Find q in a geometric infinite progression. Denominator of a geometric progression: formulas and properties

>>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 obtain:

(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 members.

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).

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… Geometric progression, in which the first member is equal to one, and $q=2$.

Example. 8,8,8,8… A geometric progression whose first term is eight,
and $q=1$.

Example. 3,-3,3,-3,3... A geometric progression whose first term is three,
and $q=-1$.

The geometric progression has the properties of monotonicity.
If $b_(1)>0$, $q>1$,
then the sequence is increasing.
If $b_(1)>0$, $0 The sequence is usually denoted as: $b_(1), b_(2), b_(3), ..., b_(n), ...$.

Just like in an arithmetic progression, if the number of elements in a geometric progression is finite, then the progression is called a finite geometric progression.

$b_(1), b_(2), b_(3), ..., b_(n-2), b_(n-1), b_(n)$.
Note that if the sequence is a geometric progression, then the sequence of squared terms is also a geometric progression. The second sequence has the first term $b_(1)^2$ and the denominator $q^2$.

Formula of the nth member of a geometric progression

Geometric progression can also be specified in analytical form. Let's see how to do it:
$b_(1)=b_(1)$.
$b_(2)=b_(1)*q$.
$b_(3)=b_(2)*q=b_(1)*q*q=b_(1)*q^2$.
$b_(4)=b_(3)*q=b_(1)*q^3$.
$b_(5)=b_(4)*q=b_(1)*q^4$.
We can easily see the pattern: $b_(n)=b_(1)*q^(n-1)$.
Our formula is called "formula of the n-th member of a geometric progression".

Let's go back to our examples.

Example. 1,2,4,8,16… A geometric progression whose first term is equal to one,
and $q=2$.
$b_(n)=1*2^(n)=2^(n-1)$.

Example. 16,8,4,2,1,1/2… A geometric progression whose first term is sixteen and $q=\frac(1)(2)$.
$b_(n)=16*(\frac(1)(2))^(n-1)$.

Example. 8,8,8,8… A geometric progression where the first term is eight and $q=1$.
$b_(n)=8*1^(n-1)=8$.

Example. 3,-3,3,-3,3… A geometric progression whose first term is three and $q=-1$.
$b_(n)=3*(-1)^(n-1)$.

Example. Given a geometric progression $b_(1), b_(2), …, b_(n), … $.
a) It is known that $b_(1)=6, q=3$. Find $b_(5)$.
b) It is known that $b_(1)=6, q=2, b_(n)=768$. Find n.
c) It is known that $q=-2, b_(6)=96$. Find $b_(1)$.
d) It is known that $b_(1)=-2, b_(12)=4096$. Find q.

Solution.
a) $b_(5)=b_(1)*q^4=6*3^4=486$.
b) $b_n=b_1*q^(n-1)=6*2^(n-1)=768$.
$2^(n-1)=\frac(768)(6)=128$ since $2^7=128 => n-1=7; n=8$.
c) $b_(6)=b_(1)*q^5=b_(1)*(-2)^5=-32*b_(1)=96 => b_(1)=-3$.
d) $b_(12)=b_(1)*q^(11)=-2*q^(11)=4096 => q^(11)=-2048 => q=-2$.

Example. The difference between the seventh and fifth members of the geometric progression is 192, the sum of the fifth and sixth members of the progression is 192. Find the tenth member of this progression.

Solution.
We know that: $b_(7)-b_(5)=192$ and $b_(5)+b_(6)=192$.
We also know: $b_(5)=b_(1)*q^4$; $b_(6)=b_(1)*q^5$; $b_(7)=b_(1)*q^6$.
Then:
$b_(1)*q^6-b_(1)*q^4=192$.
$b_(1)*q^4+b_(1)*q^5=192$.
We got a system of equations:
$\begin(cases)b_(1)*q^4(q^2-1)=192\\b_(1)*q^4(1+q)=192\end(cases)$.
Equating, our equations get:
$b_(1)*q^4(q^2-1)=b_(1)*q^4(1+q)$.
$q^2-1=q+1$.
$q^2-q-2=0$.
We got two solutions q: $q_(1)=2, q_(2)=-1$.
Substitute successively into the second equation:
$b_(1)*2^4*3=192 => b_(1)=4$.
$b_(1)*(-1)^4*0=192 =>$ no solutions.
We got that: $b_(1)=4, q=2$.
Let's find the tenth term: $b_(10)=b_(1)*q^9=4*2^9=2048$.

The sum of a finite geometric progression

Suppose we have a finite geometric progression. Let's, as well as for an arithmetic progression, calculate the sum of its members.

Let a finite geometric progression be given: $b_(1),b_(2),…,b_(n-1),b_(n)$.
Let's introduce the notation for the sum of its members: $S_(n)=b_(1)+b_(2)+⋯+b_(n-1)+b_(n)$.
In the case when $q=1$. All members of the geometric progression are equal to the first member, then it is obvious that $S_(n)=n*b_(1)$.
Consider now the case $q≠1$.
Multiply the above amount by q.
$S_(n)*q=(b_(1)+b_(2)+⋯+b_(n-1)+b_(n))*q=b_(1)*q+b_(2)*q+⋯ +b_(n-1)*q+b_(n)*q=b_(2)+b_(3)+⋯+b_(n)+b_(n)*q$.
Note:
$S_(n)=b_(1)+(b_(2)+⋯+b_(n-1)+b_(n))$.
$S_(n)*q=(b_(2)+⋯+b_(n-1)+b_(n))+b_(n)*q$.

$S_(n)*q-S_(n)=(b_(2)+⋯+b_(n-1)+b_(n))+b_(n)*q-b_(1)-(b_(2 )+⋯+b_(n-1)+b_(n))=b_(n)*q-b_(1)$.

$S_(n)(q-1)=b_(n)*q-b_(1)$.

$S_(n)=\frac(b_(n)*q-b_(1))(q-1)=\frac(b_(1)*q^(n-1)*q-b_(1)) (q-1)=\frac(b_(1)(q^(n)-1))(q-1)$.

$S_(n)=\frac(b_(1)(q^(n)-1))(q-1)$.

We have obtained the formula for the sum of a finite geometric progression.

Example.
Find the sum of the first seven terms of a geometric progression whose first term is 4 and the denominator is 3.

Solution.
$S_(7)=\frac(4*(3^(7)-1))(3-1)=2*(3^(7)-1)=4372$.

Example.
Find the fifth member of the geometric progression, which is known: $b_(1)=-3$; $b_(n)=-3072$; $S_(n)=-4095$.

Solution.
$b_(n)=(-3)*q^(n-1)=-3072$.
$q^(n-1)=1024$.
$q^(n)=1024q$.

$S_(n)=\frac(-3*(q^(n)-1))(q-1)=-4095$.
$-4095(q-1)=-3*(q^(n)-1)$.
$-4095(q-1)=-3*(1024q-1)$.
$1365q-1365=1024q-1$.
$341q=1364$.
$q=4$.
$b_5=b_1*q^4=-3*4^4=-3*256=-768$.

Characteristic property of a geometric progression

Guys, given a geometric progression. Let's consider its three consecutive members: $b_(n-1),b_(n),b_(n+1)$.
We know that:
$\frac(b_(n))(q)=b_(n-1)$.
$b_(n)*q=b_(n+1)$.
Then:
$\frac(b_(n))(q)*b_(n)*q=b_(n)^(2)=b_(n-1)*b_(n+1)$.
$b_(n)^(2)=b_(n-1)*b_(n+1)$.
If the progression is finite, then this equality holds for all terms except the first and last.
If it is not known in advance what kind of sequence the sequence has, but it is known that: $b_(n)^(2)=b_(n-1)*b_(n+1)$.
Then we can safely say that this is a geometric progression.

A number sequence is a geometric progression only when the square of each of its terms is equal to the product of its two neighboring terms of the progression. Do not forget that for a finite progression this condition is not satisfied for the first and last term.

Let's look at this identity: $\sqrt(b_(n)^(2))=\sqrt(b_(n-1)*b_(n+1))$.
$|b_(n)|=\sqrt(b_(n-1)*b_(n+1))$.
$\sqrt(a*b)$ is called the geometric mean of a and b.

The modulus of any member of a geometric progression is equal to the geometric mean of the two members adjacent to it.

Example.
Find x such that $x+2; 2x+2; 3x+3$ were three consecutive members of a geometric progression.

Solution.
Let's use the characteristic property:
$(2x+2)^2=(x+2)(3x+3)$.
$4x^2+8x+4=3x^2+3x+6x+6$.
$x^2-x-2=0$.
$x_(1)=2$ and $x_(2)=-1$.
Substitute sequentially in the original expression, our solutions:
With $x=2$, we got the sequence: 4;6;9 is a geometric progression with $q=1.5$.
With $x=-1$, we got the sequence: 1;0;0.
Answer: $x=2.$

Tasks for independent solution

1. Find the eighth first member of the geometric progression 16; -8; 4; -2 ....
2. Find the tenth member of the geometric progression 11,22,44….
3. It is known that $b_(1)=5, q=3$. Find $b_(7)$.
4. It is known that $b_(1)=8, q=-2, b_(n)=512$. Find n.
5. Find the sum of the first 11 members of the geometric progression 3;12;48….
6. Find x such that $3x+4; 2x+4; x+5$ are three consecutive members of a geometric progression.

The formula for the nth member of a geometric progression is a very simple thing. Both in meaning and in general. But there are all sorts of problems for the formula of the nth member - from very primitive to quite serious ones. And in the process of our acquaintance, we will definitely consider both of them. Well, let's meet?)

So, for starters, actually formulan

There she is:

b n = b 1 · q n -1

Formula as a formula, nothing supernatural. It looks even simpler and more compact than the similar formula for . The meaning of the formula is also simple, like a felt boot.

This formula allows you to find ANY member of a geometric progression BY ITS NUMBER " n".

As you can see, the meaning is a complete analogy with an arithmetic progression. We know the number n - we can also calculate the term under this number. What we want. Not multiplying sequentially by "q" many, many times. That's the whole point.)

I understand that at this level of work with progressions, all the quantities included in the formula should already be clear to you, but I consider it my duty to decipher each one. Just in case.

So let's go:

b 1 – the first member of a geometric progression;

q – ;

n– member number;

b n – nth (nth) member of a geometric progression.

This formula links the four main parameters of any geometric progression - bn, b 1 , q and n. And around these four key figures, all-all tasks in progression revolve.

"And how is it displayed?"- I hear a curious question ... Elementary! Look!

What is equal to second progression member? No problem! We write directly:

b 2 = b 1 q

And the third member? Not a problem either! We multiply the second term again onq.

Like this:

B 3 \u003d b 2 q

Recall now that the second term, in turn, is equal to b 1 q and substitute this expression into our equality:

B 3 = b 2 q = (b 1 q) q = b 1 q q = b 1 q 2

We get:

B 3 = b 1 q 2

Now let's read our entry in Russian: third term is equal to the first term multiplied by q in second degree. Do you get it? Not yet? Okay, one more step.

What is the fourth term? All the same! Multiply previous(i.e. the third term) on q:

B 4 \u003d b 3 q \u003d (b 1 q 2) q \u003d b 1 q 2 q \u003d b 1 q 3

Total:

B 4 = b 1 q 3

And again we translate into Russian: fourth term is equal to the first term multiplied by q in third degree.

And so on. So how is it? Did you catch the pattern? Yes! For any term with any number, the number of equal factors q (i.e. the power of the denominator) will always be one less than the number of the desired membern.

Therefore, our formula will be, without options:

b n =b 1 · q n -1

That's all.)

Well, let's solve problems, shall we?)

Solving problems on a formulanth term of a geometric progression.

Let's start, as usual, with a direct application of the formula. Here is a typical problem:

It is known exponentially that b 1 = 512 and q = -1/2. Find the tenth term of the progression.

Of course, this problem can be solved without any formulas at all. Just like a geometric progression. But we need to warm up with the formula of the nth term, right? Here we are breaking up.

Our data for applying the formula is as follows.

The first term is known. This is 512.

b 1 = 512.

The denominator of the progression is also known: q = -1/2.

It remains only to figure out what the number of the term n is equal to. No problem! Are we interested in the tenth term? So we substitute ten instead of n in the general formula.

And carefully calculate the arithmetic:

Answer: -1

As you can see, the tenth term of the progression turned out to be with a minus. No wonder: the denominator of the progression is -1/2, i.e. negative number. And this tells us that the signs of our progression alternate, yes.)

Everything is simple here. And here is a similar problem, but a little more complicated in terms of calculations.

In geometric progression, we know that:

b 1 = 3

Find the thirteenth term of the progression.

Everything is the same, only this time the denominator of the progression - irrational. Root of two. Well, no big deal. The formula is a universal thing, it copes with any numbers.

We work directly according to the formula:

The formula, of course, worked as it should, but ... this is where some will hang. What to do next with the root? How to raise a root to the twelfth power?

How-how ... You need to understand that any formula, of course, is a good thing, but the knowledge of all previous mathematics is not canceled! How to raise? Yes, remember the properties of degrees! Let's change the root to fractional degree and - by the formula of raising a power to a power.

Like this:

Answer: 192

And all things.)

What is the main difficulty in the direct application of the nth term formula? Yes! The main difficulty is work with degrees! Namely, the exponentiation of negative numbers, fractions, roots, and similar constructions. So those who have problems with this, an urgent request to repeat the degrees and their properties! Otherwise, you will slow down in this topic, yes ...)

Now let's solve typical search problems one of the elements of the formula if all the others are given. For the successful solution of such problems, the recipe is single and simple to horror - write the formulanth member in general! Right in the notebook next to the condition. And then, from the condition, we figure out what is given to us and what is not enough. And we express the desired value from the formula. Everything!

For example, such a harmless problem.

The fifth term of a geometric progression with a denominator of 3 is 567. Find the first term of this progression.

Nothing complicated. We work directly according to the spell.

We write the formula of the nth term!

b n = b 1 · q n -1

What is given to us? First, the denominator of the progression is given: q = 3.

In addition, we are given fifth member: b 5 = 567 .

Everything? Not! We are also given the number n! This is a five: n = 5.

I hope you already understand what is in the record b 5 = 567 two parameters are hidden at once - this is the fifth member itself (567) and its number (5). In a similar lesson on I already talked about this, but I think it’s not superfluous to remind here.)

Now we substitute our data into the formula:

567 = b 1 3 5-1

We consider arithmetic, simplify and get a simple linear equation:

81 b 1 = 567

We solve and get:

b 1 = 7

As you can see, there are no problems with finding the first member. But when looking for the denominator q and numbers n there may be surprises. And you also need to be prepared for them (surprises), yes.)

For example, such a problem:

The fifth term of a geometric progression with a positive denominator is 162, and the first term of this progression is 2. Find the denominator of the progression.

This time we are given the first and fifth members, and are asked to find the denominator of the progression. Here we start.

We write the formulanth member!

b n = b 1 · q n -1

Our initial data will be as follows:

b 5 = 162

b 1 = 2

n = 5

Not enough value q. No problem! Let's find it now.) We substitute everything that we know into the formula.

We get:

162 = 2q 5-1

2 q 4 = 162

q 4 = 81

A simple equation of the fourth degree. But now - carefully! At this stage of the solution, many students immediately joyfully extract the root (of the fourth degree) and get the answer q=3 .

Like this:

q4 = 81

q = 3

But in general, this is an unfinished answer. Or rather, incomplete. Why? The point is that the answer q = -3 also fits: (-3) 4 would also be 81!

This is because the power equation x n = a always has two opposite roots at evenn . Plus and minus:

Both fit.

For example, solving (i.e. second degrees)

x2 = 9

For some reason you are not surprised to see two roots x=±3? It's the same here. And with any other even degree (fourth, sixth, tenth, etc.) will be the same. Details - in the topic about

So the correct solution would be:

q 4 = 81

q= ±3

Okay, we've got the signs figured out. Which one is correct - plus or minus? Well, we read the condition of the problem again in search of additional information. It, of course, may not exist, but in this problem such information available. In our condition, it is directly stated that a progression is given with positive denominator.

So the answer is obvious:

q = 3

Everything is simple here. What do you think would happen if the problem statement were like this:

The fifth term of a geometric progression is 162, and the first term of this progression is 2. Find the denominator of the progression.

What's the Difference? Yes! In the condition nothing no mention of the denominator. Neither directly nor indirectly. And here the problem would already have two solutions!

q = 3 and q = -3

Yes Yes! And with plus and minus.) Mathematically, this fact would mean that there are two progressions that fit the task. And for each - its own denominator. For fun, practice and write down the first five terms of each.)

Now let's practice finding the member number. This is the hardest one, yes. But also more creative.

Given a geometric progression:

3; 6; 12; 24; …

What number is 768 in this progression?

The first step is the same: write the formulanth member!

b n = b 1 · q n -1

And now, as usual, we substitute the data known to us into it. Hm... it doesn't fit! Where is the first member, where is the denominator, where is everything else?!

Where, where ... Why do we need eyes? Flapping eyelashes? This time the progression is given to us directly in the form sequences. Can we see the first term? We see! This is a triple (b 1 = 3). What about the denominator? We don't see it yet, but it's very easy to count. If, of course, you understand.

Here we consider. Directly according to the meaning of a geometric progression: we take any of its members (except the first) and divide by the previous one.

At least like this:

q = 24/12 = 2

What else do we know? We also know some member of this progression, equal to 768. Under some number n:

b n = 768

We do not know his number, but our task is precisely to find him.) So we are looking for. We have already downloaded all the necessary data for substitution in the formula. Imperceptibly.)

Here we substitute:

768 = 3 2n -1

We make elementary ones - we divide both parts by three and rewrite the equation in the usual form: the unknown on the left, the known on the right.

We get:

2 n -1 = 256

Here's an interesting equation. We need to find "n". What's unusual? Yes, I do not argue. Actually, it's the simplest. It is so called because the unknown (in this case, it is the number n) stands in indicator degree.

At the stage of acquaintance with a geometric progression (this is the ninth grade), exponential equations are not taught to solve, yes ... This is a topic for high school. But there is nothing terrible. Even if you do not know how such equations are solved, let's try to find our n guided by simple logic and common sense.

We start to discuss. On the left we have a deuce to some extent. We do not yet know what exactly this degree is, but this is not scary. But on the other hand, we firmly know that this degree is equal to 256! So we remember to what extent the deuce gives us 256. Remember? Yes! AT eighth degrees!

256 = 2 8

If you didn’t remember or with the recognition of the degrees of the problem, then it’s also okay: we just successively raise the two to the square, to the cube, to the fourth power, the fifth, and so on. The selection, in fact, but at this level, is quite a ride.

One way or another, we will get:

2 n -1 = 2 8

n-1 = 8

n = 9

So 768 is ninth member of our progression. That's it, problem solved.)

Answer: 9

What? Boring? Tired of the elementary? I agree. And me too. Let's go to the next level.)

More complex tasks.

And now we solve the puzzles more abruptly. Not exactly super-cool, but on which you have to work a little to get to the answer.

For example, like this.

Find the second term of a geometric progression if its fourth term is -24 and the seventh term is 192.

This is a classic of the genre. Some two different members of the progression are known, but one more member must be found. Moreover, all members are NOT neighbors. What confuses at first, yes ...

As in , we consider two methods for solving such problems. The first way is universal. Algebraic. Works flawlessly with any source data. So that's where we'll start.)

We paint each term according to the formula nth member!

Everything is exactly the same as with an arithmetic progression. Only this time we are working with another general formula. That's all.) But the essence is the same: we take and in turn we substitute our initial data into the formula of the nth term. For each member - their own.

For the fourth term we write:

b 4 = b 1 · q 3

-24 = b 1 · q 3

There is. One equation is complete.

For the seventh term we write:

b 7 = b 1 · q 6

192 = b 1 · q 6

In total, two equations were obtained for the same progression .

We assemble a system from them:

Despite its formidable appearance, the system is quite simple. The most obvious way to solve is the usual substitution. We express b 1 from the upper equation and substitute into the lower one:

A little fiddling with the lower equation (reducing the exponents and dividing by -24) yields:

q 3 = -8

By the way, the same equation can be arrived at in a simpler way! What? Now I will show you another secret, but very beautiful, powerful and useful way to solve such systems. Such systems, in the equations of which they sit only works. At least in one. called term division method one equation to another.

So we have a system:

In both equations on the left - work, and on the right is just a number. This is a very good sign.) Let's take and ... divide, say, the lower equation by the upper one! What means, divide one equation by another? Very simple. We take left side one equation (lower) and we divide her on left side another equation (upper). The right side is similar: right side one equation we divide on the right side another.

The whole division process looks like this:

Now, reducing everything that is reduced, we get:

q 3 = -8

What is good about this method? Yes, because in the process of such a division, everything bad and inconvenient can be safely reduced and a completely harmless equation remains! That is why it is so important to have only multiplications in at least one of the equations of the system. There is no multiplication - there is nothing to reduce, yes ...

In general, this method (like many other non-trivial ways of solving systems) even deserves a separate lesson. I will definitely take a closer look at it. Someday…

However, no matter how you solve the system, in any case, now we need to solve the resulting equation:

q 3 = -8

No problem: we extract the root (cubic) and - done!

Please note that it is not necessary to put plus / minus here when extracting. We have an odd (third) degree root. And the answer is the same, yes.

So, the denominator of progression is found. Minus two. Excellent! The process is underway.)

For the first term (say from the top equation) we get:

Excellent! We know the first term, we know the denominator. And now we have the opportunity to find any member of the progression. Including the second.)

For the second member, everything is quite simple:

b 2 = b 1 · q= 3 (-2) = -6

Answer: -6

So, we have sorted out the algebraic way of solving the problem. Difficult? Not much, I agree. Long and boring? Yes, definitely. But sometimes you can significantly reduce the amount of work. For this there is graphic way. Good old and familiar to us by .)

Let's draw the problem!

Yes! Exactly. Again we depict our progression on the number axis. Not necessarily by a ruler, it is not necessary to maintain equal intervals between members (which, by the way, will not be the same, because the progression is geometric!), But simply schematically draw our sequence.

I got it like this:

Now look at the picture and think. How many equal factors "q" share fourth and seventh members? That's right, three!

Therefore, we have every right to write:

-24q 3 = 192

From here it is now easy to find q:

q 3 = -8

q = -2

That's great, the denominator is already in our pocket. And now we look at the picture again: how many such denominators sit between second and fourth members? Two! Therefore, to record the relationship between these members, we will raise the denominator squared.

Here we write:

b 2 · q 2 = -24 , where b 2 = -24/ q 2

We substitute our found denominator into the expression for b 2 , count and get:

Answer: -6

As you can see, everything is much simpler and faster than through the system. Moreover, here we didn’t even need to count the first term at all! At all.)

Here is such a simple and visual way-light. But it also has a serious drawback. Guessed? Yes! It is only good for very short pieces of progression. Those where the distances between the members of interest to us are not very large. But in all other cases it is already difficult to draw a picture, yes ... Then we solve the problem analytically, through a system.) And systems are a universal thing. Deal with any number.

Another epic one:

The second term of the geometric progression is 10 more than the first, and the third term is 30 more than the second. Find the denominator of the progression.

What's cool? Not at all! All the same. We again translate the condition of the problem into pure algebra.

1) We paint each term according to the formula nth member!

Second term: b 2 = b 1 q

Third term: b 3 \u003d b 1 q 2

2) We write down the relationship between the members from the condition of the problem.

Reading the condition: "The second term of a geometric progression is 10 more than the first." Stop, this is valuable!

So we write:

b 2 = b 1 +10

And we translate this phrase into pure mathematics:

b 3 = b 2 +30

We got two equations. We combine them into a system:

The system looks simple. But there are a lot of different indices for letters. Let's substitute instead of the second and third members of their expression through the first member and denominator! In vain, or what, we painted them?

We get:

But such a system is no longer a gift, yes ... How to solve this? Unfortunately, the universal secret spell to solve complex non-linear There are no systems in mathematics and there cannot be. It is fantastic! But the first thing that should come to your mind when trying to crack such a tough nut is to figure out But isn't one of the equations of the system reduced to a beautiful form, which makes it easy, for example, to express one of the variables in terms of another?

Let's guess. The first equation of the system is clearly simpler than the second. We will torture him.) Why not try from the first equation something express through something? Since we want to find the denominator q, then it would be most advantageous for us to express b 1 through q.

So let's try to do this procedure with the first equation, using the good old ones:

b 1 q = b 1 +10

b 1 q - b 1 \u003d 10

b 1 (q-1) = 10

Everything! Here we have expressed unnecessary us the variable (b 1) through necessary(q). Yes, not the most simple expression received. Some kind of fraction ... But our system is of a decent level, yes.)

Typical. What to do - we know.

We write ODZ (necessarily!) :

q ≠ 1

We multiply everything by the denominator (q-1) and reduce all fractions:

10 q 2 = 10 q + 30(q-1)

We divide everything by ten, open the brackets, collect everything on the left:

q 2 – 4 q + 3 = 0

We solve the resulting and get two roots:

q 1 = 1

q 2 = 3

There is only one final answer: q = 3 .

Answer: 3

As you can see, the way to solve most problems for the formula of the nth member of a geometric progression is always the same: we read carefully condition of the problem and, using the formula of the nth term, we translate all useful information into pure algebra.

Namely:

1) We write separately each member given in the problem according to the formulanth member.

2) From the condition of the problem, we translate the connection between the members into a mathematical form. We compose an equation or a system of equations.

3) We solve the resulting equation or system of equations, find the unknown parameters of the progression.

4) In case of an ambiguous answer, we carefully read the condition of the problem in search of additional information (if any). We also check the received answer with the conditions of the ODZ (if any).

And now we list the main problems that most often lead to errors in the process of solving geometric progression problems.

1. Elementary arithmetic. Operations with fractions and negative numbers.

2. If at least one of these three points is a problem, then you will inevitably be mistaken in this topic. Unfortunately... So don't be lazy and repeat what was mentioned above. And follow the links - go. Sometimes it helps.)

Modified and recurrent formulas.

And now let's look at a couple of typical exam problems with a less familiar presentation of the condition. Yes, yes, you guessed it! it modified and recurrent formulas of the nth member. We have already encountered such formulas and worked in arithmetic progression. Everything is similar here. The essence is the same.

For example, such a problem from the OGE:

The geometric progression is given by the formula b n = 3 2 n . Find the sum of the first and fourth terms.

This time the progression is given to us not quite as usual. Some kind of formula. So what? This formula is also a formulanth member! We all know that the formula of the nth term can be written both in general form, through letters, and for specific progression. FROM specific first term and denominator.

In our case, we are, in fact, given a general term formula for a geometric progression with the following parameters:

b 1 = 6

q = 2

Let's check?) Let's write the formula of the nth term in general form and substitute into it b 1 and q. We get:

b n = b 1 · q n -1

b n= 6 2n -1

We simplify, using factorization and power properties, and get:

b n= 6 2n -1 = 3 2 2n -1 = 3 2n -1+1 = 3 2n

As you can see, everything is fair. But our goal with you is not to demonstrate the derivation of a specific formula. This is so, a lyrical digression. Purely for understanding.) Our goal is to solve the problem according to the formula that is given to us in the condition. Do you catch it?) So we are working with the modified formula directly.

We count the first term. Substitute n=1 into the general formula:

b 1 = 3 2 1 = 3 2 = 6

Like this. By the way, I'm not too lazy and once again I will draw your attention to a typical blunder with the calculation of the first term. DO NOT look at the formula b n= 3 2n, immediately rush to write that the first member is a troika! It's a big mistake, yes...)

We continue. Substitute n=4 and consider the fourth term:

b 4 = 3 2 4 = 3 16 = 48

And finally, we calculate the required amount:

b 1 + b 4 = 6+48 = 54

Answer: 54

Another problem.

The geometric progression is given by the conditions:

b 1 = -7;

b n +1 = 3 b n

Find the fourth term of the progression.

Here the progression is given by the recurrent formula. Well, okay.) How to work with this formula - we also know.

Here we are acting. Step by step.

1) counting two successive member of the progression.

The first term is already given to us. Minus seven. But the next, second term, can be easily calculated using the recursive formula. If you understand how it works, of course.)

Here we consider the second term according to the famous first:

b 2 = 3 b 1 = 3 (-7) = -21

2) We consider the denominator of the progression

Also no problem. Straight, share second dick on the first.

We get:

q = -21/(-7) = 3

3) Write the formulanth member in the usual form and consider the desired member.

So, we know the first term, the denominator too. Here we write:

b n= -7 3n -1

b 4 = -7 3 3 = -7 27 = -189

Answer: -189

As you can see, working with such formulas for a geometric progression is essentially no different from that for an arithmetic progression. It is only important to understand the general essence and meaning of these formulas. Well, the meaning of geometric progression also needs to be understood, yes.) And then there will be no stupid mistakes.

Well, let's decide on our own?)

Quite elementary tasks, for warm-up:

1. Given a geometric progression in which b 1 = 243, and q = -2/3. Find the sixth term of the progression.

2. The common term of a geometric progression is given by the formula b n = 5∙2 n +1 . Find the number of the last three-digit member of this progression.

3. The geometric progression is given by the conditions:

b 1 = -3;

b n +1 = 6 b n

Find the fifth term of the progression.

A little more complicated:

4. Given a geometric progression:

b 1 =2048; q =-0,5

What is the sixth negative term of it?

What seems super difficult? Not at all. Logic and understanding of the meaning of geometric progression will save. Well, the formula of the nth term, of course.

5. The third term of the geometric progression is -14 and the eighth term is 112. Find the denominator of the progression.

6. The sum of the first and second terms of a geometric progression is 75, and the sum of the second and third terms is 150. Find the sixth term of the progression.

Answers (in disarray): 6; -3888; -one; 800; -32; 448.

That's almost all. It remains only to learn how to count the sum of the first n terms of a geometric progression yes discover infinitely decreasing geometric progression and its amount. A very interesting and unusual thing, by the way! More on that in later lessons.)

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.

Solution: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 .

Solution: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.

Solution: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.

Terms: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.

Consequently,a 3 = q 2 · a 1

When substitutingq= 4

Terms:a 2 = 6, a 3 = 12. Calculate S 6 .

Solution:To do this, it is enough to find q, the first element and substitute it into the formula.

a 3 = q· a 2 , Consequently,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.

Solution:

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.

Geometric progression no less important in mathematics than in arithmetic. A geometric progression is such a sequence of numbers b1, b2,..., b[n] each next member of which is obtained by multiplying the previous one by a constant number. This number, which also characterizes the rate of growth or decrease of the progression, is called denominator of a geometric progression and denote

For a complete assignment of a geometric progression, in addition to the denominator, it is necessary to know or determine its first term. For a positive value of the denominator, the progression is a monotone sequence, and if this sequence of numbers is monotonically decreasing and monotonically increasing when. The case when the denominator is equal to one is not considered in practice, since we have a sequence of identical numbers, and their summation is not of practical interest

General term of a geometric progression calculated according to the formula

The sum of the first n terms of a geometric progression determined by the formula

Let us consider solutions of classical geometric progression problems. Let's start with the simplest to understand.

Example 1. The first term of a geometric progression is 27, and its denominator is 1/3. Find the first six terms of a geometric progression.

Solution: We write the condition of the problem in the form

For calculations, we use the formula for the nth member of a geometric progression

Based on it, we find unknown members of the progression

As you can see, calculating the terms of a geometric progression is not difficult. The progression itself will look like this

Example 2. The first three members of a geometric progression are given: 6; -12; 24. Find the denominator and the seventh term.

Solution: We calculate the denominator of the geometric progression based on its definition

We got an alternating geometric progression whose denominator is -2. The seventh term is calculated by the formula

On this task is solved.

Example 3. A geometric progression is given by two of its members . Find the tenth term of the progression.

Solution:

Let's write the given values through the formulas

According to the rules, it would be necessary to find the denominator, and then look for the desired value, but for the tenth term we have

The same formula can be obtained on the basis of simple manipulations with the input data. We divide the sixth term of the series by another, as a result we get