For example, sequence \ (3 \); \ (6 \); \(12\); \ (24 \); \ (48 \) ... is a geometric progression, because each next element differs from the previous one twice (in other words, it can be obtained from the previous one by multiplying it by two):

Like any sequence, geometric progression is denoted by a small Latin letter. The numbers forming the progression call it members of(or elements). They are denoted by the same letter as the geometric progression, but with a numerical index equal to the number of the element in order.

For example, geometric progression \ (b_n = \ (3; 6; 12; 24; 48 ... \) \) consists of elements \ (b_1 = 3 \); \ (b_2 = 6 \); \ (b_3 = 12 \) and so on. In other words:

If you understand the above information, then you can already solve most of the problems on this topic.

Example (OGE):
Solution:

Answer : \(-686\).

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

	To continue the sequence, we need to know the denominator. Let's find it from two adjacent elements: what is to be multiplied by \ (324 \) to get \ (- 108 \)?
\ (324 q = -108 \)	From here we calculate the denominator without any problems.
\ (q = - \) \ (\ frac (108) (324) \) \ (= - \) \ (\ frac (1) (3) \)	Now we can easily find the element we need.
	The answer is ready.

Answer : \(4\).

Example: The progression is specified by the condition \ (b_n = 0.8 5 ^ n \). Which of the numbers is a member of this progression:

a) \ (- 5 \) b) \ (100 \) c) \ (25 \) d) \ (0.8 \)?

Solution: From the wording of the assignment, it is obvious that one of these numbers is definitely in our progression. Therefore, we can simply calculate its members in turn until we find the value we need. Since our progression is given by a 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. Let's continue.
\ (n = 2 \); \ (b_2 = 0.8 5 ^ 2 = 0.8 25 = 20 \) - and this is not the case either.
\ (n = 3 \); \ (b_3 = 0.8 5 ^ 3 = 0.8 125 = 100 \) - here comes our champion!

Answer: \(100\).

Example (OGE): Several members of a geometric progression are given one after the other ... \ (8 \); \ (x \); \(50\); \ (- 125 \) .... Find the value of the item denoted by \ (x \).

Solution:

Answer: \(-20\).

Example (OGE): The progression is specified 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:

	From the diagram on the left you can see that in order to "get" from \ (b_6 \) to \ (b_9 \) - we take three "steps", that is, we multiply \ (b_6 \) by the denominator of the progression three times. In other words, \ (b_9 = b_6 q q q = b_6 q ^ 3 \).
\ (b_9 = b_6 q ^ 3 \)	Let's substitute the values ​​we know.
\ (704 = (- 11) q ^ 3 \)	Let's "flip" the equation and divide it by \ ((- 11) \).
\ (q ^ 3 = \) \ (\ frac (704) (- 11) \) \ (\: \: \: ⇔ \: \: \: \) \ (q ^ 3 = - \) \ (64 \)	What number in the cube will give \ (- 64 \)? Of course \ (- 4 \)!
	The answer has been found. It can be checked by restoring the chain of numbers from \ (- 11 \) to \ (704 \).
	Everything agreed - the answer is correct.

Answer: \(-4\).

The most important formulas

As you can see, most problems on a geometric progression can be solved with pure logic, just by understanding the essence (this is generally typical for mathematics). But sometimes knowledge of some formulas and laws speeds up and greatly facilitates the solution. We will study two such formulas.

The formula for the \ (n \) -th term: \ (b_n = b_1 q ^ (n-1) \), where \ (b_1 \) is the first term of the progression; \ (n \) - number of the element being searched for; \ (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 literally one action.

Example (OGE): Geometric progression given by the conditions \ (b_1 = -2 \); \ (q = 7 \). Find \ (b_4 \).
Solution:

Answer: \(-686\).

This example was simple, so the formula didn't make the calculations too easy for us. Let's look at the problem a little more difficult.

Example: The geometric progression is specified by the conditions \ (b_1 = 20480 \); \ (q = \ frac (1) (2) \). Find \ (b_ (12) \).
Solution:

Answer: \(10\).

Of course, raising \ (\ frac (1) (2) \) to \ (11 \) - th degree is not too joyful, but it's still easier than \ (11 \) times to divide \ (20480 \) by two.

Sum of \ (n \) first members: \ (S_n = \) \ (\ frac (b_1 (q ^ n-1)) (q-1) \), where \ (b_1 \) is the first term of the progression; \ (n \) - the number of elements to be added; \ (q \) is the denominator of the progression; \ (S_n \) - the sum \ (n \) of the first members of the progression.

Example (OGE): You are given a geometric progression \ (b_n \), the denominator of which 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 "head-on" - find all six elements in turn, and then add the results. However, the number of calculations, and hence the chance of an accidental error, would increase dramatically.

For a geometric progression, there are several more formulas that we did not consider here because of their low practical value. You can find these formulas.

Ascending and Decreasing Geometric Progressions

The progression \ (b_n = \ (3; 6; 12; 24; 48 ... \) \) considered at the very beginning of the article has the denominator \ (q \) greater than one and therefore each next term is greater than the previous one. Such progressions are called increasing.

If \ (q \) is less than one, but at the same time is positive (that is, it lies in the range from zero to one), then each next element will be less than the previous one. For example, in the progression \ (4 \); \ (2 \); \(1\); \ (0.5 \); \ (0,25 \) ... the denominator \ (q \) is \ (\ frac (1) (2) \).

These progressions are called decreasing... Please note that none of the elements of such a 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 will never go beyond it. Mathematicians in such cases say "go to zero."

Note that with a negative denominator, the elements of the geometric progression will necessarily change sign. For example, in the progression \ (5 \); \(-15\); \ (45 \); \ (- 135 \); \ (675 \) ... the denominator \ (q \) is \ (- 3 \), and because of this, the element characters "blink".

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

So, for starters itself formulan

There she is:

b n = b 1 · q n -1

Formula as formula, nothing supernatural. It looks even simpler and more compact than a 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. Without multiplying sequentially by "q" many, many times. That's the whole point.)

I understand that at this level of work with progressions all the values ​​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 a member of a geometric progression;

q – ;

n- member number;

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

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

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

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

b 2 = b 1 q

And the third term? Not a problem either! We multiply the second term one more time onq.

Like this:

B 3 = b 2 q

Let us now recall that the second term, in turn, is equal to b 1 q and we 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 times 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) by q:

B 4 = b 3 q = (b 1 q 2) q = b 1 q 2 q = 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 times q in third degree.

Etc. So how? Got a pattern? Yes! For any term with any number, the number of identical factors q (i.e., the degree of the denominator) will always be one less than the number of the required termn.

Therefore, our formula will be, without options:

b n =b 1 · q n -1

That's all there is to it.)

Well, let's solve the problems, probably?)

Solving formula problemsnth member of a geometric progression.

Let's start, as usual, by applying the formula directly. Here's a typical problem:

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

Of course, this problem can be solved without any formulas at all. Directly within the meaning of a geometric progression. But we need to warm up with the formula for the nth term, right? So we warm up.

Our data for applying the formula are as follows.

The first term is known. It's 512.

b 1 = 512.

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

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

And we accurately count 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 task, but a little more complicated in terms of calculations.

It is known exponentially that:

b 1 = 3

Find the thirteenth term in the progression.

Everything is the same, only this time the denominator of the progression is irrational... The root of two. Well, that's okay. 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 freeze. What to do next with the root? How to raise the root to the twelfth power?

How-how ... You have to understand that any formula, of course, is a good thing, but knowledge of all previous mathematics is not canceled! How to build? Yes, the properties of the degrees to remember! Let's turn the root into fractional exponent and - according to the exponentiation formula.

Like this:

Answer: 192

And that's all.)

What is the main difficulty in the direct application of the n-term formula? Yes! The main difficulty is working with degrees! Namely - raising to the power of negative numbers, fractions, roots and the like. So those who have problems with this, we urge you 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 formula elements if all others are given. For the successful solution of such problems, the recipe is uniform and terribly simple - writing the formulanth member in general view! Right in the notebook next to the condition. And then, from the condition, we figure out what has been given to us and what is lacking. And we express the required value from the formula. Everything!

For example, such a harmless task.

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

Nothing complicated. We work directly by the spell.

We write the formula for the nth term!

b n = b 1 · q n -1

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

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

Everything? No! We are also given number n! This is a five: n = 5.

I hope you already understand what is in the recording b 5 = 567 two parameters are hidden at once - this is the fifth term 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 you here.)

Now we substitute our data into the formula:

567 = b 1 · 3 5-1

We count arithmetic, simplify and get a simple one 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 (for surprises), yes.)

For example, this problem:

The fifth term of the 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 terms, and we are asked to find the denominator of the progression. So let's get started.

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 meaning q... No problem! Now we will find it.) We substitute everything that we know into the formula.

We get:

162 = 2q 5-1

2 q 4 = 162

q 4 = 81

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

Like this:

q 4 = 81

q = 3

But actually, this is an unfinished answer. More precisely, incomplete. Why? The point is that the answer is q = -3 also fits: (-3) 4 will be 81 too!

This is due to the fact that the power equation x n = a always has two opposite roots at evenn . With plus and minus:

Both fit.

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

x 2 = 9

For some reason, you are not surprised at the appearance two roots x = ± 3? Here is the same thing. And with any other even degree (fourth, sixth, tenth, etc.) will be the same. Details - in the topic about

Therefore, the correct solution would be like this:

q 4 = 81

q= ± 3

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

Therefore, the answer is obvious:

q = 3

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

The fifth term of the 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 condition nothing the denominator sign is not mentioned. Neither directly nor indirectly. And here the task 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 match the condition of the problem. 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 task, yes. But also more creative.)

A geometric progression is given:

3; 6; 12; 24; …

What is the number 768 in this progression?

The first step is still the same: writing the formulanth member!

b n = b 1 · q n -1

And now, as usual, we substitute the data we know into it. Um ... not substituted! Where is the first term, where is the denominator, where is everything else ?!

Where, where ... And why do we need eyes? Clap your eyelashes? This time the progression is given to us directly in the form sequence. See the first term? We see! This is a triple (b 1 = 3). What about the denominator? We do not see it yet, but it is very easy to count. If, of course, you understand.

So we count. Directly in the sense 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 a certain 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 it.) So we are looking. We have already downloaded all the necessary data for substitution into the formula. Unbeknownst to myself.)

So we substitute:

768 = 3.2n -1

We do elementary ones - we divide both parts into 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. You need to find "n". What is unusual? Yes, I don’t argue. Actually, this is the simplest. It is so called due to the fact that 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, we will try to find our n guided by simple logic and common sense.

We begin to reason. 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 a big deal. But on the other hand, we firmly know that this degree is equal to 256! So we remember to what extent two gives us 256. Remember? Yes! V eighth degree!

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 sequentially raise the two to a square, to a cube, to the fourth degree, the fifth, and so on. The selection, in fact, but at this level is quite good.

One way or another, we get:

2 n -1 = 2 8

n-1 = 8

n = 9

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

Answer: 9

What? Boring? Tired of the elementary stuff? Agree. Me too. Let's go to the next level.)

More challenging tasks.

And now we solve problems more abruptly. Not exactly super cool, but they still have a little work to do to get to the answer.

For example, this.

Find the second term of the geometric progression if the 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 some more member must be found. Moreover, all members are NOT neighboring. Which is embarrassing at first, yes ...

As in, we will consider two ways to solve such problems. The first method is universal. Algebraic. Works flawlessly with any source data. Therefore, we will start with him.)

We write down each term according to the formula nth member!

Everything is exactly like with an arithmetic progression. Only this time we work with another general formula. That's all.) But the essence is the same: we take and one by one we substitute our initial data into the formula of the n-th term. For each member - their own.

For the fourth member, write:

b 4 = b 1 · q 3

-24 = b 1 · q 3

There is. One equation is ready.

For the seventh member, we write:

b 7 = b 1 · q 6

192 = b 1 · q 6

In total, we got two equations for the same progression .

We collect the system from them:

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

After tinkering a bit with the lower equation (by reducing the powers and dividing by -24), we get:

q 3 = -8

By the way, you can come to the same equation in a simpler way! How? Now I will demonstrate to you another secret, but very beautiful, powerful and useful way of solving such systems. Such systems in the equations of which sit only works. At least one. Called term division method one equation to another.

So, before us is the 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 divide her on left side another equation (top). The right side is similar: right side one equation divide on right side another.

The whole process of division looks like this:

Now, having reduced everything that is reduced, we get:

q 3 = -8

Why is this method good? Yes, the fact that 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 analyze it in more detail. Someday…

However, it doesn't matter how you solve the system, in any case, now we need to solve the resulting equation:

q 3 = -8

No problem: extract the root (cubic) and you're done!

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

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

For the first term (say from the upper 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 term, everything is quite simple:

b 2 = b 1 · q= 3 (-2) = -6

Answer: -6

So, we have laid out the algebraic way of solving the problem. Hard? Not really, I agree. Long and boring? Yes, absolutely. But sometimes you can significantly reduce the amount of work. For this there is graphical way. Good old and familiar to us.)

Drawing a problem!

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

I got it like this:

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

Therefore, we have every right to write down:

-24q 3 = 192

Hence, q is now easily searched for:

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 connection between these terms, the denominator will be squared.

So 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 easier and faster than through the system. Moreover, here we didn't even need to count the first term at all! At all.)

Here's a simple and intuitive way to light. But he also has a serious drawback. Have you guessed? Yes! It only works for very short slices of the 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 the system.) And systems are a universal thing. Any numbers can be dealt with.

Another epic challenge:

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 is cool? Not at all! All the same. We again translate the problem statement into pure algebra.

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

Second term: b 2 = b 1 q

Third term: b 3 = b 1 q 2

2) We write down the connection between the members from the problem statement.

We read the condition: "The second term of the exponential 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 the letters. Let's substitute instead of the second and third terms of their expression through the first term and denominator! Was it in vain that we painted them?

We get:

But such a system is no longer a gift, yes ... How to solve this? Unfortunately, a universal secret spell for solving complex nonlinear there are no systems in mathematics and cannot be. It is fantastic! But the first thing that should come to your mind when trying to bite such a tough nut is to estimate, but isn't one of the equations of the system reducible to a beautiful form that allows, for example, to easily express one of the variables in terms of the other?

So let's estimate. The first equation of the system is clearly simpler than the second. We will torture him.) Shouldn't we 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, applying the good old ones:

b 1 q = b 1 +10

b 1 q - b 1 = 10

b 1 (q-1) = 10

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

Typical. We know what to do.

We write ODZ (necessarily!) :

q ≠ 1

We multiply everything by the denominator (q-1) and cancel all the 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 result 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 term of a geometric progression is always the same: read attentively condition of the problem and, using the formula for the nth term, we translate the entire useful information into pure algebra.

Namely:

1) We write separately each term given in the problem by the formulanth member.

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

3) We solve the resulting equation or a system of equations, we 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 terms of the DLO (if any).

And now let's list the main problems that most often lead to errors in the process of solving problems on a geometric progression.

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

2. If you have problems with at least one of these three points, 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.

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

For example, such a task from the OGE:

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

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

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

b 1 = 6

q = 2

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

b n = b 1 · q n -1

b n= 6 2n -1

Simplify it using factorization and power properties to 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 a lyrical digression. Purely for understanding.) Our goal is to solve the problem according to the formula given to us in the condition. Catch?) So we work 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 will not be lazy, and once again I will draw your attention to a typical blooper with the calculation of the first member. DO NOT NEED looking at the formula b n= 3 2n, immediately rush to write that the first term is a triple! This is a gross mistake, yes ...)

Let's continue. Substitute n=4 and count 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 specified by the conditions:

b 1 = -7;

b n +1 = 3 b n

Find the fourth term in the progression.

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

So we act. Step by step.

1) Count two consecutive member of the progression.

The first term has already been assigned 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.)

So we count the second term according to the well-known first:

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

2) We consider the denominator of the progression

No problem either. Straight, divide second member on first.

We get:

q = -21/(-7) = 3

3) We write the formulan-th member in the usual form and consider the desired member.

So, we know the first term, and the denominator too. So 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 inherently 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 the geometric progression must also be understood, yes.) And then there will be no stupid mistakes.

Well, let's solve it ourselves?)

Quite basic tasks for warm-up:

1. A geometric progression is given in which b 1 = 243, and q = -2/3. Find the sixth term in the progression.

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

3. The geometric progression is set by the conditions:

b 1 = -3;

b n +1 = 6 b n

Find the fifth term in the progression.

A little more complicated:

4. A geometric progression is given:

b 1 =2048; q =-0,5

What is the sixth negative term?

What seems super difficult? Not at all. Will save the logic and understanding of the meaning of the geometric progression. Well, the formula for 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 the 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; -1; 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 this in the following lessons.)

A geometric progression is a numerical sequence, the first term of which is nonzero, and each next term is equal to the previous term multiplied by the same nonzero number. The geometric progression is denoted by b1, b2, b3,…, bn,…

Properties of a geometric progression

The ratio of any member of the geometric error to its previous term is equal to the same number, that is, b2 / b1 = b3 / b2 = b4 / b3 =… = bn / b (n-1) = b (n + 1) / bn = …. This follows directly from the definition of an arithmetic progression. This number is called the denominator of the geometric progression. Usually the denominator of a geometric progression is denoted by the letter q.

One of the ways to specify a geometric progression is to specify its first term b1 and the denominator of the geometric error q. For example, b1 = 4, q = -2. These two conditions define the geometric progression 4, -8, 16, -32,….

If q> 0 (q is not equal to 1), then the progression is a monotonic sequence. For example, the sequence, 2, 4,8,16,32, ... is a monotonically increasing sequence (b1 = 2, q = 2).

If, in the geometric error, the denominator is q = 1, then all members of the geometric progression will be equal to each other. In such cases, the progression is said to be a constant sequence.

Formula of the n-th member of the progression

In order for the numerical sequence (bn) to be a geometric progression, it is necessary that each of its terms, starting from the second, is the geometric mean of the neighboring terms. That is, it is necessary to fulfill the following equation - (b (n + 1)) ^ 2 = bn * b (n + 2), for any n> 0, where n belongs to the set of natural numbers N.

The formula for the n-th term of the geometric progression is:

bn = b1 * q ^ (n-1), where n belongs to the set of natural numbers N.

Let's take a look at a simple example:

In geometric progression b1 = 6, q = 3, n = 8 find bn.

Let's use the formula for the n-th term of the geometric progression.

So let's sit down and start writing some numbers. For example:

You can write any numbers, and there can be as many as you like (in our case, them). No matter how many numbers we write, we can always say which one is the first, which is the second, and so on to the last, that is, we can number them. This is an example of a number sequence:

Number sequence Is a set of numbers, each of which can be assigned a unique number.

For example, for our sequence:

The assigned number is specific to only one number in the sequence. In other words, there are no three second numbers in the sequence. The second number (like the -th number) is always one.

The number with the number is called the th member of the sequence.

We usually call the entire sequence some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member:.

In our case:

The most common types of progression are arithmetic and geometric. In this thread, we will talk about the second kind - geometric progression.

Why do we need a geometric progression and its history of origin.

Even in ancient times, the Italian mathematician Leonardo of Pisa (better known as Fibonacci) was engaged in solving the practical needs of trade. The monk was faced with the task of determining with the help of what least amount of weights it is possible to weigh the goods? In his writings, Fibonacci proves that such a system of weights is optimal: This is one of the first situations in which people had to face a geometric progression, which you probably already heard about and have at least a general concept. Once you fully understand the topic, think about why such a system is optimal?

Currently, in life practice, a geometric progression is manifested when investing money in a bank, when the amount of interest is charged on the amount accumulated in the account for the previous period. In other words, if you put money on a term deposit in a savings bank, then in a year the deposit will increase by more than the original amount, i.e. the new amount will be equal to the deposit multiplied by. In another year, this amount will increase by, i.e. the amount obtained at that time will be multiplied by again and so on. A similar situation is described in the problems of calculating the so-called compound interest- the percentage is taken each time from the amount on the account, taking into account the previous interest. We will talk about these tasks a little later.

There are many more simple cases where geometric progression is used. For example, the spread of influenza: one person infected a person, they, in turn, infected another person, and thus the second wave of infection is a person, and they, in turn, infected another ... and so on ...

By the way, the financial pyramid, the same MMM, is a simple and dry calculation based on the properties of a geometric progression. Interesting? Let's figure it out.

Geometric progression.

Let's say we have a numeric sequence:

You will immediately answer that it is easy and the name of such a sequence - with the difference of its members. How about this:

If you subtract the previous one from the next number, then you will see that each time a new difference is obtained (and so on), but the sequence definitely exists and it is easy to notice - each next number is times larger than the previous one!

This kind of number sequence is called geometric progression and is indicated by.

Geometric progression () is a numerical sequence, the first term of which is nonzero, and each term, starting from the second, is equal to the previous one, multiplied by the same number. This number is called the denominator of the geometric progression.

Restrictions that the first term () is not equal and not random. Let's say that there are none, and the first term is still equal, and q is equal, hmm .. let, then it turns out:

Agree that this is no longer any progression.

As you understand, we will get the same results if it is any number other than zero, and. In these cases, there will simply not be a progression, since the entire number series will be either all zeros, or one number, and all other zeros.

Now let's talk in more detail about the denominator of the geometric progression, that is, Fr.

Let's repeat: is a number, how many times does each subsequent term change geometric progression.

What do you think it can be? Correctly, positive and negative, but not zero (we talked about this a little higher).

Let's say that we have a positive one. Let in our case, as well. What is the second term and? You can easily answer that:

Everything is correct. Accordingly, if, then all subsequent members of the progression have the same sign - they positive.

What if negative? For example, a. What is the second term and?

This is a completely different story.

Try to count the term of this progression. How much did you get it? I have. Thus, if, then the signs of the members of the geometric progression alternate. That is, if you see a progression with alternating signs on its members, then its denominator is negative. This knowledge can help you test yourself when solving problems on this topic.

Now let's practice a little: try to determine which number sequences are a geometric progression, and which are arithmetic:

Understood? Let's compare our answers:

• Geometric progression - 3, 6.
• Arithmetic progression - 2, 4.
• It is neither arithmetic nor geometric progressions - 1, 5, 7.

Let's return to our last progression, and try to find its term in the same way as in arithmetic. As you might guess, there are two ways to find it.

We successively multiply each term by.

So, the th member of the described geometric progression is equal to.

As you might guess, now you yourself will deduce a formula that will help you find any member of a geometric progression. Or have you already brought it out for yourself, describing how to find the th member step by step? If so, then check the correctness of your reasoning.

Let us illustrate this by the example of finding the th member of a given progression:

In other words:

Find on your own the value of a member of a given geometric progression.

Happened? Let's compare our answers:

Pay attention that you got exactly the same number as in the previous method, when we successively multiplied by each previous term of the geometric progression.
Let's try to "depersonalize" this formula - we will bring it into a general form and get:

The derived formula is correct for all values, both positive and negative. Check it yourself by calculating the members of the geometric progression with the following conditions:, a.

Did you count it? Let's compare the results obtained:

Agree that it would be possible to find a member of the progression in the same way as a member, however, there is a possibility of incorrect counting. And if we have already found the th term of the geometric progression, then what could be easier than using the "cut off" part of the formula.

An infinitely decreasing geometric progression.

More recently, we talked about the fact that it can be either greater than or less than zero, however, there are special values ​​at which a geometric progression is called infinitely decreasing.

Why do you think such a name?
First, let's write down some geometric progression consisting of members.
Suppose, a, then:

We see that each subsequent term is less than the previous one by one factor, but will there be any number? You will immediately answer no. That is why the infinitely decreasing - decreases, decreases, and never becomes zero.

To clearly understand how it looks visually, let's try to draw a graph of our progression. So, for our case, the formula takes the following form:

It is customary for us to build dependence on charts, therefore:

The essence of the expression has not changed: in the first entry, we showed the dependence of the value of a geometric progression member on its ordinal number, and in the second entry, we simply took the value of a geometric progression term as, and the ordinal number was designated not how, but how. All that remains to be done is to build a graph.
Let's see what you get. Here's the graph I got:

See? The function decreases, tends to zero, but never crosses it, so it is infinitely decreasing. Let's mark our points on the graph, and at the same time what the coordinate and mean:

Try to schematically depict a graph of a geometric progression when, if its first term is also equal. Analyze, what's the difference with our previous chart?

Did you manage? Here's the graph I got:

Now that you have completely understood the basics of the theme of a geometric progression: you know what it is, you know how to find its term, and you also know what an infinitely decreasing geometric progression is, let's move on to its main property.

The property of a geometric progression.

Do you remember the property of the members of an arithmetic progression? Yes, yes, how to find the value of a certain number of a progression, when there are previous and subsequent values ​​of the members of a given progression. Remembered? This:

Now we are faced with exactly the same question for the members of a geometric progression. To derive a similar formula, let's start drawing and reasoning. You will see, it is very easy, and if you forget, you can bring it out on your own.

Let's take one more simple geometric progression in which we know and. How to find? With an arithmetic progression, this is easy and simple, but what about here? In fact, in geometric, too, there is nothing complicated - you just need to write down each value given to us according to the formula.

You ask, and what should we do with this now? It's very simple. To begin with, we will depict these formulas in the figure, and try to do various manipulations with them in order to arrive at a value.

We abstract from the numbers that we are given, we will focus only on expressing them through a formula. We need to find the value highlighted in orange, knowing the members adjacent to it. Let's try to perform various actions with them, as a result of which we can receive.

Addition.
Let's try to add two expressions and, we get:

From this expression, as you can see, we cannot express in any way, therefore, we will try another option - subtraction.

Subtraction.

As you can see, we also cannot express from this, therefore, we will try to multiply these expressions by each other.

Multiplication.

Now look carefully what we have, multiplying the members of the geometric progression given to us in comparison with what needs to be found:

Guess what I'm talking about? That's right, to find we need to take Square root from the geometric progression numbers multiplied by each other adjacent to the desired number:

Well. You yourself have deduced the property of a geometric progression. Try to write this formula in general terms. Happened?

Forgot the condition for? Think about why it is important, for example, try to calculate it yourself, if. What happens in this case? That's right, complete nonsense since the formula looks like this:

Accordingly, do not forget this limitation.

Now let's calculate what is equal to

Correct answer - ! If, when calculating, you did not forget the second possible value, then you are a great fellow and you can immediately proceed to training, and if you forgot, read what is discussed below and pay attention to why both roots must be written down in the answer.

Let's draw both of our geometric progressions - one with a meaning, and the other with a meaning and check if both of them have the right to exist:

In order to check whether such a geometric progression exists or not, it is necessary to see whether it is the same between all of its given members? Calculate q for the first and second cases.

See why we have to write two answers? Because the sign of the required term depends on whether it is positive or negative! And since we do not know what he is, we need to write both answers with a plus and a minus.

Now that you have mastered the main points and derived the formula for the property of a geometric progression, find, knowing and

Compare the received answers with the correct ones:

What do you think, what if we were given not the values ​​of the members of the geometric progression adjacent to the desired number, but equidistant from it. For example, we need to find, and are given and. Can we in this case use the formula we derived? Try to confirm or deny this possibility in the same way, writing out what each value consists of, as you did when initially deriving the formula, for.
What did you do?

Now look closely again.
and correspondingly:

From this we can conclude that the formula works not only with neighboring with the required terms of the geometric progression, but also with equidistant from the sought-after members.

Thus, our initial formula takes the form:

That is, if in the first case we said that, now we say that it can be equal to any natural number that is less. The main thing is to be the same for both given numbers.

Practice with specific examples, just be extremely careful!

1. ,. Find.
2. ,. Find.
3. ,. Find.

Decided? I hope you were extremely attentive and noticed a small catch.

We compare the results.

In the first two cases, we calmly apply the above formula and get the following values:

In the third case, upon careful consideration of the ordinal numbers of the numbers given to us, we understand that they are not equidistant from the number we are looking for: it is the previous number, but removed in position, so it is not possible to apply the formula.

How to solve it? It's actually not as difficult as it sounds! Let's write down with you what each number given to us and the required number consists of.

So, we have and. Let's see what we can do with them? I propose to divide by. We get:

We substitute our data into the formula:

The next step we can find - for this we need to take the cube root of the resulting number.

And now we look once again what we have. We have, but we need to find, and he, in turn, is equal to:

We found all the necessary data for the calculation. Substitute in the formula:

Our answer: .

Try to solve another similar problem yourself:
Given:,
Find:

How much did you get it? I have - .

As you can see, in fact, you need remember just one formula-. You can withdraw all the rest without any difficulty on your own at any time. To do this, just write the simplest geometric progression on a piece of paper and write down what, according to the above formula, each of its numbers is equal.

The sum of the members of a geometric progression.

Now consider the formulas that allow us to quickly calculate the sum of the members of a geometric progression in a given interval:

To derive the formula for the sum of the members of a finite geometric progression, we multiply all parts of the higher equation by. We get:

Look carefully: what do the last two formulas have in common? That's right, common members, for example, and so on, except for the first and last member. Let's try to subtract the 1st from the 2nd equation. What did you do?

Now express the term of the geometric progression through the formula and substitute the resulting expression in our last formula:

Group the expression. You should get:

All that's left to do is express:

Accordingly, in this case.

What if? What formula works then? Imagine a geometric progression at. What is she like? Correctly a series of identical numbers, respectively, the formula will look like this:

There are many legends in both arithmetic and geometric progression. One of them is the legend of Seth, the creator of chess.

Many people know that the game of chess was invented in India. When the Hindu king met her, he was delighted with her wit and the variety of possible positions in her. Upon learning that it was invented by one of his subjects, the king decided to personally reward him. He summoned the inventor to him and ordered to ask him for whatever he wanted, promising to fulfill even the most skillful desire.

Seta asked for time to think, and when the next day Seta appeared to the king, he surprised the king with the unparalleled modesty of his request. He asked to give off a grain of wheat for the first cell of the chessboard, for the second wheat grain, for the third, for the fourth, etc.

The king was angry and drove Seth away, saying that the servant's request was unworthy of the royal generosity, but promised that the servant would receive his grains for all the cells of the board.

And now the question: using the formula for the sum of the members of a geometric progression, calculate how many grains Seta should receive?

Let's start to reason. Since, according to the condition, Seth asked for a grain of wheat for the first square of the chessboard, for the second, for the third, for the fourth, etc., we see that the problem is about a geometric progression. What is equal in this case?
Right.

Total cells of the chessboard. Respectively, . We have all the data, it remains only to substitute it into the formula and calculate.

To represent at least approximately the "scales" of a given number, we transform using the properties of the degree:

Of course, if you want, you can take a calculator and calculate what number you will get in the end, but if not, you will have to take my word for it: the final value of the expression will be.
That is:

quintillion quadrillion trillion billion million thousand.

Fuh) If you want to imagine the enormity of this number, then estimate how large the barn would be required to contain the entire amount of grain.
With a barn height m and a width of m, its length would have to extend for km, i.e. twice as far as from the Earth to the Sun.

If the tsar were strong in mathematics, he could suggest that the scientist himself count the grains, because to count a million grains, he would need at least a day of tireless counting, and given that it is necessary to count quintillions, the grains would have to be counted all his life.

Now let's solve a simple problem for the sum of the members of a geometric progression.
A pupil of 5 A grade Vasya, got the flu, but continues to go to school. Every day Vasya infects two people, who, in turn, infect two more people, and so on. There are people in the class. How many days will the whole class get sick with the flu?

So, the first member of the geometric progression is Vasya, that is, a person. th member of the geometric progression, these are the two people whom he infected on the first day of his arrival. The total number of members in the progression is equal to the number of students 5A. Accordingly, we are talking about a progression in which:

Let's substitute our data into the formula for the sum of members of a geometric progression:

The whole class will get sick in days. Don't you believe in formulas and numbers? Try to portray the "infection" of the students yourself. Happened? See how it looks for me:

Calculate for yourself how many days it would take pupils to get the flu if each infect a person and there was a person in the class.

What value did you get? It turned out that everyone started to get sick after a day.

As you can see, such a task and drawing to it resembles a pyramid in which each subsequent one “brings” new people. However, sooner or later a moment comes when the latter cannot attract anyone. In our case, if we imagine that the class is isolated, the person from will close the chain (). Thus, if a person were involved in a financial pyramid, in which money was given in the event that you bring two other participants, then the person (or in the general case) would not bring anyone, respectively, they would lose everything that they invested in this financial scam.

Everything that has been said above refers to a decreasing or increasing geometric progression, but, as you remember, we have a special kind - an infinitely decreasing geometric progression. How to calculate the sum of its members? And why does this type of progression have certain features? Let's sort it out together.

So, first, let's look again at this figure of an infinitely decreasing geometric progression from our example:

Now let's look at the formula for the sum of a geometric progression, derived a little earlier:
or

What are we striving for? That's right, the graph shows that it tends to zero. That is, at, it will be almost equal, respectively, when calculating the expression, we get almost. In this regard, we believe that when calculating the sum of an infinitely decreasing geometric progression, this bracket can be neglected, since it will be equal.

- the formula is the sum of the terms of an infinitely decreasing geometric progression.

IMPORTANT! We use the formula for the sum of the terms of an infinitely decreasing geometric progression only if the condition explicitly states that we need to find the sum endless number of members.

If a specific number n is indicated, then we use the formula for the sum of n terms, even if or.

Now let's practice.

1. Find the sum of the first terms of a geometric progression with and.
2. Find the sum of the terms of an infinitely decreasing geometric progression with and.

I hope you were extremely attentive. Let's compare our answers:

Now you know everything about geometric progression, and it's time to move from theory to practice. The most common exponential problems encountered in the exam are compound interest problems. It is about them that we will talk.

Tasks for calculating compound interest.

You've probably heard of the so-called compound interest formula. Do you understand what she means? If not, let's figure it out, because having realized the process itself, you will immediately understand, and here is a geometric progression.

We all go to the bank and we know that there are different conditions on deposits: this is the term, and additional service, and interest with two different ways of calculating it - simple and complex.

WITH simple interest everything is more or less clear: interest is calculated once at the end of the term of the deposit. That is, if we say that we put 100 rubles for a year under, then they will be credited only at the end of the year. Accordingly, by the end of the deposit, we will receive rubles.

Compound interest- this is the option in which there is capitalization of interest, i.e. their addition to the amount of the deposit and the subsequent calculation of income not from the initial, but from the accumulated amount of the deposit. Capitalization does not occur constantly, but with some frequency. As a rule, such periods are equal and most often banks use a month, quarter or year.

Let's say that we put all the same rubles at annual rates, but with a monthly capitalization of the deposit. What do we get?

Do you understand everything here? If not, let's figure it out in stages.

We brought rubles to the bank. By the end of the month, our account should have an amount consisting of our rubles plus interest on them, that is:

Agree?

We can put it outside the bracket and then we get:

Agree, this formula is already more similar to the one we wrote at the beginning. It remains to deal with interest

In the problem statement, we are told about the annual. As you know, we do not multiply by - we convert interest into decimals, that is:

Right? Now you ask, where did the number come from? Very simple!
I repeat: the problem statement says about ANNUAL interest accrued MONTHLY... As you know, in a year of months, respectively, the bank will charge us a part of the annual interest per month:

Realized? Now try to write what this part of the formula will look like if I say that interest is calculated daily.
Did you manage? Let's compare the results:

Well done! Let's return to our task: write down how much will be credited to our account for the second month, taking into account that interest is charged on the accumulated amount of the deposit.
Here's what I got:

Or, in other words:

I think that you have already noticed a pattern and saw a geometric progression in all this. Write down what its member will be equal to, or, in other words, how much money we will receive at the end of the month.
Did? Checking!

As you can see, if you put money in the bank for a year at a simple interest, then you will receive rubles, and if at a complex rate - rubles. The benefit is small, but this happens only during the th year, but for a longer period, capitalization is much more profitable:

Let's consider another type of problems with compound interest. After what you figured out, it will be elementary for you. So the task:

The Zvezda company began investing in the industry in 2000, having capital in dollars. Every year since 2001, she earns a profit, which is from the capital of the previous year. How much profit will the Zvezda company receive at the end of 2003 if the profit has not been withdrawn from circulation?

Capital of the company "Zvezda" in 2000.
- the capital of the company "Zvezda" in 2001.
- the capital of the company "Zvezda" in 2002.
- the capital of the company "Zvezda" in 2003.

Or we can write briefly:

For our case:

2000, 2001, 2002 and 2003.

Respectively:
rubles
Note that in this problem we have no division either by or by, since the percentage is given ANNUALLY and it is calculated ANNUALLY. That is, when reading a problem for compound interest, pay attention to what percentage is given, and in what period it is charged, and only then proceed to the calculations.
Now you know everything about geometric progression.

Workout.

1. Find the exponential term if it is known that, and
2. Find the sum of the first terms of the geometric progression, if it is known that, and
3. MDM Capital started investing in the industry in 2003, having capital in dollars. Every year, starting in 2004, she earns a profit, which is from the capital of the previous year. The company "MSK Cash Flows" began to invest in the industry in 2005 in the amount of $ 10,000, starting to make a profit in 2006 in the amount of. How many dollars is the capital of one company more than another at the end of 2007, if the profit has not been withdrawn from circulation?

Answers:

1. Since the problem statement does not say that the progression is infinite and it is required to find the sum of a specific number of its members, the calculation is carried out according to the formula:
2. 
   MDM Capital:

   2003, 2004, 2005, 2006, 2007.
   - increases by 100%, that is, 2 times.
   Respectively:
   rubles
   MSK Cash Flows:

   2005, 2006, 2007.
   - increases by, that is, times.
   Respectively:
   rubles
   rubles

Let's summarize.

1) Geometric progression () is a numerical sequence, the first term of which is nonzero, and each term, starting from the second, is equal to the previous one, multiplied by the same number. This number is called the denominator of the geometric progression.

2) Equation of members of a geometric progression -.

3) can take any values, except for and.

• if, then all subsequent members of the progression have the same sign - they positive;
• if, then all subsequent members of the progression alternate signs;
• at - the progression is called infinitely decreasing.

4), for is the property of a geometric progression (adjacent terms)

or
, at (equidistant terms)

When finding, do not forget that there should be two answers.

For example,

5) The sum of the members of a geometric progression is calculated by the formula:
or

or

IMPORTANT! We use the formula for the sum of the terms of an infinitely decreasing geometric progression only if the condition explicitly states that it is necessary to find the sum of an infinite number of terms.

6) Problems for compound interest are also calculated according to the formula of the -th term of a geometric progression, provided that the funds have not been withdrawn from circulation:

GEOMETRIC PROGRESSION. BRIEFLY ABOUT THE MAIN

Geometric progression() is a numerical sequence, the first term of which is nonzero, and each term, starting from the second, is equal to the previous one, multiplied by the same number. This number is called the denominator of a geometric progression.

Denominator of geometric progression can take any values ​​except and.

• If, then all subsequent members of the progression have the same sign - they are positive;
• if, then all subsequent members of the progression alternate signs;
• at - the progression is called infinitely decreasing.

Equation of members of a geometric progression - .

The sum of the members of a geometric progression calculated by the formula:
or

If the progression is infinitely decreasing, then:

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Mathematics is wherebypeople control nature and themselves.

Soviet mathematician, academician A.N. Kolmogorov

Geometric progression.

Along with problems for arithmetic progressions, problems related to the concept of a geometric progression are also common in entrance examinations 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 basic properties of a geometric progression. It also provides examples of solving typical tasks., borrowed from the assignments of the entrance tests in mathematics.

First, we note the main properties of a geometric progression and recall the most important formulas and statements, related to 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 the geometric progression.

For a geometric progressionthe formulas are valid

, (1)

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

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

The above formulas (1) and (2) are generalized as follows:

, (3)

To calculate the amount the first members of a geometric progressionthe formula is applied

If we denote, then

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

In the case when and, geometric progressionis infinitely decreasing. To calculate the amountof 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, (first equality) and, (second equality).

Theorem. If, then

Proof. If, then,

The theorem is proved.

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

Example 1. Given:, and. Find .

Solution. If we apply formula (5), then

Answer: .

Example 2. Let and. Find .

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

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

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

2. If, then.

Example 3. Let, and. Find .

Solution. From formula (2) it follows that or. Since, then or.

By condition . However, therefore. Since and, then here we have the system of equations

If the second equation of the system is divided by the first, then or.

Since, then the equation has a single suitable root. In this case, it follows from the first equation of the system.

Taking into account formula (7), we obtain.

Answer: .

Example 4. Given: and. Find .

Solution. Since, then.

Since, then either

According to formula (2), we have. In this regard, from equality (10) we obtain or.

However, by condition, therefore.

Example 5. It is known that . Find .

Solution. According to the theorem, we have two equalities

Since, then or. Since, then.

Answer: .

Example 6. Given: and. Find .

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

Since, then. Since, and, then.

Example 7. Let and. 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 this and 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, you can write or.

From this we obtain the quadratic equation, whose roots are and .

Let's check if, then, and; if, then, and.

In the first case, we have and, and in the second - and.

Answer: , .

Example 10.Solve the equation

, (11)

where and.

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

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

Answer: .

Example 11. NS sequence of positive numbersforms an arithmetic progression, a - geometric progression, what does it have to do with . Find .

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

Since and, then ... In this case, the expression takes the form or. By condition , therefore from the equationwe obtain the unique solution of the considered problem, i.e. ...

Answer: .

Example 12. Calculate the amount

. (12)

Solution. We multiply both sides of equality (12) by 5 and obtain

If we subtract from the obtained expression (12), then

or .

To calculate, we substitute the values ​​in the formula (7), and we get. Since, then.

Answer: .

The examples of problem solving given here will be useful to applicants in preparation for the entrance examinations. For a deeper study of problem solving methods, exponentially related, can be used tutorials from the list of recommended literature.

1. Collection of problems in mathematics for applicants to technical colleges / Ed. M.I. Skanavi. - M .: Peace and Education, 2013. - 608 p.

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

3. Medynsky M.M. Complete course of elementary mathematics in problems and exercises. Book 2: Number sequences and progressions. - M .: Edithus, 2015 .-- 208 p.

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