Arithmetic and geometric progression formulas 9. Algebra lesson "Arithmetic and geometric progressions" (Grade 9)
- Generalization and systematization of students' knowledge on this topic.
- Familiarization of students with historical material.
Equipment: poster for the game "Progressio - moving forward."
All students are divided into five groups + the advice of the sages
The twentieth century is over.
Where does the person go?
Space and sea explored
The structure of the stars and the whole Earth.
But mathematicians are calling
famous slogan:
"Progressio - moving forward."
Today we will have a council in the class - the council of the Wise Men. Wise men are students sitting in groups in a class. And the Wise Men sitting at this table.
Do you recognize them?
Sitting at the table: Archimedes, Gauss, Magnitsky.
Who found the formula for the sum of squares?
And the right way to progress came?
Mathematician and physicist. I am Archimedes.
There are many legends about my life.
O! I am Carl Gauss! Found instantly the sum of all natural numbers 1 to 100 as an elementary school student.
Magnitsky. Lord! I have the honor to introduce myself. I am Leonty Filippovich Magnitsky, the creator of the first textbook "Arithmetic".
Teacher. Tell me, guys, why are these scientists suddenly gathered together at the same table? What math question unites them? If you haven't figured it out, then carefully watch the scene.
ancient indian legend
A Hindu king appears in the classroom with a servant.
Tsar. I, the Hindu king Sheram, have learned the game of chess and admire its wit and variety of positions. Servant, let's call the inventor Setu. I want to adequately reward you, Seth, for the wonderful game that you came up with. Name a reward that will satisfy you and you will receive it.
Seth. Lord. Order me to give me one grain of wheat for the first cell of the chessboard
Tsar. A simple grain of wheat?
Seth. Yes, lord For the second cell, order to give out 2 grains, for the third - 4, for the fourth - 8, for the fifth - 16, and so on until the 64th cell.
King Sheram laughed.
Teacher. O sages of the ninth class, let us consult. Should the king laugh?
Record on the board: 1,2,4,8,16, ... .. S 64 -?
The students decide. b 1= 1, q=2, n=64, S 64 =2 64 - 1.
Teacher. How big is this number? Who can explain it?
Archimedes. The Wisest! If the king could sow wheat on the entire surface of the Earth, counting the seas, and oceans, and mountains, and the desert, and the Arctic and Antarctica, and get a satisfactory harvest, then, perhaps, in five years he could pay off.
Gauss. Mathematics is exact science. (Writes on the board 18 446 744 073 709 551 615). 18 quintillion 446 quadrillion 744 trillion 73 billion 709 million 551 thousand 615.
Magnitsky. Lord Wise Men of the 9th grade! My contemporaries would say that S 64 18.5 10 18 . True, I confess to you that in my textbook “Arithmetic”, published 200 years ago, from which children studied for half a century, there are many problems on the topic “Progressions”, but I myself solved some of them with great difficulty, since I have not yet found all formulas relating the quantities included in them.
Under the creak of a pen on a sheet of paper.
Fill out these sheets!
May our endeavors help you!
Blank sheets are distributed to test knowledge of the theory, i.e., the basic abstract on the topic “Progressions” is restored.
Students complete the table. The following table appears on the board:
progressions |
Arithmetic a n |
Geometric b n |
|
Definition |
b n+1 =b n q (q0,q1) |
||
Formula of n first terms |
a n \u003d a 1 + (n-1) d |
||
The sum of the first n terms of the progression |
S n = |
S n = And the search for them was appreciated by us. “Mathematics is the queen of sciences, arithmetic is the queen of mathematics”
|
Understanding many topics in mathematics and physics is associated with knowledge of the properties of number series. Schoolchildren in grade 9, when studying the subject "Algebra", consider one of the important sequences of numbers - an arithmetic progression. Let's give the basic formulas of an arithmetic progression (Grade 9), as well as examples of their use for solving problems.
Algebraic or arithmetic progression
The number series that will be discussed in this article is called two different ways presented in the title of this paragraph. So, arithmetic progression in mathematics is understood as such number series, in which any two numbers standing next to each other differ by the same amount, which is called the difference. Numbers in such a series are usually denoted by letters with a lower integer index, for example, a 1 , a 2 , a 3 and so on, where the index indicates the number of the element of the series.
Given the above definition of an arithmetic progression, we can write the following equality: a 2 -a 1 =...=a n -a n-1 =d, here d is the difference of the algebraic progression and n is any integer. If d>0, then we can expect that each subsequent term of the series will be greater than the previous one, in this case we speak of an increasing progression. If d<0, тогда предыдущий член будет больше последующего, то есть ряд будет убывать. Частный случай возникает, когда d = 0, то есть ряд представляет собой последовательность, в которой a 1 =a 2 =...=a n .
Arithmetic progression formulas (grade 9)
The series of numbers under consideration, since it is ordered and obeys a certain mathematical law, has two properties that are important for its use:
- First, knowing only two numbers a 1 and d, you can find any member of the sequence. This is done using the following formula: a n = a 1 +(n-1)*d.
- Secondly, to calculate the sum of n terms of the first ones, it is not necessary to add them in order, since you can use the following formula: S n \u003d n * (a n + a 1) / 2.
The first formula is easy to understand, since it is a direct consequence of the fact that each member of the series under consideration differs from its neighbor by the same difference.
The second formula of an arithmetic progression can be obtained by paying attention to the fact that the sum a 1 +a n is equivalent to the sums a 2 +a n-1 , a 3 +a n-2 and so on. Indeed, since a 2 = d+a 1 , a n-2 = -2*d+a n , a 3 = 2*d+a 1 , and a n-1 = -d+a n , then substituting these expressions into the corresponding sums, we get that they will be the same. The factor n/2 in the 2nd formula (for S n) appears due to the fact that sums of type a i+1 +a n-i turn out to be exactly n/2, here i is an integer ranging from 0 to n/2 -1.
According to the surviving historical evidence, the formula for the sum S n was first obtained by Karl Gauss (the famous German mathematician) when he was given the task by a school teacher to add the first 100 numbers.
Sample Problem #1: Find the Difference
Tasks that pose the question as follows: knowing the formulas for an arithmetic progression, how to find q (d), are the simplest that can only be for this topic.
Here is an example: given a numerical sequence -5, -2, 1, 4, ..., it is necessary to determine its difference, that is, d.
To do this is as easy as shelling pears: you need to take two elements and subtract the smaller from the larger one. In this case, we have: d = -2 - (-5) = 3.
In order to be sure of the answer received, it is recommended to check the remaining differences, since the presented sequence may not satisfy the algebraic progression condition. We have: 1-(-2)=3 and 4-1=3. These data indicate that we got the correct result (d=3) and proved that the series of numbers in the problem statement is indeed an algebraic progression.
Sample Problem #2: Find the Difference Knowing Two Terms of the Progression
Consider another interesting problem, which is posed by the question of how to find the difference. The arithmetic progression formula in this case must be used for the nth term. So, the task: given the first and fifth numbers of a series that corresponds to all the properties of an algebraic progression, for example, these are the numbers a 1 = 8 and a 5 = -10. How to find the difference d?
You should start solving this problem by writing the general form of the formula for the n-th element: a n = a 1 + d * (-1 + n). Now you can go in two ways: either substitute the numbers right away and work with them already, or express d, and then go to specific a 1 and a 5 . Let's use the last method, we get: a 5 \u003d a 1 + d * (-1 + 5) or a 5 \u003d 4 * d + a 1, from which it follows that d \u003d (a 5 -a 1) / 4. Now you can safely substitute the known data from the condition and get the final answer: d = (-10-8)/4 = -4.5.
Note that in this case the progression difference turned out to be negative, that is, there is a decreasing sequence of numbers. It is necessary to pay attention to this fact when solving problems so as not to confuse the signs "+" and "-". All the formulas above are universal, so they should always be followed regardless of the sign of the numbers with which operations are carried out.
An example of solving problem No. 3: find a1, knowing the difference and the element
Let's change the condition of the problem a little. Let there be two numbers: the difference d=6 and the 9th element of the progression a 9 = 10. How to find a1? The formulas of the arithmetic progression remain unchanged, we will use them. For the number a 9 we have the following expression: a 1 +d*(9-1) = a 9 . From where we easily get the first element of the series: a 1 = a 9 -8 * d = 10 - 8 * 6 = -38.
An example of solving problem #4: find a1, knowing two elements
This version of the problem is a complicated version of the previous one. The essence is the same, it is necessary to calculate a 1 , but now the difference d is not known, and instead one more element of the progression is given.
An example of this type of problem is the following: find the first number in a sequence known to be an arithmetic progression and whose 15th and 23rd elements are 7 and 12, respectively.
It is necessary to solve this problem by writing an expression for the n-th member for each element known from the condition, we have: a 15 = d*(15-1)+a 1 and a 23 = d*(23-1)+a 1 . As you can see, we have received two linear equations that need to be solved with respect to a 1 and d. Let's do this: subtract the first equation from the second equation, then we get the following expression: a 23 -a 15 \u003d 22 * d - 14 * d \u003d 8 * d. In deriving the last equation, the values of a 1 have been omitted because they cancel out when subtracted. Substituting the known data, we find the difference: d \u003d (a 23 -a 15) / 8 \u003d (12-7) / 8 \u003d 0.625.
The value of d must be substituted into any formula for a known element in order to obtain the first member of the sequence: a 15 = 14*d+a 1, from where: a 1 = a 15 -14*d = 7-14*0.625 = -1.75.
Let's check the result, for this we find a 1 through the second expression: a 23 \u003d d * 22 + a 1 or a 1 \u003d a 23 -d * 22 \u003d 12 - 0.625 * 22 \u003d -1.75.
An example of solving problem No. 5: find the sum of n elements
As you can see, up to this point, only one arithmetic progression formula (Grade 9) was used for the solution. Now we give a problem for the solutions of which we need to know the second formula, that is, for the sum S n .
Given the following ordered series of numbers -1.1, -2.1, -3.1,..., you need to calculate the sum of its first 11 elements.
It can be seen from this series that it is decreasing, and a 1 \u003d -1.1. Its difference is: d = -2.1 - (-1.1) = -1. Now let's define the 11th term: a 11 \u003d 10 * d + a 1 \u003d -10 + (-1.1) \u003d -11.1. After completing the preparatory calculations, you can use the above formula for the sum, we have: S 11 \u003d 11 * (-1.1 + (-11.1)) / 2 \u003d -67.1. Since all the terms were negative numbers, their sum also has the corresponding sign.
An example of solving problem No. 6: find the sum of elements from n to m
Perhaps this type of problem is the most difficult for most students. Let's give a typical example: given a series of numbers 2, 4, 6, 8 ..., you need to find the sum from the 7th to the 13th terms.
Formulas arithmetic progression(Grade 9) are used exactly the same as in all tasks before. This task is recommended to be solved in stages:
- First, find the sum of 13 terms using the standard formula.
- Then calculate this sum for the first 6 elements.
- Then subtract the 2nd from the 1st sum.
Let's get to the decision. As in the previous case, we will carry out preparatory calculations: a 6 = 5*d+a 1 = 10+2 = 12, a 13 = 12*d+a 1 = 24+2 = 26.
Let's calculate two sums: S 13 = 13*(2+26)/2 = 182, S 6 = 6*(2+12)/2 = 42. We take the difference and get the desired answer: S 7-13 = S 13 - S 6 = 182-42 = 140. Note that when obtaining this value, it was the sum of 6 elements of the progression that was used as subtracted, since the 7th member is included in the sum S 7-13 .
Arithmetic and geometric progressions
Theoretical information
Theoretical information
Arithmetic progression |
Geometric progression |
|
Definition |
Arithmetic progression a n a sequence is called, each member of which, starting from the second, is equal to the previous member, added with the same number d (d- progression difference) |
geometric progression b n a sequence of non-zero numbers is called, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression) |
Recurrent formula |
For any natural n |
For any natural n |
nth term formula |
a n = a 1 + d (n - 1) |
b n \u003d b 1 ∙ q n - 1, b n ≠ 0 |
| characteristic property |  |
|
| Sum of the first n terms |  |
 |
Examples of tasks with comments
Exercise 1
In arithmetic progression ( a n) a 1 = -6, a 2
According to the formula of the nth term:
a 22 = a 1+ d (22 - 1) = a 1+ 21d
By condition:
a 1= -6, so a 22= -6 + 21d.
It is necessary to find the difference of progressions:
d= a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = - 48.
Answer : a 22 = -48.
Task 2
Find the fifth term of the geometric progression: -3; 6;....
1st way (using n-term formula)
According to the formula of the n-th member of a geometric progression:
b 5 \u003d b 1 ∙ q 5 - 1 = b 1 ∙ q 4.
As b 1 = -3,
2nd way (using recursive formula)
Since the denominator of the progression is -2 (q = -2), then:
b 3 = 6 ∙ (-2) = -12;
b 4 = -12 ∙ (-2) = 24;
b 5 = 24 ∙ (-2) = -48.
Answer : b 5 = -48.
Task 3
In arithmetic progression ( a n) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.
For an arithmetic progression, the characteristic property has the form 
.
Therefore:
.
Substitute the data in the formula:
Answer: 95.
Task 4
In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.
To find the sum of the first n terms of an arithmetic progression, two formulas are used:
.
Which of them is more convenient to apply in this case?
By condition, the formula of the nth member of the original progression is known ( a n) a n= 3n - 4. Can be found immediately and a 1, and a 16 without finding d . Therefore, we use the first formula.
Answer: 368.
Task 5
In arithmetic progression a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.
According to the formula of the nth term:
a 22 = a 1 + d (22 – 1) = a 1+ 21d.
By condition, if a 1= -6, then a 22= -6 + 21d. It is necessary to find the difference of progressions:
d= a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = -48.
Answer : a 22 = -48.
Task 6
Several consecutive terms of a geometric progression are recorded:
Find the term of the progression, denoted by the letter x .
When solving, we use the formula for the nth term b n \u003d b 1 ∙ q n - 1 for geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of these terms of the progression and divide by the previous one. In our example, you can take and divide by. We get that q \u003d 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.
Substituting the found values into the formula, we get:
.
Answer : .
Task 7
From the arithmetic progressions given by the formula of the nth term, choose the one for which the condition is satisfied a 27 > 9:
Since the specified condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:
.
Answer: 4.
Task 8
In arithmetic progression a 1= 3, d = -1.5. Specify the largest value of n for which the inequality holds a n > -6.
Summary of the lesson of algebra in grade 9
Lesson topic: Definition of arithmetic and geometric progression.
Formula of the nth member of the arithmetic and geometric
progressions.
Lesson type : lesson learning new material
The purpose of the lesson:
Formation of the concepts of arithmetic and geometric progression, as types of numerical sequences; derivation of the formula of the n-th member of the arithmetic and geometric sequence.
Acquaintance with the characteristic property of the members of an arithmetic and geometric progression.
Formation of students' skills to use the acquired knowledge in solving problems.
Lesson objectives:
Educational: introduce the concepts of arithmetic and geometric progression; formulas of the nth member; characteristic property that members of an arithmetic and geometric progression have.
Developing: to increase the conscious assimilation of the material through opposition; develop the ability to compare mathematical concepts, find similarities and differences, see patterns, reason by analogy, develop memory and logical thinking.
Educational: create conditions for the development of cognitive interest in the subject.
Lesson plan:
1. Organization of the beginning of the lesson, setting goals and objectives of the lesson.
2. Motivation to study the topic (“The Legend of the Chessboard”)
3. Learning new material
4. Primary fastening
5. Summing up the lesson
6. Homework
During the classes
1. Organization of the beginning of the lesson.
Name the topic of the lesson, the purpose of the lesson, the tasks.
2. Motivation to study the topic.
"The Legend of the Chessboard".
Chess is one of the most ancient games. It has existed for many centuries, and it is not surprising that legends are associated with it, the veracity of which cannot be verified due to the prescription of time. I want to tell one of these legends. To understand it, one does not need to know how to play chess at all - it is enough to know that the game takes place on a board divided into 64 cells (alternately black and white).
The game of chess was invented in India, and when the Indian king Sheram met her, he was delighted with her wit and the variety of possible positions in it. Having learned that the game was invented by one of his subjects, the king ordered to call him in order to personally reward him for a successful invention.
The inventor - his name was Seta - appeared at the throne of the ruler. He was a modestly dressed scientist who received his livelihood from his students.
I want to adequately reward you, Seth, for the wonderful game that you came up with, the king said.
The sage bowed.
I am rich enough to fulfill your most daring wish, - continued the king. - Name the reward that will satisfy you, and you will receive it.
Seth was silent.
Do not be shy, - the king encouraged him. - Express your desire. I will spare nothing to fulfill it!
Great is your kindness, my lord. But give me time to think about the answer. Tomorrow, after mature reflection, I will communicate my request to you.
When the next day, Seta again appeared at the steps of the throne, he surprised the king with the unparalleled modesty of his request.
Lord, - said Seth, - order me to give me one grain of wheat for the first cell of the chessboard.
A simple grain of wheat? - the king was amazed.
Yes, master. For the second cell, order to give out two grains, for the third - four, for the fourth - 8, for the fifth - 16, for the sixth - 32 ...
Enough! - the king interrupted him with irritation. - You will receive your grains for all 64 cells of the board, according to your desire: for each twice as much as the previous one. But know that your request is not worthy of my generosity. By asking for such an insignificant reward, you disrespectfully disregard my grace. Truly, as a teacher, you could show the best example of respect for the kindness of your sovereign. Go! My servants will bring you a sack of wheat.
Seta smiled, left the hall and waited at the gates of the palace.
At dinner, the king remembered the inventor of chess and sent to find out if the reckless Seth had already taken away his miserable reward.
Lord, - was the answer, - your order is being fulfilled. Court mathematicians calculate the number of grains to follow.
The king frowned - he was not used to his orders being carried out so slowly.
In the evening, going to bed, King Sheram once again inquired whether Seta had left the palace fence with his sack of wheat.
Lord, - they answered him, - your mathematicians work tirelessly and hope to finish counting before dawn.
Why are they delaying this? - the king exclaimed angrily. - Tomorrow, before I wake up, everything to the last grain must be given to Seth. I don't order twice!
In the morning, the king was informed that the foreman of the court mathematicians asked to listen to an important report. The king ordered to bring him in.
Before you speak of your case,” announced Sheram, “I want to hear if Seta has finally received the insignificant reward that he assigned himself.
For this reason, I dared to appear before you at such an early "hour," the old man answered. "We conscientiously counted the entire number of grains that Seth wants to receive. The number is so great ...
No matter how great it is, - the king interrupted arrogantly, - my granaries will not become scarce! A reward has been promised and must be given...
It is not in your power, lord, to fulfill such desires. In all your barns there is not such a number of grains as Seth demanded. Nor is it in the granaries of the whole kingdom. There is no such number of grains in the entire space of the Earth. And if you want to give the promised reward without fail, then order to turn the earthly kingdoms into arable fields, order to dry up the seas and oceans, order to melt the ice and snow covering the distant northern wastelands. Let all their space be completely sown with wheat. And all that is born in these fields, order to give to Seth. Then he will receive his reward.
With amazement, the king listened to the words of the elder.
Give me that monstrous number, he said thoughtfully.
Eighteen quintillion four hundred and forty-six quadrillion seven hundred and forty-four trillion seventy-three billion seven hundred nine million five hundred and fifty-one thousand six hundred and fifteen, O Lord! (18 446 744 073 709 551 615)
Such is the legend. Whether what is told here really happened is not known, but that the reward of which the tradition speaks must have been expressed in just such a number.
If you want to imagine the whole immensity of this numerical giant, estimate what size barn would be required to accommodate such a number of grains. It is known that a cubic meter of wheat contains about 15 million grains. This means that the reward for a chess inventor should have taken up approximately
12,000,000,000,000 cubic meters m, or 12,000 cubic meters. km. With a barn height of 4 m and a width of 10 m, its length would have to extend for 300,000,000 km, that is, twice as far as from the Earth to the Sun!
Of course, the Indian king was not in a position to issue such an award.
3. Presentation of new material.
Distribute to each student sheets on which the theoretical material is presented in the form of a table showing the differences in the definitions of arithmetic and geometric progressions, their characteristic properties, formulas for finding the n-th term, formulas for finding the sum of the n-first terms and for a geometric progression, the formula for the sum is infinite decreasing geometric progression.
Arithmetic progression(a/p) | Geometric progression(g/n) |
Def. An arithmetic progression is a sequence of numbers, each term of which, starting from the second, is equal to the previous one, added with the same number. For example: -6; -four; -2; 0; 2; four;… 6; = -4; = -2; =0; = 2… | Def. A geometric progression is a sequence of non-zero numbers, each term of which, starting from the second, is equal to the previous one, multiplied by the same number that is not equal to zero. For example: 5; 15; 45; 135, ... 5; =15; =45; =135; … |
d = 2 – difference a/n d = - ; d=- | q = 3 - denominator g/n q = ; Q= |
Formula of the nth member of a / p D = + 2d; D = + 3d; = + 4d; | Formula of the nth member of the g / p Q = ; Q = ; |
The formula for the middle term a / p |
The great Russian language lives and develops
Change is a small life Why school changes are needed
What is intelligence: definition, examples