Formula for the sum of an infinite geometric progression. What is geometric progression? Basic Concepts
Mathematics is whatpeople control nature and themselves.
Soviet mathematician, academician A.N. Kolmogorov
Geometric progression.
Along with problems on arithmetic progressions, problems related to the concept of geometric progression are also common in entrance examinations in mathematics. To successfully solve such problems, you need to know the properties of geometric progressions and have good skills in using them.
This article is devoted to the presentation of the basic properties of geometric progression. Examples of solving typical problems are also provided here., borrowed from the tasks of entrance examinations in mathematics.
Let us first note the basic properties of the geometric progression and recall the most important formulas and statements, related to this concept.
Definition. A number sequence is called a geometric progression if each number, starting from the second, is equal to the previous one, multiplied by the same number. The number is called the denominator of a geometric progression.
For geometric progressionthe formulas are valid
, (1)
Where . Formula (1) is called the formula of the general term of a geometric progression, and formula (2) represents the main property of a geometric progression: each term of the progression coincides with the geometric mean of its neighboring terms and .
Note, that it is precisely because of this property that the progression in question is called “geometric”.
The above formulas (1) and (2) are generalized as follows:
, (3)
To calculate the amount first members of a geometric progressionformula applies
If we denote , then
Where . Since , formula (6) is a generalization of formula (5).
In the case when and geometric progressionis infinitely decreasing. To calculate the amountof all terms of an infinitely decreasing geometric progression, the formula is used
. (7)
For example , using formula (7) we can show, What
Where . These equalities are obtained from formula (7) under the condition that , (first equality) and , (second equality).
Theorem. If , then
Proof. If , then
The theorem has been proven.
Let's move on to consider examples of solving problems on the topic “Geometric progression”.
Example 1. Given: , and . Find .
Solution. If we apply formula (5), then
Answer: .
Example 2. Let it be. Find .
Solution. Since and , we use formulas (5), (6) and obtain a system of equations
If the second equation of system (9) is divided by the first, then or . It follows from this that . Let's consider two cases.
1. If, then from the first equation of system (9) we have.
2. If , then .
Example 3. Let , and . Find .
Solution. From formula (2) it follows that or . Since , then or .
According to the condition. However, therefore. Since and then here we have a system of equations
If the second equation of the system is divided by the first, then or .
Since, the equation has a unique suitable root. In this case, it follows from the first equation of the system.
Taking into account formula (7), we obtain.
Answer: .
Example 4. Given: and . Find .
Solution. Since, then.
Since , then or
According to formula (2) we have . In this regard, from equality (10) we obtain or .
However, by condition, therefore.
Example 5. It is known that. Find .
Solution. According to the theorem, we have two equalities
Since , then or . Because , then .
Answer: .
Example 6. Given: and . Find .
Solution. Taking into account formula (5), we obtain
Since, then. Since , and , then .
Example 7. Let it be. Find .
Solution. According to formula (1) we can write
Therefore, we have or . It is known that and , therefore and .
Answer: .
Example 8. Find the denominator of an infinite decreasing geometric progression if
And .
Solution. From formula (7) it follows And . From here and from the conditions of the problem we obtain a system of equations
If the first equation of the system is squared, and then divide the resulting equation by the second equation, then we get
Or .
Answer: .
Example 9. Find all values for which the sequence , , is a geometric progression.
Solution. Let , and . According to formula (2), which defines the main property of a geometric progression, we can write or .
From here we get the quadratic equation, whose roots are And .
Let's check: if, then , and ; if , then , and .
In the first case we have and , and in the second – and .
Answer: , .
Example 10.Solve the equation
, (11)
where and .
Solution. The left side of equation (11) is the sum of an infinite decreasing geometric progression, in which and , subject to: and .
From formula (7) it follows, What . In this regard, equation (11) takes the form or . Suitable root quadratic equation is
Answer: .
Example 11. P sequence of positive numbersforms an arithmetic progression, A – geometric progression, and here. Find .
Solution. Because arithmetic sequence, That (the main property of arithmetic progression). Since, then or . It follows from this, that the geometric progression has the form. According to formula (2), then we write down that .
Since and , then . In this case, the expression takes the form or . According to the condition, so from Eq.we obtain a unique solution to the problem under consideration, i.e. .
Answer: .
Example 12. Calculate Sum
. (12)
Solution. Multiply both sides of equality (12) by 5 and get
If we subtract (12) from the resulting expression, That
or .
To calculate, we substitute the values \u200b\u200binto formula (7) and get . Since, then.
Answer: .
The examples of problem solving given here will be useful to applicants when preparing for entrance examinations. For a deeper study of problem solving methods, related to geometric progression, can be used teaching aids from the list of recommended literature.
1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Mir and Education, 2013. – 608 p.
2. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. – 216 p.
3. Medynsky M.M. Full course elementary mathematics in problems and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. – 208 p.
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The formula for the nth term of a geometric progression is very simple. Both in meaning and in general appearance. But there are all kinds of problems on the formula of the nth term - from very primitive to quite serious. And in the process of our acquaintance, we will definitely consider both. Well, let's get acquainted?)
So, to begin with, actually formulan
Here it is:
b n = b 1 · qn -1
The formula is just a formula, nothing supernatural. It looks even simpler and more compact than a similar formula for. The meaning of the formula is also as simple as felt boots.
This formula allows you to find ANY member of a geometric progression BY ITS NUMBER " n".
As you can see, the meaning is complete analogy with an arithmetic progression. We know the number n - we can also count the term under this number. Whichever one we want. Without repeatedly multiplying by "q" many, many times. That's the whole point.)
I understand that at this level of working with progressions, all the quantities included in the formula should already be clear to you, but I still consider it my duty to decipher each one. Just in case.
So, here we go:
b 1 – first term of geometric progression;
q – ;
n– member number;
b n – nth (nth) term of a geometric progression.
This formula connects the four main parameters of any geometric progression - bn, b 1 , q And n. And around these four key figures and all the problems rotate in progression.
“How is it removed?”– I hear a curious question... Elementary! Look!
What is equal to second member of the progression? No question! We write directly:
b 2 = b 1 ·q
What about the third member? Not a problem either! We multiply the second term once again onq.
Like this:
B 3 = b 2 q
Let us now remember 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 degrees. Do you get it? Not yet? Okay, one more step.
What is the fourth term? Everything is the same! Multiply previous(i.e. the third term) on q:
B 4 = b 3 q = (b 1 q 2) q = b 1 q 2 q = b 1 q 3
Total:
B 4 = b 1 ·q 3
And again we translate into Russian: fourth term is equal to the first term multiplied by q in third degrees.
And so on. So how? Did you catch the pattern? Yes! For any term with any number, the number of identical factors q (i.e. the degree of the denominator) will always be one less than the number of the desired membern.
Therefore, our formula will be, without options:
b n =b 1 · qn -1
That's all.)
Well, let’s solve the problems, I guess?)
Solving formula problemsnth term of a geometric progression.
Let's start, as usual, with the direct application of the formula. Here's a typical problem:
In geometric progression, it is known 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. Directly in the sense of geometric progression. But we need to warm up with the formula for the nth term, right? Here we are warming up.
Our data for applying the formula is as follows.
The first member is known. This is 512.
b 1 = 512.
The denominator of the progression is also known: q = -1/2.
All that remains is to figure out what the number of member n is. No question! Are we interested in the tenth term? So we substitute ten instead of n into the general formula.
And carefully calculate the arithmetic:
Answer: -1
As you can see, the tenth term of the progression turned out to be minus. Nothing surprising: our progression denominator is -1/2, i.e. negative number. And this tells us that the signs of our progression alternate, yes.)
Everything is simple here. Here is a similar problem, but a little more complicated in terms of calculations.
In geometric progression, it is known that:
b 1 = 3
Find the thirteenth term of the progression.
Everything is the same, only this time the denominator of the progression is irrational. Root of two. Well, that's okay. The formula is a universal thing; it can handle any numbers.
We work directly according to the formula:
The formula, of course, worked as it should, but... this is where some people get stuck. What to do next with the root? How to raise a root to the twelfth power?
How-how... You must understand that any formula, of course, is a good thing, but knowledge of all previous mathematics is not canceled! How to build? Yes, remember the properties of degrees! Let's turn the root into fractional degree and – according to the formula for raising a degree to a degree.
Like this:
Answer: 192
And that's all.)
What is the main difficulty in directly applying the nth term formula? Yes! The main difficulty is working with degrees! Namely, raising negative numbers, fractions, roots and similar constructions to powers. So those who have problems with this, please repeat the degrees and their properties! Otherwise, you will slow down this topic too, yes...)
Now let’s solve typical search problems one of the elements of the formula, if all others are given. To successfully solve such problems, the recipe is uniform and terribly simple - write the formulanth member in general view! Right in the notebook next to the condition. And then from the condition we figure out what is given to us and what is missing. And we express the desired value from the formula. All!
For example, such a harmless problem.
The fifth term of a geometric progression with denominator 3 is 567. Find the first term of this progression.
Nothing complicated. We work directly according to the spell.
Let's write the formula for the nth term!
b n = b 1 · qn -1
What have we been given? First, the denominator of the progression is given: q = 3.
Moreover, we are given fifth member: b 5 = 567 .
All? No! We have also been given number n! This is five: n = 5.
I hope you already understand what is in the recording b 5 = 567 two parameters are hidden at once - this is the fifth term itself (567) and its number (5). I already talked about this in a similar lesson, but I think it’s worth mentioning here too.)
Now we substitute our data into the formula:
567 = b 1 ·3 5-1
We do the 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 term. But when searching for the denominator q and numbers n There may also be surprises. And you also need to be prepared for them (surprises), yes.)
For example, this 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 terms, and are asked to find the denominator of the progression. Here we go.
We write the formulanth member!
b n = b 1 · qn -1
Our initial data will be as follows:
b 5 = 162
b 1 = 2
n = 5
Missing value q. No question! Let’s find it now.) We substitute everything 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. And 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 actually, this is an unfinished answer. More precisely, incomplete. Why? The point is that the answer q = -3 also suitable: (-3) 4 will also be 81!
This is because the power equation x n = a always has two opposite roots at evenn . With plus and minus:
Both are suitable.
For example, when deciding (i.e. second degrees)
x 2 = 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 are in the topic about
That's why the right decision will be like this:
q 4 = 81
q= ±3
Okay, we've sorted out the signs. Which one is correct - plus or minus? Well, let’s read the problem statement again in search of additional information. Of course, it may not exist, but in this problem such information available. Our condition states in plain text that a progression is given with positive denominator.
Therefore 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 condition Nothing no mention is made of the sign of the denominator. Neither directly nor indirectly. And here the problem would already have two solutions!
q = 3 And q = -3
Yes, yes! Both with a plus and with a minus.) Mathematically, this fact would mean that there are two progressions, which fit the conditions of the problem. And each has its own denominator. Just for fun, practice and write out the first five terms of each.)
Now let’s practice finding the member’s number. This problem is the most difficult, yes. But also more creative.)
Given a geometric progression:
3; 6; 12; 24; …
What number in this progression is the number 768?
The first step is still the same: write the formulanth member!
b n = b 1 · qn -1
And now, as usual, we substitute the data we know into it. Hm... it doesn't work! Where is the first term, where is the denominator, where is everything else?!
Where, where... Why do we need eyes? Flapping your eyelashes? This time the progression is given to us directly in the form sequences. Can we see the first member? 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...
So we count. Directly according to the meaning of a geometric progression: we take any of its terms (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 term of this progression, equal to 768. Under some number n:
b n = 768
We don’t know his number, but our task is precisely to find him.) So we are looking. We have already downloaded all the necessary data for substitution into the formula. Unbeknownst to yourself.)
Here we substitute:
768 = 3 2n -1
Let's do the elementary ones - divide both sides by three and rewrite the equation in the usual form: the unknown is on the left, the known is on the right.
We get:
2 n -1 = 256
This is an interesting equation. We need to find "n". What, unusual? Yes, I don't argue. Actually, this is the simplest thing. It is so called because the unknown (in this case it is the number n) costs in indicator degrees.
At the stage of learning about geometric progression (this is ninth grade), they don’t teach you how to solve exponential equations, yes... This is a topic for high school. But there's nothing scary. Even if you don’t know how such equations are solved, let’s try to find our n, guided by simple logic and common sense.
Let's start talking. On the left we have a deuce to some extent. We don’t yet know what exactly this degree is, but that’s not scary. But we know for sure that this degree is equal to 256! So we remember to what extent a two gives us 256. Do you remember? Yes! IN eighth degrees!
256 = 2 8
If you don’t remember or have problems recognizing the degrees, then that’s okay: we just successively square two, cube, fourth, fifth, and so on. Selection, in fact, but at this level will work quite well.
One way or another, we 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 elementary stuff? Agree. Me too. Let's move to the next level.)
More complex tasks.
Now let’s solve more challenging problems. Not exactly super cool, but ones that require a little work to get to the answer.
For example, this one.
Find the second term of a geometric progression if its fourth term is -24 and its seventh term is 192.
This is a classic of the genre. Some two different terms of the progression are known, but another term needs to be found. Moreover, all members are NOT neighboring. Which is confusing at first, yes...
As in, to solve such problems we will consider two methods. The first method is universal. Algebraic. Works flawlessly with any source data. That's why we'll start with that.)
We describe 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 one by one We substitute our initial data into the formula for 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
Eat. One equation is ready.
For the seventh term we write:
b 7 = b 1 · q 6
192 = b 1 · q 6
In total, we got two equations for the same progression .
We assemble a system from them:
Despite its menacing appearance, the system is quite simple. The most obvious solution is simple substitution. We express b 1 from the upper equation and substitute it into the lower one:
After fiddling around with the bottom equation a bit (reducing the powers and dividing by -24), we get:
q 3 = -8
By the way, this same equation can be arrived at in a simpler way! Which one? Now I will show you another secret, but very beautiful, powerful and useful way solutions for such systems. Such systems, the equations of which include only works. At least in one. Called division method one equation to another.
So, we have a system before us:
In both equations on the left - work, and on the right is just a number. This is very good sign.) Let's take it and... divide, say, the lower equation by the upper one! What does it mean let's divide one equation by another? Very simple. Let's take it left side one equation (lower) and divide her on left side another equation (upper). The right side is similar: right side one equation divide on right side another.
The whole division process looks like this:
Now, reducing everything that can be reduced, we get:
q 3 = -8
What's good about this method? Yes, because in the process of such division everything bad and inconvenient can be safely reduced and a completely harmless equation remains! This is why it is so important to have multiplication only 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 methods of solving systems) even deserves a separate lesson. I'll definitely look into it in more detail. Some day…
However, it doesn’t matter how exactly you solve the system, in any case, now we need to solve the resulting equation:
q 3 = -8
No problem: extract the cube root and you’re done!
Please note that there is no need to put a plus/minus here when extracting. Our root is of odd (third) degree. And the answer is also the same, yes.)
So, the denominator of the progression has been found. Minus two. Great! The process is ongoing.)
For the first term (say, from the upper equation) we get:
Great! 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 one.)
For the second term everything is quite simple:
b 2 = b 1 · q= 3·(-2) = -6
Answer: -6
So, we have broken down the algebraic method for solving the problem. Difficult? Not really, I agree. Long and tedious? Yes, definitely. But sometimes you can significantly reduce the amount of work. For this there is graphic method. Good old and familiar to us.)
Let's draw a problem!
Yes! That's right. Again we depict our progression on the number axis. It’s not necessary to follow a ruler, it’s not necessary to maintain equal intervals between the terms (which, by the way, will not be the same, since the progression is geometric!), but simply schematically Let's draw our sequence.
I got it like this:
Now look at the picture and figure it out. How many identical factors "q" separate fourth And seventh members? That's right, three!
Therefore, we have every right to write:
-24·q 3 = 192
From here it is now easy to find q:
q 3 = -8
q = -2
That’s great, we already have the denominator in our pocket. Now let’s look at the picture again: how many such denominators sit between second And fourth members? Two! Therefore, to record the connection between these terms, we will construct the denominator squared.
So we write:
b 2 · q 2 = -24 , where b 2 = -24/ q 2
We substitute our found denominator into the expression for b 2, count and get:
Answer: -6
As you can see, everything is much 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. Did you guess it? 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’s already difficult to draw a picture, yes... Then we solve the problem analytically, through the system.) And systems are universal things. They can handle any numbers.
Another epic challenge:
The second term of the geometric progression is 10 more than the first, and the third term is 30 more than the second. Find the denominator of the progression.
What, cool? Not at all! Everything is the same. Again we translate the problem statement into pure algebra.
1) We describe each term according to the formula nth member!
Second term: b 2 = b 1 q
Third term: b 3 = b 1 q 2
2) We write down the connection between the members from the problem statement.
We read the condition: "The second term of the geometric progression is 10 greater 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. Let's combine them into a system:
The system looks simple. But there are too many different indices for the letters. Let's substitute instead of the second and third terms their expressions through the first term and the denominator! Was it in vain that we painted them?
We get:
But such a system is no longer a gift, yes... How to solve this? Unfortunately, there is no universal secret spell for solving complex nonlinear There are no systems in mathematics and there cannot be. This 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 that allows, for example, to easily express one of the variables in terms of another?
Let's figure it out. The first equation of the system is clearly simpler than the second. We'll torture him.) Shouldn't we try from the first equation something express through something? Since we want to find the denominator q, then it would be most advantageous for us to express b 1 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 = 10
b 1 (q-1) = 10
All! So we expressed unnecessary give us the variable (b 1) through necessary(q). Yes, it’s not the simplest expression we got. Some kind of fraction... But our system is of a decent level, yes.)
Typical. We know what to do.
We write ODZ (Necessarily!) :
q ≠ 1
We multiply everything by the denominator (q-1) and cancel all fractions:
10 q 2 = 10 q + 30(q-1)
We divide everything by ten, open the brackets, and collect everything from the left:
q 2 – 4 q + 3 = 0
We solve the result and get two roots:
q 1 = 1
q 2 = 3
There is only one final answer: q = 3 .
Answer: 3
As you can see, the path to solving most problems involving the formula of the nth term of a geometric progression is always the same: read attentively condition of the problem and using the formula of the nth term we translate the entire useful information into pure algebra.
Namely:
1) We describe separately each term given in the problem according to the formulanth member.
2) From the conditions of the problem we translate the connection between the members into mathematical form. We compose an equation or 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 task conditions in search of additional information (if any). We also check the received response with the terms of the DL (if any).
Now let’s 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 there are problems with at least one of these three points, then you will inevitably make mistakes in this topic. Unfortunately... So don't be lazy and repeat what was mentioned above. And follow the links - go. Sometimes it helps.)
Modified and recurrent formulas.
Now let’s look at a couple of typical exam problems with a less familiar presentation of the condition. Yes, yes, you guessed it! This modified And recurrent nth term formulas. We have already encountered such formulas and worked on arithmetic progression. Everything is similar here. The essence is the same.
For example, this problem from the OGE:
The geometric progression is given by the formula b n = 3 2 n . Find the sum of its first and fourth terms.
This time the progression is not quite as usual for us. In the form of some kind of formula. So what? This formula is also a formulanth member! You and I know that the formula for the nth term can be written both in general form, using letters, and for specific progression. WITH 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 down the formula for the nth term in general form and substitute it into b 1 And q. We get:
b n = b 1 · qn -1
b n= 6 2n -1
We simplify using factorization and properties of powers, and we 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 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 given to us in the condition. Do you get it?) So we work with the modified formula directly.
We count the first term. Let's substitute n=1 into the general formula:
b 1 = 3 2 1 = 3 2 = 6
Like this. By the way, I won’t be lazy and once again draw your attention to a typical mistake with the calculation of the first term. DO NOT, looking at the formula b n= 3 2n, immediately rush to write that the first term is a three! This is a gross mistake, yes...)
Let's continue. Let's substitute n=4 and count the fourth term:
b 4 = 3 2 4 = 3 16 = 48
And finally, we calculate the required amount:
b 1 + b 4 = 6+48 = 54
Answer: 54
Another problem.
Geometric progression is specified 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 a recurrent formula. Well, okay.) How to work with this formula – we know too.
So we act. Step by step.
1) Count two consecutive member of the progression.
The first term has already been given to us. Minus seven. But the next, second term, can be easily calculated using the recurrence formula. If you understand the principle of its operation, of course.)
So we count the second term By known first:
b 2 = 3 b 1 = 3·(-7) = -21
2) Calculate the denominator of the progression
No problem either. Straight, let's divide second dick on first.
We get:
q = -21/(-7) = 3
3) Write the formulanth member in the usual form and calculate the required member.
So, we know the first term, and so do the denominator. So we write:
b n= -7·3n -1
b 4 = -7·3 3 = -7·27 = -189
Answer: -189
As you can see, working with such formulas for a geometric progression is essentially no different from that for an arithmetic progression. It is only important to understand general essence and the meaning of these formulas. Well, you also need to understand the meaning of geometric progression, yes.) And then there will be no stupid mistakes.
Well, let's decide on our own?)
Very basic tasks for warming up:
1. Given a geometric progression in which b 1 = 243, a q = -2/3. Find the sixth term of the progression.
2. The general term of the geometric progression is given by the formula b n = 5∙2 n +1 . Find the number of the last three-digit term of this progression.
3. 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 equal to?
What seems super difficult? Not at all. Logic and understanding of the meaning of geometric progression will save you. Well, the formula for the nth term, of course.
5. The third term of the geometric progression is -14, and the eighth term is 112. Find the denominator of the progression.
6. The sum of the first and second terms of the geometric progression is 75, and the sum of the second and third terms is 150. Find the sixth term of the progression.
Answers (in disarray): 6; -3888; -1; 800; -32; 448.
That's almost all. All we have to do is learn to count the sum of the first n terms of a geometric progression yes discover infinitely decreasing geometric progression and its amount. A very interesting and unusual thing, by the way! More on this in the next lessons.)
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's not hard to see that general formula nth term of the geometric progression 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 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?
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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:
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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 we're talking about, at most, 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 you sow the entire surface of the Earth, it would take at least 8 years to collect required quantity grains This legend is sometimes interpreted as indicating a practical unlimited possibilities, hidden in the chess game.
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?
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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 should this result be interpreted and why does it make any sense at all? for a long time it wasn't very clear.
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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 general theory conic sections, DC – diameter of the 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.”
A geometric progression is a numerical sequence, the first term of which is non-zero, and each subsequent term is equal to the previous term multiplied by the same non-zero number. Geometric progression is denoted b1,b2,b3, …, bn, …
Properties of geometric progression
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.
One of the ways to specify a geometric progression is to specify its first term b1 and the denominator of the geometric error q. For example, b1=4, q=-2. These two conditions define the geometric progression 4, -8, 16, -32, ….
If q>0 (q is not equal to 1), then the progression is a monotonic sequence. For example, the sequence, 2, 4,8,16,32, ... is a monotonically increasing sequence (b1=2, q=2).
If the denominator in the geometric error is q=1, then all terms of the geometric progression will be equal to each other. In such cases, the progression is said to be a constant sequence.
Formula for the nth term of the progression
In order for a number sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, be the geometric mean of 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 term of the geometric progression is:
bn=b1*q^(n-1), where n belongs to the set of natural numbers N.
Let's look at a simple example:
In geometric progression b1=6, q=3, n=8 find bn.
Let's use the formula for the nth term of a geometric progression.
Lesson and presentation on the topic: "Number sequences. Geometric progression"
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Guys, today we will get acquainted with another type of progression.
The topic of today's lesson is geometric progression.
Geometric progression
Definition. A numerical sequence in which each term, starting from the second, is equal to the product of the previous one and some fixed number is called a geometric progression.Let's define our sequence recursively: $b_(1)=b$, $b_(n)=b_(n-1)*q$,
where b and q are certain given numbers. The number q is called the denominator of the progression.
Example. 1,2,4,8,16... A geometric progression in which the first term is equal to one, and $q=2$.
Example. 8,8,8,8... A geometric progression in which the first term is equal to eight,
and $q=1$.
Example. 3,-3,3,-3,3... Geometric progression in which the first term is equal to three,
and $q=-1$.
Geometric progression has the properties of monotony.
If $b_(1)>0$, $q>1$,
then the sequence is increasing.
If $b_(1)>0$, $0 The sequence is usually denoted in the form: $b_(1), b_(2), b_(3), ..., b_(n), ...$.
Just like in an arithmetic progression, if in a geometric progression the number of elements is finite, then the progression is called a finite geometric progression.
$b_(1), b_(2), b_(3), ..., b_(n-2), b_(n-1), b_(n)$.
Note that if a sequence is a geometric progression, then the sequence of squares of terms is also a geometric progression. In the second sequence, the first term is equal to $b_(1)^2$, and the denominator is equal to $q^2$.
Formula for the nth term of a geometric progression
Geometric progression can also be specified in analytical form. Let's see how to do this:$b_(1)=b_(1)$.
$b_(2)=b_(1)*q$.
$b_(3)=b_(2)*q=b_(1)*q*q=b_(1)*q^2$.
$b_(4)=b_(3)*q=b_(1)*q^3$.
$b_(5)=b_(4)*q=b_(1)*q^4$.
We easily notice the pattern: $b_(n)=b_(1)*q^(n-1)$.
Our formula is called the "formula of the nth term of a geometric progression."
Let's return to our examples.
Example. 1,2,4,8,16... Geometric progression in which the first term is equal to one,
and $q=2$.
$b_(n)=1*2^(n)=2^(n-1)$.
Example. 16,8,4,2,1,1/2… A geometric progression in which the first term is equal to sixteen, and $q=\frac(1)(2)$.
$b_(n)=16*(\frac(1)(2))^(n-1)$.
Example. 8,8,8,8... A geometric progression in which the first term is equal to eight, and $q=1$.
$b_(n)=8*1^(n-1)=8$.
Example. 3,-3,3,-3,3... A geometric progression in which the first term is equal to three, and $q=-1$.
$b_(n)=3*(-1)^(n-1)$.
Example. Given a geometric progression $b_(1), b_(2), …, b_(n), … $.
a) It is known that $b_(1)=6, q=3$. Find $b_(5)$.
b) It is known that $b_(1)=6, q=2, b_(n)=768$. Find n.
c) It is known that $q=-2, b_(6)=96$. Find $b_(1)$.
d) It is known that $b_(1)=-2, b_(12)=4096$. Find q.
Solution.
a) $b_(5)=b_(1)*q^4=6*3^4=486$.
b) $b_n=b_1*q^(n-1)=6*2^(n-1)=768$.
$2^(n-1)=\frac(768)(6)=128$, since $2^7=128 => n-1=7; n=8$.
c) $b_(6)=b_(1)*q^5=b_(1)*(-2)^5=-32*b_(1)=96 => b_(1)=-3$.
d) $b_(12)=b_(1)*q^(11)=-2*q^(11)=4096 => q^(11)=-2048 => q=-2$.
Example. The difference between the seventh and fifth terms of the geometric progression is 192, the sum of the fifth and sixth terms of the progression is 192. Find the tenth term of this progression.
Solution.
We know that: $b_(7)-b_(5)=192$ and $b_(5)+b_(6)=192$.
We also know: $b_(5)=b_(1)*q^4$; $b_(6)=b_(1)*q^5$; $b_(7)=b_(1)*q^6$.
Then:
$b_(1)*q^6-b_(1)*q^4=192$.
$b_(1)*q^4+b_(1)*q^5=192$.
We received a system of equations:
$\begin(cases)b_(1)*q^4(q^2-1)=192\\b_(1)*q^4(1+q)=192\end(cases)$.
Equating our equations we get:
$b_(1)*q^4(q^2-1)=b_(1)*q^4(1+q)$.
$q^2-1=q+1$.
$q^2-q-2=0$.
We got two solutions q: $q_(1)=2, q_(2)=-1$.
Substitute sequentially into the second equation:
$b_(1)*2^4*3=192 => b_(1)=4$.
$b_(1)*(-1)^4*0=192 =>$ no solutions.
We got that: $b_(1)=4, q=2$.
Let's find the tenth term: $b_(10)=b_(1)*q^9=4*2^9=2048$.
Sum of a finite geometric progression
Let us have a finite geometric progression. Let's, just like for an arithmetic progression, calculate the sum of its terms.Let a finite geometric progression be given: $b_(1),b_(2),…,b_(n-1),b_(n)$.
Let us introduce the designation for the sum of its terms: $S_(n)=b_(1)+b_(2)+⋯+b_(n-1)+b_(n)$.
In the case when $q=1$. All terms of the geometric progression are equal to the first term, then it is obvious that $S_(n)=n*b_(1)$.
Let us now consider the case $q≠1$.
Let's multiply the above amount by q.
$S_(n)*q=(b_(1)+b_(2)+⋯+b_(n-1)+b_(n))*q=b_(1)*q+b_(2)*q+⋯ +b_(n-1)*q+b_(n)*q=b_(2)+b_(3)+⋯+b_(n)+b_(n)*q$.
Note:
$S_(n)=b_(1)+(b_(2)+⋯+b_(n-1)+b_(n))$.
$S_(n)*q=(b_(2)+⋯+b_(n-1)+b_(n))+b_(n)*q$.
$S_(n)*q-S_(n)=(b_(2)+⋯+b_(n-1)+b_(n))+b_(n)*q-b_(1)-(b_(2 )+⋯+b_(n-1)+b_(n))=b_(n)*q-b_(1)$.
$S_(n)(q-1)=b_(n)*q-b_(1)$.
$S_(n)=\frac(b_(n)*q-b_(1))(q-1)=\frac(b_(1)*q^(n-1)*q-b_(1)) (q-1)=\frac(b_(1)(q^(n)-1))(q-1)$.
$S_(n)=\frac(b_(1)(q^(n)-1))(q-1)$.
We have obtained the formula for the sum of a finite geometric progression.
Example.
Find the sum of the first seven terms of a geometric progression whose first term is 4 and the denominator is 3.
Solution.
$S_(7)=\frac(4*(3^(7)-1))(3-1)=2*(3^(7)-1)=4372$.
Example.
Find the fifth term of the geometric progression that is known: $b_(1)=-3$; $b_(n)=-3072$; $S_(n)=-4095$.
Solution.
$b_(n)=(-3)*q^(n-1)=-3072$.
$q^(n-1)=1024$.
$q^(n)=1024q$.
$S_(n)=\frac(-3*(q^(n)-1))(q-1)=-4095$.
$-4095(q-1)=-3*(q^(n)-1)$.
$-4095(q-1)=-3*(1024q-1)$.
$1365q-1365=1024q-1$.
$341q=$1364.
$q=4$.
$b_5=b_1*q^4=-3*4^4=-3*256=-768$.
Characteristic property of geometric progression
Guys, a geometric progression is given. Let's look at its three consecutive members: $b_(n-1),b_(n),b_(n+1)$.We know that:
$\frac(b_(n))(q)=b_(n-1)$.
$b_(n)*q=b_(n+1)$.
Then:
$\frac(b_(n))(q)*b_(n)*q=b_(n)^(2)=b_(n-1)*b_(n+1)$.
$b_(n)^(2)=b_(n-1)*b_(n+1)$.
If the progression is finite, then this equality holds for all terms except the first and last.
If it is not known in advance what form the sequence has, but it is known that: $b_(n)^(2)=b_(n-1)*b_(n+1)$.
Then we can safely say that this is a geometric progression.
A number sequence is a geometric progression only when the square of each member is equal to the product of the two adjacent members of the progression. Do not forget that for a finite progression this condition is not satisfied for the first and last terms.
Let's look at this identity: $\sqrt(b_(n)^(2))=\sqrt(b_(n-1)*b_(n+1))$.
$|b_(n)|=\sqrt(b_(n-1)*b_(n+1))$.
$\sqrt(a*b)$ is called the geometric mean of the numbers a and b.
The modulus of any term of a geometric progression is equal to the geometric mean of its two adjacent terms.
Example.
Find x such that $x+2; 2x+2; 3x+3$ were three consecutive terms of a geometric progression.
Solution.
Let's use the characteristic property:
$(2x+2)^2=(x+2)(3x+3)$.
$4x^2+8x+4=3x^2+3x+6x+6$.
$x^2-x-2=0$.
$x_(1)=2$ and $x_(2)=-1$.
Let us sequentially substitute our solutions into the original expression:
With $x=2$, we got the sequence: 4;6;9 – a geometric progression with $q=1.5$.
For $x=-1$, we get the sequence: 1;0;0.
Answer: $x=2.$
Problems to solve independently
1. Find the eighth first term of the geometric progression 16;-8;4;-2….2. Find the tenth term of the geometric progression 11,22,44….
3. It is known that $b_(1)=5, q=3$. Find $b_(7)$.
4. It is known that $b_(1)=8, q=-2, b_(n)=512$. Find n.
5. Find the sum of the first 11 terms of the geometric progression 3;12;48….
6. Find x such that $3x+4; 2x+4; x+5$ are three consecutive terms of a geometric progression.