The geometric meaning of progression. Geometric progression and its formula. Property of a 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 step. 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 mathematical model tasks.

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 from the very simple 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).

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):
Solution:

Answer : \(-686\).

Example (OGE): Given the first three terms of the progression \(324\); \(-108\); \(36\)…. Find \(b_5\).
Solution:

	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 \(-108\)?
\(324 q=-108\)	From here we can easily calculate the denominator.
\(q=-\) \(\frac(108)(324)\) \(=-\) \(\frac(1)(3)\)	Now we can easily find the element we need.
	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) \(0.8\) ?

Solution: 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 \(x\).

Solution:

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.

Solution:

Answer: \(105\).

Example (OGE): It is known that exponentially \(b_6=-11\),\(b_9=704\). Find the denominator \(q\).

Solution:

	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, \(b_9=b_6 q q q=b_6 q^3\).
\(b_9=b_6 q^3\)	Substitute the values we know.
\(704=(-11)q^3\)	“Reverse” the equation and divide it by \((-11)\).
\(q^3=\) \(\frac(704)(-11)\) \(\:\:\: ⇔ \:\:\: \)\(q^3=-\) \(64 \)	What number cubed gives \(-64\)? Of course, \(-4\)!
	Answer found. It can be checked by restoring the chain of numbers from \(-11\) to \(704\).
	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 \(b_4\).
Solution:

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 \(b_(12)\).
Solution:

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.
Solution:

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

For the progression \(b_n = \(3; 6; 12; 24; 48…\)\) considered at the very beginning of the article, the denominator \(q\) is 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 \(5\); \(-15\); \(45\); \(-135\); \(675\)... the denominator of \(q\) is \(-3\), and because of this, the signs of the elements "blink".

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

Here 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 – 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: the 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.

Etc. 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 view! 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 by the appearance 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 is 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 they don’t teach you to decide, yes ... This is the topic of the senior classes. 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 a certain degree. 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! IN 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? Agree. 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 solutions for 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 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. Some day…

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. Fine! The process is underway.)

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

Fine! 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's fantastic! But the first thing that should come to your mind when trying to crack such a tough nut is to figure out and does not one of the equations of the system reduce to beautiful view, allowing, for example, to easily express one of the variables in terms of the other?

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 across 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 the entire 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! This 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 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.)

This number is called the denominator of a geometric progression, that is, 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 of the nth member 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, they knew not only arithmetic, but also geometric progression. Here, for example, is a task from the Rhind papyrus: “Seven faces have seven cats; each cat eats seven mice, each mouse eats seven ears of corn, each ear 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 was repeated many times with different variations among other peoples at other times. For example, in written in the XIII 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 which has 7 mules, each of which has 7 bags, each of which contains 7 loaves , each of which has 7 knives, each of which is in 7 sheaths. The problem asks how many items there are.

The sum of the first n members of the geometric progression S n = b 1 (q n - 1) / (q - 1) . This formula can be proved, for example, as follows: S n \u003d b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n - 1.

Let's add the number b 1 q n to S n and get:

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

Hence S n (q - 1) = b 1 (q n - 1), and we get the necessary formula.

Already on one of the clay tablets of Ancient Babylon, dating back to the VI 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 where this fact was known to the Babylonians.

The rapid growth of a geometric progression in a number of cultures, in particular, in India, is repeatedly used as a clear symbol of the immensity of the universe. In the well-known legend about the appearance of chess, the ruler gives their inventor the opportunity to choose a reward himself, and he asks for such a number of wheat grains as will be obtained if one is placed on the first cell of the chessboard, two on the second, four on the third, eight on the fourth, and etc., each time the number is doubled. Vladyka thought that it was, at the most, a few sacks, but he miscalculated. It is easy to see that for all 64 squares of the chessboard the inventor should have received (2 64 - 1) grain, 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 number of grains. This legend is sometimes interpreted as a reference to the almost unlimited possibilities hidden in the game of chess.

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

2 64 \u003d 2 4 ∙ (2 10) 6 \u003d 16 1024 6 ≈ 16 1000 6 \u003d 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 is increasing if the denominator is greater than 1 in absolute value, or decreasing if it is less than one. In the latter case, the number q n can become arbitrarily small for sufficiently large n. While an increasing exponential increases unexpectedly fast, a decreasing exponential decreases just as quickly.

The larger n, the weaker the number q n differs from zero, and the closer the sum of n members of the geometric progression S n \u003d b 1 (1 - q n) / (1 - q) to the number S \u003d b 1 / (1 - q) . (So reasoned, for example, F. Viet). 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 the summation of the ALL 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 "Biting" and "Achilles and the tortoise". In the first case, it is clearly shown that the entire road (assume length 1) is the sum of an infinite number of segments 1/2, 1/4, 1/8, etc. This, of course, is how it is from the point of view of ideas about the finite sum infinite geometric progression. And yet - how can this be?

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

In the aporia about Achilles, the situation is a little more complicated, because here the denominator of the progression is not equal to 1/2, but to some other number. Let, for example, Achilles run at speed v, the tortoise moves at speed u, and the initial distance between them is l. Achilles will run this distance in the time l / v , the tortoise will move a distance lu / v during this time. When Achilles runs through 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 point with the turtle - is equal to l / (1 - u / v) = lv / (v - u) . But, again, how this result should be interpreted and why it makes any sense at all, was not very clear for a long time.

Rice. 3. Geometric progression with coefficient 2/3

The sum of a geometric progression was used by Archimedes when determining the area of a segment of a parabola. Let the given segment of the parabola be delimited by the chord AB and let the tangent at the 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 . Draw lines parallel to DC through points A , E , F , B ; let the tangent drawn at point D , these lines intersect at points K , L , M , N . Let's also draw segments AD and DB. Let the line EL intersect the line AD at the point G, and the parabola at the point H; line FM intersects line DB at point Q, and the parabola at point R. According to the general theory of conic sections, DC is the diameter of a 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 parabola equation is written as y 2 \u003d 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 . Since KA = 2LG , LH = HG . The area of the segment ADB of the parabola is equal to the area of the triangle ΔADB and the areas of the segments AHD and DRB combined. In turn, the area of the AHD segment is similarly equal to the area of the triangle AHD and the remaining segments AH and HD, with each of which the same operation can be performed - 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 triangle ΔAHD is equal to a quarter of the area of triangle ΔACD. Likewise, the area of triangle ΔDRB is equal to a quarter of the area of triangle ΔDFB. So, the areas of triangles ∆AHD and ∆DRB, taken together, are equal to a quarter of the area of triangle ∆ADB. Repeating this operation as applied to the segments AH , HD , DR and RB will also 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 . Etc:

Thus, Archimedes proved that "every segment enclosed between a straight line and a parabola is four-thirds of a triangle, having with it the same base and equal height."

Mathematics is whatpeople control nature and themselves.

Soviet mathematician, academician A.N. Kolmogorov

Geometric progression.

Along with tasks for arithmetic progressions, tasks related to the concept of a geometric progression are also common in entrance tests in mathematics. To successfully solve such problems, you need to know the properties of a geometric progression and have good skills in using them.

This article is devoted to the presentation of the main properties of a geometric progression. It also provides examples of solving typical problems, borrowed from the tasks of entrance tests in mathematics.

Let us preliminarily note the main properties of a geometric progression and recall the most important formulas and statements, associated with this concept.

Definition. A numerical sequence is called a geometric progression if each of its numbers, 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 a geometric progressionthe formulas are valid

, (1)

where . Formula (1) is called the formula of the general term of a geometric progression, and formula (2) is the main property of a geometric progression: each member of the progression coincides with the geometric mean of its neighboring members and .

Note, that it is precisely because of this property that the progression in question is called "geometric".

Formulas (1) and (2) above are summarized as follows:

, (3)

To calculate the sum first members of a geometric progressionthe formula applies

If we designate

where . Since , formula (6) is a generalization of formula (5).

In the case when and geometric progressionis infinitely decreasing. To calculate the sumof all members of an infinitely decreasing geometric progression, the formula is used

. (7)

For example , using formula (7), one can show, what

where . These equalities are obtained from formula (7) provided that , (the first equality) and , (the second equality).

Theorem. If , then

Proof. If , then ,

The theorem has been proven.

Let's move on to considering examples of solving problems on the topic "Geometric progression".

Example 1 Given: , and . To find .

Solution. If formula (5) is applied, then

Answer: .

Example 2 Let and . To find .

Solution. Since and , we use formulas (5), (6) and obtain the system of equations

If the second equation of system (9) is divided by the first, then or . From this it follows . Let's consider two cases.

1. If , then from the first equation of system (9) we have.

2. If , then .

Example 3 Let , and . To find .

Solution. It follows from formula (2) that or . Since , then or .

By condition . However , therefore . Because 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 single suitable root . In this case, the first equation of the system implies .

Taking into account formula (7), we obtain.

Answer: .

Example 4 Given: and . To find .

Solution. Since , then .

Because , 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 . To find .

Solution. According to the theorem, we have two equalities

Since , then or . Because , then .

Answer: .

Example 6 Given: and . To find .

Solution. Taking into account formula (5), we obtain

Since , then . Since , and , then .

Example 7 Let and . To 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 condition of the problem, we obtain the 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 10solve the equation

, (11)

where and .

Solution. The left side of equation (11) is the sum of an infinite decreasing geometric progression, in which and , provided: and .

From formula (7) it follows, what . In this regard, equation (11) takes the form or . suitable root quadratic equation is

Answer: .

Example 11. P sequence of positive numbersforms an arithmetic progression, but - geometric progression, what does it have to do with . To find .

Solution. Because arithmetic sequence, then (the main property of an arithmetic progression). Insofar as, then or . This implies , that the geometric progression is. According to formula (2), then we write that .

Since and , then . In that case, the expression takes the form or . By condition , so from the equationwe obtain the unique solution of 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, then

or .

To calculate, we substitute the values into formula (7) and obtain . Since , then .

Answer: .

The examples of problem solving given here will be useful to applicants in preparation for entrance examinations. For a deeper study of problem solving methods, associated with a geometric progression, can be used study guides from the list of recommended literature.

1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. – M.: Mir i Obrazovanie, 2013. – 608 p.

2. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. - 216 p.

3. Medynsky M.M. A complete course of elementary mathematics in tasks and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. - 208 p.

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