Lesson and presentation on the topic: "Number sequences. Geometric progression"
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Guys, today we will get acquainted with another type of progression.
The topic of today's lesson is geometric progression.
Geometric progression
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... 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 geometric mean of the numbers a and b.
The modulus of any term of a geometric progression is equal to the geometric mean of its two adjacent 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.
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 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.Solve the equation
, (11)
where and .
Solution. The left side of 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 sequence of positive numbersforms an arithmetic progression, A – geometric progression, what does it have to do with . Find .
Solution. 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), then we write down that .
Since and , then . In this case, the expression takes the form or . By condition , so from Eq.we obtain a unique solution to 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 \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, You can use tutorials 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 of the 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. This means that this series is a progression.
A geometric progression is an infinite sequence of numbers, the main feature of which is that the next number is obtained from the previous one by multiplying by a 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:S z = 2 · = 4
S z = 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.
22.09.2018 22:00Geometric progression, along with arithmetic progression, is an important number series that is studied in the school algebra course in the 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. A geometric progression is a series of rational numbers that 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 moving on to the consideration of specific problems 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 on specific numbers.
Task 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 using 2 different 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 do exactly the same 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 tortoise, is solved using the concept of an infinitely decreasing sequence of numbers.
- If you place wheat grains on each square of the chessboard so that on the 1st square you put 1 grain, on the 2nd - 2, on the 3rd - 3, and so on, then to fill all the squares 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.
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Geometric progression is a new type of number sequence that we are about to get acquainted with. For successful dating, it doesn’t hurt to at least know and understand. Then there will be no problems with geometric progression.)
What is geometric progression? The concept of geometric progression.
We start the tour, as usual, with the basics. I write an unfinished sequence of numbers:
1, 10, 100, 1000, 10000, …
Can you spot the pattern and tell which numbers will come next? The pepper is clear, then the numbers 100,000, 1,000,000 and so on will follow. Even without much mental effort, everything is clear, right?)
OK. Another example. I write this sequence:
1, 2, 4, 8, 16, …
Can you tell which numbers will come next, following the number 16, and name eighth sequence member? If you figured out that it would be the number 128, then very good. So, half the battle is in understanding meaning And key points geometric progression has already been done. You can grow further.)
And now we move again from sensations to strict mathematics.
Key points of geometric progression.
Key Point #1
Geometric progression is sequence of numbers. So is progression. Nothing fancy. Only this sequence is arranged differently. Hence, naturally, it has a different name, yes...
Key Point #2
With the second key point, the question will be trickier. Let's go back a little and remember the key property of arithmetic progression. Here it is: each member is different from the previous one by the same amount.
Is it possible to formulate a similar key property for a geometric progression? Think a little... Take a closer look at the examples given. Did you guess it? Yes! In geometric progression (any!) each of its members differs from the previous one the same number of times. Always!
In the first example, this number is ten. Whichever member of the sequence you take, it is greater than the previous one ten times.
In the second example it is a two: each term is greater than the previous one twice.
It is this key point that geometric progression differs from arithmetic progression. In an arithmetic progression, each subsequent term is obtained by adding the same value to the previous term. And here - multiplication the previous term by the same amount. That's the whole difference.)
Key Point #3
This key point is completely identical to that for an arithmetic progression. Namely: Each member of a geometric progression stands in its place. Everything is exactly the same as in the arithmetic progression and comments, I think, are unnecessary. There is the first term, there is the hundred and first, etc. Let us swap at least two terms – the pattern (and with it the geometric progression) will disappear. What will remain is just a sequence of numbers without any logic.
That's all. That's the whole point of geometric progression.
Terms and designations.
But now, having understood the meaning and key points of geometric progression, we can move on to the theory. Otherwise, what is a theory without understanding the meaning, right?
How to denote geometric progression?
How is geometric progression written in general form? No problem! Each term of the progression is also written as a letter. Only for arithmetic progression, usually the letter is used "A", for geometric – letter "b". Member number, as usual, is indicated index at the bottom right. We simply list the members of the progression themselves, separated by commas or semicolons.
Like this:
b 1,b 2 , b 3 , b 4 , b 5 , b 6 , …
Briefly, this progression is written like this: (b n) .
Or like this, for finite progressions:
b 1, b 2, b 3, b 4, b 5, b 6.
b 1, b 2, …, b 29, b 30.
Or, in short:
(b n), n=30 .
That, in fact, is all the designation. Everything is the same, only the letter is different, yes.) And now we move directly to the definition.
Definition of geometric progression.
A geometric progression is a number sequence in which the first term is non-zero, and each subsequent term is equal to the previous term multiplied by the same non-zero number.
That's the whole definition. Most words and phrases are clear and familiar to you. If, of course, you understand the meaning of geometric progression “on your fingers” and in general. But there are also a few new phrases that I would like to pay special attention to.
First, the words: "the first member of which non-zero".
This restriction on the first term was not introduced by chance. What do you think will happen if the first member b 1 will be equal to zero? What will the second term be equal to if each term is greater than the previous one? the same number of times? Let's say three times? Let's see... Multiply the first term (i.e. 0) by 3 and get... zero! What about the third member? Also zero! And the fourth term is also zero! And so on…
We just get a bag of bagels, a sequence of zeros:
0, 0, 0, 0, …
Of course, such a sequence has a right to life, but it is of no practical interest. Everything is clear. Any member of it is zero. The sum of any number of terms is also zero... What interesting things can you do with it? Nothing…
The following keywords: "multiplied by the same non-zero number."
This same number also has its own special name - denominator of geometric progression. Let's start getting acquainted.)
Denominator of a geometric progression.
Everything is as simple as shelling pears.
The denominator of a geometric progression is a non-zero number (or quantity) indicating how many timeseach term of the progression more than the previous one.
Again, similar to the arithmetic progression, the key word to look for in this definition is the word "more". It means that each term of the geometric progression is obtained multiplication to this very denominator previous member.
Let me explain.
To calculate, let's say second dick, need to take first member and multiply it to the denominator. For calculation tenth dick, need to take ninth member and multiply it to the denominator.
The denominator of the geometric progression itself can be anything. Absolutely anyone! Whole, fractional, positive, negative, irrational - everything. Except zero. This is what the word “non-zero” in the definition tells us. Why this word is needed here - more on that later.
Denominator of geometric progression most often indicated by the letter q.
How to find it q? No problem! We must take any term of the progression and divide by the previous term. Division is fraction. Hence the name - “progression denominator”. The denominator, it usually sits in a fraction, yes...) Although, logically, the value q should be called private geometric progression, similar to difference for arithmetic progression. But we agreed to call denominator. And we won’t reinvent the wheel either.)
Let us define, for example, the quantity q for this geometric progression:
2, 6, 18, 54, …
Everything is elementary. Let's take it any sequence number. We take whatever we want. Except the very first one. For example, 18. And divide by previous number. That is, at 6.
We get:
q = 18/6 = 3
That's all. This is the correct answer. For this geometric progression, the denominator is three.
Let's now find the denominator q for another geometric progression. For example, this one:
1, -2, 4, -8, 16, …
All the same. No matter what signs the members themselves have, we still take any number of the sequence (for example, 16) and divide by previous number(i.e. -8).
We get:
d = 16/(-8) = -2
And that's it.) This time the denominator of the progression turned out to be negative. Minus two. Happens.)
Let's now take this progression:
1, 1/3, 1/9, 1/27, …
And again, regardless of the type of numbers in the sequence (whether integers, even fractions, even negative, even irrational), we take any number (for example, 1/9) and divide by the previous number (1/3). According to the rules for working with fractions, of course.
We get:
That's all.) Here the denominator turned out to be fractional: q = 1/3.
What do you think of this “progression”?
3, 3, 3, 3, 3, …
Obviously here q = 1 . Formally, this is also a geometric progression, only with identical members.) But such progressions are not interesting for study and practical application. The same as progressions with solid zeros. Therefore, we will not consider them.
As you can see, the denominator of the progression can be anything - integer, fractional, positive, negative - anything! It can't just be zero. Can't guess why?
Well, let's use some specific example to see what will happen if we take as the denominator q zero.) Let us, for example, have b 1 = 2 , A q = 0 . What then will the second term be equal to?
We count:
b 2 = b 1 · q= 2 0 = 0
What about the third member?
b 3 = b 2 · q= 0 0 = 0
Types and behavior of geometric progressions.
Everything was more or less clear: if the progression difference d is positive, then the progression increases. If the difference is negative, then the progression decreases. There are only two options. There is no third.)
But with the behavior of geometric progression, everything will be much more interesting and varied!)
No matter how the terms behave here: they increase, and decrease, and indefinitely approach zero, and even change signs, alternately throwing themselves into “plus” and then into “minus”! And in all this diversity you need to be able to understand well, yes...
Let's figure it out?) Let's start with the simplest case.
The denominator is positive ( q >0)
With a positive denominator, firstly, the terms of the geometric progression can go into plus infinity(i.e. increase without limit) and can go into minus infinity(i.e., decrease without limit). We are already accustomed to this behavior of progressions.
For example:
(b n): 1, 2, 4, 8, 16, …
Everything is simple here. Each term of the progression is obtained more than previous. Moreover, each term turns out multiplication previous member on positive number +2 (i.e. q = 2 ). The behavior of such a progression is obvious: all members of the progression grow without limit, going into space. Plus infinity...
And now here's the progression:
(b n): -1, -2, -4, -8, -16, …
Here, too, each term of the progression is obtained multiplication previous member on positive number +2. But the behavior of such a progression is exactly the opposite: each term of the progression is obtained less than previous, and all its terms decrease without limit, going to minus infinity.
Now let's think: what do these two progressions have in common? That's right, denominator! Here and there q = +2 . Positive number. Two. And here behavior These two progressions are fundamentally different! Can't guess why? Yes! It's all about first member! It is he, as they say, who calls the tune.) See for yourself.
In the first case, the first term of the progression positive(+1) and, therefore, all subsequent terms obtained by multiplying by positive denominator q = +2 , will also be positive.
But in the second case, the first term negative(-1). Therefore, all subsequent terms of the progression, obtained by multiplying by positive q = +2 , will also be obtained negative. Because “minus” to “plus” always gives “minus”, yes.)
As you can see, unlike an arithmetic progression, a geometric progression can behave completely differently not only depending from the denominatorq, but also depending from the first member, Yes.)
Remember: the behavior of a geometric progression is uniquely determined by its first term b 1 and denominatorq .
And now we begin to analyze less familiar, but much more interesting cases!
Let's take, for example, this sequence:
(b n): 1, 1/2, 1/4, 1/8, 1/16, …
This sequence is also a geometric progression! Each term of this progression also turns out multiplication the previous member, by the same number. It's just a number - fractional: q = +1/2 . Or +0,5 . Moreover (important!) the number less than one:q = 1/2<1.
Why is this geometric progression interesting? Where are its members heading? Let's get a look:
1/2 = 0,5;
1/4 = 0,25;
1/8 = 0,125;
1/16 = 0,0625;
…….
What interesting things can you notice here? Firstly, the decrease in terms of the progression is immediately noticeable: each of its members less the previous one exactly 2 times. Or, according to the definition of a geometric progression, each term more previous 1/2 times, because progression denominator q = 1/2 . And when multiplied by a positive number less than one, the result usually decreases, yes...
What more can be seen in the behavior of this progression? Are its members diminishing? unlimited, going to minus infinity? No! They disappear in a special way. At first they decrease quite quickly, and then more and more slowly. And while remaining all the time positive. Albeit very, very small. And what do they themselves strive for? Didn't you guess? Yes! They strive towards zero!) Moreover, pay attention, the members of our progression are from zero never reach! Only approaching him infinitely close. It is very important.)
A similar situation will occur in the following progression:
(b n): -1, -1/2, -1/4, -1/8, -1/16, …
Here b 1 = -1 , A q = 1/2 . Everything is the same, only now the terms will approach zero from the other side, from below. Staying all the time negative.)
Such a geometric progression, the terms of which approach zero without limit(no matter from the positive or negative side), in mathematics has a special name - infinitely decreasing geometric progression. This progression is so interesting and unusual that it will even be discussed separate lesson .)
So, we have considered all possible positive the denominators are both large ones and smaller ones. We do not consider the unit itself as a denominator for the reasons stated above (remember the example with a sequence of triplets...)
Let's summarize:
positiveAnd more than one (q>1), then the terms of the progression:
a) increase without limit (ifb 1 >0);
b) decrease without limit (ifb 1 <0).
If the denominator of the geometric progression positive And less than one (0< q<1), то члены прогрессии:
a) infinitely close to zero above(Ifb 1 >0);
b) approaching infinitely close to zero from below(Ifb 1 <0).
It now remains to consider the case negative denominator.
Denominator is negative ( q <0)
We won’t go far for an example. Why, exactly, shaggy grandma?!) Let, for example, the first term of the progression be b 1 = 1 , and let’s take the denominator q = -2.
We get the following sequence:
(b n): 1, -2, 4, -8, 16, …
And so on.) Each term of the progression is obtained multiplication previous member on a negative number-2. In this case, all members standing in odd places (first, third, fifth, etc.) will be positive, and in even places (second, fourth, etc.) – negative. The signs strictly alternate. Plus-minus-plus-minus... This geometric progression is called - increasing sign alternating.
Where are its members heading? But nowhere.) Yes, in absolute value (i.e. modulo) the members of our progression increase without limit (hence the name “increasing”). But at the same time, each member of the progression alternately throws you into the heat, then into the cold. Either “plus” or “minus”. Our progression is wavering... Moreover, the scope of fluctuations is growing rapidly with each step, yes.) Therefore, the aspirations of the members of the progression are going somewhere specifically Here No. Neither to plus infinity, nor to minus infinity, nor to zero - nowhere.
Let us now consider some fractional denominator between zero and minus one.
For example, let it be b 1 = 1 , A q = -1/2.
Then we get the progression:
(b n): 1, -1/2, 1/4, -1/8, 1/16, …
And again we have an alternation of signs! But, unlike the previous example, here there is already a clear tendency for the terms to approach zero.) Only this time our terms approach zero not strictly from above or below, but again hesitating. Alternatingly taking positive and negative values. But at the same time they modules are getting closer and closer to the cherished zero.)
This geometric progression is called infinitely decreasing sign, alternating.
Why are these two examples interesting? And the fact that in both cases takes place alternation of signs! This trick is typical only for progressions with a negative denominator, yes.) Therefore, if in some task you see a geometric progression with alternating terms, you will already know for sure that its denominator is 100% negative and you will not make a mistake in the sign.)
By the way, in the case of a negative denominator, the sign of the first term does not at all affect the behavior of the progression itself. Regardless of the sign of the first term of the progression, in any case the sign of the terms will be observed. The only question is, in what places(even or odd) there will be members with specific signs.
Remember:
If the denominator of the geometric progression negative , then the signs of the terms of the progression are always alternate.
At the same time, the members themselves:
a) increase without limitmodulo, Ifq<-1;
b) approach zero infinitely if -1< q<0 (прогрессия бесконечно убывающая).
That's all. All typical cases have been analyzed.)
In the process of analyzing a variety of examples of geometric progressions, I periodically used the words: "tends to zero", "tends to plus infinity", "tends to minus infinity"... It's okay.) These figures of speech (and specific examples) are just an initial introduction to behavior a variety of number sequences. Using the example of geometric progression.
Why do we even need to know the behavior of progression? What difference does it make where she goes? Toward zero, to plus infinity, to minus infinity... What does that do to us?
The thing is that already at the university, in a course of higher mathematics, you will need the ability to work with a wide variety of numerical sequences (with any, not just progressions!) and the ability to imagine exactly how this or that sequence behaves - whether it increases whether it decreases unlimitedly, whether it tends to a specific number (and not necessarily to zero), or even does not tend to anything at all... An entire section is devoted to this topic in the course of mathematical analysis - theory of limits. And a little more specifically - the concept limit of the number sequence. A very interesting topic! It makes sense to go to college and figure it out.)
Some examples from this section (sequences having a limit) and in particular, infinitely decreasing geometric progression They begin to get used to it at school. We're getting used to it.)
Moreover, the ability to study well the behavior of sequences will greatly benefit you in the future and will be very useful in function research. The most diverse. But the ability to competently work with functions (calculate derivatives, study them in full, build their graphs) already dramatically increases your mathematical level! Do you have any doubts? No need. Also remember my words.)
Let's look at the geometric progression in life?
In the life around us, we encounter geometric progression very, very often. Even without even knowing it.)
For example, various microorganisms that surround us everywhere in huge quantities and which we cannot even see without a microscope multiply precisely in geometric progression.
Let's say one bacterium reproduces by dividing in half, giving offspring into 2 bacteria. In turn, each of them, when multiplying, also divides in half, giving a common offspring of 4 bacteria. The next generation will produce 8 bacteria, then 16 bacteria, 32, 64 and so on. With each subsequent generation, the number of bacteria doubles. A typical example of a geometric progression.)
Also, some insects – aphids and flies – multiply exponentially. And sometimes rabbits too, by the way.)
Another example of a geometric progression, closer to everyday life, is the so-called compound interest. This interesting phenomenon is often found in bank deposits and is called capitalization of interest. What it is?
You yourself are still, of course, young. You study at school, you don’t go to banks. But your parents are already adults and independent people. They go to work, earn money for their daily bread, and put part of the money in the bank, making savings.)
Let's say your dad wants to save up a certain amount of money for a family vacation in Turkey and puts 50,000 rubles in the bank at 10% per annum for a period of three years with annual interest capitalization. Moreover, during this entire period nothing can be done with the deposit. You can neither replenish the deposit nor withdraw money from the account. How much profit will he make after these three years?
Well, first of all, we need to figure out what 10% per annum is. It means that in a year The bank will add 10% to the initial deposit amount. From what? Of course, from initial deposit amount.
We calculate the size of the account after a year. If the initial deposit amount was 50,000 rubles (i.e. 100%), then after a year there will be how much interest on the account? That's right, 110%! From 50,000 rubles.
So we calculate 110% of 50,000 rubles:
50000·1.1 = 55000 rubles.
I hope you understand that finding 110% of a value means multiplying that value by the number 1.1? If you don’t understand why this is so, remember fifth and sixth grades. Namely – connection between percentages and fractions and parts.)
Thus, the increase for the first year will be 5,000 rubles.
How much money will be in the account in two years? 60,000 rubles? Unfortunately (or rather, fortunately), everything is not so simple. The whole trick of interest capitalization is that with each new interest accrual, these same interests will be considered already from the new amount! From the one who already is on the account At the moment. And the interest accrued for the previous period is added to the original deposit amount and, thus, itself participates in the calculation of new interest! That is, they become a full part of the overall account. Or general capital. Hence the name - capitalization of interest.
It's in economics. And in mathematics such percentages are called compound interest. Or percentage of interest.) Their trick is that when calculating sequentially, the percentages are calculated each time from the new value. And not from the original...
Therefore, to calculate the amount through two years, we need to calculate 110% of the amount that will be in the account in a year. That is, already from 55,000 rubles.
We count 110% of 55,000 rubles:
55000·1.1 = 60500 rubles.
This means that the percentage increase for the second year will be 5,500 rubles, and for two years – 10,500 rubles.
Now you can already guess that after three years the amount in the account will be 110% of 60,500 rubles. That is again 110% from the previous one (last year) amounts.
Here we think:
60500·1.1 = 66550 rubles.
Now we arrange our monetary amounts by year in sequence:
50000;
55000 = 50000 1.1;
60500 = 55000·1.1 = (50000·1.1)·1.1;
66550 = 60500 1.1 = ((50000 1.1) 1.1) 1.1
So how is it? Why not a geometric progression? First member b 1 = 50000 , and the denominator q = 1,1 . Each term is strictly 1.1 times larger than the previous one. Everything is in strict accordance with the definition.)
And how many additional interest bonuses will your dad “accumulate” while his 50,000 rubles have been lying in his bank account for three years?
We count:
66550 – 50000 = 16550 rubles
Not much, of course. But this is if the initial deposit amount is small. What if there is more? Let's say, not 50, but 200 thousand rubles? Then the increase over three years will be 66,200 rubles (if you do the math). Which is already very good.) What if the contribution is even greater? That's it...
Conclusion: the higher the initial deposit, the more profitable the interest capitalization becomes. That is why deposits with interest capitalization are provided by banks for long periods. Let's say for five years.
Also, all sorts of bad diseases like influenza, measles and even more terrible diseases (the same SARS in the early 2000s or the plague in the Middle Ages) like to spread exponentially. Hence the scale of epidemics, yes...) And all due to the fact that the geometric progression with whole positive denominator (q>1) – a thing that grows very quickly! Remember the reproduction of bacteria: from one bacteria two are obtained, from two - four, from four - eight, and so on... It’s the same with the spread of any infection.)
The simplest problems on geometric progression.
Let's start, as always, with a simple problem. Purely to understand the meaning.
1. It is known that the second term of the geometric progression is 6, and the denominator is -0.5. Find the first, third and fourth terms.
So we are given endless geometric progression, but known second term this progression:
b 2 = 6
In addition, we also know progression denominator:
q = -0.5
And you need to find first, third And fourth members of this progression.
So we act. We write down the sequence according to the conditions of the problem. Directly in general form, where the second term is six:
b 1, 6,b 3 , b 4 , …
Now let's start searching. We start, as always, with the simplest. You can calculate, for example, the third term b 3? Can! You and I already know (directly in the sense of geometric progression) that the third term (b 3) more than the second (b 2 ) V "q" once!
So we write:
b 3 =b 2 · q
We substitute six into this expression instead of b 2 and -0.5 instead q and we count. And we don’t ignore the minus either, of course...
b 3 = 6·(-0.5) = -3
Like this. The third term turned out to be negative. No wonder: our denominator q– negative. And multiplying a plus by a minus will, of course, be a minus.)
Now we count the next, fourth term of the progression:
b 4 =b 3 · q
b 4 = -3·(-0.5) = 1.5
The fourth term is again with a plus. The fifth term will again be minus, the sixth will be plus, and so on. The signs alternate!
So, the third and fourth terms were found. The result is the following sequence:
b 1 ; 6; -3; 1.5; ...
Now all that remains is to find the first term b 1 according to the well-known second. To do this, we step in the other direction, to the left. This means that in this case we do not need to multiply the second term of the progression by the denominator, but divide.
We divide and get:
That's all.) The answer to the problem will be like this:
-12; 6; -3; 1,5; …
As you can see, the solution principle is the same as in . We know any member and denominator geometric progression - we can find any other member of it. We’ll find the one we want.) The only difference is that addition/subtraction is replaced by multiplication/division.
Remember: if we know at least one member and denominator of a geometric progression, then we can always find any other member of this progression.
The following problem, according to tradition, is from a real version of the OGE:
2.
...; 150; X; 6; 1.2; ...
So how is it? This time there is no first term, no denominator q, just a sequence of numbers is given... Something already familiar, right? Yes! A similar problem has already been solved in arithmetic progression!
So we are not afraid. All the same. Let's turn on our heads and remember the elementary meaning of geometric progression. We look carefully at our sequence and figure out which parameters of the geometric progression of the three main ones (first term, denominator, term number) are hidden in it.
Member numbers? There are no membership numbers, yes... But there are four consecutive numbers. I don’t see any point in explaining what this word means at this stage.) Are there two neighboring known numbers? Eat! These are 6 and 1.2. So we can find progression denominator. So we take the number 1.2 and divide to the previous number. To six.
We get:
We get:
x= 150·0.2 = 30
Answer: x = 30 .
As you can see, everything is quite simple. The main difficulty is only in the calculations. It is especially difficult in the case of negative and fractional denominators. So those who have problems, repeat the arithmetic! How to work with fractions, how to work with negative numbers, and so on... Otherwise, you will slow down mercilessly here.
Now let's modify the problem a little. Now it's going to get interesting! Let's remove the last number 1.2 from it. Now let's solve this problem:
3. Several consecutive terms of the geometric progression are written out:
...; 150; X; 6; ...
Find the term of the progression indicated by the letter x.
Everything is the same, only two adjacent famous We now have no members of the progression. This is the main problem. Because the magnitude q through two neighboring terms we can easily determine we can't. Do we have a chance to cope with the task? Certainly!
Let's write down the unknown term " x"directly within the meaning of geometric progression! In general terms.
Yes Yes! Right with an unknown denominator!
On the one hand, for X we can write the following ratio:
x= 150·q
On the other hand, we have every right to describe this same X through next member, through the six! Divide six by the denominator.
Like this:
x = 6/ q
Obviously, now we can equate both of these ratios. Since we are expressing the same magnitude (x), but two different ways.
We get the equation:
Multiplying everything by q, simplifying and shortening, we get the equation:
q2 = 1/25
We solve and get:
q = ±1/5 = ±0.2
Oops! The denominator turned out to be double! +0.2 and -0.2. And which one should you choose? Dead end?
Calm! Yes, the problem really has two solutions! Nothing wrong with that. It happens.) You're not surprised when, for example, you get two roots when solving the usual problem? It's the same story here.)
For q = +0.2 we will get:
X = 150 0.2 = 30
And for q = -0,2 will:
X = 150·(-0.2) = -30
We get a double answer: x = 30; x = -30.
What does this interesting fact mean? And what exists two progressions, satisfying the conditions of the problem!
Like these ones:
…; 150; 30; 6; …
…; 150; -30; 6; …
Both are suitable.) Why do you think we had a split in answers? Just because of the elimination of a specific member of the progression (1,2), coming after six. And knowing only the previous (n-1)th and subsequent (n+1)th terms of the geometric progression, we can no longer say anything unambiguously about the nth term standing between them. There are two options – with a plus and with a minus.
But no problem. As a rule, in geometric progression tasks there is additional information that gives an unambiguous answer. Let's say the words: "alternating progression" or "progression with a positive denominator" and so on... It is these words that should serve as a clue as to which sign, plus or minus, should be chosen when preparing the final answer. If there is no such information, then yes, the task will have two solutions.)
Now we decide for ourselves.
4. Determine whether the number 20 is a member of a geometric progression:
4 ; 6; 9; …
5. The sign of an alternating geometric progression is given:
…; 5; x ; 45; …
Find the term of the progression indicated by the letter x .
6. Find the fourth positive term of the geometric progression:
625; -250; 100; …
7. The second term of the geometric progression is equal to -360, and its fifth term is equal to 23.04. Find the first term of this progression.
Answers (in disorder): -15; 900; No; 2.56.
Congratulations if everything worked out!
Something doesn't fit? Somewhere there was a double answer? Read the terms of the assignment carefully!
The last problem doesn't work out? There is nothing complicated there.) We work directly according to the meaning of geometric progression. Well, you can draw a picture. It helps.)
As you can see, everything is elementary. If the progression is short. What if it’s long? Or is the number of the required member very large? I would like, by analogy with the arithmetic progression, to somehow obtain a convenient formula that makes it easy to find any term of any geometric progression by his number. Without multiplying many, many times by q. And there is such a formula!) Details are in the next lesson.