Geometric progression formulas and examples explanation. The formula for the sum of the first n terms of the GP
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 instance , 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 an
Answer: .
Example 11. P sequence of positive numbersforms an arithmetic progression, a - geometric progression, what does it have to do with . Find .
Solution. Because arithmetic sequence, then (the main property of an arithmetic progression). Insofar as, 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, then
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 of the school curriculum. – M.: Lenand / URSS, 2014. - 216 p.
3. Medynsky M.M. A complete course of elementary mathematics in tasks and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. - 208 p.
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>>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 will be discussed in the 11th grade algebra course.
Let's return to examples 1-5 from the previous paragraph.
1) 1, 3, 9, 27, 81,... . This is a geometric progression, 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 get:
(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 a geometric progression, about which in question in the problem: 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.
Numeric, a sequence is a geometric progression if and only if the square of each of its terms, except the first Theorem (and the last, in the case of a finite sequence), is equal to the product preceding and following terms (a characteristic property of a 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
Definition. A numerical sequence in which each term, starting from the second, is equal to the product of the previous one and some fixed number, is called a geometric progression.Let's define our sequence recursively: $b_(1)=b$, $b_(n)=b_(n-1)*q$,
where b and q are certain given numbers. The number q is called the denominator of the progression.
Example. 1,2,4,8,16… Geometric progression, in which the first member is equal to one, and $q=2$.
Example. 8,8,8,8… A geometric progression whose first term is eight,
and $q=1$.
Example. 3,-3,3,-3,3... A geometric progression whose first term is three,
and $q=-1$.
The geometric progression has the properties of monotonicity.
If $b_(1)>0$, $q>1$,
then the sequence is increasing.
If $b_(1)>0$, $0 The sequence is usually denoted as: $b_(1), b_(2), b_(3), ..., b_(n), ...$.
Just like in an arithmetic progression, if the number of elements in a geometric progression is finite, then the progression is called a finite geometric progression.
$b_(1), b_(2), b_(3), ..., b_(n-2), b_(n-1), b_(n)$.
Note that if the sequence is a geometric progression, then the sequence of squared terms is also a geometric progression. The second sequence has the first term $b_(1)^2$ and the denominator $q^2$.
Formula of the nth member of a geometric progression
Geometric progression can also be specified in analytical form. Let's see how to do it:$b_(1)=b_(1)$.
$b_(2)=b_(1)*q$.
$b_(3)=b_(2)*q=b_(1)*q*q=b_(1)*q^2$.
$b_(4)=b_(3)*q=b_(1)*q^3$.
$b_(5)=b_(4)*q=b_(1)*q^4$.
We can easily see the pattern: $b_(n)=b_(1)*q^(n-1)$.
Our formula is called "formula of the n-th member of a geometric progression".
Let's go back to our examples.
Example. 1,2,4,8,16… A geometric progression whose 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 whose 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 where the first term is eight and $q=1$.
$b_(n)=8*1^(n-1)=8$.
Example. 3,-3,3,-3,3… A geometric progression whose first term is three and $q=-1$.
$b_(n)=3*(-1)^(n-1)$.
Example. 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 members of the geometric progression is 192, the sum of the fifth and sixth members of the progression is 192. Find the tenth member 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 successively into the second equation:
$b_(1)*2^4*3=192 => b_(1)=4$.
$b_(1)*(-1)^4*0=192 =>$ no solutions.
We got that: $b_(1)=4, q=2$.
Let's find the tenth term: $b_(10)=b_(1)*q^9=4*2^9=2048$.
The sum of a finite geometric progression
Suppose we have a finite geometric progression. Let's, as well as for an arithmetic progression, calculate the sum of its members.Let a finite geometric progression be given: $b_(1),b_(2),…,b_(n-1),b_(n)$.
Let's 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 member, then it is obvious that $S_(n)=n*b_(1)$.
Consider now the case $q≠1$.
Multiply the above amount 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 have obtained the formula for the sum of a finite geometric progression.
Example.
Find the sum of the first seven terms of a geometric progression whose first term is 4 and the denominator is 3.
Solution.
$S_(7)=\frac(4*(3^(7)-1))(3-1)=2*(3^(7)-1)=4372$.
Example.
Find the fifth member 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
Guys, given a geometric progression. Let's consider its three consecutive members: $b_(n-1),b_(n),b_(n+1)$.We know that:
$\frac(b_(n))(q)=b_(n-1)$.
$b_(n)*q=b_(n+1)$.
Then:
$\frac(b_(n))(q)*b_(n)*q=b_(n)^(2)=b_(n-1)*b_(n+1)$.
$b_(n)^(2)=b_(n-1)*b_(n+1)$.
If the progression is finite, then this equality holds for all terms except the first and last.
If it is not known in advance what kind of sequence the sequence has, but it is known that: $b_(n)^(2)=b_(n-1)*b_(n+1)$.
Then we can safely say that this is a geometric progression.
A number sequence is a geometric progression only when the square of each of its terms is equal to the product of its two neighboring terms of the progression. Let's not forget that for finite progression this condition is not satisfied for the first and last member.
Let's look at this identity: $\sqrt(b_(n)^(2))=\sqrt(b_(n-1)*b_(n+1))$.
$|b_(n)|=\sqrt(b_(n-1)*b_(n+1))$.
$\sqrt(a*b)$ is called the geometric mean of a and b.
The modulus of any member of a geometric progression is equal to the geometric mean of the two members adjacent to it.
Example.
Find x such that $x+2; 2x+2; 3x+3$ were three consecutive members of a geometric progression.
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 $x_(2)=-1$.
Substitute sequentially in the original expression, our solutions:
With $x=2$, we got the sequence: 4;6;9 is a geometric progression with $q=1.5$.
With $x=-1$, we got the sequence: 1;0;0.
Answer: $x=2.$
Tasks for independent solution
1. Find the eighth first member of the geometric progression 16; -8; 4; -2 ....2. Find the tenth member of the geometric progression 11,22,44….
3. It is known that $b_(1)=5, q=3$. Find $b_(7)$.
4. It is known that $b_(1)=8, q=-2, b_(n)=512$. Find n.
5. Find the sum of the first 11 members of the geometric progression 3;12;48….
6. Find x such that $3x+4; 2x+4; x+5$ are three consecutive members of a geometric progression. 22.09.2018 22:00
Geometric progression, along with arithmetic, is an important number series, which is studied in school course algebra in 9th grade. In this article, we will consider the denominator of a geometric progression, and how its value affects its properties.
Definition of geometric progression
To begin with, we give the definition of this number series. A geometric progression is a series 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 we multiply 3 (the first element) by 2, we get 6. If we multiply 6 by 2, we 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 a progression can be written in the language of mathematics 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 under consideration. Similar reasoning can be continued for large values of n.
The denominator of a geometric progression
The number b completely determines what character the entire number series will have. The denominator b can be positive, negative, and also have a value more than one or less. 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 the element a1 is negative, then the whole sequence will increase only modulo, but decrease taking into account the sign of the numbers.
- b = 1. Often such a case is not called a progression, since there is an ordinary series of identical rational numbers. For example, -4, -4, -4.
Formula for sum
Before proceeding to the consideration of specific problems using the denominator of the type of progression under consideration, one should bring important formula for the sum of its first n elements. The formula is: Sn = (bn - 1) * a1 / (b - 1).
You can get this expression yourself if you consider a recursive sequence of members of the progression. Also note that in the above formula, it is enough to know only the first element and the denominator in order to find the sum of an arbitrary number of terms.
Infinitely decreasing sequence
Above was an explanation 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 we will consider several problems, where we will show how to apply the acquired knowledge to specific numbers.
Task number 1. Calculation of unknown elements of the progression and the sum
Given a geometric progression, the denominator of the progression is 2, and its first element is 3. What will be its 7th and 10th terms, 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 the element with 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 member: a10 = 29 * 3 = 1536.
We 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.
Task number 2. Determining the sum of arbitrary elements of the progression
Let -2 be the denominator of the exponential 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 different ways. For the sake of completeness, we present both.
Method 1. Its idea is simple: you need to calculate the two corresponding sums of the first terms, and then subtract the other from one. Calculate the smaller sum: S10 = ((-2)10 - 1) * 4 / (-2 - 1) = -1364. Now we calculate the big sum: S4 = ((-2)4 - 1) * 4 / (-2 - 1) = -20. Note that in the last expression, only 4 terms were summed up, since the 5th is already included in the sum that needs to be calculated according to the condition of the problem. Finally, we take the difference: S510 = S10 - S4 = -1364 - (-20) = -1344.
Method 2. Before substituting numbers and counting, you can get a formula for the sum between the terms m and n of the series in question. We act in exactly the same way as in method 1, only we work first with the symbolic representation of the sum. 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.
Task number 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.
According to the condition of the problem, it is not difficult to guess which formula should be used to solve it. Of course, for the sum of an infinitely decreasing progression. 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 \u003d 1 - 2 / 3 \u003d -1 / 3 or -0.333 (3). We can check this result qualitatively if we remember that for this type of sequence, the modulus b must not go beyond 1. As you can see, |-1 / 3|
Task number 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 restore 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 member. We have: a5 = b4 * a1 and a10 = b9 * a1. Now we 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 degree root of the ratio of the members known from the condition of the problem, b = 1.148698. We substitute the resulting number into one of the expressions for a known element, we get: a1 = a5 / b4 = 30 / (1.148698)4 = 17.2304966.
Thus, we have found what the denominator of the progression bn is, and the geometric progression bn-1 * 17.2304966 = an, where b = 1.148698.
Where are geometric progressions used?
If there were no application of this numerical series in practice, then its study would be reduced to a purely theoretical interest. But there is such an application.
The 3 most famous examples are listed below:
- Zeno's paradox, in which the agile Achilles cannot catch up with the slow tortoise, is solved using the concept of an infinitely decreasing sequence of numbers.
- If wheat grains are placed on each cell of the chessboard so that 1 grain is placed on the 1st cell, 2 - on the 2nd, 3 - on the 3rd, and so on, then 18446744073709551615 grains will be needed to fill all the cells of the board!
- In the game "Tower of Hanoi", in order to rearrange disks from one rod to another, it is necessary to perform 2n - 1 operations, that is, their number grows exponentially from the number of disks n used.
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A geometric progression is a numerical sequence, the first term of which is non-zero, and each next term is equal to the previous term multiplied by the same non-zero number.
The geometric progression is denoted b1,b2,b3, …, bn, … .
The ratio of any term of the geometric error to its previous term is equal to the same number, that is, b2/b1 = b3/b2 = b4/b3 = … = bn/b(n-1) = b(n+1)/bn = …. This follows directly from the definition of an arithmetic progression. This number is called the denominator of a geometric progression. Usually the denominator of a geometric progression is denoted by the letter q.
Monotonic and constant sequence
One way to set a geometric progression is to set its first term b1 and the denominator of the geometric error q. For example, b1=4, q=-2. These two conditions give a geometric progression of 4, -8, 16, -32, … .
If q>0 (q is not equal to 1), then the progression is monotone sequence. For example, the sequence, 2, 4,8,16,32, ... is a monotonically increasing sequence (b1=2, q=2).
If the denominator q=1 in the geometric error, then all members of the geometric progression will be equal to each other. In such cases, the progression is said to be constant sequence.
Formula of the nth member of a geometric progression
In order for the numerical sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, be the geometric mean of the neighboring members. That is, it is necessary to fulfill the following equation
(b(n+1))^2 = bn * b(n+2), for any n>0, where n belongs to the set natural numbers N.
The formula for the nth member of a geometric progression is:
bn=b1*q^(n-1),
where n belongs to the set of natural numbers N.
The formula for the sum of the first n terms of a geometric progression
The formula for the sum of the first n terms of a geometric progression is:
Sn = (bn*q - b1)/(q-1) where q is not equal to 1.
Consider a simple example:
In geometric progression b1=6, q=3, n=8 find Sn.
To find S8, we use the formula for the sum of the first n terms of a geometric progression.
S8= (6*(3^8 -1))/(3-1) = 19680.
