Geometric progression, along with arithmetic, is an important number series that is studied in the school algebra course in the 9th grade. In this article we will look at the denominator of a geometric progression and how its value affects its properties.
Definition of geometric progression
First, let's give the definition of this number series. Such a series is called a geometric progression rational numbers, which is formed by sequentially 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 you multiply 3 (the first element) by 2, you get 6. If you multiply 6 by 2, you 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 progression can be written in mathematical language as follows: an = bn-1 * a1, where b is the denominator. It's 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 in question. Similar reasoning can be continued for larger values of n.
Denominator of 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 element a1 is negative, then the entire sequence will increase only in absolute value, but decrease depending on the sign of the numbers.
- b = 1. Often this case is not called a progression, since there is an ordinary series of identical rational numbers. For example, -4, -4, -4.
Formula for amount
Before moving on to the consideration of specific problems using the denominator of the type of progression under consideration, it is necessary to give important formula for the sum of its first n elements. The formula looks like: Sn = (bn - 1) * a1 / (b - 1).
You can obtain this expression yourself if you consider the recursive sequence of terms of the progression. Also note that in the above formula it is enough to know only the first element and the denominator to find the sum of an arbitrary number of terms.
Infinitely decreasing sequence
An explanation was given above 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 let's look at several problems where we will show how to apply the acquired knowledge on specific numbers.
Task No. 1. Calculation of unknown elements of progression and sum
Given a geometric progression, the denominator of the progression is 2, and its first element is 3. What will its 7th and 10th terms be equal to, 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 element 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 term: a10 = 29 * 3 = 1536.
Let's 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.
Problem No. 2. Determining the sum of arbitrary elements of a progression
Let -2 be equal to the denominator of the geometric 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 ways various methods. For completeness of presentation of the topic, we present both.
Method 1. The idea is simple: you need to calculate the two corresponding sums of the first terms, and then subtract the other from one. We calculate the smaller amount: S10 = ((-2)10 - 1) * 4 / (-2 - 1) = -1364. Now let's calculate a large amount: S4 = ((-2)4 - 1) * 4 / (-2 - 1) = -20. Note that in the last expression only 4 terms were summed, since the 5th is already included in the amount that needs to be calculated according to the conditions of the problem. Finally, we take the difference: S510 = S10 - S4 = -1364 - (-20) = -1344.
Method 2. Before substituting numbers and counting, you can obtain a formula for the sum between the m and n terms of the series in question. We do exactly the same as in method 1, only we first work with the symbolic representation of the amount. 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.
Problem No. 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.
Based on the conditions of the problem, it is not difficult to guess which formula should be used to solve it. Of course, for the sum of the progression infinitely decreasing. We have: S∞ = a1 / (1 - b). From where we express the denominator: b = 1 - a1 / S∞. All that remains is to substitute known values and get the required number: b = 1 - 2 / 3 = -1 / 3 or -0.333(3). We can qualitatively check this result if we remember that for this type of sequence the modulus b should not go beyond 1. As can be seen, |-1 / 3|
Task No. 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 reconstruct 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 term. We have: a5 = b4 * a1 and a10 = b9 * a1. Now 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 root of the ratio of the terms known from the problem statement, b = 1.148698. We substitute the resulting number into one of the expressions for the known element, we get: a1 = a5 / b4 = 30 / (1.148698)4 = 17.2304966.
Thus, we found the denominator of the progression bn, and the geometric progression bn-1 * 17.2304966 = an, where b = 1.148698.
Where are geometric progressions used?
If there were no practical application of this number series, then its study would be reduced to purely theoretical interest. But such an application exists.
Below are the 3 most famous examples:
- Zeno's paradox, in which the nimble Achilles cannot catch up with the slow tortoise, is solved using the concept of an infinitely decreasing sequence of numbers.
- If for each cell chessboard put wheat grains so that on the 1st cell you put 1 grain, on the 2nd - 2, on the 3rd - 3 and so on, then to fill all the cells of the board you will need 18446744073709551615 grains!
- In the game "Tower of Hanoi", in order to move disks from one rod to another, it is necessary to perform 2n - 1 operations, that is, their number grows exponentially with the number n of disks used.
NUMERIC SEQUENCES VI
§ l48. Sum of an infinitely decreasing geometric progression
Until now, when talking about 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, they say that the sum (1) exists or does not exist.
How can we find out whether sum (1) exists in each specific case? Common decision This issue goes far beyond the scope of our program. However, there is one important special case, which we now have to consider. We will talk about summing the terms of an infinitely decreasing geometric progression.
Let a 1 , a 1 q , a 1 q 2, ... is an infinitely decreasing geometric progression. This means that | q |< 1. Сумма первых P terms of this progression is equal
From the basic theorems on the limits of variables (see § 136) we obtain:
But 1 = 1, a qn = 0. Therefore
So, the sum of an infinitely decreasing geometric progression is equal to the first term of this progression divided by one minus the denominator of this progression.
1) The sum of the geometric progression 1, 1/3, 1/9, 1/27, ... is equal to
and the sum of the geometric progression is 12; -6; 3; - 3 / 2 , ... equal
2) Convert a simple periodic fraction 0.454545 ... into an ordinary one.
To solve this problem, imagine this fraction as an infinite sum:
Right part This equality is the sum of an infinitely decreasing geometric progression, the first term of which is equal to 45/100, and the denominator is 1/100. That's why
Using the described method, it can also be obtained general rule conversion of simple periodic fractions into ordinary ones (see Chapter II, § 38):
To convert a simple periodic fraction into an ordinary fraction, you need to do the following: put the period in the numerator decimal, and the denominator is a number consisting of nines taken as many times as there are digits in the period of the decimal fraction.
3) Convert the mixed periodic fraction 0.58333 .... into an ordinary fraction.
Let's imagine 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 equal to 3/1000, and the denominator is 1/10. That's why
Using the described method, a general rule for converting mixed periodic fractions into ordinary fractions can be obtained (see Chapter II, § 38). We deliberately do not present it here. There is no need to remember 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 a certain number. And the formula
for the sum of an infinitely decreasing geometric progression, you must, of course, remember.
As an exercise, we suggest that you, in addition to the problems No. 995-1000 given below, once again turn to problem No. 301 § 38.
Exercises
995. What is called the sum of an infinitely decreasing geometric progression?
996. Find the sums of infinitely decreasing geometric progressions:
997. At what values X progression
is it infinitely decreasing? Find the sum of such a progression.
998. In an equilateral triangle with 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. Square with 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. Compose 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.
Arithmetic and geometric progressions
Theoretical information
Theoretical information
Arithmetic progression |
Geometric progression |
|
Definition |
Arithmetic progression a n is a sequence in which each member, starting from the second, is equal to the previous member added to the same number d (d- progression difference) |
Geometric progression b n is a sequence of non-zero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression) |
Recurrence formula |
For any natural n |
For any natural n |
Formula nth term |
a n = a 1 + d (n – 1) |
b n = b 1 ∙ q n - 1 , b n ≠ 0 |
| Characteristic property |  |
|
| Sum of the first n terms |  |
 |
Examples of tasks with comments
Exercise 1
In arithmetic progression ( a n) a 1 = -6, a 2
According to the formula of the nth term:
a 22 = a 1+ d (22 - 1) = a 1+ 21 d
By condition:
a 1= -6, then a 22= -6 + 21 d .
It is necessary to find the difference of progressions:
d = a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = - 48.
Answer : a 22 = -48.
Task 2
Find the fifth term of the geometric progression: -3; 6;....
1st method (using the n-term formula)
According to the formula for the nth term of a geometric progression:
b 5 = b 1 ∙ q 5 - 1 = b 1 ∙ q 4.
Because b 1 = -3,
2nd method (using recurrent formula)
Since the denominator of the progression is -2 (q = -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= 156. Find the seventy-fifth term of this progression.
For an arithmetic progression, the characteristic property has the form 
.
Therefore:
.
Let's substitute the data into the formula:
Answer: 95.
Task 4
In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.
To find the sum of the first n terms of an arithmetic progression, two formulas are used:
.
Which one is in in this case more convenient to use?
By condition, the formula for the nth term of the original progression is known ( a n) a n= 3n - 4. You can find immediately and a 1, And a 16 without finding d. Therefore, we will use the first formula.
Answer: 368.
Task 5
In arithmetic progression( a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.
According to the formula of the nth term:
a 22 = a 1 + d (22 – 1) = a 1+ 21d.
By condition, if a 1= -6, then a 22= -6 + 21d . It is necessary to find the difference of progressions:
d = a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = -48.
Answer : a 22 = -48.
Task 6
Several consecutive terms of the geometric progression are written:
Find the term of the progression indicated by x.
When solving, we will use the formula for the nth term b n = b 1 ∙ q n - 1 for geometric progressions. The first term of the progression. To find the denominator of the progression q, you need to take any of the given terms of the progression and divide by the previous one. In our example, we can take and divide by. We obtain that q = 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.
Substituting the found values into the formula, we get:
.
Answer : .
Task 7
From the arithmetic progressions given by the formula of the nth term, select the one for which the condition is satisfied a 27 > 9:
Since the given condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:
.
Answer: 4.
Task 8
In arithmetic progression a 1= 3, d = -1.5. Specify highest value n for which the inequality holds a n > -6.
Instructions
10, 30, 90, 270...
You need to find the denominator of a geometric progression.
Solution:
Option 1. Let's take an arbitrary term of the progression (for example, 90) and divide it by the previous one (30): 90/30=3.
If the sum of several terms of a geometric progression or the sum of all terms 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 terms of the progression with a denominator less than one).
Example.
The first term of a decreasing geometric progression is equal to one, and the sum of all its terms is equal to two.
It is required to determine the denominator of this progression.
Solution:
Substitute the data from the problem into the formula. It will turn out:
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 a certain number q, called the denominator of the progression.
Instructions
If two adjacent geometric terms b(n+1) and b(n) are known, to obtain the denominator, you need to divide the number with the larger one by the one preceding it: q=b(n+1)/b(n). This follows from the definition of progression and its denominator. An important condition is the inequality of the first term and the denominator of the progression to zero, otherwise it is considered indefinite.
Thus, the following relationships are established between the terms of the progression: b2=b1 q, b3=b2 q, ... , b(n)=b(n-1) q. Using the formula b(n)=b1 q^(n-1), any term of the geometric progression in which the denominator q and the term b1 are known can be calculated. Also, each of the progressions is equal in modulus to the average of its neighboring members: |b(n)|=√, which is where the progression got its .
An analogue of a geometric progression is the simplest exponential function y=a^x, where x is an exponent, a is a certain number. In this case, the denominator of the progression coincides with the first term and is equal to the number a. The value of the function y can be understood as nth term progression if the argument x is taken to be natural number n (counter).
Another important property geometric progression, which gave geometric progression
Let us now consider the question of summing an infinite geometric progression. Let us call the partial sum of a given infinite progression the sum of its first terms. Let us 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 a progression, is called the sum of an infinite progression. We will prove that an infinite decreasing geometric progression always has a sum, and we will derive a formula for this sum (we can also show that if an infinite progression has no sum, it does not exist).
Let us write the expression for the partial sum as the sum of terms of the progression using formula (91.1) and consider the limit of the partial sum at
From Theorem 89 it is known that for a decreasing progression; therefore, applying the difference limit theorem, we find
(here the rule is also used: the constant factor is taken beyond the limit sign). The existence is proven, 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 in the form
Here it may seem paradoxical that the sum of an infinite number of terms is assigned a very definite finite value.
You can cite visual illustration to explain this situation. Consider a square with side equal to one(Fig. 72). Let's divide this square 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 this, we will again divide the right half of this rectangle in half with a horizontal line and attach the upper part to the lower one (as shown in Fig. 72). Continuing this process, we continually transform the original square with area equal to 1 into equal-sized figures (taking the form of a staircase with thinning steps).
With the infinite continuation of this process, the entire area of the square is decomposed into an infinite number of terms - the areas of rectangles with bases equal to 1 and heights. The areas of rectangles precisely form an infinite decreasing progression, its sum
i.e., as one would expect, equal to the area of the square.
Example. Find the sums of the following infinite progressions:
Solution, a) We notice that this progression Therefore, using formula (92.2) we find
b) Here it means that using the same formula (92.2) we have
c) We find that this progression therefore has no sum.
In paragraph 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 that if the sequence
forms an infinitely decreasing geometric progression, then the sequence
for any, it forms an infinitely decreasing geometric progression. Will this statement hold true when
Derive a formula for the product of the terms of a geometric progression.