", was introduced by the Roman author Boethius in the 6th century AD. and received further development in the works of Fibonacci, Shuke, Gauss and other scientists.

Definition. An arithmetic progression is a sequence in which each term, starting from the second, is equal to the previous term added to the same number. This number is called the difference of an arithmetic progression and is denoted d.

(a n): a 1 ; a 2 ; a 3 ; …a n … arithmetic progression.
d \u003d a 2 - a 1 \u003d a 3 - a 2 \u003d ... \u003d a n + 1 - a n

Task 3. Let a 1 = 7; d = 0.

Name the next 3 members of the sequence.

students. 7; 7; 7

Teacher. Such sequences are called constant or stationary.

Let a 1 = -12; d = 3. Name 3 members of this sequence.

student. -9; -6; -3

Teacher. Will I be right if I name the numbers: -15; -eighteen; -21?

As a rule, most students believe that this is correct. Then you should ask them to identify the number of each member. Since the number of a member of the sequence must be expressed as a natural number, the named numbers cannot be present in this sequence.

Task 4. In arithmetic progression a 1 ; a 2 ; 6; 4; a 5 find a 1 ; a 2 ; a 5 .

The task is carried out in pairs, one student, if desired, completes it with reverse side boards.

Decision:

d = 4 - 6 = -2
a 5 \u003d a 4 + d \u003d 4 - 2 \u003d 2
a 2 \u003d a 3 - d \u003d 6 - (-2) \u003d 8
a 1 \u003d a 2 - d \u003d 8 - (-2) \u003d 10

Specify for this sequence a 8 and a 126

students. and 8 \u003d -4 and 126 can be specified, but it takes too long to count.

Teacher. So it is necessary to find a way that will allow us to quickly find any member of the sequence. Try to derive the formula for the nth member of an arithmetic progression.

You can call a strong student to the board and, through clearly posed questions and help from the class, derive a formula.

Formula derivation:

a 2 = a 1 + d
a 3 = a 2 + d = a 1 + 2d
a 4 = a 3 + d = a 1 + 3d
etc.

a n = a 1 + (n – 1) d- formulan-th member of the arithmetic progression.

Teacher. So, what do you need to know to determine any member of an arithmetic progression?

students. a 1 and d

Teacher. Using this formula, find a 126.

Students. a 126 \u003d a 1 + 125d \u003d 10 \u003d 125 ∙ (- 2) \u003d 10 - 250 \u003d - 240

Task 5. Let (b n): an arithmetic progression in which b 1 is the first term and d is the difference. Find errors:

b4 = b1 + 3d		b 2k = b 1 + (2k – 1)∙d
b9 = b1 + 10d		b k-4 = b 1 + (k - 3)∙d
b -3 = b 1 - 4d		b k+7 \u003d b 1 + (k - 6) ∙d

Task 6. Consider the formula for the nth member of an arithmetic progression. Let's find out what types of problems can be solved using this formula. Formulate a direct problem.

Students. By set values a 1 and d find a n .

Teacher. What kind inverse problems can you put?

Students.

Given a 1 and a n . Find d.
Given d and a n . Find a 1 .
Given a 1 , d and a n . Find n.

Task 7. Find the difference of the arithmetic progression in which y 1 = 10; y 5 = 22

Board solution:

y 5 = y 1 + 4d
22 = 10 + 4d
4d = 12
d=3

Task 8. Does the arithmetic progression contain 2; nine; … number 156 ?

Analysis: by reasoning we come to the conclusion that since each number in the sequence has its own number, expressed as a natural number, then you need to find the number of a member of the sequence and find out if it belongs to the set of natural numbers. If it does, then the sequence contains given number otherwise, no.

Solution at the blackboard:

a n \u003d a 1 + (n - 1) d
156 = 2 + 7 (n - 1)
7(n - 1) = 154
n - 1 = 22
n=23

Answer: a 23 = 156

Task 9. Find the first three terms of the arithmetic progression in which

a 1 + a 5 = 24;
a 2 ∙ a 3 \u003d 60

We analyze the task, draw up a system of equations, which is proposed to be solved at home.

a 1 + a 1 + 4d = 24;
(a 1 + d) ∙ (a 1 + 4d) = 60.

Summing up total lesson.

What new did you learn at the lesson today? What have you learned?

Homework. Familiarize yourself with the material in paragraph 25 of the textbook. Learn the definition of an arithmetic progression and the formula for the nth term. Be able to express from the formula all the quantities included in it. Solve the system for task 9. Follow the textbook No. 575 (a, b); 576; 578(a); 579(a).

Task for an additional assessment: let a 1 ; a 2 ; a 3 ; …a n … arithmetic progression. Prove that a n+1 = (a n + a n+2) : 2

What main point formulas?

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 tasks. And don't forget to right moment, but 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 is 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 arithmetic progression recurrent formula easy to convert to normal. 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 appeared specific number: 121. Which is quite logical.) We are interested in the term 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 "trifle" often slips past the head, and without it, (without the "trifle", not the head!) The problem cannot be solved. Although ... and without a head too.)

Now we can 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 parameters? 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 Creative skills!)

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 members. 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 real version 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 in 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, the GIA or the Unified State Examination, you forgot useful formula nth 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 to be sure and right decision that's enough!) 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? The third term equals first term plus two d.

a 3 =a 1 + 2 d

Do you get it? I'm not in vain highlighting some words in bold. Okay, one more step.)

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

a 4 =a 1 + 3 d

It's time to realize that the number of gaps, i.e. d, always one less than the number of the member you are looking for n. That is, 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 is 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! It's better in the 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 member progressions.

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? It 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 element of fantasy for the fourth, and the subtle moment for the sixth, and general approaches for solving any problems on the formula of the n-th member - everything is painted. Recommend.

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