I will solve the exam geometric progression. Formula of the nth term of a geometric progression
Geometric progression is numerical sequence, the first term of which is nonzero, and each next term is equal to the previous term, multiplied by the same nonzero number. The geometric progression is denoted by b1, b2, b3,…, bn,…
Properties of a geometric progression
The ratio of any term of the geometric error to its previous term is equal to the same number, that is, b2 / b1 = b3 / b2 = b4 / b3 =… = bn / b (n-1) = b (n + 1) / bn = …. This follows directly from the definition arithmetic progression... This number is called the denominator of the geometric progression. Usually, the denominator of a geometric progression is denoted by the letter q.
One of the ways to specify a geometric progression is to specify its first term b1 and the denominator of the geometric error q. For example, b1 = 4, q = -2. These two conditions define the geometric progression 4, -8, 16, -32,….
If q> 0 (q is not equal to 1), then the progression is a monotonic sequence. For example, the sequence, 2, 4,8,16,32, ... is a monotonically increasing sequence (b1 = 2, q = 2).
If, in the geometric error, the denominator is q = 1, then all members of the geometric progression will be equal to each other. In such cases, the progression is said to be a constant sequence.
Formula of the n-th member of the progression
In order for the numerical sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, is the geometric mean of the neighboring members. That is, it is necessary to fulfill the following equation - (b (n + 1)) ^ 2 = bn * b (n + 2), for any n> 0, where n belongs to the set natural numbers N.
The formula for the n-th term of the geometric progression is:
bn = b1 * q ^ (n-1), where n belongs to the set of natural numbers N.
Let's look at a simple example:
In geometric progression b1 = 6, q = 3, n = 8 find bn.
Let's use the formula for the n-th term of the geometric progression.
Arithmetic and geometric progressions
Theoretical information
Theoretical information
Arithmetic progression |
Geometric progression |
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Definition |
Arithmetic progression a n a sequence is called, each term of which, starting from the second, is equal to the previous term added with the same number d (d- difference of progressions) |
Geometric progression b n is a sequence of nonzero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q is the denominator of the progression) |
Recurrent formula |
For any natural n |
For any natural n |
Nth term formula |
a n = a 1 + d (n - 1) |
b n = b 1 ∙ q n - 1, b n ≠ 0 |
| Characteristic property |  |
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| Sum of n-first members |  |
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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, so a 22= -6 + 21 d.
It is necessary to find the difference between the progressions:
d = a 2 - a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = - 48.
Answer : a 22 = -48.
Assignment 2
Find the fifth term of a geometric progression: -3; 6; ....
1st way (using the n-term formula)
According to the formula of the n-th member of a geometric progression:
b 5 = b 1 ∙ q 5 - 1 = b 1 ∙ q 4.
As b 1 = -3,
2nd way (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.
Assignment 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 is 
.
Therefore:
.
Let's substitute the data into the formula:
Answer: 95.
Assignment 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 more convenient to use in this case?
By condition, the formula for the nth term of the original progression is known ( a n) a n= 3n - 4. You can immediately find and a 1, and a 16 without finding d. Therefore, we will use the first formula.
Answer: 368.
Assignment 5
In arithmetic progression ( a n) a 1 = -6; a 2= -8. Find the twenty-second term in 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 between the progressions:
d = a 2 - a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = -48.
Answer : a 22 = -48.
Assignment 6
Several consecutive members of a geometric progression are written:
Find the term in the progression denoted by the letter x.
When solving, we use the formula for the nth term b n = b 1 ∙ q n - 1 for geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of the given members of the progression and divide by the previous one. In our example, you can take and divide by. We get that q = 3. Instead of n in the formula, we substitute 3, since it is necessary to find the third term given by a geometric progression.
Substituting the found values into the formula, we get:
.
Answer : .
Assignment 7
From arithmetic progressions, given by the formula n-th member, select the one for which the condition is satisfied a 27 > 9:
Since the given condition must be fulfilled 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.
Assignment 8
In arithmetic progression a 1= 3, d = -1.5. Please indicate highest value n for which the inequality a n > -6.
A geometric progression is a numerical sequence, the first term of which is nonzero, and each next term is equal to the previous term multiplied by the same nonzero number.
The geometric progression is indicated by b1, b2, b3,…, bn,….
The ratio of any member of the geometric error to its previous term is equal to the same number, that is, b2 / b1 = b3 / b2 = b4 / b3 =… = bn / b (n-1) = b (n + 1) / bn = …. This follows directly from the definition of an arithmetic progression. This number is called the denominator of the geometric progression. Usually, the denominator of a geometric progression is denoted by the letter q.
Monotonic and constant sequence
One of the ways to specify a geometric progression is to specify its first term b1 and the denominator of the geometric error q. For example, b1 = 4, q = -2. These two conditions define the geometric progression 4, -8, 16, -32,….
If q> 0 (q is not equal to 1), then the progression is monotonous sequence. For example, the sequence, 2, 4,8,16,32, ... is a monotonically increasing sequence (b1 = 2, q = 2).
If, in the geometric error, the denominator is q = 1, then all members of the geometric progression will be equal to each other. In such cases, the progression is said to be constant sequence.
Formula of the n-th term of a geometric progression
In order for the numerical sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, is the geometric mean of the neighboring members. That is, it is necessary to fulfill the following equation
(b (n + 1)) ^ 2 = bn * b (n + 2), for any n> 0, where n belongs to the set of natural numbers N.
The formula for the n-th term of the geometric progression is:
bn = b1 * q ^ (n-1),
where n belongs to the set of natural numbers N.
Formula for the sum of the first n terms of a geometric progression
The formula for the sum of the first n terms of a geometric progression is:
Sn = (bn * q - b1) / (q-1), where q is not equal to 1.
Let's look at a simple example:
Find Sn exponentially b1 = 6, q = 3, n = 8.
To find S8, we use the formula for the sum of the first n terms of a geometric progression.
S8 = (6 * (3 ^ 8 -1)) / (3-1) = 19 680.
An example of a geometric progression: 2, 6, 18, 54, 162.
Here each term after the first is 3 times larger than the previous one. That is, each subsequent term is the result of multiplying the previous term by 3:
2 3 = 6
6 3 = 18
18 3 = 54
54 3 = 162 .
In our example, when dividing the second term by the first, the third by the second, etc. we get 3. The number 3 is the denominator of this geometric progression.
Example:
Let's go back to our geometric progression 2, 6, 18, 54, 162. Take the fourth term and square it:
54 2 = 2916.
Now we multiply the terms on the left and right of the number 54:
18 162 = 2916.
As you can see, the square of the third term is equal to the product of the adjacent second and fourth terms.
Example 1: Take a certain geometric progression in which the first term is 2, and the denominator of the geometric progression is 1.5. It is necessary to find the 4th term of this progression.
Given:
b 1 = 2
q = 1,5
n = 4
————
b 4 - ?
Solution.
We apply the formula b n= b 1 q n- 1, inserting the appropriate values into it:
b 4 = 2 · 1.5 4 - 1 = 2 · 1.5 3 = 2 · 3.375 = 6.75.
Answer: The fourth term of a given geometric progression is the number 6.75.
Example 2: Let's find the fifth term of the geometric progression if the first and third terms are 12 and 192, respectively.
Given:
b 1 = 12
b 3 = 192
————
b 5 - ?
Solution.
1) First, we need to find the denominator of the geometric progression, without which it is impossible to solve the problem. As a first step, using our formula, we derive the formula for b 3:
b 3 = b 1 q 3 - 1 = b 1 q 2
Now we can find the denominator of the geometric progression:
b 3 192
q 2 = —— = —— = 16
b 1 12
q= √16 = 4 or -4.
2) It remains to find the value b 5 .
If q= 4, then
b 5 = b 1 q 5 - 1 = 12 4 4 = 12 256 = 3072.
At q= -4 the result will be the same. Thus, the problem has one solution.
Answer: The fifth term of a given geometric progression is the number 3072.
Example: Find the sum of the first five terms of the geometric progression ( b n), in which the first term is 2, and the denominator of the geometric progression is 3.
Given:
b 1 = 2
q = 3
n = 5
————
S 5 - ?
Solution.
We apply the second of the two above:
b 1 (q 5 - 1) 2 (3 5 - 1) 2 (243 - 1) 484
S 5 = ————— = ————— = ———————— = ————— = 242
q - 1 3 - 1 2 2
Answer: The sum of the first five members of a given geometric progression is 242.
The sum of an infinite geometric progression.
It is necessary to distinguish between the concepts of "the sum of an infinite geometric progression" and "the sum n members of a geometric progression ". The second concept refers to any geometric progression, and the first - only to the one where the denominator is less than 1 in absolute value.
Already in Ancient egypt knew not only arithmetic, but also geometric progression. For example, here is a problem from Rynd's papyrus: “Seven faces have seven cats each; each cat eats seven mice, each mouse eats seven ears, each ear can grow seven measures of barley. How large are the numbers of this series and their sum? "
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Rice. 1. The ancient Egyptian problem of geometric progression |
This task was repeated many times with different variations among other peoples at other times. For example, in the written in the XIII century. "The Book of the Abacus" by Leonardo of Pisa (Fibonacci) has a problem in which there are 7 old women heading to Rome (obviously pilgrims), each of whom has 7 mules, each of which has 7 sacks, each of which has 7 loaves , each of which has 7 knives, each of which is in 7 scabbards. The problem asks how many items are there.
The sum of the first n terms of the geometric progression S n = b 1 (q n - 1) / (q - 1). This formula can be proved, for example, as follows: S n = b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n - 1.
Add to S n the number b 1 q n and get:
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Hence S n (q - 1) = b 1 (q n - 1), and we obtain the required formula.
Already on one of the clay tablets of Ancient Babylon, dating back to the VI century. BC e., contains the sum 1 + 2 + 2 2 + 2 3 + ... + 2 9 = 2 10 - 1. True, as in a number of other cases, we do not know where this fact was known to the Babylonians.
The rapid growth of geometric progression in a number of cultures, in particular, in Indian, is repeatedly used as a visual symbol of the immensity of the universe. In the well-known legend about the appearance of chess, the lord gives its inventor the opportunity to choose the reward himself, and he asks for the amount of wheat grains that will be obtained if one is put on the first square of the chessboard, two on the second, four on the third, eight on the fourth and so on, each time the number doubles. Vladyka thought that it was, at most, about several sacks, but he miscalculated. It is easy to see that for all 64 squares of the chessboard, the inventor should have received (2 64 - 1) grain, which is expressed by a 20-digit number; even if the entire surface of the Earth was sown, it would take at least 8 years to collect the required amount of grains. This legend is sometimes interpreted as an indication of the almost unlimited possibilities hidden in the game of chess.
It's easy to see that this number is indeed 20 digits:
2 64 = 2 4 ∙ (2 10) 6 = 16 ∙ 1024 6 ≈ 16 ∙ 1000 6 = 1.6 ∙ 10 19 (a more accurate calculation gives 1.84 ∙ 10 19). But I wonder if you can find out what digit this number ends with?
The geometric progression is increasing if the denominator is greater than 1 in absolute value, or decreasing if it is less than one. IN the latter case the number q n for sufficiently large n can become arbitrarily small. While an increasing geometric progression increases unexpectedly quickly, a decreasing one decreases just as quickly.
The larger n, the weaker the number q n differs from zero, and the closer the sum of n terms of the geometric progression S n = b 1 (1 - q n) / (1 - q) to the number S = b 1 / (1 - q). (This is how F. Viet, for example, reasoned). The number S is called the sum of an infinitely decreasing geometric progression. Nevertheless, for many centuries the question of what is the meaning of the summation of the ENTIRE geometric progression, with its infinite number of terms, was not clear enough to mathematicians.
A diminishing geometric progression can be seen, for example, in Zeno's aporias "Halving" and "Achilles and the Turtle." In the first case, it is clearly shown that the entire road (suppose of length 1) is the sum of an infinite number of segments 1/2, 1/4, 1/8, etc. This is, of course, from the point of view of ideas about final amount endless geometric progression. And yet - how can this be?
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Rice. 2. Progression with a factor of 1/2 |
In the aporia about Achilles, the situation is a little more complicated, since here the denominator of the progression is equal not to 1/2, but to some other number. Suppose, for example, Achilles runs at speed v, the turtle moves at speed u, and the initial distance between them is equal to l. Achilles will run this distance in time l / v, the turtle will move by a distance lu / v during this time. When Achilles runs this segment, the distance between him and the turtle will become equal to l (u / v) 2, etc. It turns out that catching up with the turtle means finding the sum of an infinitely decreasing geometric progression with the first term l and the denominator u / v. This sum - the segment that Achilles will eventually run to the place of meeting with the turtle - is equal to l / (1 - u / v) = lv / (v - u). But, again, how this result should be interpreted and why it makes any sense at all was not very clear for a long time.
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Rice. 3. Geometric progression with a factor of 2/3 |
The sum of a geometric progression was used by Archimedes to determine the area of a parabola segment. Let the given segment of the parabola be delimited by the chord AB and let the tangent line at the point D of the parabola be parallel to AB. Let C be the midpoint of AB, E the midpoint of AC, F the midpoint of CB. Draw straight lines parallel to DC through points A, E, F, B; let the tangent drawn at the point D, these lines intersect at the points K, L, M, N. Let's also draw segments AD and DB. Let the line EL intersect the line AD at the point G, and the parabola at the point H; line FM intersects line DB at point Q, and parabola at point R. According to general theory conic sections, DC is the diameter of the parabola (that is, a segment parallel to its axis); it and the tangent at point D can serve as the x and y coordinate axes, in which the parabola equation is written as y 2 = 2px (x is the distance from D to any point of a given diameter, y is the length of a parallel to a given tangent line from this point of diameter to some point on the parabola itself).
By virtue of the parabola equation, DL 2 = 2 ∙ p ∙ LH, DK 2 = 2 ∙ p ∙ KA, and since DK = 2DL, then KA = 4LH. Since KA = 2LG, LH = HG. The area of the parabola ADB segment is equal to the area of the triangle ΔADB and the areas of the AHD and DRB segments combined. In turn, the area of the AHD segment is similarly equal to the area of the triangle AHD and the remaining segments AH and HD, with each of which you can perform the same operation - divide into a triangle (Δ) and two remaining segments (), etc.:
The area of triangle ΔAHD is equal to half the area of triangle ΔALD (they have a common base AD, and the heights differ by a factor of 2), which, in turn, is equal to half the area of triangle ΔAKD, and therefore half the area of triangle ΔACD. Thus, the area of the triangle ΔAHD is equal to a quarter of the area of the triangle ΔACD. Likewise, the area of the triangle ΔDRB is equal to a quarter of the area of the triangle ΔDFB. So, the areas of the triangles ΔAHD and ΔDRB, taken together, are equal to a quarter of the area of the triangle ΔADB. Repeating this operation applied to the AH, HD, DR and RB segments will also select triangles from them, the area of which, taken together, will be 4 times less than the area of the triangles ΔAHD and ΔDRB taken together, which means 16 times less. than the area of the triangle ΔADB. Etc:
Thus, Archimedes proved that "every segment enclosed between a straight line and a parabola is four-thirds of a triangle with the same base and equal height."
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