Some problems in physics and mathematics can be solved using the properties of number series. The two simplest number sequences taught in schools are algebraic and geometric. In this article, we will take a closer look at the question of how to find the sum of an infinite decreasing geometric progression.
Progression geometric
These words mean a series of real numbers whose elements a i satisfy the expression:
Here i is the number of the element in the series, r is a constant number called the denominator.
This definition shows that, knowing any member of the progression and its denominator, you can restore the entire series of numbers. For example, if the 10th element is known, then dividing it by r will get the 9th element, then dividing it again will get the 8th and so on. These simple arguments allow us to write down an expression that is valid for the series of numbers under consideration:
An example of a progression with a denominator of 2 would be the following series:
1, 2, 4, 8, 16, 32, ...
If the denominator is equal to -2, then a completely different series is obtained:
1, -2, 4, -8, 16, -32, ...
Geometric progression is much faster than algebraic progression, that is, its terms increase quickly and decrease quickly.
Sum of i terms of progression
To solve practical problems, it is often necessary to calculate the sum of several elements of the numerical sequence under consideration. For this case the following formula is valid:
S i = a 1 *(r i -1)/(r-1)
It can be seen that to calculate the sum of i terms, you need to know only two numbers: a 1 and r, which is logical, since they uniquely determine the entire sequence.
Decreasing sequence and the sum of its terms
Now let's consider special case. We will assume that the modulus of the denominator r does not exceed one, that is -1 A decreasing geometric progression is interesting to consider because the infinite sum of its terms tends to a finite real number. Let's get the formula for the sum. This is easy to do if you write out the expression for S i given in the previous paragraph. We have: S i = a 1 *(r i -1)/(r-1) Let's consider the case when i->∞. Since the modulus of the denominator is less than 1, raising it to an infinite power will give zero. This can be checked using the example of r=0.5: 0,5 2 = 0,25; 0,5 3 = 0,125; ...., 0,5 20 = 0,0000009. As a result, the sum of the terms of an infinite decreasing geometric progression will take the form: This formula is often used in practice, for example, to calculate the areas of figures. It is also used to solve the paradox of Zeno of Elea with the tortoise and Achilles. It is obvious that considering the sum of an infinite geometric increasing progression (r>1) will lead to the result S ∞ = +∞. Let us show how to apply the above formulas using an example of solving a problem. It is known that the sum of an infinite geometric progression is 11. Moreover, its 7th term is 6 times less than the third term. What is the first element for this number series? First, let's write out two expressions to determine the 7th and 3rd elements. We get: Dividing the first expression by the second and expressing the denominator, we have: a 7 /a 3 = r 4 => r = 4 √(a 7 /a 3) Since the ratio of the seventh and third terms is given in the problem statement, you can substitute it and find r: r = 4 √(a 7 /a 3) = 4 √(1/6) ≈ 0.63894 We calculated r to five decimal places. Since the resulting value is less than one, the progression is decreasing, which justifies the use of the formula for its infinite sum. Let's write the expression for the first term through the sum S ∞: We substitute known values into this formula and get the answer: a 1 = 11*(1-0.63894) = 3.97166. Zeno of Elea is a famous Greek philosopher who lived in the 5th century BC. e. A number of its apogees or paradoxes have reached the present day, in which the problem of the infinitely large and the infinitely small in mathematics is formulated. One of Zeno's famous paradoxes is the competition between Achilles and the tortoise. Zeno believed that if Achilles gave the tortoise some advantage in distance, he would never be able to catch up with it. For example, let Achilles run 10 times faster than an animal crawling, which, for example, is 100 meters in front of him. When the warrior runs 100 meters, the turtle crawls away 10 meters. Having run 10 meters again, Achilles sees that the turtle crawls another 1 meter. You can argue this way ad infinitum, the distance between the competitors will indeed decrease, but the turtle will always be in front. Led Zeno to the conclusion that movement does not exist, and all surrounding movements of objects are an illusion. Of course, the ancient Greek philosopher was wrong. The solution to the paradox lies in the fact that an infinite sum of constantly decreasing segments tends to a finite number. In the above case, for the distance that Achilles ran, we get: 100 + 10 + 1 + 0,1 + 0,01 + ... Applying the formula for the sum of an infinite geometric progression, we obtain: S ∞ = 100 /(1-0.1) ≈ 111.111 meters This result shows that Achilles will catch up with the tortoise when it crawls only 11.111 meters. The ancient Greeks did not know how to work with infinite quantities in mathematics. However, this paradox can be resolved if we pay attention not to the infinite number of gaps that Achilles must overcome, but to the finite number of steps the runner needs to reach his goal. Purpose of the lesson: to introduce students to a new type of sequence - an infinitely decreasing geometric progression. Lesson on the topic “Infinitely decreasing geometric progression” (algebra, 10th grade) The purpose of the lesson: introducing students to a new type of sequence - an infinitely decreasing geometric progression. Tasks: formulating an initial idea of the limit of a numerical sequence; acquaintance with another way to convert infinite periodic fractions into ordinary ones using the formula for the sum of an infinitely decreasing geometric progression; development of intellectual qualities of schoolchildren’s personality such as logical thinking, ability to make evaluative actions, and generalization; fostering activity, mutual assistance, collectivism, and interest in the subject. Equipment: computer class, projector, screen. Lesson type: lesson - learning a new topic. During the classes I. Org. moment. State the topic and purpose of the lesson. II. Updating students' knowledge. In 9th grade you studied arithmetic and geometric progressions. Questions 1. Definition arithmetic progression.
(An arithmetic progression is a sequence in which each member Starting from the second, it is equal to the previous term added to the same number). 2. Formula n th term of an arithmetic progression 3. Formula for the sum of the first n terms of an arithmetic progression. ( or ) 4. Definition of geometric progression. (A geometric progression is a sequence of non-zero numbers Each term of which, starting from the second, is equal to the previous term multiplied by Same number). 5. Formula n th term of the geometric progression 6. Formula for the sum of the first n members of a geometric progression. 7. What other formulas do you know? (, Where ; ; ;
,
)
Tasks 1. Arithmetic progression is given by the formula a n = 7 – 4n . Find a 10. (-33) 2. In arithmetic progression a 3 = 7 and a 5 = 1 . Find a 4 . (4) 3. In arithmetic progression a 3 = 7 and a 5 = 1 . Find a 17 . (-35) 4. In arithmetic progression a 3 = 7 and a 5 = 1 . Find S 17. (-187) 5. For geometric progressionfind the fifth term. 6. For geometric progression find the nth term. 7. Exponentially b 3 = 8 and b 5 = 2. Find b 4 . (4) 8. Exponentially b 3 = 8 and b 5 = 2. Find b 1 and q. 9. Exponentially b 3 = 8 and b 5 = 2. Find S5. (62) III. Learning a new topic(demonstration of presentation). Consider a square with a side equal to 1. Let's draw another square whose side is half the size of the first square, then another one whose side is half the second, then the next one, etc. Each time the side of the new square is equal to half of the previous one. As a result, we received a sequence of sides of squaresforming a geometric progression with the denominator.
And, what is very important, the more we build such squares, the smaller the side of the square will be. For example , Those. As the number n increases, the terms of the progression approach zero. Using this figure, you can consider another sequence. For example, the sequence of areas of squares: And, again, if n increases indefinitely, then the area approaches zero as close as you like. Let's look at another example. An equilateral triangle with sides equal to 1 cm. Let's construct the following triangle with the vertices in the midpoints of the sides of the 1st triangle, according to the theorem about the midline of the triangle - the side of the 2nd is equal to half the side of the first, the side of the 3rd is equal to half the side of the 2nd, etc. Again we obtain a sequence of lengths of the sides of triangles. At . If we consider a geometric progression with negative denominator.
Then, again, with increasing numbers n terms of the progression approach zero. Let's pay attention to the denominators of these sequences. Everywhere the denominators were less than 1 in absolute value. We can conclude: a geometric progression will be infinitely decreasing if the modulus of its denominator is less than 1. Frontal work. Definition: Geometric progression is called infinitely decreasing if the modulus of its denominator is less than one..
Using the definition, you can decide whether a geometric progression is infinitely decreasing or not. Task Is the sequence an infinitely decreasing geometric progression if it is given by the formula: Solution: Let's find q. ;
;
;
.
this geometric progression is infinitely decreasing. b) this sequence is not an infinitely decreasing geometric progression. Consider a square with a side equal to 1. Divide it in half, one of the halves in half, etc. The areas of all the resulting rectangles form an infinitely decreasing geometric progression: The sum of the areas of all rectangles obtained in this way will be equal to the area of the 1st square and equal to 1. But on the left side of this equality is the sum of an infinite number of terms. Let's consider the sum of the first n terms. According to the formula for the sum of the first n terms of a geometric progression, it is equal to.
If n increases without limit, then or . Therefore, i.e. . Sum of an infinitely decreasing geometric progressionthere is a sequence limit S 1, S 2, S 3, …, S n, …. For example, for progression,
we have Because The sum of an infinitely decreasing geometric progressioncan be found using the formula.
III. Comprehension and consolidation(completing tasks). №13; №14; №15(1,3); №16(1,3); №18(1,3); №19; №20.
IV. Summarizing. What sequence did you get acquainted with today? Define an infinitely decreasing geometric progression. How to prove that a geometric progression is infinitely decreasing? Give the formula for the sum of an infinitely decreasing geometric progression. V. Homework. 2. № 15(2,4); №16(2,4); 18(2,4).
To use presentation previews, create a Google account and log in to it: https://accounts.google.com Everyone should be able to think consistently, judge with evidence, and refute incorrect conclusions: a physicist and a poet, a tractor driver and a chemist. E. Kolman In mathematics, one should remember not the formulas, but the processes of thinking. V.P. Ermakov It is easier to find the squaring of a circle than to outwit a mathematician. Augustus de Morgan What science could be more noble, more admirable, more useful to humanity than mathematics? Franklin Infinitely decreasing geometric progression grade 10 I. Arithmetic and geometric progressions. Questions 1. Definition of arithmetic progression. 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. 2. Formula for the nth term of an arithmetic progression. 3. Formula for the sum of the first n terms of an arithmetic progression. 4. Definition of geometric progression. A geometric progression is a sequence of non-zero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number 5. Formula for the nth term of a geometric progression. 6. Formula for the sum of the first n terms of a geometric progression. II. Arithmetic progression. Tasks An arithmetic progression is given by the formula a n = 7 – 4 n Find a 10 . (-33) 2. In arithmetic progression, a 3 = 7 and a 5 = 1. Find a 4 . (4) 3. In arithmetic progression a 3 = 7 and a 5 = 1. Find a 17 . (-35) 4. In arithmetic progression, a 3 = 7 and a 5 = 1. Find S 17. (-187) II. Geometric progression. Tasks 5. For a geometric progression, find the fifth term 6. For a geometric progression, find the nth term. 7. In geometric progression b 3 = 8 and b 5 = 2. Find b 4 . (4) 8. In geometric progression b 3 = 8 and b 5 = 2. Find b 1 and q. 9. In geometric progression b 3 = 8 and b 5 = 2. Find S5. (62) definition: A geometric progression is called infinitely decreasing if the modulus of its denominator is less than one. Problem No. 1 Is the sequence an infinitely decreasing geometric progression if it is given by the formula: Solution: a) this geometric progression is infinitely decreasing. b) this sequence is not an infinitely decreasing geometric progression. The sum of an infinitely decreasing geometric progression is the limit of the sequence S 1, S 2, S 3, ..., S n, .... For example, for the progression we have Since the Sum of an infinitely decreasing geometric progression can be found using the formula Completing tasks Find the sum of an infinitely decreasing geometric progression with the first term 3, the second 0.3. 2. No. 13; No. 14; textbook, p. 138 3. No. 15(1;3); No.16(1;3) No.18(1;3); 4. No. 19; No. 20. What sequence did you get acquainted with today? Define an infinitely decreasing geometric progression. How to prove that a geometric progression is infinitely decreasing? Give the formula for the sum of an infinitely decreasing geometric progression. Questions The famous Polish mathematician Hugo Steinhaus jokingly claims that there is a law that is formulated as follows: a mathematician will do it better. Namely, if you entrust two people, one of whom is a mathematician, to perform any work unfamiliar to them, then the result will always be the following: the mathematician will do it better. Hugo Steinhaus 01/14/1887-02/25/1972 NUMERIC SEQUENCES VI § l48. Sum of an infinitely decreasing geometric progression Until now, when talking about sums, we have always assumed that the number of terms in these sums is finite (for example, 2, 15, 1000, etc.). But when solving some problems (especially higher mathematics) one has to deal with the sums of an infinite number of terms S= a
1 + a
2 + ... + a
n
+ ... . (1) What are these amounts? A-priory the sum of an infinite number of terms a
1 , a
2 , ..., a
n
, ... is called the limit of the sum S n
first P
numbers when P
-> ∞
: S=S n
= (a
1 + a
2 + ... + a
n
). (2) Limit (2), of course, may or may not exist. Accordingly, they say that the sum (1) exists or does not exist. How can we find out whether sum (1) exists in each specific case? Common decision This issue goes far beyond the scope of our program. However, there is one important special case that we must now consider. We will talk about summing the terms of an infinitely decreasing geometric progression. Let a
1 , a
1 q
, a
1 q
2, ... is an infinitely decreasing geometric progression. This means that | q
|< 1. Сумма первых P
terms of this progression is equal From the basic theorems on the limits of variables (see § 136) we obtain: But 1 = 1, a qn
= 0. Therefore So, the sum of an infinitely decreasing geometric progression is equal to the first term of this progression divided by one minus the denominator of this progression. 1) The sum of the geometric progression 1, 1/3, 1/9, 1/27, ... is equal to and the sum of the geometric progression is 12; -6; 3; - 3 / 2 , ... equal 2) Convert a simple periodic fraction 0.454545 ... into an ordinary one. To solve this problem, imagine this fraction as an infinite sum: Right part This equality is the sum of an infinitely decreasing geometric progression, the first term of which is equal to 45/100, and the denominator is 1/100. That's why Using the described method, it can also be obtained general rule conversion of simple periodic fractions into ordinary ones (see Chapter II, § 38): To convert a simple periodic fraction into an ordinary fraction, you need to do the following: in the numerator put the period of the decimal fraction, and in the denominator - a number consisting of nines taken as many times as there are digits in the period of the decimal fraction.
3) Convert the mixed periodic fraction 0.58333 .... into an ordinary fraction. Let's imagine this fraction as an infinite sum: On the right side of this equality, all terms, starting from 3/1000, form an infinitely decreasing geometric progression, the first term of which is equal to 3/1000, and the denominator is 1/10. That's why Using the described method, a general rule for converting mixed periodic fractions into ordinary fractions can be obtained (see Chapter II, § 38). We deliberately do not present it here. There is no need to remember this cumbersome rule. It is much more useful to know that any mixed periodic fraction can be represented as the sum of an infinitely decreasing geometric progression and a certain number. And the formula for the sum of an infinitely decreasing geometric progression, you must, of course, remember. As an exercise, we suggest that you, in addition to the problems No. 995-1000 given below, once again turn to problem No. 301 § 38. Exercises
995. What is called the sum of an infinitely decreasing geometric progression? 996. Find the sums of infinitely decreasing geometric progressions: 997. At what values X
progression is it infinitely decreasing? Find the sum of such a progression. 998. In an equilateral triangle with side A
a new triangle is inscribed by connecting the midpoints of its sides; a new triangle is inscribed in this triangle in the same way, and so on ad infinitum. a) the sum of the perimeters of all these triangles; b) the sum of their areas. 999. Square with side A
a new square is inscribed by connecting the midpoints of its sides; a square is inscribed in this square in the same way, and so on ad infinitum. Find the sum of the perimeters of all these squares and the sum of their areas. 1000. Compose an infinitely decreasing geometric progression such that its sum is equal to 25/4, and the sum of the squares of its terms is equal to 625/24. So, let's sit down and start writing some numbers. For example: You can write any numbers, and there can be as many of them as you like (in our case, there are them). No matter how many numbers we write, we can always say which one is first, which one is second, and so on until the last, that is, we can number them. This is an example of a number sequence: Number sequence is a set of numbers, each of which can be assigned a unique number. For example, for our sequence: The assigned number is specific to only one number in the sequence. In other words, there are no three second numbers in the sequence. The second number (like the th number) is always the same. The number with the number is called the nth member of the sequence. We usually call the entire sequence by some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member: . In our case: The most common types of progression are arithmetic and geometric. In this topic we will talk about the second type - geometric progression. Even in ancient times, the Italian mathematician monk Leonardo of Pisa (better known as Fibonacci) dealt with the practical needs of trade. The monk was faced with the task of determining what is the smallest number of weights that can be used to weigh a product? In his works, Fibonacci proves that such a system of weights is optimal: This is one of the first situations in which people had to deal with a geometric progression, which you have probably already heard about and have at least general concept. Once you fully understand the topic, think about why such a system is optimal? Currently, in life practice, geometric progression manifests itself when investing money in a bank, when the amount of interest is accrued on the amount accumulated in the account for the previous period. In other words, if you put money on a time deposit in a savings bank, then after a year the deposit will increase by the original amount, i.e. the new amount will be equal to the contribution multiplied by. In another year, this amount will increase by, i.e. the amount obtained at that time will again be multiplied by and so on. A similar situation is described in problems of calculating the so-called compound interest- the percentage is taken each time from the amount that is in the account, taking into account previous interest. We'll talk about these tasks a little later. There are many more simple cases where geometric progression is applied. For example, the spread of influenza: one person infected another person, they, in turn, infected another person, and thus the second wave of infection is a person, and they, in turn, infected another... and so on... By the way, a financial pyramid, the same MMM, is a simple and dry calculation based on the properties of a geometric progression. Interesting? Let's figure it out. Let's say we have a number sequence: You will immediately answer that this is easy and the name of such a sequence is an arithmetic progression with the difference of its terms. How about this: If you subtract the previous number from the next number, you will see that each time you get a new difference (and so on), but the sequence definitely exists and is easy to notice - each subsequent number is times larger than the previous one! This type of number sequence is called geometric progression and is designated. Geometric progression () is a numerical sequence, the first term of which is different from zero, and each term, starting from the second, is equal to the previous one, multiplied by the same number. This number is called the denominator of a geometric progression. The restrictions that the first term ( ) is not equal and are not random. Let's assume that there are none, and the first term is still equal, and q is equal to, hmm.. let it be, then it turns out: Agree that this is no longer a progression. As you understand, we will get the same results if there is any number other than zero, a. In these cases, there will simply be no progression, since the entire number series will either be all zeros, or one number, and all the rest will be zeros. Now let's talk in more detail about the denominator of the geometric progression, that is, o. Let's repeat: - this is the number how many times does each subsequent term change? geometric progression. What do you think it could be? That's right, positive and negative, but not zero (we talked about this a little higher). Let's assume that ours is positive. Let in our case, a. What is the value of the second term and? You can easily answer that: That's right. Accordingly, if, then all subsequent terms of the progression have the same sign - they are positive. What if it's negative? For example, a. What is the value of the second term and? This is a completely different story Try to count the terms of this progression. How much did you get? I have. Thus, if, then the signs of the terms of the geometric progression alternate. That is, if you see a progression with alternating signs for its members, then its denominator is negative. This knowledge can help you test yourself when solving problems on this topic. Now let's practice a little: try to determine which number sequences are a geometric progression and which are an arithmetic progression: Got it? Let's compare our answers: Let's return to our last progression and try to find its member, just like in the arithmetic one. As you may have guessed, there are two ways to find it. We successively multiply each term by. So, the th term of the described geometric progression is equal to. As you already guessed, now you yourself will derive a formula that will help you find any member of the geometric progression. Or have you already developed it for yourself, describing how to find the th member step by step? If so, then check the correctness of your reasoning. Let us illustrate this with the example of finding the th term of this progression: In other words: Find the value of the term of the given geometric progression yourself. Happened? Let's compare our answers: Please note that you got exactly the same number as in the previous method, when we sequentially multiplied by each previous term of the geometric progression. The derived formula is true for all values - both positive and negative. Check this yourself by calculating the terms of the geometric progression with the following conditions: , A. Did you count? Let's compare the results: Agree that it would be possible to find a term of a progression in the same way as a term, however, there is a possibility of calculating incorrectly. And if we have already found the th term of the geometric progression, then what could be simpler than using the “truncated” part of the formula. More recently, we talked about the fact that it can be either greater or less than zero, however, there are special values for which the geometric progression is called infinitely decreasing. Why do you think this name is given? We see that each subsequent term is less than the previous one by a factor, but will there be any number? You will immediately answer - “no”. That is why it is infinitely decreasing - it decreases and decreases, but never becomes zero. To clearly understand how this looks visually, let's try to draw a graph of our progression. So, for our case, the formula takes the following form: On graphs we are accustomed to plotting dependence on, therefore: The essence of the expression has not changed: in the first entry we showed the dependence of the value of a member of a geometric progression on its ordinal number, and in the second entry we simply took the value of a member of a geometric progression as, and designated the ordinal number not as, but as. All that remains to be done is to build a graph. Do you see? The function decreases, tends to zero, but never crosses it, so it is infinitely decreasing. Let’s mark our points on the graph, and at the same time what the coordinate and means: Try to schematically depict a graph of a geometric progression if its first term is also equal. Analyze what is the difference with our previous graph? Did you manage? Here's the graph I came up with: Now that you have fully understood the basics of the topic of geometric progression: you know what it is, you know how to find its term, and you also know what an infinitely decreasing geometric progression is, let's move on to its main property. Do you remember the property of the terms of an arithmetic progression? Yes, yes, how to find the value of a certain number of a progression when there are previous and subsequent values of the terms of this progression. Do you remember? This: Now we are faced with exactly the same question for the terms of a geometric progression. To derive such a formula, let's start drawing and reasoning. You'll see, it's very easy, and if you forget, you can get it out yourself. Let's take another simple geometric progression, in which we know and. How to find? With arithmetic progression it is easy and simple, but what about here? In fact, there is nothing complicated in geometric either - you just need to write down each value given to us according to the formula. You may ask, what should we do about it now? Yes, very simple. First, let's depict these formulas in the figure and try to do with them various manipulations to arrive at a value. Let's abstract from the numbers that are given to us, let's focus only on their expression through the formula. We need to find the value highlighted in orange, knowing the terms adjacent to it. Let's try to produce with them various actions, as a result of which we can get. Addition. From this expression, as you can see, we cannot express it in any way, therefore, we will try another option - subtraction. Subtraction. As you can see, we cannot express this either, therefore, let’s try to multiply these expressions by each other. Multiplication. Now look carefully at what we have by multiplying the terms of the geometric progression given to us in comparison with what needs to be found: Guess what I'm talking about? That's right, to find we need to take Square root from the geometric progression numbers adjacent to the desired one multiplied by each other: Here you go. You yourself derived the property of geometric progression. Try writing this formula in general view. Happened? Forgot the condition for? Think about why it is important, for example, try to calculate it yourself. What will happen in this case? That's right, complete nonsense because the formula looks like this: Accordingly, do not forget this limitation. Now let's calculate what it equals Correct answer - ! If you didn't forget the second one when calculating possible meaning, then you are a great fellow and can immediately move on to training, and if you forgot, read what is discussed below and pay attention to why it is necessary to write down both roots in the answer. Let's draw both of our geometric progressions - one with a value and the other with a value and check whether both of them have the right to exist: In order to check whether such a geometric progression exists or not, it is necessary to see whether all its given terms are the same? Calculate q for the first and second cases. See why we have to write two answers? Because the sign of the term you are looking for depends on whether it is positive or negative! And since we don’t know what it is, we need to write both answers with a plus and a minus. Now that you have mastered the main points and derived the formula for the property of geometric progression, find, knowing and Compare your answers with the correct ones: What do you think, what if we were given not the values of the terms of the geometric progression adjacent to the desired number, but equidistant from it. For example, we need to find, and given and. Can we use the formula we derived in this case? Try to confirm or refute this possibility in the same way, describing what each value consists of, as you did when you originally derived the formula, at. Now look carefully again. From this we can conclude that the formula works not only with neighboring with the desired terms of the geometric progression, but also with equidistant from what the members are looking for. Thus, our initial formula takes the form: That is, if in the first case we said that, now we say that it can be equal to any natural number, which is smaller. The main thing is that it is the same for both given numbers. Practice with specific examples, just be extremely careful! Decided? I hope you were extremely attentive and noticed a small catch. Let's compare the results. In the first two cases, we calmly apply the above formula and get the following values: In the third case, upon closer examination serial numbers numbers given to us, we understand that they are not equidistant from the number we are looking for: it is the previous number, but is removed at the position, so it is not possible to apply the formula. How to solve it? It's actually not as difficult as it seems! Let us write down what each number given to us and the number we are looking for consists of. So we have and. Let's see what we can do with them? I suggest dividing by. We get: We substitute our data into the formula: The next step we can find is - for this we need to take the cube root of the resulting number. Now let's look again at what we have. We have it, but we need to find it, and it, in turn, is equal to: We found all the necessary data for the calculation. Substitute into the formula: Our answer: . Try solving another similar problem yourself: How much did you get? I have - . As you can see, essentially you need remember just one formula- . You can withdraw all the rest yourself without any difficulty at any time. To do this, simply write the simplest geometric progression on a piece of paper and write down what each of its numbers is equal to, according to the formula described above. Now let's look at formulas that allow us to quickly calculate the sum of terms of a geometric progression in a given interval: To derive the formula for the sum of terms of a finite geometric progression, multiply all parts of the above equation by. We get: Look carefully: what do the last two formulas have in common? That's right, common members, for example, and so on, except for the first and last member. Let's try to subtract the 1st from the 2nd equation. What did you get? Now express the term of the geometric progression through the formula and substitute the resulting expression into our last formula: Group the expression. You should get: All that remains to be done is to express: Accordingly, in this case. What if? What formula works then? Imagine a geometric progression at. What is she like? A series of identical numbers is correct, so the formula will look like this: There are many legends about both arithmetic and geometric progression. One of them is the legend of Set, the creator of chess. Many people know that the game of chess was invented in India. When the Hindu king met her, he was delighted with her wit and the variety of positions possible in her. Having learned that it was invented by one of his subjects, the king decided to personally reward him. He summoned the inventor to himself and ordered him to ask him for everything he wanted, promising to fulfill even the most skillful desire. Seta asked for time to think, and when the next day Seta appeared before the king, he surprised the king with the unprecedented modesty of his request. He asked to give a grain of wheat for the first square of the chessboard, a grain of wheat for the second, a grain of wheat for the third, a fourth, etc. The king was angry and drove Seth away, saying that the servant's request was unworthy of the king's generosity, but promised that the servant would receive his grains for all the squares of the board. And now the question: using the formula for the sum of the terms of a geometric progression, calculate how many grains Seth should receive? Let's start reasoning. Since, according to the condition, Seth asked for a grain of wheat for the first square of the chessboard, for the second, for the third, for the fourth, etc., then we see that the problem is about a geometric progression. What does it equal in this case? Total squares of the chessboard. Respectively, . We have all the data, all that remains is to plug it into the formula and calculate. To imagine at least approximately the “scale” of a given number, we transform using the properties of degree: Of course, if you want, you can take a calculator and calculate what number you end up with, and if not, you’ll have to take my word for it: the final value of the expression will be. quintillion quadrillion trillion billion million thousand. Phew) If you want to imagine the enormity of this number, then estimate how large a barn would be required to accommodate the entire amount of grain. If the king were strong in mathematics, he could have invited the scientist himself to count the grains, because to count a million grains, he would need at least a day of tireless counting, and given that it is necessary to count quintillions, the grains would have to be counted throughout his life. Now let’s solve a simple problem involving the sum of terms of a geometric progression. So, the first term of the geometric progression is Vasya, that is, a person. The th term of the geometric progression is the two people he infected on the first day of his arrival. total amount members of the progression is equal to the number of students in 5A. Accordingly, we talk about a progression in which: Let's substitute our data into the formula for the sum of the terms of a geometric progression: The whole class will get sick within days. Don't believe formulas and numbers? Try to portray the “infection” of students yourself. Happened? Look how it looks for me: Calculate for yourself how many days it would take for students to get sick with the flu if each one infected a person, and there were only one person in the class. What value did you get? It turned out that everyone started getting sick after a day. As you can see, such a task and the drawing for it resemble a pyramid, in which each subsequent one “brings” new people. However, sooner or later a moment comes when the latter cannot attract anyone. In our case, if we imagine that the class is isolated, the person from closes the chain (). Thus, if a person were involved in financial pyramid, in which money was given if you bring two other participants, then the person (or general case) would not have brought anyone, and therefore would have lost everything they invested in this financial scam. Everything that was said above refers to a decreasing or increasing geometric progression, but, as you remember, we have a special type - an infinitely decreasing geometric progression. How to calculate the sum of its members? And why does this type of progression have certain characteristics? Let's figure it out together. So, first, let's look again at this drawing of an infinitely decreasing geometric progression from our example: Now let’s look at the formula for the sum of a geometric progression, derived a little earlier: What are we striving for? That's right, the graph shows that it tends to zero. That is, at, will be almost equal, respectively, when calculating the expression we will get almost. In this regard, we believe that when calculating the sum of an infinitely decreasing geometric progression, this bracket can be neglected, since it will be equal. - formula is the sum of the terms of an infinitely decreasing geometric progression. IMPORTANT! We use the formula for the sum of terms of an infinitely decreasing geometric progression only if the condition explicitly states that we need to find the sum infinite number of members. If a specific number n is specified, then we use the formula for the sum of n terms, even if or. Now let's practice. I hope you were extremely careful. Let's compare our answers: Now you know everything about geometric progression, and it’s time to move from theory to practice. The most common geometric progression problems encountered on the exam are problems calculating compound interest. These are the ones we will talk about. You've probably heard of the so-called compound interest formula. Do you understand what it means? If not, let’s figure it out, because once you understand the process itself, you will immediately understand what geometric progression has to do with it. We all go to the bank and know that there are different conditions on deposits: this is the term, and additional service, and interest with two different ways its calculations - simple and complex. WITH simple interest everything is more or less clear: interest is accrued once at the end of the deposit term. That is, if we say that we deposit 100 rubles for a year, then they will be credited only at the end of the year. Accordingly, by the end of the deposit we will receive rubles. Compound interest- this is an option in which it occurs interest capitalization, i.e. their addition to the deposit amount and subsequent calculation of income not from the initial, but from the accumulated deposit amount. Capitalization does not occur constantly, but with some frequency. As a rule, such periods are equal and most often banks use a month, quarter or year. Let’s assume that we deposit the same rubles annually, but with monthly capitalization of the deposit. What are we doing? Do you understand everything here? If not, let's figure it out step by step. We brought rubles to the bank. By the end of the month, we should have an amount in our account consisting of our rubles plus interest on them, that is: Agree? We can take it out of brackets and then we get: Agree, this formula is already more similar to what we wrote at the beginning. All that's left is to figure out the percentages In the problem statement we are told about annual rates. As you know, we don't multiply by - we convert percentages to decimals, that is: Right? Now you may ask, where did the number come from? Very simple! Realized it? Now try to write what this part of the formula would look like if I said that interest is calculated daily. Well done! Let's return to our task: write how much will be credited to our account in the second month, taking into account that interest is accrued on the accumulated deposit amount. Or, in other words: I think that you have already noticed a pattern and saw a geometric progression in all this. Write what its member will be equal to, or, in other words, what amount of money we will receive at the end of the month. As you can see, if you put money in a bank for a year at a simple interest rate, you will receive rubles, and if at a compound interest rate, you will receive rubles. The benefit is small, but this only happens during the th year, but for more a long period capitalization is much more profitable: Let's look at another type of problem involving compound interest. After what you have figured out, it will be elementary for you. So, the task: The Zvezda company began investing in the industry in 2000, with capital in dollars. Every year since 2001, it has received a profit that is equal to the previous year's capital. How much profit will the Zvezda company receive at the end of 2003 if profits were not withdrawn from circulation? Capital of the Zvezda company in 2000. Or we can write briefly: For our case: 2000, 2001, 2002 and 2003. Answers: 2003, 2004, 2005, 2006, 2007. 2005, 2006, 2007. 1) Geometric progression ( ) is a numerical sequence, the first term of which is different from zero, and each term, starting from the second, is equal to the previous one, multiplied by the same number. This number is called the denominator of a geometric progression. 2) The equation of the terms of the geometric progression is . 3) can take any values except and. 4) , with - property of geometric progression (adjacent terms) or When you find it, don’t forget that there should be two answers. For example, 5) The sum of the terms of the geometric progression is calculated by the formula: If the progression is infinitely decreasing, then: IMPORTANT! We use the formula for the sum of terms of an infinitely decreasing geometric progression only if the condition explicitly states that we need to find the sum of an infinite number of terms. 6) Problems involving compound interest are also calculated using the formula for the th term of a geometric progression, provided that cash were not withdrawn from circulation: Geometric progression( ) is a numerical sequence, the first term of which is different from zero, and each term, starting from the second, is equal to the previous one, multiplied by the same number. This number is called denominator of a geometric progression. Denominator of geometric progression can take any value except and. Equation of terms of geometric progression - . Sum of terms of a geometric progression calculated by the formula: Lesson on the topic “Infinitely decreasing geometric progression” (algebra, 10th grade)
The purpose of the lesson: introducing students to a new type of sequence - an infinitely decreasing geometric progression. Equipment: projector, screen. Lesson type: lesson - learning a new topic. During the classes I
. Org. moment. State the topic and purpose of the lesson. II
. Updating students' knowledge. In 9th grade you studied arithmetic and geometric progressions. Questions 1. Definition of arithmetic progression. (An arithmetic progression is a sequence in which each member, starting from the second, is equal to the previous member added to the same number). 2. Formula n th term of the arithmetic progression ( 3. Formula for the sum of the first n terms of an arithmetic progression. ( 4. Definition of geometric progression. (A geometric progression is a sequence of non-zero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number). 5. Formula n th term of the geometric progression ( 6. Formula for the sum of the first n members of a geometric progression. ( 7. What other formulas do you know? ( 5. For geometric progression 6. For geometric progression 7. Exponentially b
3
= 8
And b
5
= 2
. Find b
4
. (4) 8. Exponentially b
3
= 8
And b
5
= 2
. Find b
1
And
q
. 9. Exponentially b
3
= 8
And b
5
= 2
. Find S
5
. (62) III
. Learning a new topic(demonstration of presentation). Consider a square with a side equal to 1. Let's draw another square whose side is half the size of the first square, then another one whose side is half the second, then the next one, etc. Each time the side of the new square is equal to half of the previous one. As a result, we received a sequence of sides of squares And, what is very important, the more we build such squares, the smaller the side of the square will be. For example, Those. As the number n increases, the terms of the progression approach zero. Using this figure, you can consider another sequence. For example, the sequence of areas of squares: Let's look at another example. An equilateral triangle with sides equal to 1 cm. Let's construct the following triangle with the vertices in the midpoints of the sides of the 1st triangle, according to the theorem about the midline of the triangle - the side of the 2nd is equal to half the side of the first, the side of the 3rd is equal to half the side of the 2nd, etc. Again we obtain a sequence of lengths of the sides of triangles. If we consider a geometric progression with a negative denominator. Then, again, with increasing numbers n terms of the progression approach zero. Let's pay attention to the denominators of these sequences. Everywhere the denominators were less than 1 in absolute value. We can conclude: a geometric progression will be infinitely decreasing if the modulus of its denominator is less than 1. Definition:
A geometric progression is said to be infinitely decreasing if the modulus of its denominator is less than one. Using the definition, you can decide whether a geometric progression is infinitely decreasing or not. Task Is the sequence an infinitely decreasing geometric progression if it is given by the formula: Solution: this geometric progression is infinitely decreasing. b) this sequence is not an infinitely decreasing geometric progression. Consider a square with a side equal to 1. Divide it in half, one of the halves in half, etc. The areas of all the resulting rectangles form an infinitely decreasing geometric progression: The sum of the areas of all rectangles obtained in this way will be equal to the area of the 1st square and equal to 1.
The task of finding the first term of a progression
Zeno's famous paradox with the fast Achilles and the slow tortoise
Tasks:
formulating an initial idea of the limit of a numerical sequence;
acquaintance with another way to convert infinite periodic fractions into ordinary ones using the formula for the sum of an infinitely decreasing geometric progression;
development of intellectual qualities of schoolchildren’s personality such as logical thinking, ability to make evaluative actions, and generalization;
fostering activity, mutual assistance, collectivism, and interest in the subject.Download:
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Geometric progression. Comprehensive guide with examples (2019)
Number sequence
Why is geometric progression needed and its history?
Geometric progression.
Let's try to "depersonalize" this formula- Let's put it in general form and get:Infinitely decreasing geometric progression.
First, let's write down some geometric progression consisting of terms.
Let's say, then:
Let's see what you got. Here's the graph I came up with:
Property of geometric progression.
Let's try to add two expressions and we get:
What did you get?
and correspondingly:
Given: ,
Find:The sum of the terms of a geometric progression.
Right.
That is:
If the barn is m high and m wide, its length would have to extend for km, i.e. twice as far as from the Earth to the Sun.
A student of class 5A Vasya fell ill with the flu, but continues to go to school. Every day Vasya infects two people, who, in turn, infect two more people, and so on. There are only people in the class. In how many days will the whole class be sick with the flu?
or
Problems on calculating compound interest.
I repeat: the problem statement says about ANNUAL interest that accrues MONTHLY. As you know, in a year of months, accordingly, the bank will charge us a portion of the annual interest per month:
Did you manage? Let's compare the results:
Here's what I got:
Did? Let's check!
- capital of the Zvezda company in 2001.
- capital of the Zvezda company in 2002.
- capital of the Zvezda company in 2003.
Respectively:
rubles
Please note that in this problem we do not have a division either by or by, since the percentage is given ANNUALLY and it is calculated ANNUALLY. That is, when reading a problem on compound interest, pay attention to what percentage is given and in what period it is calculated, and only then proceed to calculations.
Now you know everything about geometric progression.Training.
MDM Capital Company:
- increases by 100%, that is, 2 times.
Respectively:
rubles
MSK Cash Flows company:
- increases by, that is, by times.
Respectively:
rubles
rublesLet's summarize.
, at (equidistant terms)
or
orGEOMETRIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS
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, Where 
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; 
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, 
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find the fifth term. 
find n th member. 
forming a geometric progression with the denominator .
. And, again, if n increases indefinitely, then the area approaches zero as close as you like.
at 
.
.
; 
.. We'll find q
.
; 
; 
; 
.
