Infinitely decreasing geometric progression online. The formula for the sum of the first n members of the GP. Problems for calculating compound interest

NUMERICAL SEQUENCES VI

§ l48. The sum of an infinitely decreasing geometric progression

Until now, speaking of sums, we have always assumed that the number of terms in these sums is finite (for example, 2, 15, 1000, etc.). But when solving some problems (especially higher mathematics), one has to deal with the sums of an infinite number of terms

S= a 1 + a 2 + ... + a n + ... . (1)

What are these amounts? By definition the sum of an infinite number of terms a 1 , a 2 , ..., a n , ... is called the limit of the sum S n first P numbers when P -> ∞ :

S=S n = (a 1 + a 2 + ... + a n ). (2)

Limit (2), of course, may or may not exist. Accordingly, the sum (1) is said to exist or not to exist.

How to find out whether the sum (1) exists in each particular case? A general solution to this question goes far beyond the scope of our program. However, there is one important special case that we have to consider now. We will talk about the summation of the terms of an infinitely decreasing geometric progression.

Let a 1 , a 1 q , a 1 q 2 , ... is an infinitely decreasing geometric progression. This means that | q |< 1. Сумма первых P members of this progression is equal to

From the basic theorems on the limits of variables (see § 136) we obtain:

But 1 = 1, a q n = 0. Therefore

So, the sum of an infinitely decreasing geometric progression is equal to the first term of this progress divided by one minus the denominator of this progression.

1) The sum of the geometric progression 1, 1/3, 1/9, 1/27, ... is

and the sum of a geometric progression is 12; -6; 3; - 3 / 2 , ... equals

2) A simple periodic fraction 0.454545 ... turn into an ordinary one.

To solve this problem, we represent this fraction as an infinite sum:

The right side of this equality is the sum of an infinitely decreasing geometric progression, the first term of which is 45/100, and the denominator is 1/100. That's why

In the manner described, the general rule for converting simple periodic fractions into ordinary fractions can also be obtained (see Chapter II, § 38):

To convert a simple periodic fraction into an ordinary one, you need to proceed as follows: put the period of the decimal fraction in the numerator, and in the denominator - a number consisting of nines taken as many times as there are digits in the period of the decimal fraction.

3) Mixed periodic fraction 0.58333 .... turn into an ordinary fraction.

Let's represent this fraction as an infinite sum:

On the right side of this equality, all terms, starting from 3/1000, form an infinitely decreasing geometric progression, the first term of which is 3/1000, and the denominator is 1/10. That's why

In the manner described, the general rule for the conversion of mixed periodic fractions into ordinary fractions can also be obtained (see Chapter II, § 38). We deliberately do not include it here. There is no need to memorize this cumbersome rule. It is much more useful to know that any mixed periodic fraction can be represented as the sum of an infinitely decreasing geometric progression and some number. And the formula

for the sum of an infinitely decreasing geometric progression, one must, of course, remember.

As an exercise, we invite you, in addition to the problems No. 995-1000 below, to once again turn to problem No. 301 § 38.

Exercises

995. What is called the sum of an infinitely decreasing geometric progression?

996. Find sums of infinitely decreasing geometric progressions:

997. For what values X progression

is infinitely decreasing? Find the sum of such a progression.

998. In an equilateral triangle with a side a a new triangle is inscribed by connecting the midpoints of its sides; a new triangle is inscribed in this triangle in the same way, and so on ad infinitum.

a) the sum of the perimeters of all these triangles;

b) the sum of their areas.

999. In a square with a side a a new square is inscribed by connecting the midpoints of its sides; a square is inscribed in this square in the same way, and so on ad infinitum. Find the sum of the perimeters of all these squares and the sum of their areas.

1000. Make an infinitely decreasing geometric progression, such that its sum is equal to 25 / 4, and the sum of the squares of its terms is equal to 625 / 24.

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

The concept of geometric progression

The geometric progression is denoted by b1,b2,b3, …, bn, … .

The ratio of any term 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 a geometric progression. Usually the denominator of a geometric progression is denoted by the letter q.

The sum of an infinite geometric progression for |q|<1

One way to set a geometric progression is to set its first term b1 and the denominator of the geometric error q. For example, b1=4, q=-2. These two conditions give a geometric progression of 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 the denominator q=1 in the geometric error, 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.

In order for the numerical sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, be the geometric mean of the neighboring members. 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.

Now let's put (Xn) - a geometric progression. The denominator of the geometric progression q, with |q|∞).
If we now denote by S the sum of an infinite geometric progression, then the following formula will hold:
S=x1/(1-q).

Consider a simple example:

Find the sum of an infinite geometric progression 2, -2/3, 2/9, - 2/27, ... .

To find S, we use the formula for the sum of an infinitely arithmetic progression. |-1/3|< 1. x1 = 2. S=2/(1-(-1/3)) = 3/2.

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

The geometric progression is denoted b1,b2,b3, …, bn, … .

Monotonic and constant sequence

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

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

Formula of the nth member of a geometric progression

The formula for the nth member of a geometric progression is:

bn=b1*q^(n-1),

where n belongs to the set of natural numbers N.

The formula for the sum of the first n terms of a geometric progression

The formula for the sum of the first n terms of a geometric progression is:

Sn = (bn*q - b1)/(q-1) where q is not equal to 1.

Consider a simple example:

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

To find S8, we use the formula for the sum of the first n terms of a geometric progression.

S8= (6*(3^8 -1))/(3-1) = 19680.

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

So, for starters, actually formulan

There she is:

b n = b 1 · q n -1

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

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

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

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

So let's go:

b 1 – the first member of a geometric progression;

q – ;

n– member number;

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

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

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

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

b 2 = b 1 q

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

Like this:

B 3 \u003d b 2 q

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

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

We get:

B 3 = b 1 q 2

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

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

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

Total:

B 4 = b 1 q 3

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

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

Therefore, our formula will be, without options:

b n =b 1 · q n -1

That's all.)

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

Solving problems on a formulanth term of a geometric progression.

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

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

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

Our data for applying the formula is as follows.

The first term is known. This is 512.

b 1 = 512.

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

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

And carefully calculate the arithmetic:

Answer: -1

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

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

In geometric progression, we know that:

b 1 = 3

Find the thirteenth term of the progression.

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

We work directly according to the formula:

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

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

Like this:

Answer: 192

And all things.)

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

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

For example, such a harmless problem.

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

Nothing complicated. We work directly according to the spell.

We write the formula of the nth term!

b n = b 1 · q n -1

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

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

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

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

Now we substitute our data into the formula:

567 = b 1 3 5-1

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

81 b 1 = 567

We solve and get:

b 1 = 7

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

For example, such a problem:

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

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

We write the formulanth member!

b n = b 1 · q n -1

Our initial data will be as follows:

b 5 = 162

b 1 = 2

n = 5

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

We get:

162 = 2q 5-1

2 q 4 = 162

q 4 = 81

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

Like this:

q4 = 81

q = 3

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

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

Both fit.

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

x2 = 9

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

So the correct solution would be:

q 4 = 81

q= ±3

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

So the answer is obvious:

q = 3

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

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

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

q = 3 and q = -3

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

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

Given a geometric progression:

3; 6; 12; 24; …

What number is 768 in this progression?

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

b n = b 1 · q n -1

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

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

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

At least like this:

q = 24/12 = 2

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

b n = 768

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

Here we substitute:

768 = 3 2n -1

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

We get:

2 n -1 = 256

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

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

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

256 = 2 8

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

One way or another, we will get:

2 n -1 = 2 8

n-1 = 8

n = 9

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

Answer: 9

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

More complex tasks.

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

For example, like this.

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

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

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

We paint each term according to the formula nth member!

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

For the fourth term we write:

b 4 = b 1 · q 3

-24 = b 1 · q 3

There is. One equation is complete.

For the seventh term we write:

b 7 = b 1 · q 6

192 = b 1 · q 6

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

We assemble a system from them:

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

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

q 3 = -8

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

So we have a system:

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

The whole division process looks like this:

Now, reducing everything that is reduced, we get:

q 3 = -8

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

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

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

q 3 = -8

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

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

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

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

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

For the second member, everything is quite simple:

b 2 = b 1 · q= 3 (-2) = -6

Answer: -6

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

Let's draw the problem!

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

I got it like this:

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

Therefore, we have every right to write:

-24q 3 = 192

From here it is now easy to find q:

q 3 = -8

q = -2

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

Here we write:

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

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

Answer: -6

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

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

Another epic one:

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

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

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

Second term: b 2 = b 1 q

Third term: b 3 \u003d b 1 q 2

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

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

So we write:

b 2 = b 1 +10

And we translate this phrase into pure mathematics:

b 3 = b 2 +30

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

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

We get:

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

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

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

b 1 q = b 1 +10

b 1 q - b 1 \u003d 10

b 1 (q-1) = 10

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

Typical. What to do - we know.

We write ODZ (necessarily!) :

q ≠ 1

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

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

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

q 2 – 4 q + 3 = 0

We solve the resulting and get two roots:

q 1 = 1

q 2 = 3

There is only one final answer: q = 3 .

Answer: 3

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

Namely:

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

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

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

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

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

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

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

Modified and recurrent formulas.

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

For example, such a problem from the OGE:

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

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

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

b 1 = 6

q = 2

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

b n = b 1 · q n -1

b n= 6 2n -1

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

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

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

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

b 1 = 3 2 1 = 3 2 = 6

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

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

b 4 = 3 2 4 = 3 16 = 48

And finally, we calculate the required amount:

b 1 + b 4 = 6+48 = 54

Answer: 54

Another problem.

The geometric progression is given by the conditions:

b 1 = -7;

b n +1 = 3 b n

Find the fourth term of the progression.

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

Here we are acting. Step by step.

1) counting two successive member of the progression.

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

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

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

2) We consider the denominator of the progression

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

We get:

q = -21/(-7) = 3

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

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

b n= -7 3n -1

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

Answer: -189

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

Well, let's decide on our own?)

Quite elementary tasks, for warm-up:

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

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

3. The geometric progression is given by the conditions:

b 1 = -3;

b n +1 = 6 b n

Find the fifth term of the progression.

A little more complicated:

4. Given a geometric progression:

b 1 =2048; q =-0,5

What is the sixth negative term of it?

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

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

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

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

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

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

Properties of a geometric progression

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 the denominator q=1 in the geometric error, 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 nth member of the progression

In order for the numerical sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, be the geometric mean of the neighboring members. 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 nth member of a geometric progression is:

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

Consider a simple example:

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

Let's use the formula of the n-th member of a geometric progression.