How to find the divisor of a geometric progression. The concept of geometric progression

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

Definition of geometric progression

First, let's define this number series. A geometric progression is a series rational numbers, which is formed by sequentially multiplying its first element by constant number, which is 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 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, one should bring important formula 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, when raised to great degrees tends to zero, i.e. 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 equal 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 at hand cannot be solved directly using known formulas. You can solve it with 2 various methods. 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). In the resulting expression, you can substitute known numbers 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 known values and get the required number: b = 1 - 2 / 3 = -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.

If every natural number n put in line 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 sequences , a 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 terms number sequence 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 she has finite number 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 given arithmetic progression, then the quantities a 1 , a n, d, n andS n linked by two formulas:

Therefore, if 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 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 . ◄

Instruction

10, 30, 90, 270...

It is required to find the denominator of a geometric progression.
Decision:

1 option. Let's take an arbitrary member of the progression (for example, 90) and divide it by the previous one (30): 90/30=3.

If the sum of several members of a geometric progression or the sum of all members of a decreasing geometric progression is known, then to find the denominator of the progression, use the appropriate formulas:
Sn = b1*(1-q^n)/(1-q), where Sn is the sum of the first n terms of the geometric progression and
S = b1/(1-q), where S is the sum of an infinitely decreasing geometric progression (the sum of all members of the progression with a denominator less than one).
Example.

First term of a decreasing geometric progression equal to one, and the sum of all its terms is equal to two.

It is required to determine the denominator of this progression.
Decision:

Substitute the data from the task into the formula. Get:
2=1/(1-q), whence – q=1/2.

A progression is a sequence of numbers. In a geometric progression, each subsequent term is obtained by multiplying the previous one by some number q, called the denominator of the progression.

Instruction

If two neighboring members of the geometric b(n+1) and b(n) are known, in order to get the denominator, it is necessary to divide the number with a large number by the one preceding it: q=b(n+1)/b(n). This follows from the definition of the progression and its denominator. An important condition is the inequality zero of the first term and denominator of the progression, otherwise it is considered indefinite.

Thus, the following relations are established between the members of the progression: b2=b1 q, b3=b2 q, … , b(n)=b(n-1) q. By the formula b(n)=b1 q^(n-1) any member of a geometric progression can be calculated, in which the denominator q and the member b1 are known. Also, each of the progression modulo is equal to the average of its neighboring members: |b(n)|=√, hence the progression got its .

An analogue of a geometric progression is the simplest exponential function y=a^x, where x is in the exponent, a is some number. In this case, the denominator of the progression is the same as the first term and is equal to the number a. The value of the function y can be understood as nth member progressions, if the argument x is taken as a natural number n (counter).

Exists for the sum of the first n members of a geometric progression: S(n)=b1 (1-q^n)/(1-q). This formula valid for q≠1. If q=1, then the sum of the first n terms is calculated by the formula S(n)=n b1. By the way, the progression will be called increasing for q greater than one and positive b1. When the denominator of the progression, modulo not exceeding one, the progression will be called decreasing.

special case geometric progression - an infinitely decreasing geometric progression (b.u.g.p.). The fact is that the members of a decreasing geometric progression will decrease over and over again, but will never reach zero. Despite this, it is possible to find the sum of all terms of such a progression. It is determined by the formula S=b1/(1-q). Total n members are infinite.

To visualize how you can add an infinite number of numbers and not get infinity, bake a cake. Cut off half of it. Then cut 1/2 off the half, and so on. The pieces that you will get are nothing more than members of an infinitely decreasing geometric progression with a denominator of 1/2. If you put all these pieces together, you get the original cake.

Geometry problems are special variety exercises that require spatial thinking. If you can't solve the geometric task try to follow the rules below.

Instruction

Read the condition of the problem very carefully, if you don’t remember or don’t understand something, re-read it again.

Try to figure out what kind geometric problems it is, for example: computational, when you need to find out some value, tasks that require a logical chain of reasoning, tasks to build using a compass and ruler. More tasks mixed type. Once you've figured out the type of problem, try to think logically.

Apply the necessary theorem for this problem, if there are doubts or there are no options at all, then try to remember the theory that you studied on the relevant topic.

Make a draft of the problem as well. Try to apply known ways checking the correctness of your solution.

Complete the solution of the problem neatly in a notebook, without blots and strikethroughs, and most importantly -. Perhaps it will take time and effort to solve the first geometric problems. However, once you get the hang of this process, you'll start clicking tasks like nuts and having fun doing it!

A geometric progression is a sequence of numbers b1, b2, b3, ... , b(n-1), b(n) such that b2=b1*q, b3=b2*q, ... , b(n) =b(n-1)*q, b1≠0, q≠0. In other words, each member of the progression is obtained from the previous one by multiplying it by some non-zero denominator of the progression q.

Instruction

Problems on a progression are most often solved by compiling and following a system with respect to the first term of the progression b1 and the denominator of the progression q. To write equations, it is useful to remember some formulas.

How to express the n-th member of the progression through the first member of the progression and the denominator of the progression: b(n)=b1*q^(n-1).

Consider separately the case |q|<1. Если знаменатель прогрессии по модулю меньше единицы, имеем бесконечно убывающую геометрическую . Сумма первых n членов бесконечно убывающей геометрической прогрессии ищется так же, как и для неубывающей геометрической прогрессии. Однако в случае бесконечно убывающей геометрической прогрессии можно найти также сумму всех членов этой прогрессии, поскольку при бесконечном n будет бесконечно уменьшаться значение b(n), и сумма всех членов будет стремиться к определенному пределу. Итак, сумма всех членов бесконечно убывающей геометрической прогрессии

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.

NUMERICAL SEQUENCES VI

§ l48. The sum of an infinitely decreasing geometric progression

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let's represent this fraction as an infinite sum:

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

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

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

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

Exercises

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

996. Find sums of infinitely decreasing geometric progressions:

997. For what values X progression

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

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

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

b) the sum of their areas.

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

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