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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, associated with this concept.
Definition. A number sequence is called a geometric progression if each number, starting from the second, is equal to the previous one, multiplied by the same number. The number is called the denominator of a geometric progression.
For geometric progressionthe formulas are valid
, (1)
Where . Formula (1) is called the formula of the general term of a geometric progression, and formula (2) represents the main property of a geometric progression: each term of the progression coincides with the geometric mean of its neighboring terms and .
Note, that it is precisely because of this property that the progression in question is called “geometric”.
The above formulas (1) and (2) are generalized as follows:
, (3)
To calculate the amount first members of a geometric progressionformula applies
If we denote , then
Where . Since , formula (6) is a generalization of formula (5).
In the case when and geometric progressionis infinitely decreasing. To calculate the amountof all terms of an infinitely decreasing geometric progression, the formula is used
. (7)
For example , using formula (7) we can show, What
Where . These equalities are obtained from formula (7) under the condition that , (first equality) and , (second equality).
Theorem. If , then
Proof. If , then
The theorem has been proven.
Let's move on to consider examples of solving problems on the topic “Geometric progression”.
Example 1. Given: , and . Find .
Solution. If we apply formula (5), then
Answer: .
Example 2. Let it be. Find .
Solution. Since and , we use formulas (5), (6) and obtain a system of equations
If the second equation of system (9) is divided by the first, then or . It follows from this that . Let's consider two cases.
1. If, then from the first equation of system (9) we have.
2. If , then .
Example 3. Let , and . Find .
Solution. From formula (2) it follows that or . Since , then or .
By condition . However, therefore. Since and then here we have a system of equations
If the second equation of the system is divided by the first, then or .
Since, the equation has a unique suitable root. In this case, it follows from the first equation of the system.
Taking into account formula (7), we obtain.
Answer: .
Example 4. Given: and . Find .
Solution. Since, then.
Since , then or
According to formula (2) we have . In this regard, from equality (10) we obtain or .
However, by condition, therefore.
Example 5. It is known that . Find .
Solution. According to the theorem, we have two equalities
Since , then or . Because , then .
Answer: .
Example 6. Given: and . Find .
Solution. Taking into account formula (5), we obtain
Since, then. Since , and , then .
Example 7. Let it be. Find .
Solution. According to formula (1) we can write
Therefore, we have or . It is known that and , therefore and .
Answer: .
Example 8. Find the denominator of an infinite decreasing geometric progression if
And .
Solution. From formula (7) it follows And . From here and from the conditions of the problem we obtain a system of equations
If the first equation of the system is squared, and then divide the resulting equation by the second equation, then we get
Or .
Answer: .
Example 9. Find all values for which the sequence , , is a geometric progression.
Solution. Let , and . According to formula (2), which defines the main property of a geometric progression, we can write or .
From here we get the quadratic equation, whose roots are And .
Let's check: if, then , and ;
if , then , and . In the first case we have
and , and in the second – and .
Answer: , .Example 10.
, (11)
Solve the equation
where and .
From formula (7) it follows, What Solution. The left side of equation (11) is the sum of an infinite decreasing geometric progression, in which and , subject to: and .. In this regard, equation (11) takes the form or . Suitable root quadratic equation
Answer: .
is Example 11. Psequence of positive numbers forms an arithmetic progression , A– geometric progression
Solution., what does it have to do with . Find . Because arithmetic sequence , That(the main property of arithmetic progression). Because the , then or . This implies ,that the geometric progression has the form. According to formula (2)
Since and , then . In this case, the expression takes the form or . By condition , so from Eq.we get only decision 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 arithmetic sequence
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 when 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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Geometric progression no less important in mathematics compared to arithmetic. A geometric progression is a sequence of numbers b1, b2,..., b[n], each next term of which is obtained by multiplying the previous one by a constant number. This number, which also characterizes the rate of growth or decrease of progression, is called denominator of geometric progression and denote
To completely specify a geometric progression, in addition to the denominator, it is necessary to know or determine its first term. For positive value denominator progression is monotonous sequence, and if this sequence of numbers is monotonically decreasing and if it is monotonically increasing. The case when the denominator is equal to one is not considered in practice, since we have a sequence of identical numbers, and their summation is of no practical interest
General term of geometric progression calculated by the formula
Sum of the first n terms of a geometric progression determined by the formula
Let's look at solutions to classic geometric progression problems. Let's start with the simplest ones to understand.
Example 1. The first term of a geometric progression is 27, and its denominator is 1/3. Find the first six terms of the geometric progression.
Solution: Let us write the problem condition in the form
For calculations we use the formula for the nth term of a geometric progression
Based on it, we find the unknown terms of the progression
As you can see, calculating the terms of a geometric progression is not difficult. The progression itself will look like this
Example 2. The first three terms of the geometric progression are given: 6; -12; 24. Find the denominator and its seventh term.
Solution: We calculate the denominator of the geomitric progression based on its definition
We have obtained an alternating geometric progression whose denominator is equal to -2. The seventh term is calculated using the formula
This solves the problem.
Example 3. A geometric progression is given by two of its terms 
. Find the tenth term of the progression.
Solution:
Let's write the given values using formulas
According to the rules, we would need to find the denominator and then look for the desired value, but for the tenth term we have
The same formula can be obtained based on simple manipulations with the input data. Divide the sixth term of the series by another, and as a result we get
If the resulting value is multiplied by the sixth term, we get the tenth
Thus, for such tasks, using simple transformations to quick way you can find the right solution.
Example 4. Geometric progression is given by recurrent formulas
Find the denominator of the geometric progression and the sum of the first six terms.
Solution:
Let's write the given data in the form of a system of equations
Express the denominator by dividing the second equation by the first
Let's find the first term of the progression from the first equation
Let's calculate the following five terms to find the sum of the geometric progression
Instructions
10, 30, 90, 270...
You need to find the denominator of a geometric progression.
Solution:
Option 1. Let's take an arbitrary term of the progression (for example, 90) and divide it by the previous one (30): 90/30=3.
If the sum of several terms of a geometric progression or the sum of all terms of a decreasing geometric progression is known, then to find the denominator of the progression, use the appropriate formulas:
Sn = b1*(1-q^n)/(1-q), where Sn is the sum of the first n terms of the geometric progression and
S = b1/(1-q), where S is the sum of an infinitely decreasing geometric progression (the sum of all terms of the progression with a denominator less than one).
Example.
The first term of a decreasing geometric progression is equal to one, and the sum of all its terms is equal to two.
It is required to determine the denominator of this progression.
Solution:
Substitute the data from the problem into the formula. It will turn out:
2=1/(1-q), whence – q=1/2.
A progression is a sequence of numbers. In a geometric progression, each subsequent term is obtained by multiplying the previous one by a certain number q, called the denominator of the progression.
Instructions
If two adjacent geometric terms b(n+1) and b(n) are known, to obtain the denominator, you need to divide the number with the larger one by the one preceding it: q=b(n+1)/b(n). This follows from the definition of progression and its denominator. An important condition is that the first term and the denominator of the progression are not equal to zero, otherwise it is considered undefined.
Thus, the following relationships are established between the terms of the progression: b2=b1 q, b3=b2 q, ... , b(n)=b(n-1) q. Using the formula b(n)=b1 q^(n-1), any term of the geometric progression in which the denominator q and the term b1 are known can be calculated. Also, each of the progressions is equal in modulus to the average of its neighboring members: |b(n)|=√, which is where the progression got its .
An analogue of a geometric progression is the simplest exponential function y=a^x, where x is an exponent, a is a certain number. In this case, the denominator of the progression coincides with the first term and equal to the number a. The value of the function y can be understood as nth term progression if the argument x is taken to be a natural number n (counter).
Exists for the sum of the first n terms of a geometric progression: S(n)=b1 (1-q^n)/(1-q). This formula is valid for q≠1. If q=1, then the sum of the first n terms is calculated by the formula S(n)=n b1. By the way, the progression will be called increasing when q is greater than one and b1 is positive. If the denominator of the progression does not exceed one in absolute value, the progression will be called decreasing.
A special case of a geometric progression is an infinitely decreasing geometric progression (infinitely decreasing geometric progression). The fact is that the terms of a decreasing geometric progression will decrease over and over again, but will never reach zero. Despite this, it is possible to find the sum of all terms of such a progression. It is determined by the formula S=b1/(1-q). Total n members are infinite.
To visualize how you can add an infinite number of numbers without getting infinity, bake a cake. Cut off half of it. Then cut 1/2 off half, and so on. The pieces that you will get are nothing more than members of an infinitely decreasing geometric progression with a denominator of 1/2. If you add up all these pieces, you get the original cake.
Geometry problems are a special type of exercise that requires spatial thinking. If you can't solve a geometric task, try following the rules below.
Instructions
Read the conditions of the task very carefully; if you don’t remember or don’t understand something, re-read it again.
Try to determine what type of geometric problems it is, for example: computational ones, when you need to find out some quantity, problems involving , requiring a logical chain of reasoning, problems involving construction using a compass and ruler. More tasks mixed type. Once you have figured out the type of problem, try to think logically.
Apply the necessary theorem for a given task, but if you have doubts or there are no options at all, then try to remember the theory that you studied on the relevant topic.
Also write down the solution to the problem in a draft form. Try to use known methods to check the correctness of your solution.
Fill out the solution to the problem carefully in your notebook, without erasing or crossing out, and most importantly - . It may take time and effort to solve the first geometric problems. However, as soon as you master this process, you will start clicking tasks like nuts, enjoying it!
A geometric progression is a sequence of numbers b1, b2, b3, ... , b(n-1), b(n) such that b2=b1*q, b3=b2*q, ... , b(n) =b(n-1)*q, b1≠0, q≠0. In other words, each term of the progression is obtained from the previous one by multiplying it by some non-zero denominator of the progression q.
Instructions
Progression problems are most often solved by drawing up and then following a system with respect to the first term of the progression b1 and the denominator of the progression q. To create equations, it is useful to remember some formulas.
How to express the nth term of the progression through the first term of the progression and the denominator of the progression: b(n)=b1*q^(n-1).
Let us consider separately the case |q|<1. Если знаменатель прогрессии по модулю меньше единицы, имеем бесконечно убывающую геометрическую . Сумма первых n членов бесконечно убывающей геометрической прогрессии ищется так же, как и для неубывающей геометрической прогрессии. Однако в случае бесконечно убывающей геометрической прогрессии можно найти также сумму всех членов этой прогрессии, поскольку при бесконечном n будет бесконечно уменьшаться значение b(n), и сумма всех членов будет стремиться к определенному пределу. Итак, сумма всех членов бесконечно убывающей геометрической прогрессии
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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 member, starting from the second, equal to the product the previous 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... A 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 $q=-1$.
Geometric progression has the properties of monotony.
If $b_(1)>0$, $q>1$,
then the sequence is increasing.
If $b_(1)>0$, $0 The sequence is usually denoted in the form: $b_(1), b_(2), b_(3), ..., b_(n), ...$.
Just like in an arithmetic progression, if in a geometric progression the number of elements 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 a sequence is a geometric progression, then the sequence of squares of terms is also a geometric progression. In the second sequence, the first term is equal to $b_(1)^2$, and the denominator is equal to $q^2$.
Formula for the nth term of a geometric progression
Geometric progression can also be specified in analytical form. Let's see how to do this:$b_(1)=b_(1)$.
$b_(2)=b_(1)*q$.
$b_(3)=b_(2)*q=b_(1)*q*q=b_(1)*q^2$.
$b_(4)=b_(3)*q=b_(1)*q^3$.
$b_(5)=b_(4)*q=b_(1)*q^4$.
We easily notice the pattern: $b_(n)=b_(1)*q^(n-1)$.
Our formula is called the "formula of the nth term of a geometric progression."
Let's return 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 equal to 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 equal to 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 equal to 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 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 received 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 we 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: $b_(1)=4, q=2$.
Let's find the tenth term: $b_(10)=b_(1)*q^9=4*2^9=2048$.
Sum of a finite geometric progression
Let us have a finite geometric progression. Let's, just like for an arithmetic progression, calculate the sum of its terms.Let a finite geometric progression be given: $b_(1),b_(2),…,b_(n-1),b_(n)$.
Let us introduce the designation for the sum of its terms: $S_(n)=b_(1)+b_(2)+⋯+b_(n-1)+b_(n)$.
In the case when $q=1$. All terms of the geometric progression are equal to the first term, then it is obvious that $S_(n)=n*b_(1)$.
Let us now consider the case $q≠1$.
Let's 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 term of the geometric progression that 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 geometric progression
Guys, a geometric progression is given. Let's look at 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 form 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 member is equal to the product of the two adjacent members of the progression. Do not forget that for a finite progression this condition is not satisfied for the first and last terms.
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 average geometric numbers a and b.
The modulus of any term of a geometric progression is equal to the geometric mean of its two neighboring terms.
Example.
Find x such that $x+2; 2x+2; 3x+3$ were three consecutive terms 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 $x_(2)=-1$.
Let us sequentially substitute our solutions into the original expression:
With $x=2$, we got the sequence: 4;6;9 – a geometric progression with $q=1.5$.
For $x=-1$, we get the sequence: 1;0;0.
Answer: $x=2.$
Problems to solve independently
1. Find the eighth first term of the geometric progression 16;-8;4;-2….2. Find the tenth term 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 terms of the geometric progression 3;12;48….
6. Find x such that $3x+4; 2x+4; x+5$ are three consecutive terms of a geometric progression.