>>Math: Geometric progression
For the convenience of the reader, this paragraph is constructed exactly according to the same plan that we followed in the previous paragraph.
1. Basic concepts.
Definition. A numerical sequence, all members of which are different from 0 and each member of which, starting from the second, is obtained from the previous member by multiplying it by the same number is called a geometric progression. In this case, the number 5 is called the denominator of a geometric progression.
Thus, a geometric progression is a numerical sequence (b n) defined recurrently by the relations
Is it possible to look at a number sequence and determine whether it is a geometric progression? Can. If you are convinced that the ratio of any member of the sequence to the previous member is constant, then you have a geometric progression.
Example 1.
1, 3, 9, 27, 81,... .
b 1 = 1, q = 3.
Example 2.
This is a geometric progression that has
Example 3.
This is a geometric progression that has
Example 4.
8, 8, 8, 8, 8, 8,....
This is a geometric progression in which b 1 - 8, q = 1.
Note that this sequence is also an arithmetic progression (see example 3 from § 15).
Example 5.
2,-2,2,-2,2,-2.....
This is a geometric progression in which b 1 = 2, q = -1.
Obviously, a geometric progression is an increasing sequence if b 1 > 0, q > 1 (see example 1), and a decreasing sequence if b 1 > 0, 0< q < 1 (см. пример 2).
To indicate that the sequence (b n) is a geometric progression, the following notation is sometimes convenient:
The icon replaces the phrase “geometric progression”.
Let us note one curious and at the same time quite obvious property of geometric progression:
If the sequence 
is a geometric progression, then the sequence of squares, i.e. 
is a geometric progression.
In the second geometric progression, the first term is equal to and equal to q 2.
If in a geometric progression we discard all terms following b n , we get a finite geometric progression 
In further paragraphs of this section we will consider the most important properties of geometric progression.
2. Formula for the nth term of a geometric progression.
Consider a geometric progression 
denominator q. We have:
It is not difficult to guess that for any number n the equality is true
This is the formula for the nth term of a geometric progression.
Comment.
If you have read the important remark from the previous paragraph and understood it, then try to prove formula (1) using the method of mathematical induction, just as was done for the formula for the nth term of an arithmetic progression.
Let's rewrite the formula for the nth term of the geometric progression
and introduce the notation: We get y = mq 2, or, in more detail, 
The argument x is contained in the exponent, so this function is called an exponential function. This means that a geometric progression can be considered as an exponential function defined on the set N of natural numbers. In Fig. 96a shows the graph of the function Fig. 966 - function graph 
In both cases, we have isolated points (with abscissas x = 1, x = 2, x = 3, etc.) lying on a certain curve (both figures show the same curve, only differently located and depicted in different scales). This curve is called an exponential curve. More details about the exponential function and its graph will be discussed in the 11th grade algebra course.
Let's return to examples 1-5 from the previous paragraph.
1) 1, 3, 9, 27, 81,... . This is a geometric progression for which b 1 = 1, q = 3. Let’s create the formula for the nth term 
2) 
This is a geometric progression for which Let's create a formula for the nth term 
This is a geometric progression that has 
Let's create the formula for the nth term 
4) 8, 8, 8, ..., 8, ... . This is a geometric progression for which b 1 = 8, q = 1. Let’s create the formula for the nth term 
5) 2, -2, 2, -2, 2, -2,.... This is a geometric progression in which b 1 = 2, q = -1. Let's create the formula for the nth term 
Example 6.
Given a geometric progression
In all cases, the solution is based on the formula of the nth term of the geometric progression
a) Putting n = 6 in the formula for the nth term of the geometric progression, we obtain
b) We have
Since 512 = 2 9, we get n - 1 = 9, n = 10.
d) We have
Example 7.
The difference between the seventh and fifth terms of the geometric progression is 48, the sum of the fifth and sixth terms of the progression is also 48. Find the twelfth term of this progression.
First stage. Drawing up a mathematical model.
The conditions of the problem can be briefly written as follows:
Using the formula for the nth term of a geometric progression, we get:
Then the second condition of the problem (b 7 - b 5 = 48) can be written as
The third condition of the problem (b 5 + b 6 = 48) can be written as
As a result, we obtain a system of two equations with two variables b 1 and q:
which, in combination with condition 1) written above, represents a mathematical model of the problem.
Second phase.
Working with the compiled model. Equating the left sides of both equations of the system, we obtain:
(we divided both sides of the equation by the non-zero expression b 1 q 4).
From the equation q 2 - q - 2 = 0 we find q 1 = 2, q 2 = -1. Substituting the value q = 2 into the second equation of the system, we get 
Substituting the value q = -1 into the second equation of the system, we obtain b 1 1 0 = 48; this equation has no solutions.
So, b 1 =1, q = 2 - this pair is the solution to the compiled system of equations.
Now we can write down the geometric progression discussed in the problem: 1, 2, 4, 8, 16, 32, ... .
Third stage.
Answer to the problem question. You need to calculate b 12. We have
Answer: b 12 = 2048.
3. Formula for the sum of terms of a finite geometric progression.
Let a finite geometric progression be given
Let us denote by S n the sum of its terms, i.e.
Let us derive a formula for finding this amount.
Let's start with the simplest case, when q = 1. Then the geometric progression b 1 , b 2 , b 3 ,..., bn consists of n numbers equal to b 1 , i.e. the progression looks like b 1, b 2, b 3, ..., b 4. The sum of these numbers is nb 1.
Let now q = 1 To find S n, we apply an artificial technique: we perform some transformations of the expression S n q. We have:
When performing transformations, we, firstly, used the definition of a geometric progression, according to which (see the third line of reasoning); secondly, they added and subtracted, which is why the meaning of the expression, of course, did not change (see the fourth line of reasoning); thirdly, we used the formula for the nth term of a geometric progression:
From formula (1) we find:
This is the formula for the sum of n terms of a geometric progression (for the case when q = 1).
Example 8.
Given a finite geometric progression
a) the sum of the terms of the progression; b) the sum of the squares of its terms.
b) Above (see p. 132) we have already noted that if all terms of a geometric progression are squared, then we get a geometric progression with the first term b 2 and the denominator q 2. Then the sum of the six terms of the new progression will be calculated by
Example 9.
Find the 8th term of the geometric progression for which
In fact, we have proven the following theorem.
A numerical sequence is a geometric progression if and only if the square of each of its terms, except for the first Theorem (and the last, in the case of a finite sequence), is equal to the product of the preceding and subsequent terms (a characteristic property of a geometric progression).
Arithmetic and geometric progressions
Theoretical information
Theoretical information
Arithmetic progression |
Geometric progression |
|
Definition |
Arithmetic progression a n is a sequence in which each member, starting from the second, is equal to the previous member added to the same number d (d- progression difference) |
Geometric progression b n 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 q (q- denominator of progression) |
Recurrence formula |
For any natural n |
For any natural n |
Formula nth term |
a n = a 1 + d (n – 1) |
b n = 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+ 21 d
By condition:
a 1= -6, then a 22= -6 + 21 d .
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 method (using the n-term formula)
According to the formula for the nth term of a geometric progression:
b 5 = b 1 ∙ q 5 - 1 = b 1 ∙ q 4.
Because b 1 = -3,
2nd method (using recurrent 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:
.
Let's substitute the data into 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 use in this case?
By condition, the formula for the nth term of the original progression is known ( a n) a n= 3n - 4. You can find immediately and a 1, And a 16 without finding d. Therefore, we will 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 the geometric progression are written:
Find the term of the progression indicated by x.
When solving, we will use the formula for the nth term b n = b 1 ∙ q n - 1 for geometric progressions. The first term of the progression. To find the denominator of the progression q, you need to take any of the given terms of the progression and divide by the previous one. In our example, we can take and divide by. We obtain that q = 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, select the one for which the condition is satisfied a 27 > 9:
Since the given 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.
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 tell which one is first, which 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.
Why is geometric progression needed and its history?
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 a general understanding of. 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.
Geometric progression.
Let's say we have a number sequence:
You will immediately answer that this is easy and the name of such a sequence is with the difference of its members. How about this:
If you subtract the previous number from the subsequent 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:
- Geometric progression – 3, 6.
- Arithmetic progression – 2, 4.
- It is neither an arithmetic nor a geometric progression - 1, 5, 7.
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.
Let’s try to “depersonalize” this formula - let’s put it in general form and get:
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.
Infinitely decreasing geometric progression.
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?
First, let's write down some geometric progression consisting of terms.
Let's say, then:
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.
Let's see what you got. Here's the graph I came up with:
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.
Property of geometric progression.
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 a picture and try to do various manipulations with them in order to arrive at the 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 perform various actions with them, as a result of which we can get.
Addition.
Let's try to add two expressions and we get:
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? Correctly, to find we need to take the square root of 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 to write this formula in general form. 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 possible value during the calculation, then you’re great and can immediately move on to training, and if you forgot, read what is discussed below and pay attention to why both roots must be written down 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.
What did you get?
Now look carefully again.
and correspondingly:
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 that is smaller. The main thing is that it is the same for both given numbers.
Practice with specific examples, just be extremely careful!
- , . Find.
- , . Find.
- , . Find.
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 careful examination of the serial numbers of the 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 a 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:
Given: ,
Find:
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.
The sum of the terms of a geometric progression.
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?
Right.
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.
That is:
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 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.
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.
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?
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. The total sum of the progression terms is equal to the number of 5A students. 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 a financial pyramid in which money was given if you brought two other participants, then the person (or in general) would not bring anyone, accordingly, would lose everything that 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:
or
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.
- Find the sum of the first terms of the geometric progression with and.
- Find the sum of the terms of an infinitely decreasing geometric progression with and.
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.
Problems on calculating compound interest.
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 for deposits: this is the term, and additional services, and interest with two different ways of calculating it - 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 happens 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 do not multiply by - we convert percentages to decimal fractions, that is:
Right? Now you may ask, where did the number come from? Very simple!
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:
Realized it? Now try to write what this part of the formula would look like if I said that interest is calculated daily.
Did you manage? Let's compare the results:
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.
Here's what I got:
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.
Did? Let's check!
As you can see, if you put money in the 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 a longer 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.
- capital of the Zvezda company in 2001.
- capital of the Zvezda company in 2002.
- capital of the Zvezda company in 2003.
Or we can write briefly:
For our case:
2000, 2001, 2002 and 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.
- Find the term of the geometric progression if it is known that, and
- Find the sum of the first terms of the geometric progression if it is known that, and
- The MDM Capital company began investing in the industry in 2003, with capital in dollars. Every year since 2004, it has received a profit that is equal to the previous year's capital. The MSK Cash Flows company began investing in the industry in 2005 in the amount of $10,000, starting to make a profit in 2006 in the amount of. By how many dollars is the capital of one company greater than the other at the end of 2007, if profits were not withdrawn from circulation?
Answers:
- Since the problem statement does not say that the progression is infinite and it is required to find the sum of a specific number of its terms, the calculation is carried out according to the formula:
MDM Capital Company:2003, 2004, 2005, 2006, 2007.
- increases by 100%, that is, 2 times.
Respectively:
rubles
MSK Cash Flows company:2005, 2006, 2007.
- increases by, that is, by times.
Respectively:
rubles
rubles
Let's summarize.
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.
- if, then all subsequent terms of the progression have the same sign - they are positive;
- if, then all subsequent terms of the progression alternate signs;
- when – the progression is called infinitely decreasing.
4) , at – property of geometric progression (adjacent terms)
or
, at (equidistant terms)
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:
or
or
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 on compound interest are also calculated using the formula of the th term of a geometric progression, provided that funds have not been withdrawn from circulation:
GEOMETRIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS
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.
- If, then all subsequent terms of the progression have the same sign - they are positive;
- if, then all subsequent members of the progression alternate signs;
- when – the progression is called infinitely decreasing.
Equation of terms of geometric progression - .
Sum of terms of a geometric progression calculated by the formula:
or
If the progression is infinitely decreasing, then:
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This number is called the denominator of a geometric progression, i.e. each term differs from the previous one by q times. (We will assume that q ≠ 1, otherwise everything is too trivial). It is easy to see that the general formula for the nth term of the geometric progression is b n = b 1 q n – 1 ; terms with numbers b n and b m differ by q n – m times.Already in Ancient Egypt they knew not only arithmetic, but also geometric progression. Here, for example, is a problem from the Rhind papyrus: “Seven faces have seven cats; Each cat eats seven mice, each mouse eats seven ears of corn, and each ear of barley can grow seven measures of barley. How large are the numbers in this series and their sum?
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Rice. 1. Ancient Egyptian geometric progression problem |
This task was repeated many times with different variations among other peoples at other times. For example, in written in the 13th century. “The Book of the Abacus” by Leonardo of Pisa (Fibonacci) has a problem in which 7 old women appear on their way to Rome (obviously pilgrims), each of whom has 7 mules, each of which has 7 bags, each of which contains 7 loaves , each of which has 7 knives, each of which has 7 sheaths. The problem asks how many objects there are.
The sum of the first n terms of the geometric progression S n = b 1 (q n – 1) / (q – 1) . This formula can be proven, for example, like this: S n = b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n – 1.
Add the number b 1 q n to S n and get:
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From here S n (q – 1) = b 1 (q n – 1), and we get the necessary formula.
Already on one of the clay tablets of Ancient Babylon, dating back to the 6th century. BC e., contains the sum 1 + 2 + 2 2 + 2 3 + ... + 2 9 = 2 10 – 1. True, as in a number of other cases, we do not know how this fact was known to the Babylonians.
The rapid increase in geometric progression in a number of cultures, in particular in Indian, is repeatedly used as a visual symbol of the vastness of the universe. In the famous legend about the appearance of chess, the ruler gives its inventor the opportunity to choose the reward himself, and he asks for the number of wheat grains that will be obtained if one is placed on the first square of the chessboard, two on the second, four on the third, eight on the fourth, and etc., each time the number doubles. Vladyka thought that at most we were talking about a few bags, but he miscalculated. It is easy to see that for all 64 squares of the chessboard the inventor would have to receive (2 64 - 1) grains, which is expressed as a 20-digit number; even if the entire surface of the Earth was sown, it would take at least 8 years to collect the required amount of grains. This legend is sometimes interpreted as indicating the virtually unlimited possibilities hidden in the game of chess.
It is easy to see that this number is really 20-digit:
2 64 = 2 4 ∙ (2 10) 6 = 16 ∙ 1024 6 ≈ 16 ∙ 1000 6 = 1.6∙10 19 (a more accurate calculation gives 1.84∙10 19). But I wonder if you can find out what digit this number ends with?
A geometric progression can be increasing if the denominator is greater than 1, or decreasing if it is less than one. In the latter case, the number q n for sufficiently large n can become arbitrarily small. While the increasing geometric progression increases unexpectedly quickly, the decreasing geometric progression decreases just as quickly.
The larger n, the weaker the number q n differs from zero, and the closer the sum of n terms of the geometric progression S n = b 1 (1 – q n) / (1 – q) to the number S = b 1 / (1 – q). (For example, F. Viet reasoned this way). The number S is called the sum of an infinitely decreasing geometric progression. However, for many centuries the question of what is the meaning of summing the ENTIRE geometric progression, with its infinite number of terms, was not clear enough to mathematicians.
A decreasing geometric progression can be seen, for example, in Zeno’s aporias “Half Division” and “Achilles and the Tortoise.” In the first case, it is clearly shown that the entire road (assuming length 1) is the sum of an infinite number of segments 1/2, 1/4, 1/8, etc. This is, of course, the case from the point of view of ideas about a finite sum infinite geometric progression. And yet - how can this be?
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Rice. 2. Progression with a coefficient of 1/2 |
In the aporia about Achilles, the situation is a little more complicated, because here the denominator of the progression is not 1/2, but some other number. Let, for example, Achilles run with speed v, the tortoise moves with speed u, and the initial distance between them is l. Achilles will cover this distance in time l/v, and during this time the turtle will move a distance lu/v. When Achilles runs through this segment, the distance between him and the turtle will become equal to l (u /v) 2, etc. It turns out that catching up with the turtle means finding the sum of an infinitely decreasing geometric progression with the first term l and the denominator u /v. This sum - the segment that Achilles will eventually run to the meeting place with the turtle - is equal to l / (1 – u /v) = lv / (v – u). But, again, how this result should be interpreted and why it makes any sense at all was not very clear for a long time.
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Rice. 3. Geometric progression with a coefficient of 2/3 |
Archimedes used the sum of a geometric progression to determine the area of a parabola segment. Let this segment of the parabola be delimited by the chord AB and let the tangent at point D of the parabola be parallel to AB. Let C be the midpoint of AB, E the midpoint of AC, F the midpoint of CB. Let's draw lines parallel to DC through points A, E, F, B; Let the tangent drawn at point D intersect these lines at points K, L, M, N. Let's also draw segments AD and DB. Let the line EL intersect the line AD at point G, and the parabola at point H; line FM intersects line DB at point Q, and the parabola at point R. According to the general theory of conic sections, DC is the diameter of a parabola (that is, a segment parallel to its axis); it and the tangent at point D can serve as coordinate axes x and y, in which the equation of the parabola is written as y 2 = 2px (x is the distance from D to any point of a given diameter, y is the length of a segment parallel to a given tangent from this point of diameter to some point on the parabola itself).
By virtue of the parabola equation, DL 2 = 2 ∙ p ∙ LH, DK 2 = 2 ∙ p ∙ KA, and since DK = 2DL, then KA = 4LH. Because KA = 2LG, LH = HG. The area of segment ADB of a parabola is equal to the area of triangle ΔADB and the areas of segments AHD and DRB combined. In turn, the area of the segment AHD is similarly equal to the area of the triangle AHD and the remaining segments AH and HD, with each of which you can perform the same operation - split into a triangle (Δ) and the two remaining segments (), etc.:
The area of the triangle ΔAHD is equal to half the area of the triangle ΔALD (they have a common base AD, and the heights differ by 2 times), which, in turn, is equal to half the area of the triangle ΔAKD, and therefore half the area of the triangle ΔACD. Thus, the area of the triangle ΔAHD is equal to a quarter of the area of the triangle ΔACD. Likewise, the area of triangle ΔDRB is equal to one-quarter of the area of triangle ΔDFB. So, the areas of the triangles ΔAHD and ΔDRB, taken together, are equal to a quarter of the area of the triangle ΔADB. Repeating this operation when applied to segments AH, HD, DR and RB will select triangles from them, the area of which, taken together, will be 4 times less than the area of triangles ΔAHD and ΔDRB, taken together, and therefore 16 times less, than the area of the triangle ΔADB. And so on:
Thus, Archimedes proved that “every segment contained between a straight line and a parabola constitutes four-thirds of a triangle having the same base and equal height.”
22.09.2018 22:00Geometric progression, along with arithmetic progression, is an important number series that is studied in the school algebra course in the 9th grade. In this article we will look at the denominator of a geometric progression and how its value affects its properties.
Definition of geometric progression
First, let's give the definition of this number series. A geometric progression is a series of rational numbers that is formed by sequentially multiplying its first element by a constant number called the denominator.
For example, the numbers in the series 3, 6, 12, 24, ... are a geometric progression, because if you multiply 3 (the first element) by 2, you get 6. If you multiply 6 by 2, you get 12, and so on.
The members of the sequence under consideration are usually denoted by the symbol ai, where i is an integer indicating the number of the element in the series.
The above definition of progression can be written in mathematical language as follows: an = bn-1 * a1, where b is the denominator. It's easy to check this formula: if n = 1, then b1-1 = 1, and we get a1 = a1. If n = 2, then an = b * a1, and we again come to the definition of the series of numbers in question. Similar reasoning can be continued for large values of n.
Denominator of geometric progression
The number b completely determines what character the entire number series will have. The denominator b can be positive, negative, or greater than or less than one. All of the above options lead to different sequences:
- b > 1. There is an increasing series of rational numbers. For example, 1, 2, 4, 8, ... If element a1 is negative, then the entire sequence will increase only in absolute value, but decrease depending on the sign of the numbers.
- b = 1. Often this case is not called a progression, since there is an ordinary series of identical rational numbers. For example, -4, -4, -4.
Formula for amount
Before moving on to the consideration of specific problems using the denominator of the type of progression under consideration, an important formula for the sum of its first n elements should be given. The formula looks like: Sn = (bn - 1) * a1 / (b - 1).
You can obtain this expression yourself if you consider the recursive sequence of terms of the progression. Also note that in the above formula it is enough to know only the first element and the denominator to find the sum of an arbitrary number of terms.
Infinitely decreasing sequence
An explanation was given above of what it is. Now, knowing the formula for Sn, let's apply it to this number series. Since any number whose modulus does not exceed 1 tends to zero when raised to large powers, that is, b∞ => 0 if -1
Since the difference (1 - b) will always be positive, regardless of the value of the denominator, the sign of the sum of an infinitely decreasing geometric progression S∞ is uniquely determined by the sign of its first element a1.
Now let's look at several problems where we will show how to apply the acquired knowledge on specific numbers.
Task No. 1. Calculation of unknown elements of progression and sum
Given a geometric progression, the denominator of the progression is 2, and its first element is 3. What will its 7th and 10th terms be equal to, and what is the sum of its seven initial elements?
The condition of the problem is quite simple and involves the direct use of the above formulas. So, to calculate element number n, we use the expression an = bn-1 * a1. For the 7th element we have: a7 = b6 * a1, substituting the known data, we get: a7 = 26 * 3 = 192. We do the same for the 10th term: a10 = 29 * 3 = 1536.
Let's use the well-known formula for the sum and determine this value for the first 7 elements of the series. We have: S7 = (27 - 1) * 3 / (2 - 1) = 381.
Problem No. 2. Determining the sum of arbitrary elements of a progression
Let -2 be equal to the denominator of the geometric progression bn-1 * 4, where n is an integer. It is necessary to determine the sum from the 5th to the 10th element of this series, inclusive.
The problem posed cannot be solved directly using known formulas. It can be solved using 2 different methods. For completeness of presentation of the topic, we present both.
Method 1. The idea is simple: you need to calculate the two corresponding sums of the first terms, and then subtract the other from one. We calculate the smaller amount: S10 = ((-2)10 - 1) * 4 / (-2 - 1) = -1364. Now we calculate the larger sum: S4 = ((-2)4 - 1) * 4 / (-2 - 1) = -20. Note that in the last expression only 4 terms were summed, since the 5th is already included in the amount that needs to be calculated according to the conditions of the problem. Finally, we take the difference: S510 = S10 - S4 = -1364 - (-20) = -1344.
Method 2. Before substituting numbers and counting, you can obtain a formula for the sum between the m and n terms of the series in question. We do exactly the same as in method 1, only we first work with the symbolic representation of the amount. We have: Snm = (bn - 1) * a1 / (b - 1) - (bm-1 - 1) * a1 / (b - 1) = a1 * (bn - bm-1) / (b - 1). You can substitute known numbers into the resulting expression and calculate the final result: S105 = 4 * ((-2)10 - (-2)4) / (-2 - 1) = -1344.
Problem No. 3. What is the denominator?
Let a1 = 2, find the denominator of the geometric progression, provided that its infinite sum is 3, and it is known that this is a decreasing series of numbers.
Based on the conditions of the problem, it is not difficult to guess which formula should be used to solve it. Of course, for the sum of the progression infinitely decreasing. We have: S∞ = a1 / (1 - b). From where we express the denominator: b = 1 - a1 / S∞. It remains to substitute the known values and get the required number: b = 1 - 2 / 3 = -1 / 3 or -0.333(3). We can qualitatively check this result if we remember that for this type of sequence the modulus b should not go beyond 1. As can be seen, |-1 / 3|
Task No. 4. Restoring a series of numbers
Let 2 elements of a number series be given, for example, the 5th is equal to 30 and the 10th is equal to 60. It is necessary to reconstruct the entire series from these data, knowing that it satisfies the properties of a geometric progression.
To solve the problem, you must first write down the corresponding expression for each known term. We have: a5 = b4 * a1 and a10 = b9 * a1. Now divide the second expression by the first, we get: a10 / a5 = b9 * a1 / (b4 * a1) = b5. From here we determine the denominator by taking the fifth root of the ratio of the terms known from the problem statement, b = 1.148698. We substitute the resulting number into one of the expressions for the known element, we get: a1 = a5 / b4 = 30 / (1.148698)4 = 17.2304966.
Thus, we found the denominator of the progression bn, and the geometric progression bn-1 * 17.2304966 = an, where b = 1.148698.
Where are geometric progressions used?
If there were no practical application of this number series, then its study would be reduced to purely theoretical interest. But such an application exists.
Below are the 3 most famous examples:
- Zeno's paradox, in which the nimble Achilles cannot catch up with the slow tortoise, is solved using the concept of an infinitely decreasing sequence of numbers.
- If you place wheat grains on each square of the chessboard so that on the 1st square you put 1 grain, on the 2nd - 2, on the 3rd - 3, and so on, then to fill all the squares of the board you will need 18446744073709551615 grains!
- In the game "Tower of Hanoi", in order to move disks from one rod to another, it is necessary to perform 2n - 1 operations, that is, their number grows exponentially with the number n of disks used.
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