Geometric progression, along with arithmetic, is important numerical series, which 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
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 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 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 regular row the same 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. 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 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 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. The resulting number is substituted into one of the expressions for 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.
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.
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.
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 a certain 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 that the first term and denominator of the progression are not equal to zero, 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 natural number n (counter).
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 main feature 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 the school is studying geometric progression- 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.
This number is called the denominator of a geometric progression, that is, each term differs from the previous one by q times. (We will assume that q ≠ 1, otherwise everything is too trivial). It is easy to see that the general formula of the nth member of the geometric progression is b n = b 1 q n – 1 ; terms with numbers b n and b m differ by q n – m times.Already in ancient Egypt, they knew not only arithmetic, but also geometric progression. Here, for example, is a task from the Rhind papyrus: “Seven faces have seven cats; each cat eats seven mice, each mouse eats seven ears of corn, each ear can grow seven measures of barley. How large are the numbers in this series and their sum?
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Rice. 1. Ancient Egyptian geometric progression problem |
This task was repeated many times with different variations among other peoples at other times. For example, in written in the XIII century. The "Book of the abacus" by Leonardo of Pisa (Fibonacci) has a problem in which 7 old women appear on their way to Rome (obviously pilgrims), each of which has 7 mules, each of which has 7 bags, each of which contains 7 loaves , each of which has 7 knives, each of which is in 7 sheaths. The problem asks how many items there are.
The sum of the first n members of the geometric progression S n = b 1 (q n - 1) / (q - 1) . This formula can be proved, for example, as follows: S n \u003d b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n - 1.
Let's add the number b 1 q n to S n and get:
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Hence S n (q - 1) = b 1 (q n - 1), and we get the necessary 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 India, is repeatedly used as a visual symbol of the immensity of the universe. In the well-known legend about the appearance of chess, the ruler gives their inventor the opportunity to choose a reward himself, and he asks for such a number of wheat grains as will be obtained if one is placed on the first cell of the chessboard, two on the second, four on the third, eight on the fourth, and etc., each time the number is doubled. Vladyka thought that it was, at the most, a few 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 as 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 number of grains. This legend is sometimes interpreted as a reference to the almost unlimited possibilities hidden in the game of chess.
The fact that this number is really 20-digit is easy to see:
2 64 \u003d 2 4 ∙ (2 10) 6 \u003d 16 1024 6 ≈ 16 1000 6 \u003d 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?
A 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 can become arbitrarily small for sufficiently large n. While an increasing exponential increases unexpectedly fast, a decreasing exponential decreases just as quickly.
The larger n, the weaker the number q n differs from zero, and the closer the sum of n members of the geometric progression S n \u003d b 1 (1 - q n) / (1 - q) to the number S \u003d b 1 / (1 - q) . (So reasoned, for example, F. Viet). The number S is called the sum of an infinitely decreasing geometric progression. However, for many centuries the question of what is the meaning of the summation of the ALL geometric progression, with its infinite number of terms, was not clear enough to mathematicians.
A decreasing geometric progression can be seen, for example, in Zeno's aporias "Biting" and "Achilles and the tortoise". In the first case, it is clearly shown that the entire road (assume length 1) is the sum of an infinite number of segments 1/2, 1/4, 1/8, etc. This, of course, is how it is from the point of view of ideas about the finite sum infinite 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, because here the denominator of the progression is not equal to 1/2, but to some other number. Let, for example, Achilles run at speed v, the tortoise moves at speed u, and the initial distance between them is l. Achilles will run this distance in the time l / v , the tortoise will move a distance lu / v during this time. When Achilles runs through 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 meeting point 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 coefficient 2/3 |
The sum of a geometric progression was used by Archimedes when determining the area of a segment of a parabola. Let the given segment of the parabola be delimited by the chord AB and let the tangent 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 lines parallel to DC through points A , E , F , B ; let the tangent drawn at point D , these lines intersect at 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 the parabola at point R. According to the general theory of conic sections, DC is the diameter of a parabola (that is, a segment parallel to its axis); it and the tangent at point D can serve as coordinate axes x and y, in which the parabola equation is written as y 2 \u003d 2px (x is the distance from D to any point of a given diameter, y is the length of a segment parallel to a given tangent 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 segment ADB of the parabola is equal to the area of the triangle ΔADB and the areas of the segments AHD and DRB 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 the same operation can be performed - split into a triangle (Δ) and the two remaining segments (), etc.:
The area of the triangle ΔAHD is equal to half the area of the triangle ΔALD (they have a common base AD, and the heights differ by 2 times), which, in turn, is equal to half the area of the triangle ΔAKD, and therefore half the area of the triangle ΔACD. Thus, the area of triangle ΔAHD is equal to a quarter of the area of triangle ΔACD. Likewise, the area of triangle ΔDRB is equal to a quarter of the area of triangle ΔDFB. So, the areas of triangles ∆AHD and ∆DRB, taken together, are equal to a quarter of the area of triangle ∆ADB. Repeating this operation as applied to the segments AH , HD , DR and RB will also select triangles from them, the area of which, taken together, will be 4 times less than the area of triangles ΔAHD and ΔDRB , taken together, and therefore 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, having with it the same base and equal height."
The formula for the nth member of a geometric progression is a very simple thing. Both in meaning and in general. But there are all sorts of problems for the formula of the nth member - from very primitive to quite serious ones. And in the process of our acquaintance, we will definitely consider both of them. Well, let's meet?)
So, for starters, actually formulan
Here she is:
b n = b 1 · q n -1
Formula as a formula, nothing supernatural. It looks even simpler and more compact than the similar formula for . The meaning of the formula is also simple, like a felt boot.
This formula allows you to find ANY member of a geometric progression BY ITS NUMBER " n".
As you can see, the meaning is a complete analogy with an arithmetic progression. We know the number n - we can also calculate the term under this number. What we want. Not multiplying sequentially by "q" many, many times. That's the whole point.)
I understand that at this level of work with progressions, all the quantities included in the formula should already be clear to you, but I consider it my duty to decipher each one. Just in case.
So let's go:
b 1 – first member of a geometric progression;
q – ;
n– member number;
b n – nth (nth) member of a geometric progression.
This formula links the four main parameters of any geometric progression - bn, b 1 , q and n. And around these four key figures, all-all tasks in progression revolve.
"And how is it displayed?"- I hear a curious question ... Elementary! Look!
What is equal to second progression member? No problem! We write directly:
b 2 = b 1 q
And the third member? Not a problem either! We multiply the second term again onq.
Like this:
B 3 \u003d b 2 q
Recall now that the second term, in turn, is equal to b 1 q and substitute this expression into our equality:
B 3 = b 2 q = (b 1 q) q = b 1 q q = b 1 q 2
We get:
B 3 = b 1 q 2
Now let's read our entry in Russian: the third term is equal to the first term multiplied by q in second degree. Do you get it? Not yet? Okay, one more step.
What is the fourth term? All the same! Multiply previous(i.e. the third term) on q:
B 4 \u003d b 3 q \u003d (b 1 q 2) q \u003d b 1 q 2 q \u003d b 1 q 3
Total:
B 4 = b 1 q 3
And again we translate into Russian: fourth term is equal to the first term multiplied by q in third degree.
Etc. So how? Did you catch the pattern? Yes! For any term with any number, the number of equal factors q (i.e. the power of the denominator) will always be one less than the number of the desired membern.
Therefore, our formula will be, without options:
b n =b 1 · q n -1
That's all.)
Well, let's solve problems, shall we?)
Solving problems on a formulanth term of a geometric progression.
Let's start, as usual, with a direct application of the formula. Here is a typical problem:
It is known exponentially that b 1 = 512 and q = -1/2. Find the tenth term of the progression.
Of course, this problem can be solved without any formulas at all. Just like a geometric progression. But we need to warm up with the formula of the nth term, right? Here we are breaking up.
Our data for applying the formula is as follows.
The first term is known. This is 512.
b 1 = 512.
The denominator of the progression is also known: q = -1/2.
It remains only to figure out what the number of the term n is equal to. No problem! Are we interested in the tenth term? So we substitute ten instead of n in the general formula.
And carefully calculate the arithmetic:
Answer: -1
As you can see, the tenth term of the progression turned out to be with a minus. No wonder: the denominator of the progression is -1/2, i.e. negative number. And this tells us that the signs of our progression alternate, yes.)
Everything is simple here. And here is a similar problem, but a little more complicated in terms of calculations.
In geometric progression, we know that:
b 1 = 3
Find the thirteenth term of the progression.
Everything is the same, only this time the denominator of the progression - irrational. Root of two. Well, no big deal. The formula is a universal thing, it copes with any numbers.
We work directly according to the formula:
The formula, of course, worked as it should, but ... this is where some will hang. What to do next with the root? How to raise a root to the twelfth power?
How-how ... You need to understand that any formula, of course, is a good thing, but the knowledge of all previous mathematics is not canceled! How to raise? Yes, remember the properties of degrees! Let's change the root to fractional degree and - by the formula of raising a power to a power.
Like this:
Answer: 192
And all things.)
What is the main difficulty in the direct application of the nth term formula? Yes! The main difficulty is work with degrees! Namely, the exponentiation of negative numbers, fractions, roots, and similar constructions. So those who have problems with this, an urgent request to repeat the degrees and their properties! Otherwise, you will slow down in this topic, yes ...)
Now let's solve typical search problems one of the elements of the formula if all the others are given. For the successful solution of such problems, the recipe is single and simple to horror - write the formulanth member in general! Right in the notebook next to the condition. And then, from the condition, we figure out what is given to us and what is not enough. And we express the desired value from the formula. Everything!
For example, such a harmless problem.
The fifth term of a geometric progression with a denominator of 3 is 567. Find the first term of this progression.
Nothing complicated. We work directly according to the spell.
We write the formula of the nth term!
b n = b 1 · q n -1
What is given to us? First, the denominator of the progression is given: q = 3.
In addition, we are given fifth term: b 5 = 567 .
Everything? Not! We are also given the number n! This is a five: n = 5.
I hope you already understand what is in the record b 5 = 567 two parameters are hidden at once - this is the fifth member itself (567) and its number (5). In a similar lesson on I already talked about this, but I think it’s not superfluous to remind here.)
Now we substitute our data into the formula:
567 = b 1 3 5-1
We consider arithmetic, simplify and get a simple linear equation:
81 b 1 = 567
We solve and get:
b 1 = 7
As you can see, there are no problems with finding the first member. But when looking for the denominator q and numbers n there may be surprises. And you also need to be prepared for them (surprises), yes.)
For example, such a problem:
The fifth term of a geometric progression with a positive denominator is 162, and the first term of this progression is 2. Find the denominator of the progression.
This time we are given the first and fifth members, and are asked to find the denominator of the progression. Here we start.
We write the formulanth member!
b n = b 1 · q n -1
Our initial data will be as follows:
b 5 = 162
b 1 = 2
n = 5
Not enough value q. No problem! Let's find it now.) We substitute everything that we know into the formula.
We get:
162 = 2q 5-1
2 q 4 = 162
q 4 = 81
A simple equation of the fourth degree. But now - carefully! At this stage of the solution, many students immediately joyfully extract the root (of the fourth degree) and get the answer q=3 .
Like this:
q4 = 81
q = 3
But in general, this is an unfinished answer. Or rather, incomplete. Why? The point is that the answer q = -3 also fits: (-3) 4 would also be 81!
This is because the power equation x n = a always has two opposite roots at evenn . Plus and minus:
Both fit.
For example, solving (i.e. second degrees)
x2 = 9
For some reason you are not surprised to see two roots x=±3? It's the same here. And with any other even degree (fourth, sixth, tenth, etc.) will be the same. Details - in the topic about
So the correct solution would be:
q 4 = 81
q= ±3
Okay, we've got the signs figured out. Which one is correct - plus or minus? Well, we read the condition of the problem again in search of additional information. It, of course, may not exist, but in this problem such information available. In our condition, it is directly stated that a progression is given with positive denominator.
So the answer is obvious:
q = 3
Everything is simple here. What do you think would happen if the problem statement were like this:
The fifth term of a geometric progression is 162, and the first term of this progression is 2. Find the denominator of the progression.
What's the Difference? Yes! In the condition nothing no mention of the denominator. Neither directly nor indirectly. And here the problem would already have two solutions!
q = 3 and q = -3
Yes Yes! And with plus and minus.) Mathematically, this fact would mean that there are two progressions that fit the task. And for each - its own denominator. For fun, practice and write down the first five terms of each.)
Now let's practice finding the member number. This is the hardest one, yes. But also more creative.
Given a geometric progression:
3; 6; 12; 24; …
What number is 768 in this progression?
The first step is the same: write the formulanth member!
b n = b 1 · q n -1
And now, as usual, we substitute the data known to us into it. Hm... it doesn't fit! Where is the first member, where is the denominator, where is everything else?!
Where, where ... Why do we need eyes? Flapping eyelashes? This time the progression is given to us directly in the form sequences. Can we see the first term? We see! This is a triple (b 1 = 3). What about the denominator? We don't see it yet, but it's very easy to count. If, of course, you understand.
Here we consider. Directly according to the meaning of a geometric progression: we take any of its members (except the first) and divide by the previous one.
At least like this:
q = 24/12 = 2
What else do we know? We also know some member of this progression, equal to 768. Under some number n:
b n = 768
We do not know his number, but our task is precisely to find him.) So we are looking for. We have already downloaded all the necessary data for substitution in the formula. Imperceptibly.)
Here we substitute:
768 = 3 2n -1
We make elementary ones - we divide both parts by three and rewrite the equation in the usual form: the unknown on the left, the known on the right.
We get:
2 n -1 = 256
Here's an interesting equation. We need to find "n". What's unusual? Yes, I do not argue. Actually, it's the simplest. It is so called because the unknown (in this case, it is the number n) stands in indicator degree.
At the stage of acquaintance with a geometric progression (this is the ninth grade), exponential equations are not taught to solve, yes ... This is a topic for high school. But there is nothing terrible. Even if you do not know how such equations are solved, let's try to find our n guided by simple logic and common sense.
We start to discuss. On the left we have a deuce to a certain degree. We do not yet know what exactly this degree is, but this is not scary. But on the other hand, we firmly know that this degree is equal to 256! So we remember to what extent the deuce gives us 256. Remember? Yes! AT eighth degrees!
256 = 2 8
If you didn’t remember or with the recognition of the degrees of the problem, then it’s also okay: we just successively raise the two to the square, to the cube, to the fourth power, the fifth, and so on. The selection, in fact, but at this level, is quite a ride.
One way or another, we will get:
2 n -1 = 2 8
n-1 = 8
n = 9
So 768 is ninth member of our progression. That's it, problem solved.)
Answer: 9
What? Boring? Tired of the elementary? I agree. Me too. Let's go to the next level.)
More complex tasks.
And now we solve the puzzles more abruptly. Not exactly super-cool, but on which you have to work a little to get to the answer.
For example, like this.
Find the second term of a geometric progression if its fourth term is -24 and the seventh term is 192.
This is a classic of the genre. Some two different members of the progression are known, but one more member must be found. Moreover, all members are NOT neighbors. What confuses at first, yes ...
As in , we consider two methods for solving such problems. The first way is universal. Algebraic. Works flawlessly with any source data. So that's where we'll start.)
We paint each term according to the formula nth member!
Everything is exactly the same as with an arithmetic progression. Only this time we are working with another general formula. That's all.) But the essence is the same: we take and in turn we substitute our initial data into the formula of the nth term. For each member - their own.
For the fourth term we write:
b 4 = b 1 · q 3
-24 = b 1 · q 3
There is. One equation is complete.
For the seventh term we write:
b 7 = b 1 · q 6
192 = b 1 · q 6
In total, two equations were obtained for the same progression .
We assemble a system from them:
Despite its formidable appearance, the system is quite simple. The most obvious way to solve is the usual substitution. We express b 1 from the upper equation and substitute into the lower one:
A little fiddling with the lower equation (reducing the exponents and dividing by -24) yields:
q 3 = -8
By the way, the same equation can be arrived at in a simpler way! What? Now I will show you another secret, but very beautiful, powerful and useful way to solve such systems. Such systems, in the equations of which they sit only works. At least in one. called term division method one equation to another.
So we have a system:
In both equations on the left - work, and on the right is just a number. This is a very good sign.) Let's take and ... divide, say, the lower equation by the upper one! What means, divide one equation by another? Very simple. We take left side one equation (lower) and we divide her on left side another equation (upper). The right side is similar: right side one equation we divide on the right side another.
The whole division process looks like this:
Now, reducing everything that is reduced, we get:
q 3 = -8
What is good about this method? Yes, because in the process of such a division, everything bad and inconvenient can be safely reduced and a completely harmless equation remains! That is why it is so important to have only multiplications in at least one of the equations of the system. There is no multiplication - there is nothing to reduce, yes ...
In general, this method (like many other non-trivial ways of solving systems) even deserves a separate lesson. I will definitely take a closer look at it. Someday…
However, no matter how you solve the system, in any case, now we need to solve the resulting equation:
q 3 = -8
No problem: we extract the root (cubic) and - done!
Please note that it is not necessary to put plus / minus here when extracting. We have an odd (third) degree root. And the answer is the same, yes.
So, the denominator of progression is found. Minus two. Fine! The process is underway.)
For the first term (say from the top equation) we get:
Fine! We know the first term, we know the denominator. And now we have the opportunity to find any member of the progression. Including the second.)
For the second member, everything is quite simple:
b 2 = b 1 · q= 3 (-2) = -6
Answer: -6
So, we have sorted out the algebraic way of solving the problem. Complicated? Not much, I agree. Long and boring? Yes, definitely. But sometimes you can significantly reduce the amount of work. For this there is graphic way. Good old and familiar to us by .)
Let's draw the problem!
Yes! Exactly. Again we depict our progression on the number axis. Not necessarily by a ruler, it is not necessary to maintain equal intervals between members (which, by the way, will not be the same, because the progression is geometric!), But simply schematically draw our sequence.
I got it like this:
Now look at the picture and think. How many equal factors "q" share fourth and seventh members? That's right, three!
Therefore, we have every right to write:
-24q 3 = 192
From here it is now easy to find q:
q 3 = -8
q = -2
That's great, the denominator is already in our pocket. And now we look at the picture again: how many such denominators sit between second and fourth members? Two! Therefore, to record the relationship between these members, we will raise the denominator squared.
Here we write:
b 2 · q 2 = -24 , where b 2 = -24/ q 2
We substitute our found denominator into the expression for b 2 , count and get:
Answer: -6
As you can see, everything is much simpler and faster than through the system. Moreover, here we didn’t even need to count the first term at all! At all.)
Here is such a simple and visual way-light. But it also has a serious drawback. Guessed? Yes! It is only good for very short pieces of progression. Those where the distances between the members of interest to us are not very large. But in all other cases it is already difficult to draw a picture, yes ... Then we solve the problem analytically, through a system.) And systems are a universal thing. Deal with any number.
Another epic one:
The second term of the geometric progression is 10 more than the first, and the third term is 30 more than the second. Find the denominator of the progression.
What's cool? Not at all! All the same. We again translate the condition of the problem into pure algebra.
1) We paint each term according to the formula nth member!
Second term: b 2 = b 1 q
Third term: b 3 \u003d b 1 q 2
2) We write down the relationship between the members from the condition of the problem.
Reading the condition: "The second term of a geometric progression is 10 more than the first." Stop, this is valuable!
So we write:
b 2 = b 1 +10
And we translate this phrase into pure mathematics:
b 3 = b 2 +30
We got two equations. We combine them into a system:
The system looks simple. But there are a lot of different indices for letters. Let's substitute instead of the second and third members of their expression through the first member and denominator! In vain, or what, we painted them?
We get:
But such a system is no longer a gift, yes ... How to solve this? Unfortunately, the universal secret spell to solve complex non-linear There are no systems in mathematics and there cannot be. It's fantastic! But the first thing that should come to your mind when trying to crack such a tough nut is to figure out But isn't one of the equations of the system reduced to a beautiful form, which makes it easy, for example, to express one of the variables in terms of another?
Let's guess. The first equation of the system is clearly simpler than the second. We will torture him.) Why not try from the first equation something express through something? Since we want to find the denominator q, then it would be most advantageous for us to express b 1 through q.
So let's try to do this procedure with the first equation, using the good old ones:
b 1 q = b 1 +10
b 1 q - b 1 \u003d 10
b 1 (q-1) = 10
Everything! Here we have expressed unnecessary us the variable (b 1) through necessary(q). Yes, not the most simple expression received. Some kind of fraction ... But our system is of a decent level, yes.)
Typical. What to do - we know.
We write ODZ (necessarily!) :
q ≠ 1
We multiply everything by the denominator (q-1) and reduce all fractions:
10 q 2 = 10 q + 30(q-1)
We divide everything by ten, open the brackets, collect everything on the left:
q 2 – 4 q + 3 = 0
We solve the resulting and get two roots:
q 1 = 1
q 2 = 3
There is only one final answer: q = 3 .
Answer: 3
As you can see, the way to solve most problems for the formula of the nth member of a geometric progression is always the same: we read attentively condition of the problem and, using the formula of the nth term, we translate all useful information into pure algebra.
Namely:
1) We write separately each member given in the problem according to the formulanth member.
2) From the condition of the problem, we translate the connection between the members into a mathematical form. We compose an equation or a system of equations.
3) We solve the resulting equation or system of equations, find the unknown parameters of the progression.
4) In case of an ambiguous answer, we carefully read the condition of the problem in search of additional information (if any). We also check the received answer with the conditions of the ODZ (if any).
And now we list the main problems that most often lead to errors in the process of solving geometric progression problems.
1. Elementary arithmetic. Operations with fractions and negative numbers.
2. If at least one of these three points is a problem, then you will inevitably be mistaken in this topic. Unfortunately... So don't be lazy and repeat what was mentioned above. And follow the links - go. Sometimes it helps.)
Modified and recurrent formulas.
And now let's look at a couple of typical exam problems with a less familiar presentation of the condition. Yes, yes, you guessed it! This is modified and recurrent formulas of the nth member. We have already encountered such formulas and worked in arithmetic progression. Everything is similar here. The essence is the same.
For example, such a problem from the OGE:
The geometric progression is given by the formula b n = 3 2 n . Find the sum of the first and fourth terms.
This time the progression is given to us not quite as usual. Some kind of formula. So what? This formula is also a formulanth member! We all know that the formula of the nth term can be written both in general form, through letters, and for specific progression. With specific first term and denominator.
In our case, we are, in fact, given a general term formula for a geometric progression with the following parameters:
b 1 = 6
q = 2
Let's check?) Let's write the formula of the nth term in general form and substitute into it b 1 and q. We get:
b n = b 1 · q n -1
b n= 6 2n -1
We simplify, using factorization and power properties, and get:
b n= 6 2n -1 = 3 2 2n -1 = 3 2n -1+1 = 3 2n
As you can see, everything is fair. But our goal with you is not to demonstrate the derivation of a specific formula. This is so, a lyrical digression. Purely for understanding.) Our goal is to solve the problem according to the formula that is given to us in the condition. Do you catch it?) So we are working with the modified formula directly.
We count the first term. Substitute n=1 into the general formula:
b 1 = 3 2 1 = 3 2 = 6
Like this. By the way, I'm not too lazy and once again I will draw your attention to a typical blunder with the calculation of the first term. DO NOT look at the formula b n= 3 2n, immediately rush to write that the first member is a troika! It's a big mistake, yes...)
We continue. Substitute n=4 and consider the fourth term:
b 4 = 3 2 4 = 3 16 = 48
And finally, we calculate the required amount:
b 1 + b 4 = 6+48 = 54
Answer: 54
Another problem.
The geometric progression is given by the conditions:
b 1 = -7;
b n +1 = 3 b n
Find the fourth term of the progression.
Here the progression is given by the recurrent formula. Well, okay.) How to work with this formula - we also know.
Here we are acting. Step by step.
1) counting two successive member of the progression.
The first term is already given to us. Minus seven. But the next, second term, can be easily calculated using the recursive formula. If you understand how it works, of course.)
Here we consider the second term according to the famous first:
b 2 = 3 b 1 = 3 (-7) = -21
2) We consider the denominator of the progression
Also no problem. Straight, share second dick on first.
We get:
q = -21/(-7) = 3
3) Write the formulanth member in the usual form and consider the desired member.
So, we know the first term, the denominator too. Here we write:
b n= -7 3n -1
b 4 = -7 3 3 = -7 27 = -189
Answer: -189
As you can see, working with such formulas for a geometric progression is essentially no different from that for an arithmetic progression. It is only important to understand the general essence and meaning of these formulas. Well, the meaning of geometric progression also needs to be understood, yes.) And then there will be no stupid mistakes.
Well, let's decide on our own?)
Quite elementary tasks, for warm-up:
1. Given a geometric progression in which b 1 = 243, and q = -2/3. Find the sixth term of the progression.
2. The common term of a geometric progression is given by the formula b n = 5∙2 n +1 . Find the number of the last three-digit member of this progression.
3. The geometric progression is given by the conditions:
b 1 = -3;
b n +1 = 6 b n
Find the fifth term of the progression.
A little more complicated:
4. Given a geometric progression:
b 1 =2048; q =-0,5
What is the sixth negative term of it?
What seems super difficult? Not at all. Logic and understanding of the meaning of geometric progression will save. Well, the formula of the nth term, of course.
5. The third term of the geometric progression is -14 and the eighth term is 112. Find the denominator of the progression.
6. The sum of the first and second terms of a geometric progression is 75, and the sum of the second and third terms is 150. Find the sixth term of the progression.
Answers (in disarray): 6; -3888; -one; 800; -32; 448.
That's almost all. It remains only to learn how to count the sum of the first n terms of a geometric progression yes discover infinitely decreasing geometric progression and its amount. A very interesting and unusual thing, by the way! More on that in later lessons.)
