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 next number obtained from the previous one by multiplying by some specific number. This is expressed by the following formula.

a z +1 =a z q, where z is the number of the selected element.

Accordingly, z ∈ N.

The period when a geometric progression is studied at school is grade 9. Examples will help you understand the concept:

0.25 0.125 0.0625...

Based on this formula, the denominator of the progression can be found as follows:

Neither q nor b z can be zero. Also, each of the elements of the progression should not be equal to zero.

Accordingly, to find out the next number in the series, you need to multiply the last one by q.

To specify this progression, you must specify its first element and denominator. After that, it is possible to find any of the subsequent terms and their sum.

Varieties

Depending on q and a 1, this progression is divided into several types:

  • If both a 1 and q are greater than one, then such a sequence is a geometric progression increasing with each next element. An example of such is presented below.

Example: a 1 =3, q=2 - both parameters are greater than one.

Then the numerical sequence can be written like this:

3 6 12 24 48 ...

  • If |q| less than one, that is, multiplication by it is equivalent to division, then a progression with similar conditions is a decreasing geometric progression. An example of such is presented below.

Example: a 1 =6, q=1/3 - a 1 is greater than one, q is less.

Then the numerical sequence can be written as follows:

6 2 2/3 ... - any element is 3 times greater than the element following it.

  • Sign-variable. If q<0, то знаки у чисел последовательности постоянно чередуются вне зависимости от a 1 , а элементы ни возрастают, ни убывают.

Example: a 1 = -3 , q = -2 - both parameters are less than zero.

Then the sequence can be written like this:

3, 6, -12, 24,...

Formulas

For convenient use of geometric progressions, there are many formulas:

  • Formula of the z-th member. Allows you to calculate the element under a specific number without calculating the previous numbers.

Example:q = 3, a 1 = 4. It is required to calculate the fourth element of the progression.

Solution: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 .

Solution: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.

Solution: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 next element more than the previousq 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 .

Solution: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.

Solution:

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.

22.09.2018 22:00

Geometric progression, along with arithmetic, is an important number series that 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 of rational numbers that is formed by successively 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 ​​of n.

The denominator of a geometric progression


The number b completely determines what character the entire number series will have. The denominator b can be positive, negative, or greater than or less than one. All of the above options lead to different sequences:

  • b > 1. There is an increasing series of rational numbers. For example, 1, 2, 4, 8, ... If the element a1 is negative, then the whole sequence will increase only modulo, but decrease taking into account the sign of the numbers.
  • b = 1. Often such a case is not called a progression, since there is an ordinary series of identical rational numbers. For example, -4, -4, -4.

Formula for sum

Before proceeding to the consideration of specific problems using the denominator of the type of progression under consideration, 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 to find the sum arbitrary number members.

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 the last expression, only 4 terms were summed up, since the 5th is already included in the sum that needs to be calculated according to the condition of the problem. Finally, we take the difference: S510 = S10 - S4 = -1364 - (-20) = -1344.

Method 2. Before substituting numbers and counting, you can get a formula for the sum between the terms m and n of the series in question. We act in exactly the same way as in method 1, only we work first with the symbolic representation of the sum. We have: Snm = (bn - 1) * a1 / (b - 1) - (bm-1 - 1) * a1 / (b - 1) = a1 * (bn - bm-1) / (b - 1). 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 the known values ​​​​and get the required number: b \u003d 1 - 2 / 3 \u003d -1 / 3 or -0.333 (3). We can check this result qualitatively if we remember that for this type of sequence, the modulus b must not go beyond 1. As you can see, |-1 / 3|

Task number 4. Restoring a series of numbers

Let 2 elements of a number series be given, for example, the 5th is equal to 30 and the 10th is equal to 60. It is necessary to restore the entire series from these data, knowing that it satisfies the properties of a geometric progression.

To solve the problem, you must first write down the corresponding expression for each known member. We have: a5 = b4 * a1 and a10 = b9 * a1. Now we divide the second expression by the first, we get: a10 / a5 = b9 * a1 / (b4 * a1) = b5. From here we determine the denominator by taking the fifth degree root of the ratio of the members known from the condition of the problem, b = 1.148698. We substitute the resulting number into one of the expressions for a known element, we get: a1 = a5 / b4 = 30 / (1.148698)4 = 17.2304966.

Thus, we have found what the denominator of the progression bn is, and the geometric progression bn-1 * 17.2304966 = an, where b = 1.148698.

Where are geometric progressions used?


If there were no application of this numerical series in practice, then its study would be reduced to a purely theoretical interest. But there is such an application.


The 3 most famous examples are listed below:

  • Zeno's paradox, in which the agile Achilles cannot catch up with the slow tortoise, is solved using the concept of an infinitely decreasing sequence of numbers.
  • If wheat grains are placed on each cell of the chessboard so that 1 grain is placed on the 1st cell, 2 - on the 2nd, 3 - on the 3rd, and so on, then 18446744073709551615 grains will be needed to fill all the cells of the board!
  • In the game "Tower of Hanoi", in order to rearrange disks from one rod to another, it is necessary to perform 2n - 1 operations, that is, their number grows exponentially from the number of disks n used.

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For example, sequence \(3\); \(6\); \(12\); \(24\); \(48\)… is a geometric progression, because each next element differs from the previous one by a factor of two (in other words, it can be obtained from the previous one by multiplying it by two):

Like any sequence, a geometric progression is denoted by a small Latin letter. The numbers that form a progression are called it members(or elements). They are denoted by the same letter as the geometric progression, but with a numerical index equal to the element number in order.

For example, the geometric progression \(b_n = \(3; 6; 12; 24; 48…\)\) consists of the elements \(b_1=3\); \(b_2=6\); \(b_3=12\) and so on. In other words:

If you understand the above information, you will already be able to solve most of the problems on this topic.

Example (OGE):
Solution:

Answer : \(-686\).

Example (OGE): Given the first three terms of the progression \(324\); \(-108\); \(36\)…. Find \(b_5\).
Solution:


To continue the sequence, we need to know the denominator. Let's find it from two neighboring elements: what should \(324\) be multiplied by to get \(-108\)?

\(324 q=-108\)

From here we can easily calculate the denominator.

\(q=-\) \(\frac(108)(324)\) \(=-\) \(\frac(1)(3)\)

Now we can easily find the element we need.


Answer ready.

Answer : \(4\).

Example: The progression is given by the condition \(b_n=0.8 5^n\). Which number is a member of this progression:

a) \(-5\) b) \(100\) c) \(25\) d) \(0.8\) ?

Solution: From the wording of the task, it is obvious that one of these numbers is definitely in our progression. Therefore, we can simply calculate its members one by one until we find the value we need. Since our progression is given by the formula , we calculate the values ​​of the elements by substituting different \(n\):
\(n=1\); \(b_1=0.8 5^1=0.8 5=4\) – there is no such number in the list. We continue.
\(n=2\); \(b_2=0.8 5^2=0.8 25=20\) - and this is not there either.
\(n=3\); \(b_3=0.8 5^3=0.8 125=100\) – and here is our champion!

Answer: \(100\).

Example (OGE): Several successive members of the geometric progression …\(8\) are given; \(x\); \(50\); \(-125\)…. Find the value of the element denoted by the letter \(x\).

Solution:

Answer: \(-20\).

Example (OGE): The progression is given by the conditions \(b_1=7\), \(b_(n+1)=2b_n\). Find the sum of the first \(4\) terms of this progression.

Solution:

Answer: \(105\).

Example (OGE): It is known that exponentially \(b_6=-11\),\(b_9=704\). Find the denominator \(q\).

Solution:


It can be seen from the diagram on the left that in order to “get” from \ (b_6 \) to \ (b_9 \) - we take three “steps”, that is, we multiply \ (b_6 \) three times by the denominator of the progression. In other words, \(b_9=b_6 q q q=b_6 q^3\).

\(b_9=b_6 q^3\)

Substitute the values ​​we know.

\(704=(-11)q^3\)

“Reverse” the equation and divide it by \((-11)\).

\(q^3=\) \(\frac(704)(-11)\) \(\:\:\: ⇔ \:\:\: \)\(q^3=-\) \(64 \)

What number cubed gives \(-64\)?
Of course, \(-4\)!

Answer found. It can be checked by restoring the chain of numbers from \(-11\) to \(704\).


All agreed - the answer is correct.

Answer: \(-4\).

The most important formulas

As you can see, most geometric progression problems can be solved with pure logic, simply by understanding the essence (this is generally characteristic of mathematics). But sometimes the knowledge of certain formulas and patterns speeds up and greatly facilitates the decision. We will study two such formulas.

The formula for the \(n\)th member is: \(b_n=b_1 q^(n-1)\), where \(b_1\) is the first member of the progression; \(n\) – number of the required element; \(q\) is the denominator of the progression; \(b_n\) is a member of the progression with the number \(n\).

Using this formula, you can, for example, solve the problem from the very first example in just one step.

Example (OGE): The geometric progression is given by the conditions \(b_1=-2\); \(q=7\). Find \(b_4\).
Solution:

Answer: \(-686\).

This example was simple, so the formula did not make the calculations easier for us too much. Let's look at the problem a little more complicated.

Example: The geometric progression is given by the conditions \(b_1=20480\); \(q=\frac(1)(2)\). Find \(b_(12)\).
Solution:

Answer: \(10\).

Of course, raising \(\frac(1)(2)\) to the \(11\)th power is not very joyful, but still easier than \(11\) dividing \(20480\) into two.

The sum \(n\) of the first terms: \(S_n=\)\(\frac(b_1 (q^n-1))(q-1)\) , where \(b_1\) is the first term of the progression; \(n\) – the number of summed elements; \(q\) is the denominator of the progression; \(S_n\) is the sum \(n\) of the first members of the progression.

Example (OGE): Given a geometric progression \(b_n\), whose denominator is \(5\), and the first term \(b_1=\frac(2)(5)\). Find the sum of the first six terms of this progression.
Solution:

Answer: \(1562,4\).

And again, we could solve the problem “on the forehead” - find all six elements in turn, and then add the results. However, the number of calculations, and hence the chance of a random error, would increase dramatically.

For a geometric progression, there are several more formulas that we did not consider here because of their low practical use. You can find these formulas.

Increasing and decreasing geometric progressions

The progression \(b_n = \(3; 6; 12; 24; 48…\)\) considered at the very beginning of the article has a denominator \(q\) greater than one, and therefore each next term is greater than the previous one. Such progressions are called increasing.

If \(q\) is less than one, but is positive (that is, lies between zero and one), then each next element will be less than the previous one. For example, in the progression \(4\); \(2\); \(1\); \(0.5\); \(0.25\)… the denominator of \(q\) is \(\frac(1)(2)\).


These progressions are called decreasing. Note that none of the elements of this progression will be negative, they just get smaller and smaller with each step. That is, we will gradually approach zero, but we will never reach it and we will not go beyond it. Mathematicians in such cases say "to tend to zero."

Note that with a negative denominator, the elements of a geometric progression will necessarily change sign. For example, the progression \(5\); \(-15\); \(45\); \(-135\); \(675\)... the denominator of \(q\) is \(-3\), and because of this, the signs of the elements "blink".

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 general formula n -th member of the geometric progression 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 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?

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 has 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:

S n + b 1 q n = b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n – 1 + b 1 q n = b 1 + (b 1 + b 1 q + b 1 q 2 + b 1 q 3 + ... + b 1 q n –1) q = b 1 + S n q .

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 a geometric progression in a number of cultures, in particular, in India, is repeatedly used as a clear 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 grains of wheat 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. The lord thought that we are talking, at most, about a few bags, 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 one were to sow the entire surface of the Earth, it would take at least 8 years to harvest required amount 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 the case from the point of view of ideas about the finite sum infinite geometric progression. And yet - how can this be?

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 to interpret this result and why it makes any sense at all, for a long time it wasn't very clear.

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 segment AHD 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 ​​the triangles ΔAHD and ΔDRB , taken together, and therefore 16 times less, than the area of ​​the triangle ΔADB . And so on:

Thus, Archimedes proved that "every segment enclosed between a straight line and a parabola is four-thirds of a triangle having the same base and equal height with it."

An example of a geometric progression: 2, 6, 18, 54, 162.

Here, each term after the first is 3 times 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, and so on. 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. Let's take the fourth term and square it:
54 2 = 2916.

Now we multiply the terms to 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 neighboring second and fourth terms.

Example 1: Let's take some geometric progression, in which the first term is equal to 2, and the denominator of the geometric progression is equal to 1.5. We need to find the 4th term of this progression.

Given:
b 1 = 2

q = 1,5
n = 4

————
b 4 - ?

Solution.

Applying the formula b n= b 1 q n- 1 , inserting the appropriate values ​​into it:
b 4 \u003d 2 1.5 4 - 1 \u003d 2 1.5 3 \u003d 2 3.375 \u003d 6.75.

Answer: The fourth term of a given geometric progression is the number 6.75.

Example 2: Find the fifth member of the geometric progression if the first and third members are 12 and 192, respectively.

Given:
b 1 = 12
b 3 = 192
————
b 5 - ?

Solution.

1) First we need to find the denominator of a 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 a 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 the 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 equal to 2, and the denominator of a geometric progression is 3.

Given:

b 1 = 2

q = 3

n = 5
————
S 5 - ?

Solution.

We apply the second formula 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 terms 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 one where the denominator is less than 1 modulo.