Geometric 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 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 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 we look at specific tasks 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 in 2 ways various 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 let's calculate a large amount: 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 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∞. All that remains is to substitute 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.
Geometric progression no less important in mathematics compared to arithmetic. A geometric progression is a sequence of numbers b1, b2,..., b[n], each next term of which is obtained by multiplying the previous one by a constant number. This number, which also characterizes the rate of growth or decrease of progression, is called denominator of geometric progression and denote
To completely specify a geometric progression, in addition to the denominator, it is necessary to know or determine its first term. For positive value denominator progression is a monotonic sequence, and if this sequence of numbers is monotonically decreasing and if it is monotonically increasing. The case when the denominator is equal to one is not considered in practice, since we have a sequence of identical numbers, and their summation is of no practical interest
General term of geometric progression calculated by the formula
Sum of the first n terms of a geometric progression determined by the formula
Let's look at solutions to classic geometric progression problems. Let's start with the simplest ones to understand.
Example 1. The first term of a geometric progression is 27, and its denominator is 1/3. Find the first six terms of the geometric progression.
Solution: Let us write the problem condition in the form
For calculations we use the formula for the nth term of a geometric progression
Based on it, we find the unknown terms of the progression
As you can see, calculating the terms of a geometric progression is not difficult. The progression itself will look like this
Example 2. The first three terms of the geometric progression are given: 6; -12; 24. Find the denominator and its seventh term.
Solution: We calculate the denominator of the geomitric progression based on its definition
We have obtained an alternating geometric progression whose denominator is equal to -2. The seventh term is calculated using the formula
This solves the problem.
Example 3. A geometric progression is given by two of its terms 
. Find the tenth term of the progression.
Solution:
Let's write the given values using formulas
According to the rules, we would need to find the denominator and then look for the desired value, but for the tenth term we have
The same formula can be obtained based on simple manipulations with the input data. Divide the sixth term of the series by another, and as a result we get
If the resulting value is multiplied by the sixth term, we get the tenth
Thus, for such tasks, using simple transformations to quick way you can find the right solution.
Example 4. Geometric progression is given by recurrent formulas
Find the denominator of the geometric progression and the sum of the first six terms.
Solution:
Let's write the given data in the form of a system of equations
Express the denominator by dividing the second equation by the first
Let's find the first term of the progression from the first equation
Let us calculate the following five terms to find the sum of the geometric progression
Mathematics is whatpeople control nature and themselves.
Soviet mathematician, academician A.N. Kolmogorov
Geometric progression.
Along with problems on arithmetic progressions, problems related to the concept of geometric progression are also common in entrance examinations in mathematics. To successfully solve such problems, you need to know the properties of geometric progressions and have good skills in using them.
This article is devoted to the presentation of the basic properties of geometric progression. Examples of solving typical problems are also provided here., borrowed from the tasks of entrance examinations in mathematics.
Let us first note the basic properties of the geometric progression and recall the most important formulas and statements, related to this concept.
Definition. A number sequence is called a geometric progression if each number, starting from the second, is equal to the previous one, multiplied by the same number. The number is called the denominator of a geometric progression.
For geometric progressionthe formulas are valid
, (1)
Where . Formula (1) is called the formula of the general term of a geometric progression, and formula (2) represents the main property of a geometric progression: each term of the progression coincides with the geometric mean of its neighboring terms and .
Note, that it is precisely because of this property that the progression in question is called “geometric”.
The above formulas (1) and (2) are generalized as follows:
, (3)
To calculate the amount first members of a geometric progressionformula applies
If we denote , then
Where . Since , formula (6) is a generalization of formula (5).
In the case when and geometric progressionis infinitely decreasing. To calculate the amountof all terms of an infinitely decreasing geometric progression, the formula is used
. (7)
For example , using formula (7) we can show, What
Where . These equalities are obtained from formula (7) under the condition that , (first equality) and , (second equality).
Theorem. If , then
Proof. If , then
The theorem has been proven.
Let's move on to consider examples of solving problems on the topic “Geometric progression”.
Example 1. Given: , and . Find .
Solution. If we apply formula (5), then
Answer: .
Example 2. Let it be. Find .
Solution. Since and , we use formulas (5), (6) and obtain a system of equations
If the second equation of system (9) is divided by the first, then or . It follows from this that . Let's consider two cases.
1. If, then from the first equation of system (9) we have.
2. If , then .
Example 3. Let , and . Find .
Solution. From formula (2) it follows that or . Since , then or .
By condition . However, therefore. Since and then here we have a system of equations
If the second equation of the system is divided by the first, then or .
Since, the equation has a unique suitable root. In this case, it follows from the first equation of the system.
Taking into account formula (7), we obtain.
Answer: .
Example 4. Given: and . Find .
Solution. Since, then.
Since , then or
According to formula (2) we have . In this regard, from equality (10) we obtain or .
However, by condition, therefore.
Example 5. It is known that . Find .
Solution. According to the theorem, we have two equalities
Since , then or . Because , then .
Answer: .
Example 6. Given: and . Find .
Solution. Taking into account formula (5), we obtain
Since, then. Since , and , then .
Example 7. Let it be. Find .
Solution. According to formula (1) we can write
Therefore, we have or . It is known that and , therefore and .
Answer: .
Example 8. Find the denominator of an infinite decreasing geometric progression if
And .
Solution. From formula (7) it follows And . From here and from the conditions of the problem we obtain a system of equations
If the first equation of the system is squared, and then divide the resulting equation by the second equation, then we get
Or .
Answer: .
Example 9. Find all values for which the sequence , , is a geometric progression.
Solution. Let , and . According to formula (2), which defines the main property of a geometric progression, we can write or .
From here we get the quadratic equation, whose roots are And .
Let's check: if, then , and ; if , then , and .
In the first case we have and , and in the second – and .
Answer: , .
Example 10.Solve the equation
, (11)
where and .
Solution. The left side of equation (11) is the sum of an infinite decreasing geometric progression, in which and , subject to: and .
From formula (7) it follows, What . In this regard, equation (11) takes the form or . Suitable root quadratic equation is
Answer: .
Example 11. P sequence of positive numbersforms an arithmetic progression, A – geometric progression, what does it have to do with . Find .
Solution. Because arithmetic sequence, That (main property arithmetic progression). Because the, then or . This implies , that the geometric progression has the form. According to formula (2), then we write down that .
Since and , then . In this case, the expression takes the form or . By condition , so from Eq.we obtain a unique solution to the problem under consideration, i.e. .
Answer: .
Example 12. Calculate Sum
. (12)
Solution. Multiply both sides of equality (12) by 5 and get
If we subtract (12) from the resulting expression, That
or .
To calculate, we substitute the values \u200b\u200binto formula (7) and get . Since, then.
Answer: .
The examples of problem solving given here will be useful to applicants in preparing for entrance examinations. For a deeper study of problem solving methods, related to geometric progression, can be used teaching aids from the list of recommended literature.
1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Mir and Education, 2013. – 608 p.
2. Suprun V.P. Mathematics for high school students: additional sections school curriculum. – M.: Lenand / URSS, 2014. – 216 p.
3. Medynsky M.M. Full course elementary mathematics in problems and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. – 208 p.
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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? The general solution to 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:
The right side of 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.
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Number sequence
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.
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 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.
Geometric progression.
Let's say we have 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 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:
- 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 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 orange, knowing the members 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? 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.
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, which is smaller. The main thing is that it is the same for both given numbers.
Practice on 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 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:
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” given number, transform using the properties of the 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 financial pyramid, in which money was given if you brought two other participants, then the person (or in the general case) would not have brought anyone, and accordingly 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 special kind- 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 includes a term, additional services, 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!
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 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 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. MSK company Cash flows"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) , with - 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:
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
Now the most important thing.
You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough...
For what?
For successful completion Unified State Exam, for admission to college on a budget and, MOST IMPORTANTLY, for life.
I won’t convince you of anything, I’ll just say one thing...
People who received a good education, earn much more than those who did not receive it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?
GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
You won't be asked for theory during the exam.
You will need solve problems against time.
And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.
It's like in sports - you need to repeat it many times to win for sure.
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