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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
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:
$b_(1)=b_(1)$.
$b_(2)=b_(1)*q$.
$b_(5)=b_(4)*q=b_(1)*q^4$.
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.
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, … .
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.
Monotonous and constant sequence
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 monotonous 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 they say that progression is constant sequence.
$b_(1), b_(2), b_(3), ..., b_(n-2), b_(n-1), b_(n)$.
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 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.
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 has the form:
Sn = (bn*q - b1)/(q-1), where q is not equal to 1.
Let's look at 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) = 19,680.
Geometric progression is the new kind numerical 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 it turns out 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, by analogy with arithmetic progression, keyword The thing to pay attention to 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 for study and practical application not interesting. 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 exactly the previous one 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 this 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 go to school, 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.
>>Math: Geometric progression
For the convenience of the reader, this paragraph is constructed exactly according to the same plan that we followed in the previous paragraph.
1. Basic concepts.
Definition. A numerical sequence, all members of which are different from 0 and each member of which, starting from the second, is obtained from the previous member by multiplying it by the same number is called a geometric progression. In this case, the number 5 is called the denominator of a geometric progression.
Thus, a geometric progression is a numerical sequence (b n) defined recurrently by the relations
Is it possible, looking at number sequence, determine whether it is a geometric progression? Can. If you are convinced that the ratio of any member of the sequence to the previous member is constant, then you have a geometric progression.
Example 1.
1, 3, 9, 27, 81,... .
b 1 = 1, q = 3.
Example 2.
This is a geometric progression that has
Example 3.
This is a geometric progression that has
Example 4.
8, 8, 8, 8, 8, 8,....
This is a geometric progression in which b 1 - 8, q = 1.
Note that this sequence is also an arithmetic progression (see example 3 from § 15).
Example 5.
2,-2,2,-2,2,-2.....
This is a geometric progression in which b 1 = 2, q = -1.
Obviously, a geometric progression is an increasing sequence if b 1 > 0, q > 1 (see example 1), and a decreasing sequence if b 1 > 0, 0< q < 1 (см. пример 2).
To indicate that the sequence (b n) is a geometric progression, the following notation is sometimes convenient:
The icon replaces the phrase “geometric progression”.
Let us note one curious and at the same time quite obvious property of geometric progression:
If the sequence 
is a geometric progression, then the sequence of squares, i.e. 
is a geometric progression.
In the second geometric progression, the first term is equal to and equal to q 2.
If in a geometric progression we discard all terms following b n , we get a finite geometric progression 
In further paragraphs of this section we will consider the most important properties of geometric progression.
2. Formula for the nth term of a geometric progression.
Consider a geometric progression 
denominator q. We have:
It is not difficult to guess that for any number n the equality is true
This is the formula for the nth term of a geometric progression.
Comment.
If you have read the important remark from the previous paragraph and understood it, then try to prove formula (1) using the method mathematical induction in the same way as was done for the formula for the nth term of an arithmetic progression.
Let's rewrite the formula for the nth term of the geometric progression
and introduce the notation: We get y = mq 2, or, in more detail, 
The argument x is contained in the exponent, so this function is called an exponential function. This means that a geometric progression can be considered as an exponential function defined on the set N of natural numbers. In Fig. 96a shows the graph of the function Fig. 966 - function graph 
In both cases we have isolated points(with abscissas x = 1, x = 2, x = 3, etc.) lying on a certain curve (both figures show the same curve, only differently located and depicted on different scales). This curve is called an exponential curve. Read more about exponential function and its graphics will be discussed in the 11th grade algebra course.
Let's return to examples 1-5 from the previous paragraph.
1) 1, 3, 9, 27, 81,... . This is a geometric progression for which b 1 = 1, q = 3. Let’s create the formula for the nth term 
2) 
This is a geometric progression for which Let's create a formula for the nth term 
This is a geometric progression that has 
Let's create the formula for the nth term 
4) 8, 8, 8, ..., 8, ... . This is a geometric progression for which b 1 = 8, q = 1. Let’s create the formula for the nth term 
5) 2, -2, 2, -2, 2, -2,.... This is a geometric progression in which b 1 = 2, q = -1. Let's create the formula for the nth term 
Example 6.
Given a geometric progression
In all cases, the solution is based on the formula of the nth term of the geometric progression
a) Putting n = 6 in the formula for the nth term of the geometric progression, we obtain
b) We have
Since 512 = 2 9, we get n - 1 = 9, n = 10.
d) We have
Example 7.
The difference between the seventh and fifth terms of the geometric progression is 48, the sum of the fifth and sixth terms of the progression is also 48. Find the twelfth term of this progression.
First stage. Drawing up a mathematical model.
The conditions of the problem can be briefly written as follows:
Using the formula for the nth term of a geometric progression, we get:
Then the second condition of the problem (b 7 - b 5 = 48) can be written as
The third condition of the problem (b 5 + b 6 = 48) can be written as
As a result, we obtain a system of two equations with two variables b 1 and q:
which, in combination with condition 1) written above, represents a mathematical model of the problem.
Second phase.
Working with the compiled model. Equating the left sides of both equations of the system, we obtain:
(we divided both sides of the equation by the non-zero expression b 1 q 4).
From the equation q 2 - q - 2 = 0 we find q 1 = 2, q 2 = -1. Substituting the value q = 2 into the second equation of the system, we get 
Substituting the value q = -1 into the second equation of the system, we obtain b 1 1 0 = 48; this equation has no solutions.
So, b 1 =1, q = 2 - this pair is the solution to the compiled system of equations.
Now we can write down the geometric progression discussed in the problem: 1, 2, 4, 8, 16, 32, ... .
Third stage.
Answer to the problem question. You need to calculate b 12. We have
Answer: b 12 = 2048.
3. Formula for the sum of terms of a finite geometric progression.
Let a finite geometric progression be given
Let us denote by S n the sum of its terms, i.e.
Let us derive a formula for finding this amount.
Let's start with the simplest case, when q = 1. Then the geometric progression b 1 , b 2 , b 3 ,..., bn consists of n numbers equal to b 1 , i.e. the progression looks like b 1, b 2, b 3, ..., b 4. The sum of these numbers is nb 1.
Let now q = 1 To find S n, we apply an artificial technique: we perform some transformations of the expression S n q. We have:
When performing transformations, we, firstly, used the definition of a geometric progression, according to which (see the third line of reasoning); secondly, they added and subtracted, which is why the meaning of the expression, of course, did not change (see the fourth line of reasoning); thirdly, we used the formula for the nth term of a geometric progression:
From formula (1) we find:
This is the formula for the sum of n terms of a geometric progression (for the case when q = 1).
Example 8.
Given a finite geometric progression
a) the sum of the terms of the progression; b) the sum of the squares of its terms.
b) Above (see p. 132) we have already noted that if all terms of a geometric progression are squared, then we get a geometric progression with the first term b 2 and the denominator q 2. Then the sum of the six terms of the new progression will be calculated by
Example 9.
Find the 8th term of the geometric progression for which
In fact, we have proven the following theorem.
A numerical sequence is a geometric progression if and only if the square of each of its terms, except the first Theorem (and the last, in the case of a finite sequence), is equal to the product of the preceding and subsequent terms (a characteristic property of a geometric progression).
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 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?
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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 to interpret this result and why it makes any sense at all was not very clear for a long time.
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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 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 common ground AD, and the heights differ by a factor of 2), 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.”