Mathematics is whatpeople control nature and themselves.
Soviet mathematician, academician A.N. Kolmogorov
Geometric progression.
Along with tasks on arithmetic progressions, tasks related to the concept of geometric progression. To successfully solve such problems, you need to know the properties of a geometric progression and have good skills in using them.
This article is devoted to the presentation of the main properties of a geometric progression. It also provides examples of solving typical problems, borrowed from the tasks of entrance tests in mathematics.
Let us preliminarily note the main properties of a geometric progression and recall the most important formulas and statements, associated with this concept.
Definition. A numerical sequence is called a geometric progression if each of its numbers, 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 a geometric progressionthe formulas are valid
, (1)
Where . Formula (1) is called the formula of the general term of a geometric progression, and formula (2) is the main property of a geometric progression: each member of the progression coincides with the geometric mean of its neighboring members and .
Note, that it is precisely because of this property that the progression in question is called "geometric".
Formulas (1) and (2) above are summarized as follows:
, (3)
To calculate the sum first members of a geometric progressionthe formula applies
If we designate
Where . Since , formula (6) is a generalization of formula (5).
In the case when and geometric progressionis infinitely decreasing. To calculate the sumof all members of an infinitely decreasing geometric progression, the formula is used
. (7)
For example , using formula (7), one can show, What
Where . These equalities are obtained from formula (7) provided that , (the first equality) and , (the second equality).
Theorem. If , then
Proof. If , then ,
The theorem has been proven.
Let's move on to considering examples of solving problems on the topic "Geometric progression".
Example 1 Given: , and . Find .
Solution. If formula (5) is applied, then
Answer: .
Example 2 Let and . Find .
Solution. Since and , we use formulas (5), (6) and obtain the system of equations
If the second equation of system (9) is divided by the first, then or . From this it follows . 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. It follows from formula (2) that or . Since , then or .
By condition . However , therefore . Because 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 single suitable root . In this case, the first equation of the system implies .
Taking into account formula (7), we obtain.
Answer: .
Example 4 Given: and . Find .
Solution. Since , then .
Because , 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 and . 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 condition of the problem, we obtain the 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 10solve the equation
, (11)
where and .
Solution. The left side of equation (11) is the sum of an infinite decreasing geometric progression, in which and , provided: 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 positive numbers forms an arithmetic progression, A - geometric progression, what does it have to do with . Find .
Solution. Because arithmetic sequence, That (the main property of an arithmetic progression). Because the, then or . This implies , that the geometric progression is. According to formula (2), then we write that .
Since and , then . In that case, the expression takes the form or . By condition , so from the equationwe obtain the unique solution of 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 into formula (7) and obtain . Since , then .
Answer: .
The examples of problem solving given here will be useful to applicants in preparation for entrance examinations. For a deeper study of problem solving methods, associated with a geometric progression, can be used study guides from the list of recommended literature.
1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. – M.: Mir i Obrazovanie, 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 tasks and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. - 208 p.
Do you have any questions?
To get the help of a tutor - register.
site, with full or partial copying of the material, a link to the source is required.
Arithmetic and geometric progressions
Theoretical information
Theoretical information
Arithmetic progression |
Geometric progression |
|
Definition |
Arithmetic progression a n a sequence is called, each member of which, starting from the second, is equal to the previous member, added with the same number d (d- progression difference) |
geometric progression b n a sequence of non-zero numbers is called, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression) |
Recurrent formula |
For any natural n |
For any natural n |
nth term formula |
a n = a 1 + d (n - 1) |
b n \u003d b 1 ∙ q n - 1, b n ≠ 0 |
| characteristic property |  |
|
| Sum of the first n terms |  |
 |
Examples of tasks with comments
Exercise 1
In arithmetic progression ( a n) a 1 = -6, a 2
According to the formula of the nth term:
a 22 = a 1+ d (22 - 1) = a 1+ 21d
By condition:
a 1= -6, so a 22= -6 + 21d.
It is necessary to find the difference of progressions:
d= a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = - 48.
Answer : a 22 = -48.
Task 2
Find the fifth term of the geometric progression: -3; 6;....
1st way (using n-term formula)
According to the formula of the n-th member of a geometric progression:
b 5 \u003d b 1 ∙ q 5 - 1 = b 1 ∙ q 4.
Because b 1 = -3,
2nd way (using recursive formula)
Since the denominator of the progression is -2 (q = -2), then:
b 3 = 6 ∙ (-2) = -12;
b 4 = -12 ∙ (-2) = 24;
b 5 = 24 ∙ (-2) = -48.
Answer : b 5 = -48.
Task 3
In arithmetic progression ( a n) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.
For an arithmetic progression, the characteristic property has the form 
.
Therefore:
.
Substitute the data in the formula:
Answer: 95.
Task 4
In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.
To find the sum of the first n terms of an arithmetic progression, two formulas are used:
.
Which of them is more convenient to apply in this case?
By condition, the formula of the nth member of the original progression is known ( a n) a n= 3n - 4. Can be found immediately and a 1, And a 16 without finding d . Therefore, we use the first formula.
Answer: 368.
Task 5
In arithmetic progression a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.
According to the formula of the nth term:
a 22 = a 1 + d (22 – 1) = a 1+ 21d.
By condition, if a 1= -6, then a 22= -6 + 21d. It is necessary to find the difference of progressions:
d= a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = -48.
Answer : a 22 = -48.
Task 6
Several consecutive terms of a geometric progression are recorded:
Find the term of the progression, denoted by the letter x .
When solving, we use the formula for the nth term b n \u003d 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 these terms of the progression and divide by the previous one. In our example, you can take and divide by. We get that q \u003d 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.
Substituting the found values into the formula, we get:
.
Answer : .
Task 7
From arithmetic progressions, given by the formula nth term, select the one for which the condition is met a 27 > 9:
Since the specified condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:
.
Answer: 4.
Task 8
In arithmetic progression a 1= 3, d = -1.5. Specify highest value n , for which the inequality a n > -6.
An example of a geometric progression: 2, 6, 18, 54, 162.
Here, each term after the first is 3 times the previous one. That is, each subsequent term is the result of multiplying the previous term by 3:
2 3 = 6
6 3 = 18
18 3 = 54
54 3 = 162 .
In our example, when dividing the second term by the first, the third by the second, and so on. we get 3. The number 3 is the denominator of this geometric progression.
Example:
Let's go back to our geometric progression 2, 6, 18, 54, 162. Let's take the fourth term and square it:
54 2 = 2916.
Now we multiply the terms to the left and right of the number 54:
18 162 = 2916.
As you can see, the square of the third term is equal to the product of the neighboring second and fourth terms.
Example 1: Let's take some geometric progression, in which the first term is equal to 2, and the denominator of the geometric progression is equal to 1.5. We need to find the 4th term of this progression.
Given:
b 1 = 2
q = 1,5
n = 4
————
b 4 - ?
Solution.
Applying the formula b n= b 1 q n- 1 , inserting the appropriate values into it:
b 4 \u003d 2 1.5 4 - 1 \u003d 2 1.5 3 \u003d 2 3.375 \u003d 6.75.
Answer: The fourth term of a given geometric progression is the number 6.75.
Example 2: Find the fifth member of the geometric progression if the first and third members are 12 and 192, respectively.
Given:
b 1 = 12
b 3 = 192
————
b 5 - ?
Solution.
1) First we need to find the denominator of a geometric progression, without which it is impossible to solve the problem. As a first step, using our formula, we derive the formula for b 3:
b 3 = b 1 q 3 - 1 = b 1 q 2
Now we can find the denominator of a geometric progression:
b 3 192
q 2 = —— = —— = 16
b 1 12
q= √16 = 4 or -4.
2) It remains to find the value b 5 .
If q= 4, then
b 5 = b 1 q 5-1 = 12 4 4 = 12 256 = 3072.
At q= -4 the result will be the same. Thus, the problem has one solution.
Answer: The fifth term of the given geometric progression is the number 3072.
Example: Find the sum of the first five terms of the geometric progression ( b n), in which the first term is equal to 2, and the denominator of a geometric progression is 3.
Given:
b 1 = 2
q = 3
n = 5
————
S 5 - ?
Solution.
We apply the second formula of the two above:
b 1 (q 5 - 1) 2 (3 5 - 1) 2 (243 - 1) 484
S 5 = ————— = ————— = ———————— = ————— = 242
q - 1 3 - 1 2 2
Answer: The sum of the first five terms of a given geometric progression is 242.
The sum of an infinite geometric progression.
It is necessary to distinguish between the concepts of "the sum of an infinite geometric progression" and "the sum n members of a geometric progression. The second concept refers to any geometric progression, and the first - only to one where the denominator is less than 1 modulo.
Already in Ancient Egypt knew not only arithmetic, but also geometric progression. Here, for example, is a task from the Rhind papyrus: “Seven faces have seven cats; each cat eats seven mice, each mouse eats seven ears of corn, each ear can grow seven measures of barley. How large are the numbers in this series and their sum?
 |
|
Rice. 1. Ancient Egyptian geometric progression problem |
This task was repeated many times with different variations among other peoples at other times. For example, in written in the XIII century. The "Book of the abacus" by Leonardo of Pisa (Fibonacci) has a problem in which 7 old women appear on their way to Rome (obviously pilgrims), each of which has 7 mules, each of which has 7 bags, each of which has 7 loaves , each of which has 7 knives, each of which is in 7 sheaths. The problem asks how many items there are.
The sum of the first n members of the geometric progression S n = b 1 (q n - 1) / (q - 1) . This formula can be proved, for example, as follows: S n \u003d b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n - 1.
Let's add the number b 1 q n to S n and get:
|
Hence S n (q - 1) = b 1 (q n - 1), and we get the necessary formula.
Already on one of the clay tablets Ancient Babylon relating 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 a geometric progression in a number of cultures, in particular, in India, is repeatedly used as a clear symbol of the immensity of the universe. In the well-known legend about the appearance of chess, the ruler gives their inventor the opportunity to choose a reward himself, and he asks for such a number of grains of wheat as will be obtained if one is placed on the first cell of the chessboard, two on the second, four on the third, eight on the fourth, and etc., each time the number is doubled. Vladyka thought that it was, at the most, a few 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 as a 20-digit number; even if the entire surface of the Earth was sown, it would take at least 8 years to collect the required number of grains. This legend is sometimes interpreted as a reference to the almost unlimited possibilities hidden in the game of chess.
The fact that this number is really 20-digit is easy to see:
2 64 \u003d 2 4 ∙ (2 10) 6 \u003d 16 1024 6 ≈ 16 1000 6 \u003d 1.6 10 19 (a more accurate calculation gives 1.84 10 19). But I wonder if you can find out what digit this number ends with?
A geometric progression 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 can become arbitrarily small for sufficiently large n. While an increasing exponential increases unexpectedly fast, a decreasing exponential decreases just as quickly.
The larger n, the weaker the number q n differs from zero, and the closer the sum of n members of the geometric progression S n \u003d b 1 (1 - q n) / (1 - q) to the number S \u003d b 1 / (1 - q) . (So reasoned, for example, F. Viet). The number S is called the sum of an infinitely decreasing geometric progression. However, for many centuries the question of what is the meaning of the summation of the ALL geometric progression, with its infinite number of terms, was not clear enough to mathematicians.
A decreasing geometric progression can be seen, for example, in Zeno's aporias "Biting" and "Achilles and the tortoise". In the first case, it is clearly shown that the entire road (assume length 1) is the sum of an infinite number of segments 1/2, 1/4, 1/8, etc. This, of course, is the case from the point of view of ideas about the finite sum infinite geometric progression. And yet - how can this be?
|
Rice. 2. Progression with a factor of 1/2 |
In the aporia about Achilles, the situation is a little more complicated, because here the denominator of the progression is not equal to 1/2, but to some other number. Let, for example, Achilles run at speed v, the tortoise moves at speed u, and the initial distance between them is l. Achilles will run this distance in the time l / v , the tortoise will move a distance lu / v during this time. When Achilles runs through this segment, the distance between him and the turtle will become equal to l (u / v) 2, etc. It turns out that catching up with the turtle means finding the sum of an infinitely decreasing geometric progression with the first term l and the denominator u / v. This sum - the segment that Achilles will eventually run to the meeting point 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.
|
Rice. 3. Geometric progression with coefficient 2/3 |
The sum of a geometric progression was used by Archimedes when determining the area of a segment of a parabola. Let this segment the parabola is delimited by the chord AB and let the tangent 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 lines parallel to DC through points A , E , F , B ; let the tangent drawn at point D , these lines intersect at 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 the parabola at point R. According to general theory conic sections, DC is the diameter of the parabola (that is, a segment parallel to its axis); it and the tangent at point D can serve as coordinate axes x and y, in which the parabola equation is written as y 2 \u003d 2px (x is the distance from D to any point of a given diameter, y is the length of a segment parallel to a given tangent from this point of diameter to some point on the parabola itself).
By virtue of the parabola equation, DL 2 = 2 ∙ p ∙ LH , DK 2 = 2 ∙ p ∙ KA , and since DK = 2DL , then KA = 4LH . Since KA = 2LG , LH = HG . The area of the segment ADB of the parabola is equal to the area of the triangle ΔADB and the areas of the segments AHD and DRB combined. In turn, the area of the segment AHD is similarly equal to the area of the triangle AHD and the remaining segments AH and HD, with each of which the same operation can be performed - split into a triangle (Δ) and the two remaining segments (), etc.:
The area of the triangle ΔAHD is equal to half the area of the triangle ΔALD (they have a common base AD, and the heights differ by 2 times), which, in turn, is equal to half the area of the triangle ΔAKD, and therefore half the area of the triangle ΔACD. Thus, the area of triangle ΔAHD is equal to a quarter of the area of triangle ΔACD. Likewise, the area of triangle ΔDRB is equal to a quarter of the area of triangle ΔDFB. So, the areas of triangles ∆AHD and ∆DRB, taken together, are equal to a quarter of the area of triangle ∆ADB. Repeating this operation as 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 the triangles ΔAHD and ΔDRB , taken together, and therefore 16 times less, than the area of the triangle ΔADB . And so on:
Thus, Archimedes proved that "every segment enclosed between a straight line and a parabola is four-thirds of a triangle having the same base and equal height with it."
>>Math: Geometric progression
For the convenience of the reader, this section follows exactly the same plan as we followed in the previous section.
1. Basic concepts.
Definition. A numerical sequence, all members of which are different from 0 and each member of which, starting from the second, is obtained from the previous member by multiplying it by the same number is called a geometric progression. In this case, the number 5 is called the denominator of a geometric progression.
Thus, a geometric progression is a numerical sequence (b n) given recursively by the relations
Is it possible, by looking at a number sequence, to determine whether it is a geometric progression? Can. If you are convinced that the ratio of any member of the sequence to the previous member is constant, then you have a geometric progression.
Example 1
1, 3, 9, 27, 81,... .
b 1 = 1, q = 3.
Example 2
This is a geometric progression that
Example 3
This is a geometric progression that
Example 4
8, 8, 8, 8, 8, 8,....
This is a geometric progression where b 1 - 8, q = 1.
Note that this sequence is also an arithmetic progression (see Example 3 from § 15).
Example 5
2,-2,2,-2,2,-2.....
This is a geometric progression, in which b 1 \u003d 2, q \u003d -1.
Obviously, a geometric progression is an increasing sequence if b 1 > 0, q > 1 (see Example 1), and a decreasing sequence if b 1 > 0, 0< q < 1 (см. пример 2).
To indicate that the sequence (b n) is a geometric progression, the following notation is sometimes convenient:
The icon replaces the phrase "geometric progression".
We note one curious and at the same time quite obvious property of a geometric progression:
If the sequence 
is a geometric progression, then the sequence of squares, i.e. 
is a geometric progression.
In the second geometric progression, the first term is equal to a equal to q 2.
If we discard all the terms following b n exponentially, then we get a finite geometric progression 
In the following paragraphs of this section, we will consider the most important properties geometric progression.
2. Formula of the n-th term of a geometric progression.
Consider a geometric progression 
denominator q. We have:
It is not difficult to guess that for any number n the equality
This is the formula for the nth term of a geometric progression.
Comment.
If you have read the important remark from the previous paragraph and understood it, then try to prove formula (1) by mathematical induction, just as it was done for the formula of the nth term of an arithmetic progression.
Let's rewrite the formula of the nth term of the geometric progression
and introduce the notation: We get y \u003d mq 2, or, in more detail, 
The argument x is contained in the exponent, so such a function is called an exponential function. This means that a geometric progression can be considered as an exponential function given on the set N of natural numbers. On fig. 96a shows a graph of the function of Fig. 966 - function graph 
In both cases, we have isolated points (with abscissas x = 1, x = 2, x = 3, etc.) lying on some curve (both figures show the same curve, only differently located and depicted in different scales). This curve is called the exponent. More about exponential function and her graphics we will talk in the 11th grade algebra course.
Let's return to examples 1-5 from the previous paragraph.
1) 1, 3, 9, 27, 81,... . This is a geometric progression, in which b 1 \u003d 1, q \u003d 3. Let's make a formula for the nth term 
2) 
This is a geometric progression, in which Let's formulate the n-th term 
This is a geometric progression that 
Compose the formula for the nth term 
4) 8, 8, 8, ..., 8, ... . This is a geometric progression, in which b 1 \u003d 8, q \u003d 1. Let's make a formula for the nth term 
5) 2, -2, 2, -2, 2, -2,.... This is a geometric progression, in which b 1 = 2, q = -1. Compose the formula for the nth term 
Example 6
Given a geometric progression
In all cases, the solution is based on the formula of the nth member of a geometric progression
a) Putting n = 6 in the formula of the nth term of the geometric progression, we get
b) We have
Since 512 \u003d 2 9, we get n - 1 \u003d 9, n \u003d 10.
d) We have
Example 7
The difference between the seventh and fifth members of the geometric progression is 48, the sum of the fifth and sixth members of the progression is also 48. Find the twelfth member of this progression.
First stage. Drawing up a mathematical model.
The conditions of the task can be briefly written as follows:
Using the formula of the n-th member of a geometric progression, we get:
Then the second condition of the problem (b 7 - b 5 = 48) can be written as
The third condition of the problem (b 5 +b 6 = 48) can be written as
As a result, we obtain a system of two equations with two variables b 1 and q:
which, in combination with condition 1) written above, is mathematical model tasks.
Second phase.
Working with the compiled model. Equating the left parts of both equations of the system, we obtain:
(we have divided both sides of the equation into the expression b 1 q 4 , which is different from zero).
From the equation q 2 - q - 2 = 0 we find q 1 = 2, q 2 = -1. Substituting the value q = 2 into the second equation of the system, we obtain 
Substituting the value q = -1 into the second equation of the system, we get b 1 1 0 = 48; this equation has no solutions.
So, b 1 \u003d 1, q \u003d 2 - this pair is the solution to the compiled system of equations.
Now we can write down the geometric progression in question: 1, 2, 4, 8, 16, 32, ... .
Third stage.
The answer to the problem question. It is required to calculate b 12 . We have
Answer: b 12 = 2048.
3. The formula for the sum of members of a finite geometric progression.
Let there be a finite geometric progression
Denote by S n the sum of its terms, i.e.
Let's derive a formula for finding this sum.
Let's start with the simplest case, when q = 1. Then the geometric progression b 1 ,b 2 , b 3 ,..., bn consists of n numbers equal to b 1 , i.e. the progression is b 1 , b 2 , b 3 , ..., b 4 . The sum of these numbers is nb 1 .
Let now q = 1 To find S n we use an artificial method: let's perform some transformations of the expression S n q. We have:
Performing transformations, we, firstly, used the definition of a geometric progression, according to which (see the third line of reasoning); secondly, they added and subtracted why the meaning of the expression, of course, did not change (see the fourth line of reasoning); thirdly, we used the formula of the n-th member of a geometric progression:
From formula (1) we find:
This is the formula for the sum of n members of a geometric progression (for the case when q = 1).
Example 8
Given a finite geometric progression
a) the sum of the members of the progression; b) the sum of the squares of its terms.
b) Above (see p. 132) we have already noted that if all members of a geometric progression are squared, then a geometric progression with the first member b 2 and the denominator q 2 will be obtained. Then the sum of the six terms of the new progression will be calculated by
Example 9
Find the 8th term of a geometric progression for which
In fact, we have proved the following theorem.
A numerical sequence is a geometric progression if and only if the square of each of its terms, except for the first one (and the last one, in the case of a finite sequence), is equal to the product of the previous and subsequent terms (a characteristic property of a geometric progression).
