How to find g in geometric progression formula. Geometric progression

This number is called the denominator of a geometric progression, i.e. each term differs from the previous one by q times. (We will assume that q ≠ 1, otherwise everything is too trivial). It is easy to see that the general formula for the nth term of the geometric progression is b n = b 1 q n – 1 ; terms with numbers b n and b m differ by q n – m times.

Already in Ancient Egypt knew not only arithmetic, but also geometric progression. Here, for example, is a problem from the Rhind papyrus: “Seven faces have seven cats; Each cat eats seven mice, each mouse eats seven ears of corn, and each ear of barley can grow seven measures of barley. How large are the numbers in this series and their sum?

Rice. 1. Ancient Egyptian geometric progression problem

This task many times different variations was repeated among other peoples at other times. For example, in written in the 13th century. “The Book of the Abacus” by Leonardo of Pisa (Fibonacci) has a problem in which 7 old women appear on their way to Rome (obviously pilgrims), each of whom has 7 mules, each of which has 7 bags, each of which contains 7 loaves , each of which has 7 knives, each of which has 7 sheaths. The problem asks how many objects there are.

The sum of the first n terms of the geometric progression S n = b 1 (q n – 1) / (q – 1) . This formula can be proven, for example, like this: S n = b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n – 1.

Add the number b 1 q n to S n and get:

S n + b 1 q n = b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n – 1 + b 1 q n = b 1 + (b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n –1) q = b 1 + S n q .

From here S n (q – 1) = b 1 (q n – 1), and we get the necessary formula.

Already on one of clay tablets Ancient Babylon dating back to the 6th century. BC e., contains the sum 1 + 2 + 2 2 + 2 3 + ... + 2 9 = 2 10 – 1. True, as in a number of other cases, we do not know how this fact was known to the Babylonians.

The rapid increase in geometric progression in a number of cultures, in particular in Indian, is repeatedly used as a visual symbol of the vastness of the universe. In the famous legend about the appearance of chess, the ruler gives its inventor the opportunity to choose the reward himself, and he asks for the number of wheat grains that would be obtained if one were placed on the first square chessboard, two for the second, four for the third, eight for the fourth, etc., each time the number doubles. Vladyka thought that we're talking about, at most, about a few bags, but he miscalculated. It is easy to see that for all 64 squares of the chessboard the inventor would have to receive (2 64 - 1) grains, which is expressed as a 20-digit number; even if the entire surface of the Earth was sown, it would take at least 8 years to collect the required amount of grains. This legend is sometimes interpreted as indicating the virtually unlimited possibilities hidden in the game of chess.

It is easy to see that this number is really 20-digit:

2 64 = 2 4 ∙ (2 10) 6 = 16 ∙ 1024 6 ≈ 16 ∙ 1000 6 = 1.6∙10 19 (a more accurate calculation gives 1.84∙10 19). But I wonder if you can find out what digit this number ends with?

A geometric progression can be increasing if the denominator is greater than 1, or decreasing if it is less than one. In the latter case, the number q n for sufficiently large n can become arbitrarily small. While the increasing geometric progression increases unexpectedly quickly, the decreasing geometric progression decreases just as quickly.

The larger n, the weaker the number q n differs from zero, and the closer the sum of n terms of the geometric progression S n = b 1 (1 – q n) / (1 – q) to the number S = b 1 / (1 – q). (For example, F. Viet reasoned this way). The number S is called the sum of an infinitely decreasing geometric progression. However, for many centuries the question of what is the meaning of summing the ENTIRE geometric progression, with its infinite number of terms, was not clear enough to mathematicians.

A decreasing geometric progression can be seen, for example, in Zeno’s aporias “Half Division” and “Achilles and the Tortoise.” In the first case, it is clearly shown that the entire road (assuming length 1) is the sum of an infinite number of segments 1/2, 1/4, 1/8, etc. This is, of course, the case from the point of view of ideas about a finite sum infinite geometric progression. And yet - how can this be?

Rice. 2. Progression with a coefficient of 1/2

In the aporia about Achilles, the situation is a little more complicated, because here the denominator of the progression is not 1/2, but some other number. Let, for example, Achilles run with speed v, the tortoise moves with speed u, and the initial distance between them is l. Achilles will cover this distance in time l/v, and during this time the turtle will move a distance lu/v. When Achilles runs this segment, the distance between him and the turtle will become equal to l (u /v) 2, etc. It turns out that catching up with the turtle means finding the sum of an infinitely decreasing geometric progression with the first term l and the denominator u /v. This sum - the segment that Achilles will eventually run to the meeting place with the turtle - is equal to l / (1 – u /v) = lv / (v – u). But, again, how should this result be interpreted and why does it make any sense at all? for a long time it wasn't very clear.

Rice. 3. Geometric progression with a coefficient of 2/3

Archimedes used the sum of a geometric progression to determine the area of ​​a parabola segment. Let this segment of the parabola be delimited by the chord AB and let the tangent at point D of the parabola be parallel to AB. Let C be the midpoint of AB, E the midpoint of AC, F the midpoint of CB. Let's draw lines parallel to DC through points A, E, F, B; Let the tangent drawn at point D intersect these lines at points K, L, M, N. Let's also draw segments AD and DB. Let the line EL intersect the line AD at point G, and the parabola at point H; line FM intersects line DB at point Q, and the parabola at point R. According to general theory conic sections, DC – diameter of the parabola (that is, a segment parallel to its axis); it and the tangent at point D can serve as coordinate axes x and y, in which the equation of the parabola is written as y 2 = 2px (x is the distance from D to any point of a given diameter, y is the length of a segment parallel to a given tangent from this point of diameter to some point on the parabola itself).

By virtue of the parabola equation, DL 2 = 2 ∙ p ∙ LH, DK 2 = 2 ∙ p ∙ KA, and since DK = 2DL, then KA = 4LH. Because KA = 2LG, LH = HG. The area of ​​segment ADB of a parabola is equal to the area of ​​triangle ΔADB and the areas of segments AHD and DRB combined. In turn, the area of ​​the segment AHD is similarly equal to the area of ​​the triangle AHD and the remaining segments AH and HD, with each of which you can perform the same operation - split into a triangle (Δ) and the two remaining segments (), etc.:

The area of ​​the triangle ΔAHD is equal to half the area of ​​the triangle ΔALD (they have a common base AD, and the heights differ by 2 times), which, in turn, is equal to half the area of ​​the triangle ΔAKD, and therefore half the area of ​​the triangle ΔACD. Thus, the area of ​​the triangle ΔAHD is equal to a quarter of the area of ​​the triangle ΔACD. Likewise, the area of ​​triangle ΔDRB is equal to one-quarter of the area of ​​triangle ΔDFB. So, the areas of the triangles ΔAHD and ΔDRB, taken together, are equal to a quarter of the area of ​​the triangle ΔADB. Repeating this operation when applied to segments AH, HD, DR and RB will select triangles from them, the area of ​​which, taken together, will be 4 times less than the area of ​​triangles ΔAHD and ΔDRB, taken together, and therefore 16 times less, than the area of ​​the triangle ΔADB. And so on:

Thus, Archimedes proved that “every segment contained between a straight line and a parabola constitutes four-thirds of a triangle having the same base and equal height.”

Let us now consider the question of summing an infinite geometric progression. Let us call the partial sum of a given infinite progression the sum of its first terms. Let us denote the partial sum by the symbol

For every infinite progression

one can compose a (also infinite) sequence of its partial sums

Let a sequence with unlimited increase have a limit

In this case, the number S, i.e., the limit of partial sums of a progression, is called the sum of an infinite progression. We will prove that an infinite decreasing geometric progression always has a sum, and we will derive a formula for this sum (we can also show that if an infinite progression has no sum, it does not exist).

Let us write the expression for the partial sum as the sum of terms of the progression using formula (91.1) and consider the limit of the partial sum at

From Theorem 89 it is known that for a decreasing progression; therefore, applying the difference limit theorem, we find

(here the rule is also used: the constant factor is taken beyond the limit sign). The existence is proven, and at the same time the formula for the sum of an infinitely decreasing geometric progression is obtained:

Equality (92.1) can also be written in the form

Here it may seem paradoxical that the sum of an infinite number of terms is assigned a very definite finite value.

A clear illustration can be given to explain this situation. Consider a square with a side equal to one (Fig. 72). Divide this square with a horizontal line into two equal parts and attach the upper part to the lower one so that a rectangle is formed with sides 2 and . After this, we will again divide the right half of this rectangle in half with a horizontal line and attach the upper part to the lower one (as shown in Fig. 72). Continuing this process, we continually transform the original square with area equal to 1 into equal-sized figures (taking the form of a staircase with thinning steps).

With the infinite continuation of this process, the entire area of ​​the square is decomposed into an infinite number of terms - the areas of rectangles with bases equal to 1 and heights. The areas of rectangles precisely form an infinite decreasing progression, its sum

i.e., as one would expect, equal to the area of ​​the square.

Example. Find the sums of the following infinite progressions:

Solution, a) We notice that this progression Therefore, using formula (92.2) we find

b) Here it means that using the same formula (92.2) we have

c) We find that this progression therefore has no sum.

In paragraph 5, we showed the application of the formula for the sum of terms of an infinitely decreasing progression to the inversion of a periodic decimal into a common fraction.

Exercises

1. The sum of an infinitely decreasing geometric progression is 3/5, and the sum of its first four terms is 13/27. Find the first term and denominator of the progression.

2. Find four numbers that form an alternating geometric progression, in which the second term is less than the first by 35, and the third is greater than the fourth by 560.

3. Show that if the sequence

forms an infinitely decreasing geometric progression, then the sequence

for any, it forms an infinitely decreasing geometric progression. Will this statement hold true when

Derive a formula for the product of the 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. 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. Left side 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 consistency positive numbers forms an arithmetic progression, A – geometric progression, what does it have to do with . Find .

Solution. Because arithmetic sequence, That (main property 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 in 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. Full course elementary mathematics in problems and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. – 208 p.

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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 the inequality of the first term and the denominator of the progression to zero, otherwise it is considered indefinite.

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 is 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 natural number n (counter).

Another important property of geometric progression, which gave geometric progression

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 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.