Geometric progression, along with arithmetic, is an important number series that is studied in school course algebra in 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. Such a series is called a geometric progression rational numbers, which is formed by sequentially multiplying its first element by a constant number called the denominator.

For example, the numbers in the series 3, 6, 12, 24, ... are a geometric progression, because if you multiply 3 (the first element) by 2, you get 6. If you multiply 6 by 2, you get 12, and so on.

The members of the sequence under consideration are usually denoted by the symbol ai, where i is an integer indicating the number of the element in the series.

The above definition of progression can be written in mathematical language as follows: an = bn-1 * a1, where b is the denominator. It is easy to check this formula: if n = 1, then b1-1 = 1, and we get a1 = a1. If n = 2, then an = b * a1, and we again come to the definition of the series of numbers in question. Similar reasoning can be continued for large values ​​of n.

Denominator of geometric progression

The number b completely determines what character the entire number series will have. The denominator b can be positive, negative, or greater than or less than one. All of the above options lead to different sequences:

• b > 1. There is an increasing series of rational numbers. For example, 1, 2, 4, 8, ... If element a1 is negative, then the entire sequence will increase only in absolute value, but decrease depending on the sign of the numbers.
• b = 1. Often this case is not called a progression, since there is an ordinary series of identical rational numbers. For example, -4, -4, -4.

Formula for amount

Before 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 to specific numbers.

Problem 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 we calculate the larger sum: S4 = ((-2)4 - 1) * 4 / (-2 - 1) = -20. Note that in the last expression only 4 terms were summed, since the 5th is already included in the amount that needs to be calculated according to the conditions of the problem. Finally, we take the difference: S510 = S10 - S4 = -1364 - (-20) = -1344.

Method 2. Before substituting numbers and counting, you can obtain a formula for the sum between the m and n terms of the series in question. We proceed in exactly the same way as in method 1, only we first work with the symbolic representation of the amount. We have: Snm = (bn - 1) * a1 / (b - 1) - (bm-1 - 1) * a1 / (b - 1) = a1 * (bn - bm-1) / (b - 1). You can substitute known numbers into the resulting expression and calculate the final result: S105 = 4 * ((-2)10 - (-2)4) / (-2 - 1) = -1344.

Problem No. 3. What is the denominator?

Let a1 = 2, find the denominator of the geometric progression, provided that its infinite sum is 3, and it is known that this is a decreasing series of numbers.

Based on the conditions of the problem, it is not difficult to guess which formula should be used to solve it. Of course, for the sum of the progression infinitely decreasing. We have: S∞ = a1 / (1 - b). From where we express the denominator: b = 1 - a1 / S∞. It remains to substitute the known values ​​and get the required number: b = 1 - 2 / 3 = -1 / 3 or -0.333(3). We can qualitatively check this result if we remember that for this type of sequence the modulus b should not go beyond 1. As can be seen, |-1 / 3|

Task No. 4. Restoring a series of numbers

Let 2 elements of a number series be given, for example, the 5th is equal to 30 and the 10th is equal to 60. It is necessary to reconstruct the entire series from these data, knowing that it satisfies the properties of a geometric progression.

To solve the problem, you must first write down the corresponding expression for each known term. We have: a5 = b4 * a1 and a10 = b9 * a1. Now divide the second expression by the first, we get: a10 / a5 = b9 * a1 / (b4 * a1) = b5. From here we determine the denominator by taking the fifth root of the ratio of the terms known from the problem statement, b = 1.148698. We substitute the resulting number into one of the expressions for the known element, we get: a1 = a5 / b4 = 30 / (1.148698)4 = 17.2304966.

Thus, we found the denominator of the progression bn, and the geometric progression bn-1 * 17.2304966 = an, where b = 1.148698.

Where are geometric progressions used?

If there were no practical application of this number series, then its study would be reduced to purely theoretical interest. But such an application exists.

Below are the 3 most famous examples:

• Zeno's paradox, in which the nimble Achilles cannot catch up with the slow turtle, is solved using the concept of an infinitely decreasing sequence of numbers.
• If for each cell chessboard put wheat grains so that on the 1st cell you put 1 grain, on the 2nd - 2, on the 3rd - 3 and so on, then to fill all the cells 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 increase 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 the sequence 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's 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 terms of geometric progressionformula applies

If we denote , then

Where . Since , formula (6) is a generalization of formula (5).

In the case when and geometric progression is 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 .

According to the 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. Left side 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 consistency positive numbers forms an arithmetic progression, A – geometric progression, and here. Find .

Solution. Because arithmetic sequence, That (main property arithmetic progression). Since, then or . It follows from this, 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 . According to the condition, so from Eq.we obtain a unique solution to the problem under consideration, i.e. .

Answer: .

Example 12. Calculate Sum

. (12)

Solution. Let's 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. A complete course of elementary mathematics in problems and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. – 208 p.

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Let's consider a certain series.

7 28 112 448 1792...

It is absolutely clear that the value of any of its elements is exactly four times greater than the previous one. Means, this series is a progression.

A geometric progression is an infinite sequence of numbers. main feature which is that the next number is obtained from the previous one by multiplying by some specific number. This is expressed by the following formula.

a z +1 =a z ·q, where z is the number of the selected element.

Accordingly, z ∈ N.

The period when geometric progression is studied at school is 9th grade. Examples will help you understand the concept:

0.25 0.125 0.0625...

Based on this formula, the denominator of the progression can be found as follows:

Neither q nor b z can be zero. Also, each of the elements of the progression should not be equal to zero.

Accordingly, to find out the next number in a series, you need to multiply the last one by q.

To set this progression, you must specify its first element and denominator. After this, it is possible to find any of the subsequent terms and their sum.

Varieties

Depending on q and a 1, this progression is divided into several types:

• If both a 1 and q are greater than one, then such a sequence is a geometric progression increasing with each subsequent element. An example of this is presented below.

Example: a 1 =3, q=2 - both parameters are greater than one.

Then the number sequence can be written like this:

3 6 12 24 48 ...

• If |q| is less than one, that is, multiplication by it is equivalent to division, then a progression with similar conditions is a decreasing geometric progression. An example of this is presented below.

Example: a 1 =6, q=1/3 - a 1 is greater than one, q is less.

Then the number sequence can be written as follows:

6 2 2/3 ... - any element is 3 times larger than the element following it.

• Alternating sign. If q<0, то знаки у чисел последовательности постоянно чередуются вне зависимости от a 1 , а элементы ни возрастают, ни убывают.

Example: a 1 = -3, q = -2 - both parameters are less than zero.

Then the number sequence can be written like this:

3, 6, -12, 24,...

Formulas

There are many formulas for convenient use of geometric progressions:

• Z-term formula. Allows you to calculate an element under a specific number without calculating previous numbers.

Example:q = 3, a 1 = 4. It is required to count the fourth element of the progression.

Solution:a 4 = 4 · 3 4-1 = 4 · 3 3 = 4 · 27 = 108.

• The sum of the first elements whose quantity is equal to z. Allows you to calculate the sum of all elements of a sequence up toa zinclusive.

Since (1-q) is in the denominator, then (1 - q)≠ 0, therefore q is not equal to 1.

Note: if q=1, then the progression would be a series of infinitely repeating numbers.

Sum of geometric progression, examples:a 1 = 2, q= -2. Calculate S5.

Solution:S 5 = 22 - calculation using the formula.

• Amount if |q| < 1 и если z стремится к бесконечности.

Example:a 1 = 2 , q= 0.5. Find the amount.

Solution:Sz = 2 · = 4

Sz = 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 3.9375 4

Some properties:

• Characteristic property. If the following condition works for anyz, then the given number series is a geometric progression:

a z 2 = a z -1 · az+1

• Also, the square of any number in a geometric progression is found by adding the squares of any two other numbers in a given series, if they are equidistant from this element.

a z 2 = a z - t 2 + a z + t 2 , Wheret- the distance between these numbers.

• Elementsdiffer in qonce.
• The logarithms of the elements of a progression also form a progression, but an arithmetic one, that is, each of them is greater than the previous one by a certain number.

Examples of some classic problems

To better understand what a geometric progression is, examples with solutions for class 9 can help.

• Conditions:a 1 = 3, a 3 = 48. Findq.

Solution: each subsequent element is greater than the previous one inq once.It is necessary to express some elements in terms of others using a denominator.

Hence,a 3 = q 2 · a 1

When substitutingq= 4

• Conditions:a 2 = 6, a 3 = 12. Calculate S 6.

Solution:To do this, just find q, the first element and substitute it into the formula.

a 3 = q· a 2 , hence,q= 2

a 2 = q · a 1 ,That's why a 1 = 3

S 6 = 189

• · a 1 = 10, q= -2. Find the fourth element of the progression.

Solution: to do this, it is enough to express the fourth element through the first and through the denominator.

a 4 = q 3· a 1 = -80

Application example:

• A bank client made a deposit in the amount of 10,000 rubles, under the terms of which every year the client will have 6% of it added to the principal amount. How much money will be in the account after 4 years?

Solution: The initial amount is 10 thousand rubles. This means that a year after the investment the account will have an amount equal to 10,000 + 10,000 · 0.06 = 10000 1.06

Accordingly, the amount in the account after another year will be expressed as follows:

(10000 · 1.06) · 0.06 + 10000 · 1.06 = 1.06 · 1.06 · 10000

That is, every year the amount increases by 1.06 times. This means that to find the amount of funds in the account after 4 years, it is enough to find the fourth element of the progression, which is given by the first element equal to 10 thousand and the denominator equal to 1.06.

S = 1.06 1.06 1.06 1.06 10000 = 12625

Examples of sum calculation problems:

Geometric progression is used in various problems. An example for finding the sum can be given as follows:

a 1 = 4, q= 2, calculateS 5.

Solution: all the data necessary for the calculation is known, you just need to substitute them into the formula.

S 5 = 124

• a 2 = 6, a 3 = 18. Calculate the sum of the first six elements.

Solution:

In geom. progression, each next element is q times greater than the previous one, that is, to calculate the sum you need to know the elementa 1 and denominatorq.

a 2 · q = a 3

q = 3

Similarly, you need to finda 1 , knowinga 2 Andq.

a 1 · q = a 2

a 1 =2

S 6 = 728.