How the geometric progression is solved. What is a geometric progression? Basic concepts

Arithmetic and geometric progressions

Theoretical information

	Arithmetic progression	Geometric progression
Definition	Arithmetic progression a n a sequence is called, each term of which, starting from the second, is equal to the previous term added with the same number d (d- difference of progressions)	Geometric progression b n is a sequence of nonzero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q is the denominator of the progression)
Recurrent formula	For any natural n a n + 1 = a n + d	For any natural n b n + 1 = b n ∙ q, b n ≠ 0
Nth term formula	a n = a 1 + d (n - 1)	b n = b 1 ∙ q n - 1, b n ≠ 0
Characteristic property
Sum of n-first members

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, so a 22= -6 + 21 d.

It is necessary to find the difference between the progressions:

d = a 2 - a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = - 48.

Answer : a 22 = -48.

Assignment 2

Find the fifth term of a geometric progression: -3; 6; ....

1st way (using the n-term formula)

According to the formula of the n-th member of a geometric progression:

b 5 = b 1 ∙ q 5 - 1 = b 1 ∙ q 4.

Because b 1 = -3,

2nd way (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.

Assignment 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 is .

Therefore:

.

Let's substitute the data into the formula:

Answer: 95.

Assignment 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 immediately find and a 1, and a 16 without finding d. Therefore, we will use the first formula.

Answer: 368.

Assignment 5

In arithmetic progression ( a n) a 1 = -6; a 2= -8. Find the twenty-second term in 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 between the progressions:

d = a 2 - a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = -48.

Answer : a 22 = -48.

Assignment 6

Several consecutive members of a geometric progression are written:

Find the term in the progression denoted by the letter x.

When solving, we use the formula for the nth term b n = b 1 ∙ q n - 1 for geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of the given members of the progression and divide by the previous one. In our example, you can take and divide by. We get that q = 3. Instead of n in the formula, we substitute 3, since it is necessary to find the third term given by a geometric progression.

Substituting the found values ​​into the formula, we get:

.

Answer : .

Assignment 7

From the arithmetic progressions given by the formula of the nth term, select the one for which the condition a 27 > 9:

Since the given condition must be fulfilled 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.

Assignment 8

In arithmetic progression a 1= 3, d = -1.5. Specify the largest n-value that satisfies the inequality a n > -6.

Let us now consider the question of summation of an infinite geometric progression. Let's call the partial sum of a given infinite progression the sum of its first terms. We denote the partial sum by the symbol

For every endless progression

one can compose an (also infinite) sequence of its partial sums

Let the sequence with unbounded increase have a limit

In this case, the number S, that is, the limit of the partial sums of the progression, is called the sum of the infinite progression. We will prove that an infinite decreasing geometric progression always has a sum, and derive a formula for this sum (you can also show that for, an infinite progression has no sum, it does not exist).

We write the expression for the partial sum as the sum of the terms of the progression by formula (91.1) and consider the limit of the partial sum for

From Theorem 89 it is known that for a decreasing progression; therefore, applying the theorem on the limit of the difference, we find

(the rule is also used here: the constant factor is taken out of the limit sign). The existence is proved, and at the same time a formula for the sum of an infinitely decreasing geometric progression is obtained:

Equality (92.1) can also be written in the form

It may seem paradoxical here that the sum of an infinite set of terms is assigned a well-defined finite value.

You can give a visual illustration to clarify this position. Consider a square with a side equal to one (Fig. 72). We divide this square with a horizontal line into two equal parts and attach the upper part to the lower one so that a rectangle with sides 2 and is formed. After that, we again divide the right half of this rectangle by a horizontal line in half and attach the upper part to the lower one (as shown in Fig. 72). Continuing this process, we are constantly transforming the original square with an area of ​​1 into equal-sized figures (taking the form of a staircase with thinning steps).

With the infinite continuation of this process, the entire area of ​​the square is decomposed into an infinite number of terms - the areas of rectangles with bases equal to 1, and the heights of the Squares of the rectangles just form an infinite decreasing progression, its sum

i.e., as expected, it is equal to the area of ​​the square.

Example. Find the sums of the following infinite progressions:

Solution, a) Note that for this progression, therefore, using formula (92.2), we find

b) Here, therefore, by the same formula (92.2) we have

c) We find that this progression has Therefore, this progression has no sum.

In Section 5, the application of the formula for the sum of the terms of an infinitely decreasing progression to the conversion of a periodic decimal fraction into an ordinary fraction was shown.

Exercises

1. The sum of the infinitely decreasing geometric progression is 3/5, and the sum of its first four terms is 13/27. Find the first term and denominator of the progression.

2. Find four numbers that form an alternating geometric progression, in which the second term is less than the first by 35, and the third is greater than the fourth by 560.

3. Show that if the sequence

forms an infinitely decreasing geometric progression, then the sequence

for any it forms an infinitely decreasing geometric progression. Will this statement hold for

Derive a formula for the product of the members of a geometric progression.

Lesson and presentation on the topic: "Number sequences. Geometric progression"

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Guys, today we will get acquainted with another type of progression.
The topic of today's lesson is geometric progression.

Geometric progression

Example. 1,2,4,8,16 ... Geometric progression, in which the first term is equal to one, and $ q = 2 $.

Example. 8,8,8,8 ... A geometric progression, in which the first term is equal to eight,
and $ q = 1 $.

Example. 3, -3.3, -3.3 ... Geometric progression, in which the first term is equal to three,
and $ q = -1 $.

The geometric progression has the properties of monotony.
If $ b_ (1)> 0 $, $ q> 1 $,
then the sequence is ascending.
If $ b_ (1)> 0 $, $ 0 The sequence is usually denoted as: $ b_ (1), b_ (2), b_ (3), ..., b_ (n), ... $.

As well as in an arithmetic progression, if the number of elements is finite in a geometric progression, then the progression is called a finite geometric progression.

$ b_ (1), b_ (2), b_ (3), ..., b_ (n-2), b_ (n-1), b_ (n) $.
Note, if the sequence is a geometric progression, then the sequence of squares of members is also a geometric progression. For the second sequence, the first term is $ b_ (1) ^ 2 $, and the denominator is $ q ^ 2 $.

Formula of the n-th term of a geometric progression

Let's go back to our examples.

Example. 1,2,4,8,16 ... Geometric progression, in which the first term is equal to one,
and $ q = 2 $.
$ b_ (n) = 1 * 2 ^ (n) = 2 ^ (n-1) $.

Example. 16,8,4,2,1,1 / 2 ... A geometric progression in which the first term is sixteen and $ q = \ frac (1) (2) $.
$ b_ (n) = 16 * (\ frac (1) (2)) ^ (n-1) $.

Example. 8,8,8,8 ... A geometric progression in which the first term is eight and $ q = 1 $.
$ b_ (n) = 8 * 1 ^ (n-1) = 8 $.

Example. 3, -3.3, -3.3 ... A geometric progression in which the first term is three and $ q = -1 $.
$ b_ (n) = 3 * (- 1) ^ (n-1) $.

Example. You are given a geometric progression $ b_ (1), b_ (2),…, b_ (n),… $.
a) It is known that $ b_ (1) = 6, q = 3 $. Find $ b_ (5) $.
b) It is known that $ b_ (1) = 6, q = 2, b_ (n) = 768 $. Find n.
c) It is known that $ q = -2, b_ (6) = 96 $. Find $ b_ (1) $.
d) It is known that $ b_ (1) = - 2, b_ (12) = 4096 $. Find q.

Solution.
a) $ b_ (5) = b_ (1) * q ^ 4 = 6 * 3 ^ 4 = 486 $.
b) $ b_n = b_1 * q ^ (n-1) = 6 * 2 ^ (n-1) = 768 $.
$ 2 ^ (n-1) = \ frac (768) (6) = 128 $ since $ 2 ^ 7 = 128 => n-1 = 7; n = 8 $.
c) $ b_ (6) = b_ (1) * q ^ 5 = b_ (1) * (- 2) ^ 5 = -32 * b_ (1) = 96 => b_ (1) = - 3 $.
d) $ b_ (12) = b_ (1) * q ^ (11) = - 2 * q ^ (11) = 4096 => q ^ (11) = - 2048 => q = -2 $.

Example. The difference between the seventh and fifth terms of the geometric progression is 192, the sum of the fifth and sixth terms of the progression is 192. Find the tenth term of this progression.

Solution.
We know that: $ b_ (7) -b_ (5) = 192 $ and $ b_ (5) + b_ (6) = 192 $.
We also know: $ b_ (5) = b_ (1) * q ^ 4 $; $ b_ (6) = b_ (1) * q ^ 5 $; $ b_ (7) = b_ (1) * q ^ 6 $.
Then:
$ b_ (1) * q ^ 6-b_ (1) * q ^ 4 = 192 $.
$ b_ (1) * q ^ 4 + b_ (1) * q ^ 5 = 192 $.
We got a system of equations:
$ \ begin (cases) b_ (1) * q ^ 4 (q ^ 2-1) = 192 \\ b_ (1) * q ^ 4 (1 + q) = 192 \ end (cases) $.
Equating, our equations get:
$ b_ (1) * q ^ 4 (q ^ 2-1) = b_ (1) * q ^ 4 (1 + q) $.
$ q ^ 2-1 = q + 1 $.
$ q ^ 2-q-2 = 0 $.
We got two solutions q: $ q_ (1) = 2, q_ (2) = - 1 $.
Substitute sequentially into the second equation:
$ b_ (1) * 2 ^ 4 * 3 = 192 => b_ (1) = 4 $.
$ b_ (1) * (- 1) ^ 4 * 0 = 192 => $ no solutions.
We got that: $ b_ (1) = 4, q = 2 $.
Find the tenth term: $ b_ (10) = b_ (1) * q ^ 9 = 4 * 2 ^ 9 = 2048 $.

Sum of a finite geometric progression

Let a finite geometric progression be given: $ b_ (1), b_ (2),…, b_ (n-1), b_ (n) $.
Let us introduce the notation for the sum of its members: $ S_ (n) = b_ (1) + b_ (2) + ⋯ + b_ (n-1) + b_ (n) $.
In the case when $ q = 1 $. All members of the geometric progression are equal to the first term, then it is obvious that $ S_ (n) = n * b_ (1) $.
Consider now the case $ q ≠ 1 $.
Multiply the above sum by q.
$ S_ (n) * q = (b_ (1) + b_ (2) + ⋯ + b_ (n-1) + b_ (n)) * q = b_ (1) * q + b_ (2) * q + ⋯ + b_ (n-1) * q + b_ (n) * q = b_ (2) + b_ (3) + ⋯ + b_ (n) + b_ (n) * q $.
Note:
$ S_ (n) = b_ (1) + (b_ (2) + ⋯ + b_ (n-1) + b_ (n)) $.
$ S_ (n) * q = (b_ (2) + ⋯ + b_ (n-1) + b_ (n)) + b_ (n) * q $.

$ S_ (n) * q-S_ (n) = (b_ (2) + ⋯ + b_ (n-1) + b_ (n)) + b_ (n) * q-b_ (1) - (b_ (2 ) + ⋯ + b_ (n-1) + b_ (n)) = b_ (n) * q-b_ (1) $.

$ S_ (n) (q-1) = b_ (n) * q-b_ (1) $.

$ S_ (n) = \ frac (b_ (n) * q-b_ (1)) (q-1) = \ frac (b_ (1) * q ^ (n-1) * q-b_ (1)) (q-1) = \ frac (b_ (1) (q ^ (n) -1)) (q-1) $.

$ S_ (n) = \ frac (b_ (1) (q ^ (n) -1)) (q-1) $.

We got the formula for the sum of a finite geometric progression.

Solution.
$ S_ (7) = \ frac (4 * (3 ^ (7) -1)) (3-1) = 2 * (3 ^ (7) -1) = 4372 $.

Example.
Find the fifth term of the geometric progression, which is known: $ b_ (1) = - 3 $; $ b_ (n) = - 3072 $; $ S_ (n) = - 4095 $.

Solution.
$ b_ (n) = (- 3) * q ^ (n-1) = - 3072 $.
$ q ^ (n-1) = 1024 $.
$ q ^ (n) = 1024q $.

$ S_ (n) = \ frac (-3 * (q ^ (n) -1)) (q-1) = - 4095 $.
$ -4095 (q-1) = - 3 * (q ^ (n) -1) $.
$ -4095 (q-1) = - 3 * (1024q-1) $.
$ 1365q-1365 = 1024q-1 $.
$ 341q = $ 1364.
$ q = 4 $.
$ b_5 = b_1 * q ^ 4 = -3 * 4 ^ 4 = -3 * 256 = -768 $.

Characteristic property of a geometric progression

A numerical sequence is a geometric progression only when the square of each of its members is equal to the product of two adjacent members of the progression. Do not forget that for a finite progression, this condition is not met for the first and last members.

The modulus of any member of a geometric progression is equal to the geometric mean of two members adjacent to it.

Solution.
Let's use the characteristic property:
$ (2x + 2) ^ 2 = (x + 2) (3x + 3) $.
$ 4x ^ 2 + 8x + 4 = 3x ^ 2 + 3x + 6x + 6 $.
$ x ^ 2-x-2 = 0 $.
$ x_ (1) = 2 $ and $ x_ (2) = - 1 $.
Substituting sequentially into the original expression, our solutions:
With $ x = 2 $, we got the sequence: 4; 6; 9 - a geometric progression, in which $ q = 1.5 $.
With $ x = -1 $, we got the sequence: 1; 0; 0.
Answer: $ x = 2. $

Tasks for independent solution

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Numerical sequence

So let's sit down and start writing some numbers. For example:

You can write any numbers, and there can be as many as you like (in our case, them). No matter how many numbers we write, we can always say which one is the first, which is the second, and so on to 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 one.

The number with the number is called the th member of the sequence.

We usually call the entire sequence 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 thread, we will talk about the second kind - geometric progression.

Why do we need a geometric progression and its history of origin.

Even in ancient times, the Italian mathematician Leonardo of Pisa (better known as Fibonacci) was engaged in solving the practical needs of trade. The monk was faced with the task of determining with the help of what least amount of weights it is possible to weigh the goods? In his writings, Fibonacci proves that such a system of weights is optimal: This is one of the first situations in which people had to face a geometric progression, which you probably already heard about and have at least a general concept. Once you fully understand the topic, think about why such a system is optimal?

Currently, in life practice, a geometric progression is manifested when investing money in a bank, when the amount of interest is charged on the amount accumulated in the account for the previous period. In other words, if you put money on a term deposit in a savings bank, then in a year the deposit will increase by more than the original amount, i.e. the new amount will be equal to the deposit multiplied by. In another year, this amount will increase by, i.e. the amount obtained at that time will be multiplied by again and so on. A similar situation is described in the problems of calculating the so-called compound interest- the percentage is taken each time from the amount on the account, taking into account the previous interest. We will talk about these tasks a little later.

There are many more simple cases where geometric progression is used. For example, the spread of influenza: one person infected a 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, the 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 numeric sequence:

You will immediately answer that it is easy and the name of such a sequence - with the difference of its members. How about this:

If you subtract the previous one from the next number, then you will see that each time a new difference is obtained (and so on), but the sequence definitely exists and it is easy to notice - each next number is times larger than the previous one!

This kind of number sequence is called geometric progression and is indicated by.

Geometric progression () is a numerical sequence, the first term of which is nonzero, 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 the geometric progression.

Restrictions that the first term () is not equal and not random. Let's say that there are none, and the first term is still equal, and q is equal, hmm .. let, then it turns out:

Agree that this is no longer any progression.

As you can imagine, we will get the same results if it is any number other than zero, and. In these cases, there will simply not be a progression, since the entire number series will be either all zeros, or one number, and all other zeros.

Now let's talk in more detail about the denominator of the geometric progression, that is, Fr.

Let's repeat: is a number, how many times does each subsequent term change geometric progression.

What do you think it can be? Correctly, positive and negative, but not zero (we talked about this just above).

Let's say that we have a positive one. Let in our case, as well. What is the second term and? You can easily answer that:

Everything is correct. Accordingly, if, then all subsequent members of the progression have the same sign - they positive.

What if negative? For example, a. What is the second term and?

This is a completely different story.

Try to count the term of this progression. How much did you get? I have. Thus, if, then the signs of the members of the geometric progression alternate. That is, if you see a progression with alternating signs on 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 arithmetic:

Understood? Let's compare our answers:

• Geometric progression - 3, 6.
• Arithmetic progression - 2, 4.
• It is neither arithmetic nor geometric progressions - 1, 5, 7.

Let's return to our last progression, and try to find its term in the same way as in arithmetic. As you might guess, there are two ways to find it.

We successively multiply each term by.

So, the th member of the described geometric progression is equal to.

As you might guess, now you yourself will derive a formula that will help you find any member of a geometric progression. Or have you already brought it out 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 by the example of finding the th member of a given progression:

In other words:

Find on your own the value of a member of a given geometric progression.

Happened? Let's compare our answers:

Please note that you get exactly the same number as in the previous method, when we successively multiplied by each previous term of the geometric progression.
Let's try to "depersonalize" this formula - we will bring it into general form and get:

The derived formula is correct for all values, both positive and negative. Check it yourself by calculating the members of the geometric progression with the following conditions:, a.

Have you counted? Let's compare the results obtained:

Agree that it would be possible to find a member of the progression in the same way as a member, however, there is a possibility of incorrect counting. And if we have already found the th term of the geometric progression, then what could be easier than using the "cut off" part of the formula.

An infinitely decreasing geometric progression.

More recently, we talked about the fact that it can be either greater than or less than zero, however, there are special values ​​at which a geometric progression is called infinitely decreasing.

Why do you think such a name?
First, let's write down some geometric progression consisting of members.
Suppose, a, then:

We see that each subsequent term is less than the previous one by one factor, but will there be any number? You will immediately answer no. That is why the infinitely decreasing - decreases, decreases, and never becomes zero.

To clearly understand how it looks visually, let's try to draw a graph of our progression. So, for our case, the formula takes the following form:

It is customary for us to build dependence on charts, therefore:

The essence of the expression has not changed: in the first entry, we showed the dependence of the value of the geometric progression member on its ordinal number, and in the second entry, we simply took the value of the geometrical progression term as, and the ordinal number was designated not how, but how. All that remains to be done is to build a graph.
Let's see what you get. Here's the graph I got:

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 mean:

Try to schematically depict a graph of a geometric progression when, if its first term is also equal. Analyze, what's the difference with our previous chart?

Did you manage? Here's the graph I got:

Now that you have completely understood the basics of the theme of a 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.

Geometric progression property.

Remember the property of the members 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 members of a given progression. Remembered? This:

Now we are faced with exactly the same question for the members of a geometric progression. To derive a similar formula, let's start drawing and reasoning. You will see, it is very easy, and if you forget, you can bring it out on your own.

Let's take another simple geometric progression in which we know and. How to find? With an arithmetic progression, this 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 using a formula.

You ask, what should we do with this now? It's very simple. To begin with, we will depict these formulas in the figure, and try to do various manipulations with them in order to arrive at a value.

We abstract from the numbers that we are given, we will focus only on expressing them through a formula. We need to find the value highlighted in orange, knowing the members adjacent to it. Let's try to perform various actions with them, as a result of which we can receive.

Addition.
Let's try to add two expressions and, we get:

From this expression, as you can see, we cannot express in any way, therefore, we will try another option - subtraction.

Subtraction.

As you can see, we also cannot express from this, therefore, we will try to multiply these expressions by each other.

Multiplication.

Now look carefully what we have, multiplying the members 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 number multiplied by each other:

Well. You yourself have deduced the property of a geometric progression. Try to write this formula in general terms. Happened?

Forgot the condition for? Think about why it is important, for example, try to calculate it yourself, if. What happens in this case? That's right, complete nonsense since the formula looks like this:

Accordingly, do not forget this limitation.

Now let's count what is equal to

Correct answer - ! If, when calculating, you did not forget the second possible value, then you are a great fellow and you can immediately proceed 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 meaning, and the other with a meaning and check if 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 it is the same between all its given members? Calculate q for the first and second cases.

See why we have to write two answers? Because the sign of the required term depends on whether it is positive or negative! And since we do not know what he is, we need to write both answers with plus and minus.

Now that you have mastered the main points and derived the formula for the property of a geometric progression, find, knowing and

Compare the received answers with the correct ones:

What do you think, what if we were given not the values ​​of the members of the geometric progression adjacent to the desired number, but equidistant from it. For example, we need to find, and are given and. Can we in this case use the formula we derived? Try to confirm or deny this possibility in the same way, writing down what each value consists of, as you did when initially deriving the formula, for.
What did you do?

Now look closely again.
and correspondingly:

From this we can conclude that the formula works not only with neighboring with the required terms of the geometric progression, but also with equidistant from the sought-after members.

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 less. The main thing is to be the same for both given numbers.

Practice with specific examples, just be extremely careful!

1. ,. Find.
2. ,. Find.
3. ,. Find.

Decided? I hope you were extremely attentive and noticed a small catch.

We 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 consideration of the ordinal 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 removed in position, so it is not possible to apply the formula.

How can we solve it? It's actually not as difficult as it sounds! Let's write down with you what each number given to us and the required number consists of.

So, we have and. Let's see what we can do with them? I propose to divide by. We get:

We substitute our data into the formula:

The next step we can find - for this we need to take the cube root of the resulting number.

And now we 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 in the formula:

Our answer: .

Try to solve another similar problem yourself:
Given:,
Find:

How much did you get? I have - .

As you can see, in fact, you need remember just one formula-. You can withdraw all the rest without any difficulty on your own at any time. To do this, simply write the simplest geometric progression on a piece of paper and write down what, according to the above formula, each of its numbers is equal.

The sum of the members of a geometric progression.

Now consider the formulas that allow us to quickly calculate the sum of the members of a geometric progression in a given interval:

To derive the formula for the sum of the members of a finite geometric progression, we multiply all parts of the higher 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 do?

Now express the term of the geometric progression through the formula and substitute the resulting expression in our last formula:

Group the expression. You should get:

All that's left to do is express:

Accordingly, in this case.

What if? What formula works then? Imagine a geometric progression at. What is she like? Correctly a series of identical numbers, respectively, the formula will look like this:

There are many legends in both arithmetic and geometric progression. One of them is the legend of Seth, the creator of chess.

Many people know that the chess game was invented in India. When the Hindu king met her, he was delighted with her wit and the variety of possible positions in her. Upon learning that it was invented by one of his subjects, the king decided to personally reward him. He summoned the inventor to him and ordered him to ask him for whatever he wanted, promising to fulfill even the most skillful desire.

Seta asked for time to think, and when the next day Seth appeared to the king, he surprised the king with the unparalleled modesty of his request. He asked to give out a grain of wheat for the first square of the chessboard, for the second one for wheat grains, for the third, for the fourth, etc.

The king was angry and drove Seth away, saying that the servant's request was unworthy of the royal generosity, but promised that the servant would receive his grains for all the cells of the board.

And now the question: using the formula for the sum of the members of a geometric progression, calculate how many grains Seta should receive?

Let's start to reason. Since, according to the condition, Seta asked for a grain of wheat for the first square of the chessboard, for the second, for the third, for the fourth, etc., we see that the problem is about a geometric progression. What is equal in this case?
Right.

Total cells of the chessboard. Respectively, . We have all the data, it remains only to substitute it into the formula and calculate.

To represent at least approximately the "scales" of a given number, we transform using the properties of the degree:

Of course, if you want, you can take a calculator and calculate what number you will get in the end, but if not, you will have to take my word for it: the final value of the expression will be.
That is:

quintillion quadrillion trillion billion million thousand.

Fuh) If you want to imagine the enormity of this number, then estimate how large the barn would be required to contain the entire amount of grain.
With a barn height m and a width of m, its length would have to extend for km, i.e. twice as far as from the Earth to the Sun.

If the tsar were strong in mathematics, he could suggest that the scientist himself 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 all his life.

Now let's solve a simple problem for the sum of the members of a geometric progression.
Vasya, a pupil of grade 5 A, has 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 people in the class. In how many days will the whole class get sick with the flu?

So, the first member of the geometric progression is Vasya, that is, a person. th member of the geometric progression, these are the two people he infected on the first day of his arrival. The total number of members in the progression is equal to the number of students 5A. Accordingly, we are talking about a progression in which:

Let's substitute our data into the formula for the sum of members of a geometric progression:

The whole class will get sick in days. Don't you believe in formulas and numbers? Try to portray the "infection" of the students yourself. Happened? See how it looks for me:

Calculate for yourself how many days it would take pupils to get the flu if each infect a person and there was a person in the class.

What value did you get? It turned out that everyone started to get sick after a day.

As you can see, such a task and drawing to it resembles 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 will close the chain (). Thus, if a person were involved in a financial pyramid, in which money was given in the event that you bring two other participants, then the person (or in the general case) would not bring anyone, respectively, they 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 kind - an infinitely decreasing geometric progression. How to calculate the sum of its members? And why does this type of progression have certain features? Let's sort it out together.

So, first, let's look again at this figure 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, when, it will be almost equal, respectively, when calculating the expression, we 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.

- the formula is the sum of the terms of an infinitely decreasing geometric progression.

IMPORTANT! We use the formula for the sum of the terms of an infinitely decreasing geometric progression only if the condition explicitly states that we need to find the sum endless number of members.

If a specific number n is indicated, then we use the formula for the sum of n terms, even if or.

Now let's practice.

1. Find the sum of the first terms of a geometric progression with and.
2. Find the sum of the terms of an infinitely decreasing geometric progression with and.

I hope you were extremely attentive. 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 in the exam are compound interest problems. It is about them that we will talk.

Tasks for calculating compound interest.

You've probably heard of the so-called compound interest formula. Do you understand what she means? If not, let's figure it out, because having realized the process itself, you will immediately understand, and here is a geometric progression.

We all go to the bank and know that there are different conditions for deposits: this is the term, and additional service, and interest with two different ways of calculating it - simple and complex.

WITH simple interest everything is more or less clear: interest is charged once at the end of the deposit term. That is, if we say that we put 100 rubles for a year under, 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 there is capitalization of interest, i.e. their addition to the amount of the deposit and the subsequent calculation of income not from the initial, but from the accumulated amount of the deposit. 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 say that we put all the same rubles at annual rates, but with a monthly capitalization of the deposit. What do we get?

Do you understand everything here? If not, let's figure it out in stages.

We brought rubles to the bank. By the end of the month, our account should have an amount consisting of our rubles plus interest on them, that is:

Agree?

We can put it outside the bracket and then we get:

Agree, this formula is already more similar to the one we wrote at the beginning. It remains to deal with interest

In the problem statement, we are told about the annual. As you know, we do not multiply by - we convert percentages to decimal fractions, that is:

Right? Now you ask, where did the number come from? Very simple!
I repeat: the problem statement says about ANNUAL interest accrued MONTHLY... As you know, in a year of months, respectively, the bank will charge us a part of the annual interest per month:

Realized? Now try to write what this part of the formula will look like if I say that interest is calculated daily.
Did you manage? Let's compare the results:

Well done! Let's return to our problem: write down how much will be credited to our account for the second month, taking into account that interest is charged on the accumulated amount of the deposit.
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 down what its member will be equal to, or, in other words, how much money we will receive at the end of the month.
Did? Checking!

As you can see, if you put money in the bank for a year at a simple interest, then you will receive rubles, and if at a complex rate - rubles. The benefit is small, but this happens only during the th year, but for a longer period, capitalization is much more profitable:

Let's consider another type of problems with compound interest. After what you figured out, it will be elementary for you. So the task:

The Zvezda company began investing in the industry in 2000, having capital in dollars. Every year since 2001, she earns a profit, which is from the capital of the previous year. How much profit will the Zvezda company receive at the end of 2003 if the profit has not been withdrawn from circulation?

Capital of the company "Zvezda" in 2000.
- the capital of the company "Zvezda" in 2001.
- the capital of the company "Zvezda" in 2002.
- the capital of the company "Zvezda" in 2003.

Or we can write briefly:

For our case:

2000, 2001, 2002 and 2003.

Respectively:
rubles
Note that in this problem we have no division either by or by, since the percentage is given ANNUALLY and it is calculated ANNUALLY. That is, when reading a problem for compound interest, pay attention to what percentage is given, and in what period it is charged, and only then proceed to the calculations.
Now you know everything about geometric progression.

Workout.

1. Find the exponential term if it is known that, and
2. Find the sum of the first terms of the geometric progression, if it is known that, and
3. MDM Capital started investing in the industry in 2003, having capital in dollars. Every year, starting in 2004, she earns a profit, which is from the capital of the previous year. The company "MSK Cash Flows" began to invest in the industry in 2005 in the amount of $ 10,000, starting to make a profit in 2006 in the amount of. How many dollars is the capital of one company more than another at the end of 2007, if the profit has not been withdrawn from circulation?

Answers:

1. 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 members, the calculation is carried out according to the formula:
2. 
   MDM Capital:

   2003, 2004, 2005, 2006, 2007.
   - increases by 100%, that is, 2 times.
   Respectively:
   rubles
   MSK Cash Flows:

   2005, 2006, 2007.
   - increases by, that is, times.
   Respectively:
   rubles
   rubles

Let's summarize.

1) Geometric progression () is a numerical sequence, the first term of which is nonzero, 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 the geometric progression.

2) Equation of members of a geometric progression -.

3) can take any values, except for and.

• if, then all subsequent members of the progression have the same sign - they positive;
• if, then all subsequent members of the progression alternate signs;
• at - the progression is called infinitely decreasing.

4), for is the property of a geometric progression (adjacent terms)

or
, at (equidistant terms)

When finding, do not forget that there should be two answers.

For example,

5) The sum of the members of a geometric progression is calculated by the formula:
or

or

IMPORTANT! We use the formula for the sum of the terms of an infinitely decreasing geometric progression only if the condition explicitly states that it is necessary to find the sum of an infinite number of terms.

6) Problems for compound interest are also calculated according to the formula of the -th term of a geometric progression, provided that the funds have not been withdrawn from circulation:

GEOMETRIC PROGRESSION. BRIEFLY ABOUT THE MAIN

Geometric progression() is a numerical sequence, the first term of which is nonzero, 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.

Denominator of geometric progression can take any values ​​except and.

• If, then all subsequent members of the progression have the same sign - they are positive;
• if, then all subsequent members of the progression alternate signs;
• at - the progression is called infinitely decreasing.

Equation of members of a geometric progression - .

The sum of the members 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, then 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 that 5%!

Now comes the most important thing.

You figured out the theory on this topic. And, again, this is ... it's 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 the successful passing of the exam, for admission to the institute on the budget and, MOST IMPORTANT, for life.

I will not convince you of anything, I will just say one thing ...

People who have received a good education earn much more than those who have not received it. These are statistics.

But this is not the main thing either.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because there are so many more opportunities open up before them and life becomes brighter? Do not know...

But think for yourself ...

What does it take to be for sure better than others on the exam and ultimately be ... more happy?

GET A HAND SOLVING PROBLEMS ON THIS TOPIC.

On the exam, you will not be asked for theory.

You will need solve problems for a while.

And, if you didn’t solve them (A LOT!), You are sure to go somewhere stupidly mistaken or simply will not be in time.

It's like in sports - you have to repeat it over and over to win for sure.

Find a collection where you want, necessarily with solutions, detailed analysis and decide, decide, decide!

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Find problems and solve!

Already in Ancient Egypt, they knew not only arithmetic, but also geometric progression. For example, here is a problem from Rind's papyrus: “Seven faces have seven cats each; each cat eats seven mice, each mouse eats seven ears, each ear can grow seven measures of barley. How large are the numbers of this series and their sum? "

Rice. 1. The ancient Egyptian problem of geometric progression

This task was repeated many times with different variations among other peoples at other times. For example, in the written in the XIII century. "The Book of the Abacus" by Leonardo of Pisa (Fibonacci) has a problem in which there are 7 old women heading to Rome (obviously pilgrims), each of whom has 7 mules, each of which has 7 sacks, each of which has 7 loaves , each of which has 7 knives, each of which is in 7 scabbards. The problem asks how many items are there.

The sum of the first n terms of the geometric progression S n = b 1 (q n - 1) / (q - 1). This formula can be proved, for example, as follows: S n = b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n - 1.

Add to S n the number b 1 q n and get:

Hence S n (q - 1) = b 1 (q n - 1), and we obtain the required formula.

Already on one of the clay tablets of Ancient Babylon, dating back to the VI 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 where this fact was known to the Babylonians.

The rapid growth of geometric progression in a number of cultures, in particular, in Indian, is repeatedly used as a visual symbol of the immensity of the universe. In the well-known legend about the appearance of chess, the lord gives its inventor the opportunity to choose the reward himself, and he asks for the amount of wheat grains that will be obtained if one is put on the first square of the chessboard, two on the second, four on the third, eight on the fourth and so on, each time the number doubles. Vladyka thought that it was, at most, about several sacks, but he miscalculated. It is easy to see that for all 64 squares of the chessboard, the inventor should have received (2 64 - 1) grain, which is expressed by 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 pointing to the almost unlimited possibilities hidden in the game of chess.

It's easy to see that this number is indeed 20 digits:

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?

The geometric progression is increasing if the denominator is greater than 1 in absolute value, 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 an increasing geometric progression increases unexpectedly quickly, a decreasing one 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). (This is how F. Viet, for example, reasoned). The number S is called the sum of an infinitely decreasing geometric progression. Nevertheless, for many centuries the question of what is the meaning of the summation of the ENTIRE geometric progression, with its infinite number of terms, was not clear enough to mathematicians.

A diminishing geometric progression can be seen, for example, in Zeno's aporias "Halving" and "Achilles and the Turtle." In the first case, it is clearly shown that the entire road (suppose of length 1) is the sum of an infinite number of segments 1/2, 1/4, 1/8, etc. This is, of course, from the point of view of the concept of a finite sum endless geometric progression. And yet - how can this be?

In the aporia about Achilles, the situation is a little more complicated, since here the denominator of the progression is equal not to 1/2, but to some other number. Suppose, for example, Achilles runs at speed v, the turtle moves at speed u, and the initial distance between them is equal to l. Achilles will run this distance in time l / v, the turtle will move by a distance lu / v during this time. When Achilles runs 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 place of meeting 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.

The sum of a geometric progression was used by Archimedes to determine the area of ​​a parabola segment. Let the given segment of the parabola be delimited by the chord AB and let the tangent line at the 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. Draw straight lines parallel to DC through points A, E, F, B; let the tangent drawn at the point D, these lines intersect at the points K, L, M, N. Let's also draw segments AD and DB. Let the line EL intersect the line AD at the point G, and the parabola at the point H; line FM intersects line DB at point Q, and 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 the x and y coordinate axes, in which the parabola equation 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 parallel to a given tangent line 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. Since KA = 2LG, LH = HG. The area of ​​the parabola ADB segment is equal to the area of ​​the triangle ΔADB and the areas of the AHD and DRB segments combined. In turn, the area of ​​the AHD segment 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 - divide into a triangle (Δ) and two remaining segments (), etc.:

The area of ​​triangle ΔAHD is equal to half the area of ​​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 ​​triangle ΔAKD, and therefore half the area of ​​triangle ΔACD. Thus, the area of ​​the triangle ΔAHD is equal to a quarter of the area of ​​the triangle ΔACD. Similarly, the area of ​​the triangle ΔDRB is equal to a quarter of the area of ​​the 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 applied to the segments AH, HD, DR and RB will also 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, which means 16 times less. than the area of ​​the triangle ΔADB. Etc:

Thus, Archimedes proved that "every segment enclosed between a straight line and a parabola is four-thirds of a triangle with the same base and equal height."