How to calculate the sum of a geometric progression. Arithmetic and geometric progressions

If every natural number n match a real number a n , then they say that given number sequence :

a 1 , a 2 , a 3 , . . . , a n , . . . .

So, a numerical sequence is a function of a natural argument.

Number a 1 called the first member of the sequence , number a 2 — the second member of the sequence , number a 3 — third etc. Number a n called nth member of the sequence , and the natural number n — his number .

From two neighboring members a n and a n +1 member sequences a n +1 called subsequent (towards a n ), a a n — previous (towards a n +1 ).

To specify a sequence, you must specify a method that allows you to find a sequence member with any number.

Often the sequence is given with nth term formulas , that is, a formula that allows you to determine a sequence member by its number.

For example,

the sequence of positive odd numbers can be given by the formula

a n= 2n- 1,

and the sequence of alternating 1 and -1 - formula

b n = (-1)n +1 . ◄

The sequence can be determined recurrent formula, that is, a formula that expresses any member of the sequence, starting with some, through the previous (one or more) members.

For example,

if a 1 = 1 , a a n +1 = a n + 5

a 1 = 1,

a 2 = a 1 + 5 = 1 + 5 = 6,

a 3 = a 2 + 5 = 6 + 5 = 11,

a 4 = a 3 + 5 = 11 + 5 = 16,

a 5 = a 4 + 5 = 16 + 5 = 21.

If a a 1= 1, a 2 = 1, a n +2 = a n + a n +1 , then the first seven members of the numerical sequence are set as follows:

a 1 = 1,

a 2 = 1,

a 3 = a 1 + a 2 = 1 + 1 = 2,

a 4 = a 2 + a 3 = 1 + 2 = 3,

a 5 = a 3 + a 4 = 2 + 3 = 5,

a 6 = a 4 + a 5 = 3 + 5 = 8,

a 7 = a 5 + a 6 = 5 + 8 = 13. ◄

Sequences can be final and endless .

The sequence is called ultimate if it has a finite number of members. The sequence is called endless if it has infinitely many members.

For example,

sequence of two-digit natural numbers:

10, 11, 12, 13, . . . , 98, 99

final.

Prime number sequence:

2, 3, 5, 7, 11, 13, . . .

endless. ◄

The sequence is called increasing , if each of its members, starting from the second, is greater than the previous one.

The sequence is called waning , if each of its members, starting from the second, is less than the previous one.

For example,

2, 4, 6, 8, . . . , 2n, . . . is an ascending sequence;

1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . is a descending sequence. ◄

A sequence whose elements do not decrease with increasing number, or, conversely, do not increase, is called monotonous sequence .

Monotonic sequences, in particular, are increasing sequences and decreasing sequences.

Arithmetic progression

Arithmetic progression a sequence is called, each member of which, starting from the second, is equal to the previous one, to which the same number is added.

a 1 , a 2 , a 3 , . . . , a n, . . .

is an arithmetic progression if for any natural number n condition is met:

a n +1 = a n + d,

where d - some number.

Thus, the difference between the next and the previous members of a given arithmetic progression is always constant:

a 2 - a 1 = a 3 - a 2 = . . . = a n +1 - a n = d.

Number d called the difference of an arithmetic progression.

To set an arithmetic progression, it is enough to specify its first term and difference.

For example,

if a 1 = 3, d = 4 , then the first five terms of the sequence are found as follows:

a 1 =3,

a 2 = a 1 + d = 3 + 4 = 7,

a 3 = a 2 + d= 7 + 4 = 11,

a 4 = a 3 + d= 11 + 4 = 15,

a 5 = a 4 + d= 15 + 4 = 19. ◄

For an arithmetic progression with the first term a 1 and difference d her n

a n = a 1 + (n- 1)d.

For example,

find the thirtieth term of an arithmetic progression

1, 4, 7, 10, . . .

a 1 =1, d = 3,

a 30 = a 1 + (30 - 1)d= 1 + 29· 3 = 88. ◄

a n-1 = a 1 + (n- 2)d,

a n= a 1 + (n- 1)d,

a n +1 = a 1 + nd,

then obviously

a n=	a n-1 + a n+1
	2

each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the previous and subsequent members.

numbers a, b and c are consecutive members of some arithmetic progression if and only if one of them is equal to the arithmetic mean of the other two.

For example,

a n = 2n- 7 , is an arithmetic progression.

Let's use the statement above. We have:

a n = 2n- 7,

a n-1 = 2(n- 1) - 7 = 2n- 9,

a n+1 = 2(n+ 1) - 7 = 2n- 5.

Hence,

a n+1 + a n-1	=	2n- 5 + 2n- 9	= 2n- 7 = a n,
2		2

◄

Note that n -th member of an arithmetic progression can be found not only through a 1 , but also any previous a k

a n = a k + (n- k)d.

For example,

for a 5 can be written

a 5 = a 1 + 4d,

a 5 = a 2 + 3d,

a 5 = a 3 + 2d,

a 5 = a 4 + d. ◄

a n = a n-k + kd,

a n = a n+k - kd,

then obviously

a n=	a n-k + a n+k
	2

any member of an arithmetic progression, starting from the second, is equal to half the sum of the members of this arithmetic progression equally spaced from it.

In addition, for any arithmetic progression, the equality is true:

a m + a n = a k + a l,

m + n = k + l.

For example,

in arithmetic progression

1) a 10 = 28 = (25 + 31)/2 = (a 9 + a 11 )/2;

2) 28 = a 10 = a 3 + 7d= 7 + 7 3 = 7 + 21 = 28;

3) a 10= 28 = (19 + 37)/2 = (a 7 + a 13)/2;

4) a 2 + a 12 = a 5 + a 9, as

a 2 + a 12= 4 + 34 = 38,

a 5 + a 9 = 13 + 25 = 38. ◄

S n= a 1 + a 2 + a 3 + . . .+ a n,

first n members of an arithmetic progression is equal to the product of half the sum of the extreme terms by the number of terms:

From this, in particular, it follows that if it is necessary to sum the terms

a k, a k +1 , . . . , a n,

then the previous formula retains its structure:

For example,

in arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, . . .

S 10 = 1 + 4 + . . . + 28 = (1 + 28) · 10/2 = 145;

10 + 13 + 16 + 19 + 22 + 25 + 28 = S 10 - S 3 = (10 + 28 ) · (10 - 4 + 1)/2 = 133. ◄

If an arithmetic progression is given, then the quantities a 1 , a n, d, n andS n linked by two formulas:

Therefore, if the values of three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.

An arithmetic progression is a monotonic sequence. Wherein:

if d > 0 , then it is increasing;
if d < 0 , then it is decreasing;
if d = 0 , then the sequence will be stationary.

Geometric progression

geometric progression a sequence is called, each term of which, starting from the second, is equal to the previous one, multiplied by the same number.

b 1 , b 2 , b 3 , . . . , b n, . . .

is a geometric progression if for any natural number n condition is met:

b n +1 = b n · q,

where q ≠ 0 - some number.

Thus, the ratio of the next term of this geometric progression to the previous one is a constant number:

b 2 / b 1 = b 3 / b 2 = . . . = b n +1 / b n = q.

Number q called denominator of a geometric progression.

To set a geometric progression, it is enough to specify its first term and denominator.

For example,

if b 1 = 1, q = -3 , then the first five terms of the sequence are found as follows:

b 1 = 1,

b 2 = b 1 · q = 1 · (-3) = -3,

b 3 = b 2 · q= -3 · (-3) = 9,

b 4 = b 3 · q= 9 · (-3) = -27,

b 5 = b 4 · q= -27 · (-3) = 81. ◄

b 1 and denominator q her n -th term can be found by the formula:

b n = b 1 · q n -1 .

For example,

find the seventh term of a geometric progression 1, 2, 4, . . .

b 1 = 1, q = 2,

b 7 = b 1 · q 6 = 1 2 6 = 64. ◄

bn-1 = b 1 · q n -2 ,

b n = b 1 · q n -1 ,

b n +1 = b 1 · q n,

then obviously

b n 2 = b n -1 · b n +1 ,

each member of the geometric progression, starting from the second, is equal to the geometric mean (proportional) of the previous and subsequent members.

Since the converse is also true, the following assertion holds:

numbers a, b and c are consecutive members of some geometric progression if and only if the square of one of them is equal to the product of the other two, that is, one of the numbers is the geometric mean of the other two.

For example,

let us prove that the sequence given by the formula b n= -3 2 n , is a geometric progression. Let's use the statement above. We have:

b n= -3 2 n,

b n -1 = -3 2 n -1 ,

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

Hence,

b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) (-3 2 n +1 ) = b n -1 · b n +1 ,

which proves the required assertion. ◄

Note that n th term of a geometric progression can be found not only through b 1 , but also any previous term b k , for which it suffices to use the formula

b n = b k · q n - k.

For example,

for b 5 can be written

b 5 = b 1 · q 4 ,

b 5 = b 2 · q 3,

b 5 = b 3 · q2,

b 5 = b 4 · q. ◄

b n = b k · q n - k,

b n = b n - k · q k,

then obviously

b n 2 = b n - k· b n + k

the square of any member of a geometric progression, starting from the second, is equal to the product of the members of this progression equidistant from it.

In addition, for any geometric progression, the equality is true:

b m· b n= b k· b l,

m+ n= k+ l.

For example,

exponentially

1) b 6 2 = 32 2 = 1024 = 16 · 64 = b 5 · b 7 ;

2) 1024 = b 11 = b 6 · q 5 = 32 · 2 5 = 1024;

3) b 6 2 = 32 2 = 1024 = 8 · 128 = b 4 · b 8 ;

4) b 2 · b 7 = b 4 · b 5 , as

b 2 · b 7 = 2 · 64 = 128,

b 4 · b 5 = 8 · 16 = 128. ◄

S n= b 1 + b 2 + b 3 + . . . + b n

first n members of a geometric progression with a denominator q ≠ 0 calculated by the formula:

And when q = 1 - according to the formula

S n= n.b. 1

Note that if we need to sum the terms

b k, b k +1 , . . . , b n,

then the formula is used:

S n- Sk -1 = b k + b k +1 + . . . + b n = b k ·	1 - q n - k +1	.
	1 - q

For example,

exponentially 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, . . .

S 10 = 1 + 2 + . . . + 512 = 1 · (1 - 2 10) / (1 - 2) = 1023;

64 + 128 + 256 + 512 = S 10 - S 6 = 64 · (1 - 2 10-7+1) / (1 - 2) = 960. ◄

If a geometric progression is given, then the quantities b 1 , b n, q, n and S n linked by two formulas:

Therefore, if the values of any three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.

For a geometric progression with the first term b 1 and denominator q the following take place monotonicity properties :

the progression is increasing if one of the following conditions is met:

b 1 > 0 and q> 1;

b 1 < 0 and 0 < q< 1;

A progression is decreasing if one of the following conditions is met:

b 1 > 0 and 0 < q< 1;

b 1 < 0 and q> 1.

If a q< 0 , then the geometric progression is sign-alternating: its odd-numbered terms have the same sign as its first term, and even-numbered terms have the opposite sign. It is clear that an alternating geometric progression is not monotonic.

Product of the first n terms of a geometric progression can be calculated by the formula:

P n= b 1 · b 2 · b 3 · . . . · b n = (b 1 · b n) n / 2 .

For example,

1 · 2 · 4 · 8 · 16 · 32 · 64 · 128 = (1 · 128) 8/2 = 128 4 = 268 435 456;

3 · 6 · 12 · 24 · 48 = (3 · 48) 5/2 = (144 1/2) 5 = 12 5 = 248 832.◄

Infinitely decreasing geometric progression

Infinitely decreasing geometric progression is called an infinite geometric progression whose denominator modulus is less than 1 , i.e

|q| < 1 .

Note that an infinitely decreasing geometric progression may not be a decreasing sequence. This fits the case

1 < q< 0 .

With such a denominator, the sequence is sign-alternating. For example,

1, - 1 / 2 , 1 / 4 , - 1 / 8 , . . . .

The sum of an infinitely decreasing geometric progression name the number to which the sum of the first n terms of the progression with an unlimited increase in the number n . This number is always finite and is expressed by the formula

S= b 1 + b 2 + b 3 + . . . =	b 1	.
	1 - q

For example,

10 + 1 + 0,1 + 0,01 + . . . = 10 / (1 - 0,1) = 11 1 / 9 ,

10 - 1 + 0,1 - 0,01 + . . . = 10 / (1 + 0,1) = 9 1 / 11 . ◄

Relationship between arithmetic and geometric progressions

Arithmetic and geometric progressions are closely related. Let's consider just two examples.

a 1 , a 2 , a 3 , . . . d , then

b a 1 , b a 2 , b a 3 , . . . b d .

For example,

1, 3, 5, . . . — arithmetic progression with difference 2 and

7 1 , 7 3 , 7 5 , . . . is a geometric progression with a denominator 7 2 . ◄

b 1 , b 2 , b 3 , . . . is a geometric progression with a denominator q , then

log a b 1, log a b 2, log a b 3, . . . — arithmetic progression with difference log aq .

For example,

2, 12, 72, . . . is a geometric progression with a denominator 6 and

lg 2, lg 12, lg 72, . . . — arithmetic progression with difference lg 6 . ◄

A geometric progression is a new kind of number sequence that we have to get acquainted with. For a successful acquaintance, it does not hurt to at least know and understand. Then there will be no problem with geometric progression.)

What is a geometric progression? The concept of geometric progression.

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

1, 10, 100, 1000, 10000, …

Can you catch a pattern and tell which numbers will go next? The pepper is clear, the numbers 100000, 1000000 and so on will go further. Even without much mental stress, everything is clear, right?)

OK. Another example. I write the following sequence:

1, 2, 4, 8, 16, …

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

And now we turn again from sensations to rigorous mathematics.

Key moments of a geometric progression.

Key moment #1

The geometric progression is sequence of numbers. As is progression. Nothing tricky. Just arranged this sequence differently. Hence, of course, it has another name, yes ...

Key moment #2

With the second key point, the question will be trickier. Let's go back a little and remember the key property of an 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 look at the examples given. Guessed? Yes! In a geometric progression (any!) each of its members differs from the previous one in the same number of times. Always!

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

In the second example, this is a two: each member is greater than the previous one. twice.

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

Key moment #3

This key point is completely identical to that for an arithmetic progression. Namely: each member of the geometric progression is in its place. Everything is exactly the same as in arithmetic progression and comments, I think, are unnecessary. There is the first term, there is a hundred and first, and so on. Let's rearrange at least two members - the pattern (and with it the geometric progression) will disappear. What remains is just a sequence of numbers without any logic.

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

Terms and designations.

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

What is a geometric progression?

How is a geometric progression written in general terms? No problem! Each member of the progression is also written as a letter. For arithmetic progression only, the letter is usually used "a", for geometric - letter "b". Member number, as usual, is indicated lower right index. The members of the progression themselves are simply listed separated by commas or semicolons.

Like this:

b1,b 2 , b 3 , b 4 , b 5 , b 6 , …

Briefly, such a progression is written as follows: (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 designations. Everything is the same, only the letter is different, yes.) And now we go directly to the definition.

Definition 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.

That's the whole definition. Most of the words and phrases are clear and familiar to you. Unless, of course, you understand the meaning of a geometric progression "on the fingers" and in general. But there are also a few new phrases to which I would like to draw special attention.

First, the words: "the first term of which different from zero".

This restriction on the first term was not introduced by chance. What do you think will happen if the first term b 1 turns out to be zero? What will be the second term if each term is greater than the previous 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! And the third member? Zero too! And the fourth term is also zero! Etc…

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

0, 0, 0, 0, …

Of course, such a sequence has the right to life, but it is of no practical interest. Everything is so clear. Any of its members is zero. The sum of any number of members 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 a geometric progression. Let's start dating.)

The denominator of a geometric progression.

Everything is simple.

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

Again, by analogy with the arithmetic progression, the key word to pay attention to in this definition is the word "more". It means that each term of a geometric progression is obtained multiplication to this very denominator previous member.

I explain.

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

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

Denominator of a geometric progression usually denoted by a letter q.

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

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

2, 6, 18, 54, …

Everything is elementary. We take any sequence number. What we want is what we take. 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 a given geometric progression, the denominator is three.

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

1, -2, 4, -8, 16, …

All the same. Whatever signs the members themselves have, we still take any sequence number (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. It happens.)

Let's take this progression:

1, 1/3, 1/9, 1/27, …

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

We get:

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

But such a "progression" as you?

3, 3, 3, 3, 3, …

Obviously here q = 1 . Formally, this is also a geometric progression, only with same members.) But such progressions are not interesting for study and practical application. Just like 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. Didn't guess why?

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

We believe:

b 2 = b 1 · q= 2 0 = 0

And the third member?

b 3 = b 2 · q= 0 0 = 0

Types and behavior of geometric progressions.

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

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

As soon as the members behave here: they increase and decrease, and indefinitely approach zero, and even change signs, alternately rushing either to "plus" or to "minus"! And in all this diversity one must be able to understand well, yes ...

We understand?) Let's start with the simplest case.

The denominator is positive ( q >0)

With a positive denominator, firstly, the members of a geometric progression can go into plus infinity(i.e. increase indefinitely) and can go into minus infinity(i.e. decrease indefinitely). We have already got used to such behavior of progressions.

For example:

(b n): 1, 2, 4, 8, 16, …

Everything is simple here. Each member of the progression is more than the previous. And each member gets 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 indefinitely, going into space. Plus infinity...

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 already directly opposite: each member of the progression is obtained less than previous, and all its terms decrease indefinitely, 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. Deuce. And here behavior These two progressions are fundamentally different! Didn't guess why? Yes! It's all about first member! It is he, as they say, who orders the music.) 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 positive.

But in the second case, the first term negative(-one). Therefore, all subsequent members of the progression obtained by multiplying by positive q = +2 , will also be obtained negative. For "minus" to "plus" always gives "minus", yes.)

As you can see, unlike an arithmetic progression, a geometric progression can behave in completely different ways, 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 member b 1 and denominatorq .

And now we begin the analysis of less familiar, but much more interesting cases!

Take, for example, the following sequence:

(b n): 1, 1/2, 1/4, 1/8, 1/16, …

This sequence is also a geometric progression! Each member of this progression is also obtained multiplication the previous term, by the same number. Only the number is fractional: q = +1/2 . Or +0,5 . And (important!) number, smaller one:q = 1/2<1.

What is interesting about this geometric progression? Where are its members going? Let's get a look:

1/2 = 0,5;

1/4 = 0,25;

1/8 = 0,125;

1/16 = 0,0625;

…….

What is interesting here? First, the decrease in the members of the progression is immediately striking: each of its members smaller the previous exactly 2 times. Or, according to the definition of a geometric progression, each term more previous 1/2 times, because progression denominator q = 1/2 . And from multiplying by a positive number less than one, the result usually decreases, yes ...

What more can be seen in the behavior of this progression? Do its members disappear? unlimited, going to minus infinity? Not! They disappear in a special way. At first they decrease quite quickly, and then more and more slowly. And all the while staying positive. Albeit very, very small. And what are they striving for? Didn't guess? Yes! They tend to zero!) And, pay attention, the members of our progression never reach! Only infinitely close to him. It is very important.)

A similar situation will be in such a 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 members will approach zero from the other side, from below. Staying all the time negative.)

Such a geometric progression, the members of which approaching zero indefinitely.(it doesn’t matter, on the positive or negative side), in mathematics it has a special name - infinitely decreasing geometric progression. This progression is so interesting and unusual that it will even be separate lesson .)

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

To summarize:

positiveand more than one (q>1), then the members of the progression:

a) increase indefinitely (ifb 1 >0);

b) decrease indefinitely (ifb 1 <0).

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

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

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

It remains now to consider the case negative denominator.

The denominator is negative ( q <0)

We won't go far for an example. Why, in fact, shaggy grandmother ?!) Let, for example, the first member of the progression be b 1 = 1 , and 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 in odd places (first, third, fifth, etc.) will be positive, and in even places (second, fourth, etc.) - negative. Signs are strictly interleaved. Plus-minus-plus-minus ... Such a geometric progression is called - increasing sign alternating.

Where are its members going? And nowhere.) Yes, in absolute value (i.e. modulo) the terms of our progression increase indefinitely (hence the name "increasing"). But at the same time, each member of the progression alternately throws it into the heat, then into the cold. Either plus or minus. Our progression fluctuates... Moreover, the range of fluctuations grows rapidly with each step, yes.) Therefore, the aspirations of the members of the progression to go somewhere specifically here no. Neither to plus infinity, nor to minus infinity, nor to zero - nowhere.

Consider now 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 terms to approach zero.) Only this time our terms approach zero not strictly from above or below, but again hesitating. Alternately taking either positive or 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 alternating sign.

Why are these two examples interesting? And the fact that in both cases takes place alternating characters! Such a chip is typical only for progressions with a negative denominator, yes.) Therefore, if in some task you see a geometric progression with alternating members, then you will already firmly know that its denominator is 100% negative and you will not be mistaken in the sign.)

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

Remember:

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

At the same time, the members themselves:

a) increase indefinitelymodulo, ifq<-1;

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

That's all. All typical cases are analyzed.)

In the process of parsing 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 speech turns (and specific examples) are just an initial acquaintance with behavior various number sequences. An example of a geometric progression.

Why do we even need to know the progression behavior? What difference does it make where she goes? To zero, to plus infinity, to minus infinity ... What do we care about this?

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

Some examples from this section (sequences that have a limit) and in particular, infinitely decreasing geometric progression begin to learn at school. Getting used.)

Moreover, the ability to study the behavior of sequences well in the future will greatly play into the hands and will be very useful in function research. The most varied. But the ability to competently work with functions (calculate derivatives, explore them in full, build their graphs) already dramatically increases your mathematical level! Doubt? No need. Also remember my words.)

Let's look at a geometric progression in life?

In the life around us, we encounter exponential progression very, very often. Without even knowing it.)

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

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

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

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

You yourself are still, of course, young. You study at school, you don't apply to banks. But your parents are adults and independent people. They go to work, earn money for their daily bread, and put some 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 put 50,000 rubles in the bank at 10% per annum for a period of three years with annual interest capitalization. Moreover, nothing can be done with the deposit during this entire period. You can neither replenish the deposit nor withdraw money from the account. What profit will he make in these three years?

Well, firstly, you need to figure out what 10% per annum is. It means that in a year 10% will be added to the initial deposit amount by the bank. From what? Of course, from initial deposit amount.

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

So we consider 110% of 50,000 rubles:

50,000 1.1 \u003d 55,000 rubles.

I hope you understand that finding 110% of the value means multiplying this value by the number 1.1? If you do not understand why this is so, remember the fifth and sixth grades. Namely - the relationship of percentages with fractions and parts.)

Thus, the increase for the first year will be 5000 rubles.

How much money will be in the account after two years? 60,000 rubles? Unfortunately (or rather, fortunately), it's not that simple. The whole trick of interest capitalization is that with each new interest accrual, these same interest will be considered already from the new amount! From the one who already is on account At the moment. And the interest accrued for the previous term is added to the initial amount of the deposit and, thus, they themselves participate in the calculation of new interest! That is, they become a full part of the total account. or general capital. Hence the name - interest capitalization.

It's in the economy. And in mathematics, such percentages are called compound interest. Or percent of percent.) Their trick is that in sequential calculation, the percentages are calculated each time from the new value. Not from the original...

Therefore, in order to calculate the sum 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 consider 110% of 55,000 rubles:

55000 1.1 \u003d 60500 rubles.

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

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

Here we consider:

60500 1.1 \u003d 66550 rubles.

And now we build our monetary amounts by years 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? Why not a geometric progression? First member b 1 = 50000 , and the denominator q = 1,1 . Each term is strictly 1.1 times greater than the previous one. Everything is in strict accordance with the definition.)

And how many additional percentage bonuses will your dad "drop in" while his 50,000 rubles were in the bank account for three years?

We believe:

66550 - 50000 = 16550 rubles

It's bad, of course. But this is if the initial amount of the contribution is small. What if there's more? Say, not 50, but 200 thousand rubles? Then the increase for three years will already be 66,200 rubles (if you count). Which is already very good.) And if the contribution is even greater? That's what it is...

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

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

The simplest problems in 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 a geometric progression is 6, and the denominator is -0.5. Find the first, third and fourth terms.

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

b2 = 6

In addition, we also know progression denominator:

q = -0.5

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

Here we are acting. We write down the sequence according to the condition of the problem. Directly in general terms, where the second member is the six:

b1,6,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! We already know (directly in the sense of a geometric progression) that the third term (b 3) more than a second (b 2 ) in "q" once!

So we write:

b 3 =b 2 · q

We substitute the six in this expression instead of b 2 and -0.5 instead q and we think. And the minus is also not ignored, of course ...

b 3 \u003d 6 (-0.5) \u003d -3

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

We now consider the next, fourth term of the progression:

b 4 =b 3 · q

b 4 \u003d -3 (-0.5) \u003d 1.5

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

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

b1; 6; -3; 1.5; …

It remains now 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 share.

We divide and get:

That's all.) The answer to the problem will be as follows:

-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 term. Whatever we want, we will find one.) 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 task, according to tradition, is from the real version of the OGE:

…; 150; X; 6; 1.2; …

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

Here we are not afraid. All the same. Turn on your head and remember the elementary meaning of a geometric progression. We look carefully at our sequence and figure out which parameters of the geometric progression of the three main ones (the first member, denominator, member number) are hidden in it.

Member numbers? There are no member numbers, yes ... But there are four successive numbers. What this word means, I don’t see the point in explaining at this stage.) Are there two neighboring known numbers? There is! 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. For six.

We get:

x= 150 0.2 = 30

Answer: x = 30 .

As you can see, everything is quite simple. The main difficulty lies 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 change the problem a bit. Now it will get interesting! Let's remove the last number 1.2 in it. Let's solve this problem now:

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

…; 150; X; 6; …

Find the term of the progression, denoted by the letter x.

Everything is the same, only two neighboring famous we no longer have members of the progression. This is the main problem. Because the magnitude q through two neighboring terms, we can already easily determine we can't. Do we have a chance to meet the challenge? Certainly!

Let's write the unknown term " x"Directly in the sense of a geometric progression! In general terms.

Yes Yes! Directly with an unknown denominator!

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

x= 150q

On the other hand, we have every right to paint the 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 value (x), but two different ways.

We get the equation:

Multiplying everything by q, simplifying, reducing, we get the equation:

q 2 \u003d 1/25

We solve and get:

q = ±1/5 = ±0.2

Oops! The denominator is double! +0.2 and -0.2. And which one to choose? Dead end?

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

For q = +0.2 we'll get:

X \u003d 150 0.2 \u003d 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 condition of the problem!

Like these ones:

…; 150; 30; 6; …

…; 150; -30; 6; …

Both are suitable.) What do you think is the reason for the bifurcation of answers? Just because of the elimination of a specific member of the progression (1,2), coming after the six. And knowing only the previous (n-1)-th and subsequent (n+1)-th members of the geometric progression, we can no longer unequivocally say anything about the n-th member standing between them. There are two options - plus and minus.

But it doesn't matter. As a rule, in tasks for a geometric progression there is additional information that gives an unambiguous answer. Let's say the words: "sign-alternating progression" or "progression with a positive denominator" and so on... It is these words that should serve as a clue, which sign, plus or minus, should be chosen when making the final answer. If there is no such information, then - yes, the task will have two solutions.)

And now we decide on our own.

4. Determine if the number 20 will be a member of a geometric progression:

4 ; 6; 9; …

5. 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 -360, and its fifth term is 23.04. Find the first term of this progression.

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

Congratulations if everything worked out!

Something doesn't fit? Is there a double answer somewhere? We read the conditions of the assignment carefully!

The last puzzle doesn't work? Nothing complicated there.) We work directly according to the meaning of a 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 desired member very large? I would like, by analogy with an arithmetic progression, to somehow get a convenient formula that makes it easy to find any member of any geometric progression by his number. Without multiplying many, many times by q. And there is such a formula!) Details - in the next lesson.

Let's consider a series.

7 28 112 448 1792...

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

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

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

Accordingly, z ∈ N.

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

0.25 0.125 0.0625...

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

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

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

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

Varieties

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

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

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

Then the numerical sequence can be written like this:

3 6 12 24 48 ...

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

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

Then the numerical sequence can be written as follows:

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

Sign-variable. If q<0, то знаки у чисел последовательности постоянно чередуются вне зависимости от a 1 , а элементы ни возрастают, ни убывают.

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

Then the sequence can be written like this:

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

Formulas

For convenient use of geometric progressions, there are many formulas:

Formula of the z-th member. Allows you to calculate the element under a specific number without calculating the previous numbers.

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

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

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

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

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

The sum of a geometric progression, examples:a 1 = 2, q= -2. Calculate S 5 .

Decision:S 5 = 22 - calculation by formula.

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

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

Decision:Sz = 2 · = 4

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

Some properties:

characteristic property. If the following condition performed for anyz, then the given number series is a geometric progression:

a z 2 = a z -1 · az+1

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

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

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

Examples of some classical problems

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

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

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

Hence,a 3 = q 2 · a 1

When substitutingq= 4

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

Decision:To do this, it is enough to find q, the first element and substitute it into the formula.

a 3 = q· a 2 , hence,q= 2

a 2 = q a 1 ,That's why a 1 = 3

S 6 = 189

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

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

a 4 = q 3· a 1 = -80

Application example:

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

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

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

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

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

S = 1.06 1.06 1.06 1.06 10000 = 12625

Examples of tasks for calculating the sum:

In various problems, a geometric progression is used. An example for finding the sum can be given as follows:

a 1 = 4, q= 2, calculateS5.

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

S 5 = 124

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

Decision:

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

a 2 · q = a 3

q = 3

Similarly, we need to finda 1 , knowinga 2 andq.

a 1 · q = a 2

a 1 =2

S 6 = 728.

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

Definition of geometric progression

To begin with, we give the definition of this number series. A geometric progression is a series of rational numbers that is formed by successively multiplying its first element by a constant number called the denominator.

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

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

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

The denominator of a geometric progression

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

b > 1. There is an increasing series of rational numbers. For example, 1, 2, 4, 8, ... If the element a1 is negative, then the whole sequence will increase only modulo, but decrease taking into account the sign of the numbers.
b = 1. Often such a case is not called a progression, since there is an ordinary series of identical rational numbers. For example, -4, -4, -4.

Formula for sum

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

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

Infinitely decreasing sequence

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

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

Now we will consider several problems, where we will show how to apply the acquired knowledge to specific numbers.

Task number 1. Calculation of unknown elements of the progression and the sum

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

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

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

Task number 2. Determining the sum of arbitrary elements of the progression

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

The problem posed cannot be solved directly using known formulas. It can be solved in 2 different ways. For the sake of completeness, we present both.

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

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

Task number 3. What is the denominator?

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

According to the condition of the problem, it is not difficult to guess which formula should be used to solve it. Of course, for the sum of an infinitely decreasing progression. We have: S∞ = a1 / (1 - b). From where we express the denominator: b = 1 - a1 / S∞. It remains to substitute the known values and get the required number: b \u003d 1 - 2 / 3 \u003d -1 / 3 or -0.333 (3). We can check this result qualitatively if we remember that for this type of sequence, the modulus b must not go beyond 1. As you can see, |-1 / 3|

Task number 4. Restoring a series of numbers

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

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

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

Where are geometric progressions used?

If there were no application of this numerical series in practice, then its study would be reduced to a purely theoretical interest. But there is such an application.

The 3 most famous examples are listed below:

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

Consider now the question of summation of an infinite geometric progression. Let us call the partial sum of a given infinite progression the sum of its first terms. Denote the partial sum by the symbol

For every infinite progression

one can compose a (also infinite) sequence of its partial sums

Let a sequence with unlimited increase have a limit

In this case, the number S, i.e., the limit of partial sums of the progression, is called the sum of an infinite progression. We will prove that an infinite decreasing geometric progression always has a sum, and derive a formula for this sum (we can also show that for an infinite progression has no sum, does not exist).

We write the expression for the partial sum as the sum of the members of the progression according to formula (91.1) and consider the limit of the partial sum at

From the theorem of item 89 it is known that for a decreasing progression ; therefore, applying the difference limit theorem, we find

(the rule is also used here: the constant factor is taken out of the sign of the limit). The existence is proved, and at the same time the formula for the sum of an infinitely decreasing geometric progression is obtained:

Equality (92.1) can also be written as

Here it may seem paradoxical that a well-defined finite value is assigned to the sum of an infinite set of terms.

A clear illustration can be given to explain this situation. Consider a square with a side equal to one (Fig. 72). Let us divide this square by a horizontal line into two equal parts and apply the upper part to the lower one so that a rectangle is formed with sides 2 and . After that, we again divide the right half of this rectangle in half by a horizontal line and attach the upper part to the lower one (as shown in Fig. 72). Continuing this process, we are constantly transforming the original square with area equal to 1 into equal-sized figures (taking the form of a staircase with thinner steps).

With an infinite continuation of this process, the entire area of the square decomposes into an infinite number of terms - the areas of rectangles with bases equal to 1 and heights. The areas of the rectangles just form an infinite decreasing progression, its sum

i.e., as expected, is equal to the area of the square.

Example. Find the sums of the following infinite progressions:

Solution, a) We note that this progression Therefore, by the formula (92.2) we find

b) Here it means that by the same formula (92.2) we have

c) We find that this progression Therefore, this progression has no sum.

In Section 5, the application of the formula for the sum of terms of an infinitely decreasing progression to the conversion of a periodic decimal fraction into an ordinary fraction was shown.

Exercises

1. The sum of an infinitely decreasing geometric progression is 3/5, and the sum of its first four terms is 13/27. Find the first term and denominator of the progression.

2. Find four numbers that form an alternating geometric progression, in which the second term is less than the first by 35, and the third is greater than the fourth by 560.

3. Show what if sequence

forms an infinitely decreasing geometric progression, then the sequence

for any form an infinitely decreasing geometric progression. Does this assertion hold for

Derive a formula for the product of the terms of a geometric progression.