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 of 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.
for example, sequence \(3\); \(6\); \(12\); \(24\); \(48\)… is a geometric progression, because each next element differs from the previous one by a factor of two (in other words, it can be obtained from the previous one by multiplying it by two):
Like any sequence, a geometric progression is denoted by a small Latin letter. The numbers that form a progression are called it members(or elements). They are denoted by the same letter as the geometric progression, but with a numerical index equal to the element number in order.
for example, the geometric progression \(b_n = \(3; 6; 12; 24; 48…\)\) consists of the elements \(b_1=3\); \(b_2=6\); \(b_3=12\) and so on. In other words:
If you understand the above information, you will already be able to solve most of the problems on this topic.
Example (OGE):
Decision:
Answer : \(-686\).
Example (OGE):
Given the first three terms of the progression \(324\); \(-108\); \(36\)…. Find \(b_5\).
Decision:
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To continue the sequence, we need to know the denominator. Let's find it from two neighboring elements: what should \(324\) be multiplied by to get \(-108\)? |
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\(324 q=-108\) |
From here we can easily calculate the denominator. |
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\(q=-\) \(\frac(108)(324)\) \(=-\) \(\frac(1)(3)\) |
Now we can easily find the element we need. |
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Answer ready. |
Answer : \(4\).
Example: The progression is given by the condition \(b_n=0.8 5^n\). Which number is a member of this progression:
a) \(-5\) b) \(100\) c) \(25\) d) \(0.8\) ?
Decision:
From the wording of the task, it is obvious that one of these numbers is definitely in our progression. Therefore, we can simply calculate its members one by one until we find the value we need. Since our progression is given by the formula , we calculate the values of the elements by substituting different \(n\):
\(n=1\); \(b_1=0.8 5^1=0.8 5=4\) – there is no such number in the list. We continue.
\(n=2\); \(b_2=0.8 5^2=0.8 25=20\) - and this is not there either.
\(n=3\); \(b_3=0.8 5^3=0.8 125=100\) – and here is our champion!
Answer: \(100\).
Example (OGE): Several successive members of the geometric progression …\(8\) are given; \(x\); \(fifty\); \(-125\)…. Find the value of the element denoted by the letter \(x\).
Decision:
Answer: \(-20\).
Example (OGE): The progression is given by the conditions \(b_1=7\), \(b_(n+1)=2b_n\). Find the sum of the first \(4\) terms of this progression.
Decision:
Answer: \(105\).
Example (OGE): It is known that exponentially \(b_6=-11\),\(b_9=704\). Find the denominator \(q\).
Decision:
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It can be seen from the diagram on the left that in order to “get” from \ (b_6 \) to \ (b_9 \) - we take three “steps”, that is, we multiply \ (b_6 \) three times by the denominator of the progression. In other words, \(b_9=b_6 q q q=b_6 q^3\). |
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\(b_9=b_6 q^3\) |
Substitute the values we know. |
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\(704=(-11)q^3\) |
“Reverse” the equation and divide it by \((-11)\). |
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\(q^3=\) \(\frac(704)(-11)\) \(\:\:\: ⇔ \:\:\: \)\(q^3=-\) \(64 \) |
What number cubed gives \(-64\)? |
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Answer found. It can be checked by restoring the chain of numbers from \(-11\) to \(704\). |
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All agreed - the answer is correct. |
Answer: \(-4\).
The most important formulas
As you can see, most geometric progression problems can be solved with pure logic, simply by understanding the essence (this is generally characteristic of mathematics). But sometimes the knowledge of certain formulas and patterns speeds up and greatly facilitates the solution. We will study two such formulas.
The formula for the \(n\)th member is: \(b_n=b_1 q^(n-1)\), where \(b_1\) is the first member of the progression; \(n\) – number of the required element; \(q\) is the denominator of the progression; \(b_n\) is a member of the progression with the number \(n\).
Using this formula, you can, for example, solve the problem from the very first example in just one step.
Example (OGE):
The geometric progression is given by the conditions \(b_1=-2\); \(q=7\). Find \(b_4\).
Decision:
Answer: \(-686\).
This example was simple, so the formula did not make the calculations easier for us too much. Let's look at the problem a little more complicated.
Example:
The geometric progression is given by the conditions \(b_1=20480\); \(q=\frac(1)(2)\). Find \(b_(12)\).
Decision:
Answer: \(10\).
Of course, raising \(\frac(1)(2)\) to the \(11\)th power is not very joyful, but still easier than \(11\) dividing \(20480\) into two.
The sum \(n\) of the first terms: \(S_n=\)\(\frac(b_1 (q^n-1))(q-1)\) , where \(b_1\) is the first term of the progression; \(n\) – the number of summed elements; \(q\) is the denominator of the progression; \(S_n\) is the sum \(n\) of the first members of the progression.
Example (OGE):
Given a geometric progression \(b_n\), whose denominator is \(5\), and the first term \(b_1=\frac(2)(5)\). Find the sum of the first six terms of this progression.
Decision:
Answer: \(1562,4\).
And again, we could solve the problem “on the forehead” - find all six elements in turn, and then add the results. However, the number of calculations, and hence the chance of a random error, would increase dramatically.
For a geometric progression, there are several more formulas that we did not consider here because of their low practical use. You can find these formulas.
Increasing and decreasing geometric progressions
The progression \(b_n = \(3; 6; 12; 24; 48…\)\) considered at the very beginning of the article has a denominator \(q\) greater than one, and therefore each next term is greater than the previous one. Such progressions are called increasing.
If \(q\) is less than one, but is positive (that is, lies between zero and one), then each next element will be less than the previous one. For example, in the progression \(4\); \(2\); \(one\); \(0.5\); \(0.25\)… the denominator of \(q\) is \(\frac(1)(2)\).
These progressions are called decreasing. Note that none of the elements of this progression will be negative, they just get smaller and smaller with each step. That is, we will gradually approach zero, but we will never reach it and we will not go beyond it. Mathematicians in such cases say "to tend to zero."
Note that with a negative denominator, the elements of a geometric progression will necessarily change sign. for example, the progression \(5\); \(-fifteen\); \(45\); \(-135\); \(675\)... the denominator of \(q\) is \(-3\), and because of this, the signs of the elements "blink".
Let's consider a 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. So 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 some specific number. This is expressed by the following formula.
a z +1 =a z q, where z is the number of the selected element.
Accordingly, z ∈ N.
The period when a geometric progression is studied at school is grade 9. 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 the series, you need to multiply the last one by q.
To specify this progression, you must specify its first element and denominator. After that, 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 next element. An example of such is presented below.
Example: a 1 =3, q=2 - both parameters are greater than one.
Then the numerical sequence can be written like this:
3 6 12 24 48 ...
- If |q| 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 such is presented below.
Example: a 1 =6, q=1/3 - a 1 is greater than one, q is less.
Then the numerical sequence can be written as follows:
6 2 2/3 ... - any element is 3 times greater than the element following it.
- Sign-variable. 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 sequence can be written like this:
3, 6, -12, 24,...
Formulas
For convenient use of geometric progressions, there are many formulas:
- Formula of the z-th member. Allows you to calculate the element under a specific number without calculating the previous numbers.
Example:q = 3, a 1 = 4. It is required to calculate the fourth element of the progression.
Decision:a 4 = 4 · 3 4-1 = 4 · 3 3 = 4 · 27 = 108.
- The sum of the first elements whose number is 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, hence q is not equal to 1.
Note: if q=1, then the progression would be a series of an infinitely repeating number.
The sum of a geometric progression, examples:a 1 = 2, q= -2. Calculate S 5 .
Decision:S 5 = 22 - calculation by formula.
- Amount if |q| < 1 и если z стремится к бесконечности.
Example:a 1 = 2 , q= 0.5. Find the amount.
Decision:Sz = 2 · = 4
Sz = 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 3.9375 4
Some properties:
- characteristic property. If the following condition performed 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 of a geometric progression is found by adding the squares of any other two numbers in a given series, if they are equidistant from this element.
a z 2 = a z - t 2 + a z + t 2 , wheretis the distance between these numbers.
- Elementsdiffer in qonce.
- The logarithms of the progression elements also form a progression, but already arithmetic, that is, each of them is greater than the previous one by a certain number.
Examples of some classical problems
To better understand what a geometric progression is, examples with a solution for grade 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 through others using a denominator.
Hence,a 3 = q 2 · a 1
When substitutingq= 4
- Conditions:a 2 = 6, a 3 = 12. Calculate S 6 .
Decision:To do this, it is enough to 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:
- The client of the bank made a deposit in the amount of 10,000 rubles, under the terms of which every year the client will add 6% of it to the principal amount. How much money will be in the account after 4 years?
Solution: The initial amount is 10 thousand rubles. So, 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 in order 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 tasks for calculating the sum:
In various problems, a geometric progression is used. An example for finding the sum can be given as follows:
a 1 = 4, q= 2, calculateS5.
Solution: all the data necessary for the calculation are 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.
Decision:
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, we need to finda 1 , knowinga 2 andq.
a 1 · q = a 2
a 1 =2
S 6 = 728.
An example of a geometric progression: 2, 6, 18, 54, 162.
Here, each term after the first is 3 times the previous one. That is, each subsequent term is the result of multiplying the previous term by 3:
2 3 = 6
6 3 = 18
18 3 = 54
54 3 = 162 .
In our example, when dividing the second term by the first, the third by the second, and so on. we get 3. The number 3 is the denominator of this geometric progression.
Example:
Let's go back to our geometric progression 2, 6, 18, 54, 162. Let's take the fourth term and square it:
54 2 = 2916.
Now we multiply the terms to the left and right of the number 54:
18 162 = 2916.
As you can see, the square of the third term is equal to the product of the neighboring second and fourth terms.
Example 1: Let's take some geometric progression, in which the first term is equal to 2, and the denominator of the geometric progression is equal to 1.5. We need to find the 4th term of this progression.
Given:
b 1 = 2
q = 1,5
n = 4
————
b 4 - ?
Decision.
Applying the formula b n= b 1 q n- 1 , inserting the appropriate values into it:
b 4 \u003d 2 1.5 4 - 1 \u003d 2 1.5 3 \u003d 2 3.375 \u003d 6.75.
Answer: The fourth term of a given geometric progression is the number 6.75.
Example 2: Find the fifth member of the geometric progression if the first and third members are 12 and 192, respectively.
Given:
b 1 = 12
b 3 = 192
————
b 5 - ?
Decision.
1) First we need to find the denominator of a geometric progression, without which it is impossible to solve the problem. As a first step, using our formula, we derive the formula for b 3:
b 3 = b 1 q 3 - 1 = b 1 q 2
Now we can find the denominator of a geometric progression:
b 3 192
q 2 = —— = —— = 16
b 1 12
q= √16 = 4 or -4.
2) It remains to find the value b 5 .
If a q= 4, then
b 5 = b 1 q 5-1 = 12 4 4 = 12 256 = 3072.
At q= -4 the result will be the same. Thus, the problem has one solution.
Answer: The fifth term of the given geometric progression is the number 3072.
Example: Find the sum of the first five terms of the geometric progression ( b n), in which the first term is equal to 2, and the denominator of a geometric progression is 3.
Given:
b 1 = 2
q = 3
n = 5
————
S 5 - ?
Decision.
We apply the second formula of the two above:
b 1 (q 5 - 1) 2 (3 5 - 1) 2 (243 - 1) 484
S 5 = ————— = ————— = ———————— = ————— = 242
q - 1 3 - 1 2 2
Answer: The sum of the first five terms of a given geometric progression is 242.
The sum of an infinite geometric progression.
It is necessary to distinguish between the concepts of "the sum of an infinite geometric progression" and "the sum n members of a geometric progression. The second concept refers to any geometric progression, and the first - only to one where the denominator is less than 1 modulo.
>>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 of a 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 the exponential function and its graph 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 the mathematical model of the problem.
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 the geometric progression in question: 1, 2, 4, 8, 16, 32, ... .
Third stage.
The answer to the problem question. It is required to calculate b 12 . We have
Answer: b 12 = 2048.
3. The formula for the sum of members of a finite geometric progression.
Let there be a finite geometric progression
Denote by S n the sum of its terms, i.e.
Let's derive a formula for finding this sum.
Let's start with the simplest case, when q = 1. Then the geometric progression b 1 ,b 2 , b 3 ,..., bn consists of n numbers equal to b 1 , i.e. the progression is b 1 , b 2 , b 3 , ..., b 4 . The sum of these numbers is nb 1 .
Let now q = 1 To find S n we use an artificial method: let's perform some transformations of the expression S n q. We have:
Performing transformations, we, firstly, used the definition of a geometric progression, according to which (see the third line of reasoning); secondly, they added and subtracted why the meaning of the expression, of course, did not change (see the fourth line of reasoning); thirdly, we used the formula of the n-th member of a geometric progression:
From formula (1) we find:
This is the formula for the sum of n members of a geometric progression (for the case when q = 1).
Example 8
Given a finite geometric progression
a) the sum of the members of the progression; b) the sum of the squares of its terms.
b) Above (see p. 132) we have already noted that if all members of a geometric progression are squared, then a geometric progression with the first member b 2 and the denominator q 2 will be obtained. Then the sum of the six terms of the new progression will be calculated by
Example 9
Find the 8th term of a geometric progression for which
In fact, we have proved the following theorem.
A numerical sequence is a geometric progression if and only if the square of each of its terms, except for the first one (and the last one, in the case of a finite sequence), is equal to the product of the previous and subsequent terms (a characteristic property of a geometric progression).
