Geometric progression is the numerical sequence, the first term of which is different from zero, and each next term is equal to the previous member, multiplied by the same one and the same non-zero number. Geometric progression is denoted by B1, B2, B3, ..., BN, ...

Properties of geometric progression

The attitude of any member of the geometric error towards its previous member is equal to the same number, that is, B2 / B1 \u003d B3 / B2 \u003d B4 / B3 \u003d ... \u003d BN / B (N-1) \u003d B (n + 1) / BN \u003d .... This should be directly from the determination of arithmetic progression. This number is called a denominator of geometric progression. Typically, the denominator of geometric progression is denoted by the letter Q.

One way to set the geometric progression is to specify its first term B1 and denominator of the geometric error Q. For example, B1 \u003d 4, Q \u003d -2. These two conditions set the geometric progression of 4, -8, 16, -32, ....

If q\u003e 0 (q is not 1), then the progression is a monotonous sequence. For example, a sequence, 2, 4,8,16.32, ... is a monotonously increasing sequence (B1 \u003d 2, Q \u003d 2).

If in the geometric error of the denominator Q \u003d 1, then all members of the geometric progression will be equal to each other. In such cases, it is said that the progression is a constant sequence.

Formula N-th member of progression

In order for the numerical sequence (BN) to be a geometric progression, it is necessary that each of its member starting from the second is the average geometric neighboring members. That is, it is necessary to perform the following equation - (b (n + 1)) ^ 2 \u003d bn * b (n + 2), for any n\u003e 0, where N belongs to the set of natural numbers N.

The formula of the NO member of the geometric progression has the form:

bN \u003d B1 * Q ^ (n - 1), where N belongs to the set of natural numbers N.

Consider a simple example:

In the geometrical progression of B1 \u003d 6, Q \u003d 3, N \u003d 8 to find BN.

We use the formula of the NO member of the geometric progression.

Arithmetic and geometric progression

Theoretical information

	Arithmetic progression	Geometric progression
Definition	Arithmetic progression a N. The sequence is called, each member of which, starting from the second, is equal to the previous member, folded with the same number d. (d. - progression difference)	Geometric progression b N. The sequence of non-zero numbers is called, each member of which, starting from the second, is the previous member, multiplied by the same number q. (q. - denominator of progression)
Recurrent formula	For any natural n. a n + 1 \u003d a n + d	For any natural n. b n + 1 \u003d b n ∙ q, b n ≠ 0
NO Formula	a n \u003d a 1 + d (n - 1)	b n \u003d b 1 ∙ q n - 1, b n ≠ 0
Characteristic property
N-first members

Examples of tasks with comments

Exercise 1

In arithmetic progression ( a N.) a 1. = -6, a 2.

According to the formula of the NO member:

a 22. = a 1. + d (22 - 1) \u003d a 1. + 21 D.

By condition:

a 1. \u003d -6, then a 22. \u003d -6 + 21 d.

It is necessary to find the difference in progression:

d \u003d a 2 - A 1 = -8 – (-6) = -2

a 22. = -6 + 21 ∙ (-2) = - 48.

Answer: a 22. = -48.

Task 2.

Find the fifth member of the geometric progression: -3; 6; ....

1st method (using the N formula)

According to the formula of the NO member of the geometric progression:

b 5 \u003d B 1 ∙ Q 5 - 1 = b 1 ∙ Q 4.

As b 1. = -3,

2nd method (using a recurrent formula)

Since the progression denominator is -2 (Q \u003d -2), then:

b 3. = 6 ∙ (-2) = -12;

b 4. = -12 ∙ (-2) = 24;

b 5. = 24 ∙ (-2) = -48.

Answer: b 5. = -48.

Task 3.

In arithmetic progression ( a N) A 74 = 34; a 76. \u003d 156. Find a seventy-fifth member of this progression.

For arithmetic progression, the characteristic property has the form .

Therefore:

.

Substitute data in the formula:

Answer: 95.

Task 4.

In arithmetic progression ( a n) a n \u003d 3N - 4. Find the sum of seventeen first members.

To find the sum of the N first members of arithmetic progression, two formulas are used:

.

What of them are more convenient to apply?

Under the condition is known for the formula of the N-WHO member of the initial progression ( a N.) a N. \u003d 3n - 4 can be found immediately and a 1., I. a 16. without finding d. Therefore, we use the first formula.

Answer: 368.

Task 5.

In arithmetic progression ( a N.) a 1. = -6; a 2. \u003d -8. Find a twenty-second progression member.

According to the formula of the NO member:

a 22 \u003d A 1 + D (22 – 1) = a 1. + 21d.

Under the condition if a 1. \u003d -6 then a 22. \u003d -6 + 21d. It is necessary to find the difference in progression:

d \u003d a 2 - A 1 = -8 – (-6) = -2

a 22. = -6 + 21 ∙ (-2) = -48.

Answer: a 22. = -48.

Task 6.

Several consecutive members of the geometric progression are recorded:

Find a member of the progression indicated by the letter x.

When solving, we use the formula of the N-th member b n \u003d b 1 ∙ q n - 1 For geometric progressions. The first member of the progression. To find a denominator of the progression of Q, you must take any of the data of the progression of progression and divide into the previous one. In our example, you can take and divide on. We obtain that Q \u003d 3. Instead of N in the formula, we substitute 3, since it is necessary to find a third term given by geometric progression.

Substitting the found values \u200b\u200bin the formula, we get:

.

Answer:.

Task 7.

From the arithmetic progress given to the formula of the N-th member, select the one for which the condition is performed a 27. > 9:

Since the specified condition should be performed for the 27th member of the progression, we will substitute 27 instead of n into each of the four progressions. In the 4th progression we get:

.

Answer: 4.

Task 8.

In arithmetic progression a 1. \u003d 3, d \u003d -1.5. Specify the highest value of n for which inequality is performed a N. > -6.

Numeric sequences VI

§ L48. The amount of infinitely decreasing geometric progression

So far, speaking of sums, we always assumed that the number of components in these amounts of course (for example, 2, 15, 1000, etc.). But when solving some tasks (especially higher mathematics), it is necessary to face and with the sums of the infinite number of terms

S \u003d. a. 1 + a. 2 + ... + a. n. + ... . (1)

What are the amounts of themselves? A-priory the sum of the infinite number of terms a. 1 , a. 2 , ..., a. n. , ... called the sum of the sum s n. First p numbers when p -> ∞ :

S \u003d S. n. = (a. 1 + a. 2 + ... + a. n. ). (2)

The limit (2), of course, may exist, and may not exist. Accordingly, it is said that the sum (1) exists or does not exist.

How to find out if the amount (1) exists in each particular case? The general solution of this issue goes far beyond our program. However, there is one important private case that we have to consider now. It will be about the summation of members of infinitely decreasing geometric progression.

Let be a. 1 , a. 1 q. , a. 1 q. 2, ...- Infinitely decreasing geometric progression. This means that | q. |< 1. Сумма первых p members of this progression is equal

Of the main theorems about the limits of variable values \u200b\u200b(see § 136) we get:

But 1 \u003d 1, a q N. \u003d 0. Therefore

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

1) the amount of geometric progression 1, 1/3, 1/9, 1/2, ... equal

and the amount of geometric progression 12; -6; 3; - 3/2, ... equal

2) a simple periodic fraction 0,454545 ... to turn into an ordinary one.

To solve this problem, we will present this fraction in the form of an infinite amount:

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

The general rule of treatment of simple periodic fractions to ordinary (see Chapter II, § 38) can be obtained by the method described.

To appeal the simple periodic fraction in the ordinary, you need to do as follows: in the numerator put a period of decimal fraction, and in the denominator - a number consisting of nine times, taken as many times as signs in the period of decimal fraction.

3) Mixed periodic fraction 0,58333 .... turn into an ordinary one.

Imagine this fraction in the form of an infinite amount:

In the right-hand side of this equality, all the components, starting with 3/1000, form infinitely decreasing geometric progression, the first term of which is 3/1000, and the denominator 1/10. therefore

The described method can also obtain a general rule of circulation of mixed periodic fractions into ordinary (see ch. II, § 38). We consciously do not bring it here. Memorize this cumbersome rule is not necessary. It is much more useful to know that any mixed periodic fraction can be represented as a sum of infinitely decreasing geometric progression and some number. A formula

for the sum of infinitely decreasing geometric progression, it is necessary to remember.

We offer you as an exercise, in addition to the following tasks No. 995-1000, refer to the problem number 301 § 38 again.

Exercises

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

996. Find the sums of infinitely decreasing geometric progressions:

997. Under what values h. progression

is infinitely decreasing? Find the sum of such progression.

998. In the equilateral triangle of the party but entered by connecting a new triangle sideways; In this triangle, the same method entered a new triangle and so on to infinity.

a) the amount of perimeters of all these triangles;

b) the sum of their squares.

999. Square but entered by connecting the middle of its sides a new square; In this square, the square was inscribed in the same way and so on to infinity. Find the amount of perimeters of all these squares and the sum of their area.

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

Do you know an amazing legend of grains on a chessboard?

Legend of grains on a chessboard

When the Creator of Chess (Ancient Indian Mathematics named Sissy) showed his invention to the ruler of the country, so liked the game that he allowed the inventor to choose the reward himself. The sage asked the King for the first cell of the chessboard to pay him one wheat grain, for the second - two, for the third - four, etc., double the amount of grain on each of the next cell. The ruler who did not dealt in mathematics quickly agreed, even somewhat offended by such a low estimate of the invention, and ordered the treasurer to calculate and issue the necessary amount of grain to the inventor. However, when a week later, the treasurer was still unable to calculate how much the grain, the ruler asked what the cause of such a delay was. The treasurer showed him the calculations and said that it was impossible to pay it. With amazement, the king did not listen to the words of the elder.

Call me this is a monstrous number, "he said.

18 quintillas 446 quadrillions 744 trillion 73 billion 709 million 551 thousand 615, O Lady!

If we assume that one wheat grains has a lot of 0.065 grams, then the total weight of wheat on a chessboard will be 1,200 trillion tons, which exceeds the entire volume of wheat harvest, collected in the entire history of mankind!

Definition

Geometric progression - sequence of numbers ( members of Progression), in which each subsequent number, starting from the second, is obtained from the previous multiplication of it to a certain number ( denominator progression):

For example, a sequence of 1, 2, 4, 8, 16, ... - geometric ()

Geometric progression

Denominator geometric progression

Characteristic property of geometric progression

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The sequence is geometric if and only if, for any N\u003e 1, the specified ratio is performed.

In particular, for geometric progression with positive members, right:

Formula N-th member of geometric progression

Amount n first members of geometric progression

(if, then)

Infinitely decreasing geometric progression

When, geometric progression is called infinitely descending . The sum of infinitely decreasing geometric progression is the number and

Examples

Example 1..

Sequence () -Gometric progression.

Find if

Decision:

According to the formula, we have:

Prix \u200b\u200b2.

Find a denominator of geometric progression () in which

Geometric progression A sequence of numbers in which each member (starting from the second) is obtained from the previous one by multiplying it to the same number Q ≠ 0. The number Q is called denominator geometric progression. In order to set the geometric progression, it is necessary to specify its first term A 1 and the denominator Q.

Geometric progression increases with Q\u003e 1, decreases at 0< q < 1.

Examples of geometric progressions:

1. 2, 4, 8, 16 .... Here the first term is 1, and the denominator is 2.

81, 27, 9, 3, 1, 1/3 .... Here, the first term is 81, and the denominator is 1/3.

So, the first term of the progression is equal to A 1, the second - a 1 Q, the third a 1 Q * Q \u003d A 1 Q 2, the fourth A 1 Q 2 * Q \u003d A 1 Q 3 .... In this way, the n-th member of the progression is calculated by the formula A n \u003d A 1 Q n-1.

Statement: The sum of n members of geometric progression is calculated by the formula

S n \u003d a 1 + A 1 Q + A 1 Q 2 + A 1 Q 3 + ... + A 1 Q n-1.

Multiply on, we get:

S N Q \u003d A 1 Q + A 1 Q 2 + A 1 Q 3 + ... A 1 Q n.

Now read S n q from s n.

Examples of tasks for geometric progression.

1. Find the amount of the first 10 members of the geometric progression, if it is known that A 1 \u003d 3, Q \u003d 4.

2. For one minute, biomass increases by 2 times. What weight she will have in 5 minutes if its weight is 3 kg.

We are dealing with geometric progress, in which A 1 \u003d 3, and Q \u003d 2 To solve the task, we need to find the sixth member of this progression.