The sum of the first 7 numbers of a geometric progression. Geometric progression. Example with solution

Geometric progression is a new type of number sequence that we are about to get acquainted with. For successful dating, it doesn’t hurt to at least know and understand. Then there will be no problems with geometric progression.)

What is geometric progression? The concept of geometric progression.

We start the tour, as usual, with the basics. I write an unfinished sequence of numbers:

1, 10, 100, 1000, 10000, …

Can you spot the pattern and tell which numbers will come next? The pepper is clear, then the numbers 100,000, 1,000,000 and so on will follow. Even without much mental effort, everything is clear, right?)

OK. Another example. I write this sequence:

1, 2, 4, 8, 16, …

Can you tell which numbers will come next, following the number 16, and name eighth sequence member? If you figured out that it would be the number 128, then very good. So, half the battle is in understanding sense And key points geometric progression has already been done. You can grow further.)

And now we move again from sensations to strict mathematics.

Key points of geometric progression.

Key Point #1

Geometric progression is sequence of numbers. So is progression. Nothing fancy. Only this sequence is arranged differently. Hence, naturally, it has a different name, yes...

Key Point #2

With the second key point, the question will be trickier. Let's go back a little and remember the key property of arithmetic progression. Here it is: each member is different from the previous one by the same amount.

Is it possible to formulate a similar key property for a geometric progression? Think a little... Take a closer look at the examples given. Did you guess it? Yes! In geometric progression (any!) each of its members differs from the previous one the same number of times. Always!

In the first example, this number is ten. Whichever member of the sequence you take, it is greater than the previous one ten times.

In the second example it is a two: each term is greater than the previous one twice.

It is this key point that geometric progression differs from arithmetic progression. In an arithmetic progression, each subsequent term is obtained by adding the same value to the previous term. And here - multiplication the previous term by the same amount. That's the whole difference.)

Key Point #3

This key point is completely identical to that for an arithmetic progression. Namely: Each member of a geometric progression stands in its place. Everything is exactly the same as in the arithmetic progression and comments, I think, are unnecessary. There is the first term, there is the hundred and first, etc. Let us swap at least two terms – the pattern (and with it the geometric progression) will disappear. What will remain is just a sequence of numbers without any logic.

That's all. That's the whole point of geometric progression.

Terms and designations.

But now, having understood the meaning and key points of geometric progression, we can move on to the theory. Otherwise, what is a theory without understanding the meaning, right?

How to denote geometric progression?

How is geometric progression written in general form? No problem! Each term of the progression is also written as a letter. Only for arithmetic progression, usually the letter is used "A", for geometric – letter "b". Member number, as usual, is indicated index at the bottom right. We simply list the members of the progression themselves, separated by commas or semicolons.

Like this:

b 1,b 2 , b 3 , b 4 , b 5 , b 6 , …

Briefly, this progression is written like this: (b n) .

Or like this, for finite progressions:

b 1, b 2, b 3, b 4, b 5, b 6.

b 1, b 2, …, b 29, b 30.

Or, in short:

(b n), n=30 .

That, in fact, is all the designation. Everything is the same, only the letter is different, yes.) And now we move directly to the definition.

Definition of geometric progression.

A geometric progression is a number sequence in which the first term is non-zero, and each subsequent term is equal to the previous term multiplied by the same non-zero number.

That's the whole definition. Most words and phrases are clear and familiar to you. If, of course, you understand the meaning of geometric progression “on your fingers” and in general. But there are also a few new phrases that I would like to pay special attention to.

First, the words: "the first member of which non-zero".

This restriction on the first term was not introduced by chance. What do you think will happen if the first member b 1 will be equal to zero? What will the second term be equal to if each term is greater than the previous one? the same number of times? Let's say three times? Let's see... Multiply the first term (i.e. 0) by 3 and get... zero! What about the third member? Also zero! And the fourth term is also zero! And so on…

We just get a bag of bagels, a sequence of zeros:

0, 0, 0, 0, …

Of course, such a sequence has a right to life, but it is of no practical interest. Everything is clear. Any member of it is zero. The sum of any number of terms is also zero... What interesting things can you do with it? Nothing…

The following keywords: "multiplied by the same non-zero number."

This same number also has its own special name - denominator of geometric progression. Let's start getting acquainted.)

Denominator of a geometric progression.

Everything is as simple as shelling pears.

The denominator of a geometric progression is a non-zero number (or quantity) indicating how many timeseach term of the progression more than the previous one.

Again, similar to the arithmetic progression, the key word to look for in this definition is the word "more". It means that each term of the geometric progression is obtained multiplication to this very denominator previous member.

Let me explain.

To calculate, let's say second dick, need to take first member and multiply it to the denominator. For calculation tenth dick, need to take ninth member and multiply it to the denominator.

The denominator of the geometric progression itself can be anything. Absolutely anyone! Whole, fractional, positive, negative, irrational - everything. Except zero. This is what the word “non-zero” in the definition tells us. Why this word is needed here - more on that later.

Denominator of geometric progression most often indicated by the letter q.

How to find it q? No problem! We must take any term of the progression and divide by the previous term. Division is fraction. Hence the name - “progression denominator”. The denominator, it usually sits in a fraction, yes...) Although, logically, the value q should be called private geometric progression, similar to difference for arithmetic progression. But we agreed to call denominator. And we won’t reinvent the wheel either.)

Let us define, for example, the quantity q for this geometric progression:

2, 6, 18, 54, …

Everything is elementary. Let's take it any sequence number. We take whatever we want. Except the very first one. For example, 18. And divide by previous number. That is, at 6.

We get:

q = 18/6 = 3

That's all. This is the correct answer. For this geometric progression, the denominator is three.

Let's now find the denominator q for another geometric progression. For example, this one:

1, -2, 4, -8, 16, …

All the same. No matter what signs the members themselves have, we still take any number of the sequence (for example, 16) and divide by previous number(i.e. -8).

We get:

d = 16/(-8) = -2

And that's it.) This time the denominator of the progression turned out to be negative. Minus two. Happens.)

Let's now take this progression:

1, 1/3, 1/9, 1/27, …

And again, regardless of the type of numbers in the sequence (whether integers, even fractions, even negative, even irrational), we take any number (for example, 1/9) and divide by the previous number (1/3). According to the rules for working with fractions, of course.

We get:

That's all.) Here the denominator turned out to be fractional: q = 1/3.

What do you think of this “progression”?

3, 3, 3, 3, 3, …

Obviously here q = 1 . Formally, this is also a geometric progression, only with identical members.) But such progressions are not interesting for study and practical application. The same as progressions with solid zeros. Therefore, we will not consider them.

As you can see, the denominator of the progression can be anything - integer, fractional, positive, negative - anything! It can't just be zero. Can't guess why?

Well, let's use some specific example to see what will happen if we take as the denominator q zero.) Let us, for example, have b 1 = 2 , A q = 0 . What then will the second term be equal to?

We count:

b 2 = b 1 · q= 2 0 = 0

What about the third member?

b 3 = b 2 · q= 0 0 = 0

Types and behavior of geometric progressions.

Everything was more or less clear: if the progression difference d is positive, then the progression increases. If the difference is negative, then the progression decreases. There are only two options. There is no third.)

But with the behavior of geometric progression, everything will be much more interesting and varied!)

No matter how the terms behave here: they increase, and decrease, and indefinitely approach zero, and even change signs, alternately throwing themselves into “plus” and then into “minus”! And in all this diversity you need to be able to understand well, yes...

Let's figure it out?) Let's start with the simplest case.

The denominator is positive ( q >0)

With a positive denominator, firstly, the terms of the geometric progression can go into plus infinity(i.e. increase without limit) and can go into minus infinity(i.e., decrease without limit). We are already accustomed to this behavior of progressions.

For example:

(b n): 1, 2, 4, 8, 16, …

Everything is simple here. Each term of the progression is obtained more than previous. Moreover, each term turns out multiplication previous member on positive number +2 (i.e. q = 2 ). The behavior of such a progression is obvious: all members of the progression grow without limit, going into space. Plus infinity...

And now here's the progression:

(b n): -1, -2, -4, -8, -16, …

Here, too, each term of the progression is obtained multiplication previous member on positive number +2. But the behavior of such a progression is exactly the opposite: each term of the progression is obtained less than previous, and all its terms decrease without limit, going to minus infinity.

Now let's think: what do these two progressions have in common? That's right, denominator! Here and there q = +2 . Positive number. Two. And here behavior These two progressions are fundamentally different! Can't guess why? Yes! It's all about first member! It is he, as they say, who calls the tune.) See for yourself.

In the first case, the first term of the progression positive(+1) and, therefore, all subsequent terms obtained by multiplying by positive denominator q = +2 , will also be positive.

But in the second case, the first term negative(-1). Therefore, all subsequent terms of the progression, obtained by multiplying by positive q = +2 , will also be obtained negative. Because “minus” to “plus” always gives “minus”, yes.)

As you can see, unlike an arithmetic progression, a geometric progression can behave completely differently not only depending from the denominatorq, but also depending from the first member, Yes.)

Remember: the behavior of a geometric progression is uniquely determined by its first term b 1 and denominatorq .

And now we begin to analyze less familiar, but much more interesting cases!

Let's take, for example, this sequence:

(b n): 1, 1/2, 1/4, 1/8, 1/16, …

This sequence is also a geometric progression! Each term of this progression also turns out multiplication the previous member, by the same number. It's just a number - fractional: q = +1/2 . Or +0,5 . Moreover (important!) the number less than one:q = 1/2<1.

Why is this geometric progression interesting? Where are its members heading? Let's get a look:

1/2 = 0,5;

1/4 = 0,25;

1/8 = 0,125;

1/16 = 0,0625;

…….

What interesting things can you notice here? Firstly, the decrease in terms of the progression is immediately noticeable: each of its members less the previous one exactly 2 times. Or, according to the definition of a geometric progression, each term more previous 1/2 times, because progression denominator q = 1/2 . And when multiplied by a positive number less than one, the result usually decreases, yes...

What more can be seen in the behavior of this progression? Are its members diminishing? unlimited, going to minus infinity? No! They disappear in a special way. At first they decrease quite quickly, and then more and more slowly. And while remaining all the time positive. Albeit very, very small. And what do they themselves strive for? Didn't you guess? Yes! They strive towards zero!) Moreover, pay attention, the members of our progression are from zero never reach! Only approaching him infinitely close. It is very important.)

A similar situation will occur in the following progression:

(b n): -1, -1/2, -1/4, -1/8, -1/16, …

Here b 1 = -1 , A q = 1/2 . Everything is the same, only now the terms will approach zero from the other side, from below. Staying all the time negative.)

Such a geometric progression, the terms of which approach zero without limit(no matter from the positive or negative side), in mathematics has a special name - infinitely decreasing geometric progression. This progression is so interesting and unusual that it will even be discussed separate lesson .)

So, we have considered all possible positive the denominators are both large ones and smaller ones. We do not consider the unit itself as a denominator for the reasons stated above (remember the example with a sequence of triplets...)

Let's summarize:

positiveAnd more than one (q>1), then the terms of the progression:

a) increase without limit (ifb 1 >0);

b) decrease without limit (ifb 1 <0).

If the denominator of the geometric progression positive And less than one (0< q<1), то члены прогрессии:

a) infinitely close to zero above(Ifb 1 >0);

b) approaching infinitely close to zero from below(Ifb 1 <0).

It now remains to consider the case negative denominator.

Denominator is negative ( q <0)

We won’t go far for an example. Why, exactly, shaggy grandma?!) Let, for example, the first term of the progression be b 1 = 1 , and let’s take the denominator q = -2.

We get the following sequence:

(b n): 1, -2, 4, -8, 16, …

And so on.) Each term of the progression is obtained multiplication previous member on a negative number-2. In this case, all members standing in odd places (first, third, fifth, etc.) will be positive, and in even places (second, fourth, etc.) – negative. The signs strictly alternate. Plus-minus-plus-minus... This geometric progression is called - increasing sign alternating.

Where are its members heading? But nowhere.) Yes, in absolute value (i.e. modulo) the members of our progression increase without limit (hence the name “increasing”). But at the same time, each member of the progression alternately throws you into the heat, then into the cold. Either “plus” or “minus”. Our progression is wavering... Moreover, the scope of fluctuations is growing rapidly with each step, yes.) Therefore, the aspirations of the members of the progression are going somewhere specifically Here No. Neither to plus infinity, nor to minus infinity, nor to zero - nowhere.

Let us now consider some fractional denominator between zero and minus one.

For example, let it be b 1 = 1 , A q = -1/2.

Then we get the progression:

(b n): 1, -1/2, 1/4, -1/8, 1/16, …

And again we have an alternation of signs! But, unlike the previous example, here there is already a clear tendency for the terms to approach zero.) Only this time our terms approach zero not strictly from above or below, but again hesitating. Alternatingly taking positive and negative values. But at the same time they modules are getting closer and closer to the cherished zero.)

This geometric progression is called infinitely decreasing sign, alternating.

Why are these two examples interesting? And the fact that in both cases takes place alternation of signs! This trick is typical only for progressions with a negative denominator, yes.) Therefore, if in some task you see a geometric progression with alternating terms, you will already know for sure that its denominator is 100% negative and you will not make a mistake in the sign.)

By the way, in the case of a negative denominator, the sign of the first term does not at all affect the behavior of the progression itself. Regardless of the sign of the first term of the progression, in any case the sign of the terms will be observed. The only question is, in what places(even or odd) there will be members with specific signs.

Remember:

If the denominator of the geometric progression negative , then the signs of the terms of the progression are always alternate.

At the same time, the members themselves:

a) increase without limitmodulo, Ifq<-1;

b) approach zero infinitely if -1< q<0 (прогрессия бесконечно убывающая).

That's all. All typical cases have been analyzed.)

In the process of analyzing a variety of examples of geometric progressions, I periodically used the words: "tends to zero", "tends to plus infinity", "tends to minus infinity"... It's okay.) These figures of speech (and specific examples) are just an initial introduction to behavior a variety of number sequences. Using the example of geometric progression.

Why do we even need to know the behavior of progression? What difference does it make where she goes? Toward zero, to plus infinity, to minus infinity... What does that do to us?

The thing is that already at the university, in a course of higher mathematics, you will need the ability to work with a wide variety of numerical sequences (with any, not just progressions!) and the ability to imagine exactly how this or that sequence behaves - whether it increases whether it decreases unlimitedly, whether it tends to a specific number (and not necessarily to zero), or even does not tend to anything at all... An entire section is devoted to this topic in the course of mathematical analysis - theory of limits. And a little more specifically - the concept limit of the number sequence. A very interesting topic! It makes sense to go to college and figure it out.)

Some examples from this section (sequences having a limit) and in particular, infinitely decreasing geometric progression They begin to get used to it at school. We're getting used to it.)

Moreover, the ability to study well the behavior of sequences will greatly benefit you in the future and will be very useful in function research. The most diverse. But the ability to competently work with functions (calculate derivatives, study them in full, build their graphs) already dramatically increases your mathematical level! Do you have any doubts? No need. Also remember my words.)

Let's look at the geometric progression in life?

In the life around us, we encounter geometric progression very, very often. Even without even knowing it.)

For example, various microorganisms that surround us everywhere in huge quantities and which we cannot even see without a microscope multiply precisely in geometric progression.

Let's say one bacterium reproduces by dividing in half, giving offspring into 2 bacteria. In turn, each of them, when multiplying, also divides in half, giving a common offspring of 4 bacteria. The next generation will produce 8 bacteria, then 16 bacteria, 32, 64 and so on. With each subsequent generation, the number of bacteria doubles. A typical example of a geometric progression.)

Also, some insects – aphids and flies – multiply exponentially. And sometimes rabbits too, by the way.)

Another example of a geometric progression, closer to everyday life, is the so-called compound interest. This interesting phenomenon is often found in bank deposits and is called capitalization of interest. What it is?

You yourself are still, of course, young. You study at school, you don’t go to banks. But your parents are already adults and independent people. They go to work, earn money for their daily bread, and put part of the money in the bank, making savings.)

Let's say your dad wants to save up a certain amount of money for a family vacation in Turkey and puts 50,000 rubles in the bank at 10% per annum for a period of three years with annual interest capitalization. Moreover, during this entire period nothing can be done with the deposit. You can neither replenish the deposit nor withdraw money from the account. How much profit will he make after these three years?

Well, first of all, we need to figure out what 10% per annum is. It means that in a year The bank will add 10% to the initial deposit amount. From what? Of course, from initial deposit amount.

We calculate the size of the account after a year. If the initial deposit amount was 50,000 rubles (i.e. 100%), then after a year there will be how much interest on the account? That's right, 110%! From 50,000 rubles.

So we calculate 110% of 50,000 rubles:

50000·1.1 = 55000 rubles.

I hope you understand that finding 110% of a value means multiplying that value by the number 1.1? If you don’t understand why this is so, remember fifth and sixth grades. Namely – connection between percentages and fractions and parts.)

Thus, the increase for the first year will be 5,000 rubles.

How much money will be in the account in two years? 60,000 rubles? Unfortunately (or rather, fortunately), everything is not so simple. The whole trick of interest capitalization is that with each new interest accrual, these same interests will be considered already from the new amount! From the one who already is on the account At the moment. And the interest accrued for the previous period is added to the original deposit amount and, thus, itself participates in the calculation of new interest! That is, they become a full part of the overall account. Or general capital. Hence the name - capitalization of interest.

It's in economics. And in mathematics such percentages are called compound interest. Or percentage of interest.) Their trick is that when calculating sequentially, the percentages are calculated each time from the new value. And not from the original...

Therefore, to calculate the amount through two years, we need to calculate 110% of the amount that will be in the account in a year. That is, already from 55,000 rubles.

We count 110% of 55,000 rubles:

55000·1.1 = 60500 rubles.

This means that the percentage increase for the second year will be 5,500 rubles, and for two years – 10,500 rubles.

Now you can already guess that after three years the amount in the account will be 110% of 60,500 rubles. That is again 110% from the previous one (last year) amounts.

Here we think:

60500·1.1 = 66550 rubles.

Now we arrange our monetary amounts by year in sequence:

50000;

55000 = 50000 1.1;

60500 = 55000 1.1 = (50000 1.1) 1.1;

66550 = 60500 1.1 = ((50000 1.1) 1.1) 1.1

So how is it? Why not a geometric progression? First member b 1 = 50000 , and the denominator q = 1,1 . Each term is strictly 1.1 times larger than the previous one. Everything is in strict accordance with the definition.)

And how many additional interest bonuses will your dad “accumulate” while his 50,000 rubles have been lying in his bank account for three years?

We count:

66550 – 50000 = 16550 rubles

Not much, of course. But this is if the initial deposit amount is small. What if there is more? Let's say, not 50, but 200 thousand rubles? Then the increase over three years will be 66,200 rubles (if you do the math). Which is already very good.) What if the contribution is even greater? That's it...

Conclusion: the higher the initial deposit, the more profitable the interest capitalization becomes. That is why deposits with interest capitalization are provided by banks for long periods. Let's say for five years.

Also, all sorts of bad diseases like influenza, measles and even more terrible diseases (the same SARS in the early 2000s or the plague in the Middle Ages) like to spread exponentially. Hence the scale of epidemics, yes...) And all due to the fact that the geometric progression with whole positive denominator (q>1) – a thing that grows very quickly! Remember the reproduction of bacteria: from one bacteria two are obtained, from two - four, from four - eight, and so on... It’s the same with the spread of any infection.)

The simplest problems on geometric progression.

Let's start, as always, with a simple problem. Purely to understand the meaning.

1. It is known that the second term of the geometric progression is equal to 6, and the denominator is equal to -0.5. Find the first, third and fourth terms.

So we are given endless geometric progression, but known second term this progression:

b 2 = 6

In addition, we also know progression denominator:

q = -0.5

And you need to find first, third And fourth members of this progression.

So we act. We write down the sequence according to the conditions of the problem. Directly in general form, where the second term is six:

b 1, 6,b 3 , b 4 , …

Now let's start searching. We start, as always, with the simplest. You can calculate, for example, the third term b 3? Can! You and I already know (directly in the sense of geometric progression) that the third term (b 3) more than the second (b 2 ) V "q" once!

So we write:

b 3 =b 2 · q

We substitute six into this expression instead of b 2 and -0.5 instead q and we count. And we don’t ignore the minus either, of course...

b 3 = 6·(-0.5) = -3

Like this. The third term turned out to be negative. No wonder: our denominator q– negative. And multiplying a plus by a minus will, of course, be a minus.)

Now we count the next, fourth term of the progression:

b 4 =b 3 · q

b 4 = -3·(-0.5) = 1.5

The fourth term is again with a plus. The fifth term will again be minus, the sixth will be plus, and so on. The signs alternate!

So, the third and fourth terms were found. The result is the following sequence:

b 1 ; 6; -3; 1.5; ...

Now all that remains is to find the first term b 1 according to the well-known second. To do this, we step in the other direction, to the left. This means that in this case we do not need to multiply the second term of the progression by the denominator, but divide.

We divide and get:

That's all.) The answer to the problem will be like this:

-12; 6; -3; 1,5; …

As you can see, the solution principle is the same as in . We know any member and denominator geometric progression - we can find any other member of it. We’ll find the one we want.) The only difference is that addition/subtraction is replaced by multiplication/division.

Remember: if we know at least one member and denominator of a geometric progression, then we can always find any other member of this progression.

The following problem, according to tradition, is from a real version of the OGE:

2.

...; 150; X; 6; 1.2; ...

So how is it? This time there is no first term, no denominator q, just a sequence of numbers is given... Something already familiar, right? Yes! A similar problem has already been solved in arithmetic progression!

So we are not afraid. All the same. Let's turn on our heads and remember the elementary meaning of geometric progression. We look carefully at our sequence and figure out which parameters of the geometric progression of the three main ones (first term, denominator, term number) are hidden in it.

Member numbers? There are no membership numbers, yes... But there are four consecutive numbers. I don’t see any point in explaining what this word means at this stage.) Are there two neighboring known numbers? Eat! These are 6 and 1.2. So we can find progression denominator. So we take the number 1.2 and divide to the previous number. To six.

We get:

We get:

x= 150·0.2 = 30

Answer: x = 30 .

As you can see, everything is quite simple. The main difficulty is only in the calculations. It is especially difficult in the case of negative and fractional denominators. So those who have problems, repeat the arithmetic! How to work with fractions, how to work with negative numbers, and so on... Otherwise, you will slow down mercilessly here.

Now let's modify the problem a little. Now it's going to get interesting! Let's remove the last number 1.2 from it. Now let's solve this problem:

3. Several consecutive terms of the geometric progression are written out:

...; 150; X; 6; ...

Find the term of the progression indicated by the letter x.

Everything is the same, only two adjacent famous We now have no members of the progression. This is the main problem. Because the magnitude q through two neighboring terms we can easily determine we can't. Do we have a chance to cope with the task? Certainly!

Let's write down the unknown term " x"directly within the meaning of geometric progression! In general terms.

Yes Yes! Right with an unknown denominator!

On the one hand, for X we can write the following ratio:

x= 150·q

On the other hand, we have every right to describe this same X through next member, through the six! Divide six by the denominator.

Like this:

x = 6/ q

Obviously, now we can equate both of these ratios. Since we are expressing the same magnitude (x), but two different ways.

We get the equation:

Multiplying everything by q, simplifying and shortening, we get the equation:

q2 = 1/25

We solve and get:

q = ±1/5 = ±0.2

Oops! The denominator turned out to be double! +0.2 and -0.2. And which one should you choose? Dead end?

Calm! Yes, the problem really has two solutions! Nothing wrong with that. It happens.) You're not surprised when, for example, you get two roots when solving the usual problem? It's the same story here.)

For q = +0.2 we will get:

X = 150 0.2 = 30

And for q = -0,2 will:

X = 150·(-0.2) = -30

We get a double answer: x = 30; x = -30.

What does this interesting fact mean? And what exists two progressions, satisfying the conditions of the problem!

Like these ones:

…; 150; 30; 6; …

…; 150; -30; 6; …

Both are suitable.) Why do you think we had a split in answers? Just because of the elimination of a specific member of the progression (1,2), coming after six. And knowing only the previous (n-1)th and subsequent (n+1)th terms of the geometric progression, we can no longer say anything unambiguously about the nth term standing between them. There are two options – with a plus and with a minus.

But no problem. As a rule, in geometric progression tasks there is additional information that gives an unambiguous answer. Let's say the words: "alternating progression" or "progression with a positive denominator" and so on... It is these words that should serve as a clue as to which sign, plus or minus, should be chosen when preparing the final answer. If there is no such information, then yes, the task will have two solutions.)

Now we decide for ourselves.

4. Determine whether the number 20 is a member of a geometric progression:

4 ; 6; 9; …

5. The sign of an alternating geometric progression is given:

…; 5; x ; 45; …

Find the term of the progression indicated by the letter x .

6. Find the fourth positive term of the geometric progression:

625; -250; 100; …

7. The second term of the geometric progression is equal to -360, and its fifth term is equal to 23.04. Find the first term of this progression.

Answers (in disorder): -15; 900; No; 2.56.

Congratulations if everything worked out!

Something doesn't fit? Somewhere there was a double answer? Read the terms of the assignment carefully!

The last problem doesn't work out? There is nothing complicated there.) We work directly according to the meaning of geometric progression. Well, you can draw a picture. It helps.)

As you can see, everything is elementary. If the progression is short. What if it’s long? Or is the number of the required member very large? I would like, by analogy with the arithmetic progression, to somehow obtain a convenient formula that makes it easy to find any term of any geometric progression by his number. Without multiplying many, many times by q. And there is such a formula!) Details are in the next lesson.

Let's consider a certain series.

7 28 112 448 1792...

It is absolutely clear that the value of any of its elements is exactly four times greater than the previous one. This means that this series is a progression.

A geometric progression is an infinite sequence of numbers, the main feature of which is that the next number is obtained from the previous one by multiplying by a specific number. This is expressed by the following formula.

a z +1 =a z ·q, where z is the number of the selected element.

Accordingly, z ∈ N.

The period when geometric progression is studied at school is 9th grade. Examples will help you understand the concept:

0.25 0.125 0.0625...

Based on this formula, the denominator of the progression can be found as follows:

Neither q nor b z can be zero. Also, each of the elements of the progression should not be equal to zero.

Accordingly, to find out the next number in a series, you need to multiply the last one by q.

To set this progression, you must specify its first element and denominator. After this, it is possible to find any of the subsequent terms and their sum.

Varieties

Depending on q and a 1, this progression is divided into several types:

• If both a 1 and q are greater than one, then such a sequence is a geometric progression increasing with each subsequent element. An example of this is presented below.

Example: a 1 =3, q=2 - both parameters are greater than one.

Then the number sequence can be written like this:

3 6 12 24 48 ...

• If |q| is less than one, that is, multiplication by it is equivalent to division, then a progression with similar conditions is a decreasing geometric progression. An example of this is presented below.

Example: a 1 =6, q=1/3 - a 1 is greater than one, q is less.

Then the number sequence can be written as follows:

6 2 2/3 ... - any element is 3 times larger than the element following it.

• Alternating sign. If q<0, то знаки у чисел последовательности постоянно чередуются вне зависимости от a 1 , а элементы ни возрастают, ни убывают.

Example: a 1 = -3, q = -2 - both parameters are less than zero.

Then the number sequence can be written like this:

3, 6, -12, 24,...

Formulas

There are many formulas for convenient use of geometric progressions:

• Z-term formula. Allows you to calculate an element under a specific number without calculating previous numbers.

Example:q = 3, a 1 = 4. It is required to count the fourth element of the progression.

Solution:a 4 = 4 · 3 4-1 = 4 · 3 3 = 4 · 27 = 108.

• The sum of the first elements whose quantity is equal to z. Allows you to calculate the sum of all elements of a sequence up toa zinclusive.

Since (1-q) is in the denominator, then (1 - q)≠ 0, therefore q is not equal to 1.

Note: if q=1, then the progression would be a series of infinitely repeating numbers.

Sum of geometric progression, examples:a 1 = 2, q= -2. Calculate S5.

Solution:S 5 = 22 - calculation using the formula.

• Amount if |q| < 1 и если z стремится к бесконечности.

Example:a 1 = 2 , q= 0.5. Find the amount.

Solution:Sz = 2 · = 4

Sz = 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 3.9375 4

Some properties:

• Characteristic property. If the following condition works for anyz, then the given number series is a geometric progression:

a z 2 = a z -1 · az+1

• Also, the square of any number in a geometric progression is found by adding the squares of any two other numbers in a given series, if they are equidistant from this element.

a z 2 = a z - t 2 + a z + t 2 , Wheret- the distance between these numbers.

• Elementsdiffer in qonce.
• The logarithms of the elements of a progression also form a progression, but an arithmetic one, that is, each of them is greater than the previous one by a certain number.

Examples of some classic problems

To better understand what a geometric progression is, examples with solutions for class 9 can help.

• Conditions:a 1 = 3, a 3 = 48. Findq.

Solution: each subsequent element is greater than the previous one inq once.It is necessary to express some elements in terms of others using a denominator.

Hence,a 3 = q 2 · a 1

When substitutingq= 4

• Conditions:a 2 = 6, a 3 = 12. Calculate S 6.

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

a 3 = q· a 2 , hence,q= 2

a 2 = q · a 1 ,That's why a 1 = 3

S 6 = 189

• · a 1 = 10, q= -2. Find the fourth element of the progression.

Solution: to do this, it is enough to express the fourth element through the first and through the denominator.

a 4 = q 3· a 1 = -80

Application example:

• A bank client made a deposit in the amount of 10,000 rubles, under the terms of which every year the client will have 6% of it added to the principal amount. How much money will be in the account after 4 years?

Solution: The initial amount is 10 thousand rubles. This means that a year after the investment the account will have an amount equal to 10,000 + 10,000 · 0.06 = 10000 1.06

Accordingly, the amount in the account after another year will be expressed as follows:

(10000 · 1.06) · 0.06 + 10000 · 1.06 = 1.06 · 1.06 · 10000

That is, every year the amount increases by 1.06 times. This means that to find the amount of funds in the account after 4 years, it is enough to find the fourth element of the progression, which is given by the first element equal to 10 thousand and the denominator equal to 1.06.

S = 1.06 1.06 1.06 1.06 10000 = 12625

Examples of sum calculation problems:

Geometric progression is used in various problems. An example for finding the sum can be given as follows:

a 1 = 4, q= 2, calculateS 5.

Solution: all the data necessary for the calculation is known, you just need to substitute them into the formula.

S 5 = 124

• a 2 = 6, a 3 = 18. Calculate the sum of the first six elements.

Solution:

In geom. progression, each next element is q times greater than the previous one, that is, to calculate the sum you need to know the elementa 1 and denominatorq.

a 2 · q = a 3

q = 3

Similarly, you need to finda 1 , knowinga 2 Andq.

a 1 · q = a 2

a 1 =2

S 6 = 728.

Geometric progression, along with arithmetic progression, is an important number series that is studied in the school algebra course in the 9th grade. In this article we will look at the denominator of a geometric progression and how its value affects its properties.

Definition of geometric progression

First, let's give the definition of this number series. A geometric progression is a series of rational numbers that is formed by sequentially multiplying its first element by a constant number called the denominator.

For example, the numbers in the series 3, 6, 12, 24, ... are a geometric progression, because if you multiply 3 (the first element) by 2, you get 6. If you multiply 6 by 2, you get 12, and so on.

The members of the sequence under consideration are usually denoted by the symbol ai, where i is an integer indicating the number of the element in the series.

The above definition of progression can be written in mathematical language as follows: an = bn-1 * a1, where b is the denominator. It's easy to check this formula: if n = 1, then b1-1 = 1, and we get a1 = a1. If n = 2, then an = b * a1, and we again come to the definition of the series of numbers in question. Similar reasoning can be continued for large values ​​of n.

Denominator of geometric progression

The number b completely determines what character the entire number series will have. The denominator b can be positive, negative, or greater than or less than one. All of the above options lead to different sequences:

• b > 1. There is an increasing series of rational numbers. For example, 1, 2, 4, 8, ... If element a1 is negative, then the entire sequence will increase only in absolute value, but decrease depending on the sign of the numbers.
• b = 1. Often this case is not called a progression, since there is an ordinary series of identical rational numbers. For example, -4, -4, -4.

Formula for amount

Before moving on to the consideration of specific problems using the denominator of the type of progression under consideration, an important formula for the sum of its first n elements should be given. The formula looks like: Sn = (bn - 1) * a1 / (b - 1).

You can obtain this expression yourself if you consider the recursive sequence of terms of the progression. Also note that in the above formula it is enough to know only the first element and the denominator to find the sum of an arbitrary number of terms.

Infinitely decreasing sequence

An explanation was given above of what it is. Now, knowing the formula for Sn, let's apply it to this number series. Since any number whose modulus does not exceed 1 tends to zero when raised to large powers, that is, b∞ => 0 if -1

Since the difference (1 - b) will always be positive, regardless of the value of the denominator, the sign of the sum of an infinitely decreasing geometric progression S∞ is uniquely determined by the sign of its first element a1.

Now let's look at several problems where we will show how to apply the acquired knowledge on specific numbers.

Task No. 1. Calculation of unknown elements of progression and sum

Given a geometric progression, the denominator of the progression is 2, and its first element is 3. What will its 7th and 10th terms be equal to, and what is the sum of its seven initial elements?

The condition of the problem is quite simple and involves the direct use of the above formulas. So, to calculate element number n, we use the expression an = bn-1 * a1. For the 7th element we have: a7 = b6 * a1, substituting the known data, we get: a7 = 26 * 3 = 192. We do the same for the 10th term: a10 = 29 * 3 = 1536.

Let's use the well-known formula for the sum and determine this value for the first 7 elements of the series. We have: S7 = (27 - 1) * 3 / (2 - 1) = 381.

Problem No. 2. Determining the sum of arbitrary elements of a progression

Let -2 be equal to the denominator of the geometric progression bn-1 * 4, where n is an integer. It is necessary to determine the sum from the 5th to the 10th element of this series, inclusive.

The problem posed cannot be solved directly using known formulas. It can be solved using 2 different methods. For completeness of presentation of the topic, we present both.

Method 1. The idea is simple: you need to calculate the two corresponding sums of the first terms, and then subtract the other from one. We calculate the smaller amount: S10 = ((-2)10 - 1) * 4 / (-2 - 1) = -1364. Now we calculate the larger sum: S4 = ((-2)4 - 1) * 4 / (-2 - 1) = -20. Note that in the last expression only 4 terms were summed, since the 5th is already included in the amount that needs to be calculated according to the conditions of the problem. Finally, we take the difference: S510 = S10 - S4 = -1364 - (-20) = -1344.

Method 2. Before substituting numbers and counting, you can obtain a formula for the sum between the m and n terms of the series in question. We do exactly the same as in method 1, only we first work with the symbolic representation of the amount. We have: Snm = (bn - 1) * a1 / (b - 1) - (bm-1 - 1) * a1 / (b - 1) = a1 * (bn - bm-1) / (b - 1). You can substitute known numbers into the resulting expression and calculate the final result: S105 = 4 * ((-2)10 - (-2)4) / (-2 - 1) = -1344.

Problem No. 3. What is the denominator?

Let a1 = 2, find the denominator of the geometric progression, provided that its infinite sum is 3, and it is known that this is a decreasing series of numbers.

Based on the conditions of the problem, it is not difficult to guess which formula should be used to solve it. Of course, for the sum of the progression infinitely decreasing. We have: S∞ = a1 / (1 - b). From where we express the denominator: b = 1 - a1 / S∞. It remains to substitute the known values ​​and get the required number: b = 1 - 2 / 3 = -1 / 3 or -0.333(3). We can qualitatively check this result if we remember that for this type of sequence the modulus b should not go beyond 1. As can be seen, |-1 / 3|

Task No. 4. Restoring a series of numbers

Let 2 elements of a number series be given, for example, the 5th is equal to 30 and the 10th is equal to 60. It is necessary to reconstruct the entire series from these data, knowing that it satisfies the properties of a geometric progression.

To solve the problem, you must first write down the corresponding expression for each known term. We have: a5 = b4 * a1 and a10 = b9 * a1. Now divide the second expression by the first, we get: a10 / a5 = b9 * a1 / (b4 * a1) = b5. From here we determine the denominator by taking the fifth root of the ratio of the terms known from the problem statement, b = 1.148698. We substitute the resulting number into one of the expressions for the known element, we get: a1 = a5 / b4 = 30 / (1.148698)4 = 17.2304966.

Thus, we found the denominator of the progression bn, and the geometric progression bn-1 * 17.2304966 = an, where b = 1.148698.

Where are geometric progressions used?

If there were no practical application of this number series, then its study would be reduced to purely theoretical interest. But such an application exists.

Below are the 3 most famous examples:

• Zeno's paradox, in which the nimble Achilles cannot catch up with the slow tortoise, is solved using the concept of an infinitely decreasing sequence of numbers.
• If you place wheat grains on each square of the chessboard so that on the 1st square you put 1 grain, on the 2nd - 2, on the 3rd - 3, and so on, then to fill all the squares of the board you will need 18446744073709551615 grains!
• In the game "Tower of Hanoi", in order to move disks from one rod to another, it is necessary to perform 2n - 1 operations, that is, their number grows exponentially with the number n of disks used.

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

You can write any numbers, and there can be as many of them as you like (in our case, there are them). No matter how many numbers we write, we can always say which one is first, which one is second, and so on until 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 the same.

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

We usually call the entire sequence by 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 topic we will talk about the second type - geometric progression.

Why is geometric progression needed and its history?

Even in ancient times, the Italian mathematician monk Leonardo of Pisa (better known as Fibonacci) dealt with the practical needs of trade. The monk was faced with the task of determining what is the smallest number of weights that can be used to weigh a product? In his works, Fibonacci proves that such a system of weights is optimal: This is one of the first situations in which people had to deal with a geometric progression, which you have probably already heard about and have at least a general understanding of. Once you fully understand the topic, think about why such a system is optimal?

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

There are many more simple cases where geometric progression is applied. For example, the spread of influenza: one person infected another 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, a 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 number sequence:

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

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

This type of number sequence is called geometric progression and is designated.

Geometric progression () is a numerical sequence, the first term of which is different from zero, 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.

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

Agree that this is no longer a progression.

As you understand, we will get the same results if there is any number other than zero, a. In these cases, there will simply be no progression, since the entire number series will either be all zeros, or one number, and all the rest will be zeros.

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

Let's repeat: - this is the number how many times does each subsequent term change? geometric progression.

What do you think it could be? That's right, positive and negative, but not zero (we talked about this a little higher).

Let's assume that ours is positive. Let in our case, a. What is the value of the second term and? You can easily answer that:

That's right. Accordingly, if, then all subsequent terms of the progression have the same sign - they are positive.

What if it's negative? For example, a. What is the value of the second term and?

This is a completely different story

Try to count the terms of this progression. How much did you get? I have. Thus, if, then the signs of the terms of the geometric progression alternate. That is, if you see a progression with alternating signs for 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 an arithmetic progression:

Got it? Let's compare our answers:

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

Let's return to our last progression and try to find its member, just like in the arithmetic one. As you may have guessed, there are two ways to find it.

We successively multiply each term by.

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

As you already guessed, now you yourself will derive a formula that will help you find any member of the geometric progression. Or have you already developed it 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 with the example of finding the th term of this progression:

In other words:

Find the value of the term of the given geometric progression yourself.

Happened? Let's compare our answers:

Please note that you got exactly the same number as in the previous method, when we sequentially multiplied by each previous term of the geometric progression.
Let’s try to “depersonalize” this formula - let’s put it in general form and get:

The derived formula is true for all values ​​- both positive and negative. Check this yourself by calculating the terms of the geometric progression with the following conditions: , a.

Did you count? Let's compare the results:

Agree that it would be possible to find a term of a progression in the same way as a term, however, there is a possibility of calculating incorrectly. And if we have already found the th term of the geometric progression, then what could be simpler than using the “truncated” part of the formula.

Infinitely decreasing geometric progression.

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

Why do you think this name is given?
First, let's write down some geometric progression consisting of terms.
Let's say, then:

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

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

On graphs we are accustomed to plotting dependence on, therefore:

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

Do you 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 means:

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

Did you manage? Here's the graph I came up with:

Now that you have fully understood the basics of the topic of 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.

Property of geometric progression.

Do you remember the property of the terms 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 terms of this progression. Do you remember? This:

Now we are faced with exactly the same question for the terms of a geometric progression. To derive such a formula, let's start drawing and reasoning. You'll see, it's very easy, and if you forget, you can get it out yourself.

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

You may ask, what should we do about it now? Yes, very simple. First, let's depict these formulas in a picture and try to do various manipulations with them in order to arrive at the value.

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

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

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

Subtraction.

As you can see, we cannot express this either, therefore, let’s try to multiply these expressions by each other.

Multiplication.

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

Guess what I'm talking about? Correctly, to find we need to take the square root of the geometric progression numbers adjacent to the desired one multiplied by each other:

Here you go. You yourself derived the property of geometric progression. Try to write this formula in general form. Happened?

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

Accordingly, do not forget this limitation.

Now let's calculate what it equals

Correct answer - ! If you didn’t forget the second possible value during the calculation, then you’re great and can immediately move on to training, and if you forgot, read what is discussed below and pay attention to why it is necessary to write down both roots in the answer.

Let's draw both of our geometric progressions - one with a value and the other with a value and check whether 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 all its given terms are the same? Calculate q for the first and second cases.

See why we have to write two answers? Because the sign of the term you are looking for depends on whether it is positive or negative! And since we don’t know what it 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 geometric progression, find, knowing and

Compare your answers with the correct ones:

What do you think, what if we were given not the values ​​of the terms of the geometric progression adjacent to the desired number, but equidistant from it. For example, we need to find, and given and. Can we use the formula we derived in this case? Try to confirm or refute this possibility in the same way, describing what each value consists of, as you did when you originally derived the formula, at.
What did you get?

Now look carefully again.
and correspondingly:

From this we can conclude that the formula works not only with neighboring with the desired terms of the geometric progression, but also with equidistant from what the members are looking for.

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 smaller. The main thing is that it is 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.

Let's 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 examination of the serial 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 is removed at a position, so it is not possible to apply the formula.

How to solve it? It's actually not as difficult as it seems! Let us write down what each number given to us and the number we are looking for consists of.

So we have and. Let's see what we can do with them? I suggest dividing by. We get:

We substitute our data into the formula:

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

Now let's look again at what we have. We have it, but we need to find it, and it, in turn, is equal to:

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

Our answer: .

Try solving another similar problem yourself:
Given: ,
Find:

How much did you get? I have - .

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

The sum of the terms of a geometric progression.

Now let's look at formulas that allow us to quickly calculate the sum of terms of a geometric progression in a given interval:

To derive the formula for the sum of terms of a finite geometric progression, multiply all parts of the above 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 get?

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

Group the expression. You should get:

All that remains to be done is to express:

Accordingly, in this case.

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

There are many legends about both arithmetic and geometric progression. One of them is the legend of Set, 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 positions possible in her. Having learned that it was invented by one of his subjects, the king decided to personally reward him. He summoned the inventor to himself and ordered him to ask him for everything he wanted, promising to fulfill even the most skillful desire.

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

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

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

Let's start reasoning. 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., then we see that the problem is about a geometric progression. What does it equal in this case?
Right.

Total squares of the chessboard. Respectively, . We have all the data, all that remains is to plug it into the formula and calculate.

To imagine at least approximately the “scale” of a given number, we transform using the properties of degree:

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

quintillion quadrillion trillion billion million thousand.

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

If the king were strong in mathematics, he could have invited the scientist himself to 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 throughout his life.

Now let’s solve a simple problem involving the sum of terms of a geometric progression.
A student of class 5A Vasya fell ill with 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 only people in the class. In how many days will the whole class be sick with the flu?

So, the first term of the geometric progression is Vasya, that is, a person. The th term of the geometric progression is the two people he infected on the first day of his arrival. The total sum of the progression terms is equal to the number of 5A students. Accordingly, we talk about a progression in which:

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

The whole class will get sick within days. Don't believe formulas and numbers? Try to portray the “infection” of students yourself. Happened? Look how it looks for me:

Calculate for yourself how many days it would take for students to get sick with the flu if each one infected a person, and there were only one person in the class.

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

As you can see, such a task and the drawing for it resemble 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 closes the chain (). Thus, if a person were involved in a financial pyramid in which money was given if you brought two other participants, then the person (or in general) would not bring anyone, accordingly, would lose everything that they invested in this financial scam.

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

So, first, let's look again at this drawing 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, will be almost equal, respectively, when calculating the expression we will 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.

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

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

If a specific number n is specified, 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 the geometric progression with and.
2. Find the sum of the terms of an infinitely decreasing geometric progression with and.

I hope you were extremely careful. 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 geometric progression problems encountered on the exam are problems calculating compound interest. These are the ones we will talk about.

Problems on calculating compound interest.

You've probably heard of the so-called compound interest formula. Do you understand what it means? If not, let’s figure it out, because once you understand the process itself, you will immediately understand what geometric progression has to do with it.

We all go to the bank and know that there are different conditions for deposits: this includes a term, additional services, and interest with two different ways of calculating it - simple and complex.

WITH simple interest everything is more or less clear: interest is accrued once at the end of the deposit term. That is, if we say that we deposit 100 rubles for a year, 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 an option in which it occurs interest capitalization, i.e. their addition to the deposit amount and subsequent calculation of income not from the initial, but from the accumulated deposit amount. 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 assume that we deposit the same rubles annually, but with monthly capitalization of the deposit. What are we doing?

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

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

Agree?

We can take it out of brackets and then we get:

Agree, this formula is already more similar to what we wrote at the beginning. All that's left is to figure out the percentages

In the problem statement we are told about annual rates. As you know, we do not multiply by - we convert percentages to decimal fractions, that is:

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

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

Well done! Let's return to our task: write how much will be credited to our account in the second month, taking into account that interest is accrued on the accumulated deposit amount.
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 what its member will be equal to, or, in other words, what amount of money we will receive at the end of the month.
Did? Let's check!

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

Let's look at another type of problem involving compound interest. After what you have figured out, it will be elementary for you. So, the task:

The Zvezda company began investing in the industry in 2000, with capital in dollars. Every year since 2001, it has received a profit that is equal to the previous year's capital. How much profit will the Zvezda company receive at the end of 2003 if profits were not withdrawn from circulation?

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

Or we can write briefly:

For our case:

2000, 2001, 2002 and 2003.

Respectively:
rubles
Please note that in this problem we do not have a division either by or by, since the percentage is given ANNUALLY and it is calculated ANNUALLY. That is, when reading a problem on compound interest, pay attention to what percentage is given and in what period it is calculated, and only then proceed to calculations.
Now you know everything about geometric progression.

Training.

1. Find the term of the geometric progression 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. The MDM Capital company began investing in the industry in 2003, with capital in dollars. Every year since 2004, it has received a profit that is equal to the previous year's capital. The MSK Cash Flows company began investing in the industry in 2005 in the amount of $10,000, starting to make a profit in 2006 in the amount of. By how many dollars is the capital of one company greater than the other at the end of 2007, if profits were not 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 terms, the calculation is carried out according to the formula:
2. 
   MDM Capital Company:

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

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

Let's summarize.

1) Geometric progression ( ) is a numerical sequence, the first term of which is different from zero, 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.

2) The equation of the terms of the geometric progression is .

3) can take any values ​​except and.

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

4) , with - property of geometric progression (adjacent terms)

or
, at (equidistant terms)

When you find it, don’t forget that there should be two answers.

For example,

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

or

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

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

GEOMETRIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

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

Denominator of geometric progression can take any value except and.

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

Equation of terms of geometric progression - .

Sum of terms of a geometric progression calculated by the formula:
or

If the progression is infinitely decreasing, then:

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough...

For what?

For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who have received a good education earn much more than those who have not received it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

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

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

Rice. 1. Ancient Egyptian geometric progression problem

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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