A geometric progression is a numerical sequence, the first term of which is different from zero, and each subsequent term is equal to the previous term multiplied by the same equal to zero number.
Concept of geometric progression
Geometric progression is denoted b1,b2,b3, …, bn, ….
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 a geometric progression. Usually the denominator of a geometric progression is denoted by the letter q.
The sum of an infinite geometric progression for |q|<1
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 the denominator in the geometric error is q=1, then all terms of the geometric progression will be equal to each other. In such cases, the progression is said to be a constant sequence.
In order for a number sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, be the geometric mean of 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.
Now let's put (Xn) - geometric progression. The denominator of the geometric progression q, and |q|∞).
If we now denote by S the sum of an infinite geometric progression, then we will have following formula:
S=x1/(1-q).
Let's look at a simple example:
Find the sum of the infinite geometric progression 2, -2/3, 2/9, - 2/27, ….
To find S, we use the formula for the sum of an infinite arithmetic progression. |-1/3|< 1. x1 = 2. S=2/(1-(-1/3)) = 3/2.
If for every natural number n match a real number a n , then they say that it is given number sequence :
a 1 , a 2 , a 3 , . . . , a n , . . . .
So, the number sequence is a function of the natural argument.
Number a 1 called first term of the sequence , number a 2 — second term of the sequence , number a 3 — third and so on. Number a n called nth term sequences , and a natural number n — his number .
From two adjacent members a n And a n +1 sequence member a n +1 called subsequent (towards a n ), A a n — previous (towards a n +1 ).
To define a sequence, you need to specify a method that allows you to find a member of the sequence with any number.
Often the sequence is specified using nth term formulas , that is, a formula that allows you to determine a member of a sequence by its number.
For example,
sequence of positive odd numbers can be given by the formula
a n= 2n- 1,
and the sequence of alternating 1 And -1 - formula
b n = (-1)n +1 . ◄
The sequence can be determined recurrent formula, that is, a formula that expresses any member of the sequence, starting with some, through the previous (one or more) members.
For example,
If a 1 = 1 , A a n +1 = a n + 5
a 1 = 1,
a 2 = a 1 + 5 = 1 + 5 = 6,
a 3 = a 2 + 5 = 6 + 5 = 11,
a 4 = a 3 + 5 = 11 + 5 = 16,
a 5 = a 4 + 5 = 16 + 5 = 21.
If a 1= 1, a 2 = 1, a n +2 = a n + a n +1 , then the first seven members number sequence install as follows:
a 1 = 1,
a 2 = 1,
a 3 = a 1 + a 2 = 1 + 1 = 2,
a 4 = a 2 + a 3 = 1 + 2 = 3,
a 5 = a 3 + a 4 = 2 + 3 = 5,
a 6 = a 4 + a 5 = 3 + 5 = 8,
a 7 = a 5 + a 6 = 5 + 8 = 13. ◄
Sequences can be final And endless .
The sequence is called ultimate if she has final number members. The sequence is called endless , if it has infinitely many members.
For example,
sequence of two-digit natural numbers:
10, 11, 12, 13, . . . , 98, 99
final.
Sequence of prime numbers:
2, 3, 5, 7, 11, 13, . . .
endless. ◄
The sequence is called increasing , if each of its members, starting from the second, is greater than the previous one.
The sequence is called decreasing , if each of its members, starting from the second, is less than the previous one.
For example,
2, 4, 6, 8, . . . , 2n, . . . — increasing sequence;
1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . — decreasing sequence. ◄
A sequence whose elements do not decrease as the number increases, or, conversely, do not increase, is called monotonous sequence .
Monotonic sequences, in particular, are increasing sequences and decreasing sequences.
Arithmetic progression
Arithmetic progression is a sequence in which each member, starting from the second, is equal to the previous one, to which the same number is added.
a 1 , a 2 , a 3 , . . . , a n, . . .
is an arithmetic progression if for any natural number n the condition is met:
a n +1 = a n + d,
Where d - a certain number.
Thus, the difference between the subsequent and previous terms of a given arithmetic progression is always constant:
a 2 - a 1 = a 3 - a 2 = . . . = a n +1 - a n = d.
Number d called difference of arithmetic progression.
To define an arithmetic progression, it is enough to indicate its first term and difference.
For example,
If a 1 = 3, d = 4 , then we find the first five terms of the sequence as follows:
a 1 =3,
a 2 = a 1 + d = 3 + 4 = 7,
a 3 = a 2 + d= 7 + 4 = 11,
a 4 = a 3 + d= 11 + 4 = 15,
a 5 = a 4 + d= 15 + 4 = 19. ◄
For an arithmetic progression with the first term a 1 and the difference d her n
a n = a 1 + (n- 1)d.
For example,
find the thirtieth term of the arithmetic progression
1, 4, 7, 10, . . .
a 1 =1, d = 3,
a 30 = a 1 + (30 - 1)d = 1 + 29· 3 = 88. ◄
a n-1 = a 1 + (n- 2)d,
a n= a 1 + (n- 1)d,
a n +1 = a 1 + nd,
then obviously
| a n=
| a n-1 + a n+1
|
| 2
|
Each member of an arithmetic progression, starting from the second, is equal to the arithmetic mean of the preceding and subsequent members.
the numbers a, b and c are successive terms of some arithmetic progression if and only if one of them is equal to the arithmetic mean of the other two.
For example,
a n = 2n- 7 , is an arithmetic progression.
Let's use the above statement. We have:
a n = 2n- 7,
a n-1 = 2(n- 1) - 7 = 2n- 9,
a n+1 = 2(n+ 1) - 7 = 2n- 5.
Hence,
| a n+1 + a n-1
| =
| 2n- 5 + 2n- 9
| = 2n- 7 = a n,
|
| 2
| 2
|
◄
Note that n The th term of an arithmetic progression can be found not only through a 1 , but also any previous a k
a n = a k + (n- k)d.
For example,
For a 5 can be written down
a 5 = a 1 + 4d,
a 5 = a 2 + 3d,
a 5 = a 3 + 2d,
a 5 = a 4 + d. ◄
a n = a n-k + kd,
a n = a n+k - kd,
then obviously
| a n=
| a n-k
+ a n+k
|
| 2
|
any member of an arithmetic progression, starting from the second, is equal to half the sum of the equally spaced members of this arithmetic progression.
In addition, for any arithmetic progression the following equality holds:
a m + a n = a k + a l,
m + n = k + l.
For example,
in arithmetic progression
1) a 10 = 28 = (25 + 31)/2 = (a 9 + a 11 )/2;
2) 28 = a 10 = a 3 + 7d= 7 + 7 3 = 7 + 21 = 28;
3) a 10= 28 = (19 + 37)/2 = (a 7 + a 13)/2;
4) a 2 + a 12 = a 5 + a 9, because
a 2 + a 12= 4 + 34 = 38,
a 5 + a 9 = 13 + 25 = 38. ◄
S n= a 1 + a 2 + a 3 + . . .+ a n,
first n terms of an arithmetic progression is equal to the product of half the sum of the extreme terms and the number of terms:
From here, in particular, it follows that if you need to sum the terms
a k, a k +1 , . . . , a n,
then the previous formula retains its structure:
For example,
in arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, . . .
S 10 = 1 + 4 + . . . + 28 = (1 + 28) · 10/2 = 145;
10 + 13 + 16 + 19 + 22 + 25 + 28 = S 10 - S 3 = (10 + 28 ) · (10 - 4 + 1)/2 = 133. ◄
If an arithmetic progression is given, then the quantities a 1 , a n, d, n AndS n connected by two formulas:
Therefore, if meanings of three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas, combined into a system of two equations with two unknowns.
An arithmetic progression is a monotonic sequence. Wherein:
- If d > 0 , then it is increasing;
- If d < 0 , then it is decreasing;
- If d = 0 , then the sequence will be stationary.
Geometric progression
Geometric progression is a sequence in which each member, starting from the second, is equal to the previous one multiplied by the same number.
b 1 , b 2 , b 3 , . . . , b n, . . .
is a geometric progression if for any natural number n the condition is met:
b n +1 = b n · q,
Where q ≠ 0 - a certain number.
Thus, the ratio of the subsequent term of a given geometric progression to the previous one is a constant number:
b 2 / b 1 = b 3 / b 2 = . . . = b n +1 / b n = q.
Number q called denominator of geometric progression.
To define a geometric progression, it is enough to indicate its first term and denominator.
For example,
If b 1 = 1, q = -3 , then we find the first five terms of the sequence as follows:
b 1 = 1,
b 2 = b 1 · q = 1 · (-3) = -3,
b 3 = b 2 · q= -3 · (-3) = 9,
b 4 = b 3 · q= 9 · (-3) = -27,
b 5 = b 4 · q= -27 · (-3) = 81. ◄
b 1 and denominator q her n The th term can be found using the formula:
b n = b 1 · qn -1 .
For example,
find the seventh term of the geometric progression 1, 2, 4, . . .
b 1 = 1, q = 2,
b 7 = b 1 · q 6 = 1 2 6 = 64. ◄
b n-1 = b 1 · qn -2 ,
b n = b 1 · qn -1 ,
b n +1 = b 1 · qn,
then obviously
b n 2 = b n -1 · b n +1 ,
each member of the geometric progression, starting from the second, is equal to the geometric mean (proportional) of the preceding and subsequent members.
Since the converse is also true, the following statement holds:
numbers a, b and c are consecutive terms of some geometric progression if and only if the square of one of them equal to the product the other two, that is, one of the numbers is the geometric mean of the other two.
For example,
Let us prove that the sequence given by the formula b n= -3 2 n , is a geometric progression. Let's use the above statement. We have:
b n= -3 2 n,
b n -1 = -3 2 n -1 ,
b n +1 = -3 2 n +1 .
Hence,
b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) · (-3 · 2 n +1 ) = b n -1 · b n +1 ,
which proves the desired statement. ◄
Note that n The th term of a geometric progression can be found not only through b 1 , but also any previous member b k , for which it is enough to use the formula
b n = b k · qn - k.
For example,
For b 5 can be written down
b 5 = b 1 · q 4 ,
b 5 = b 2 · q 3,
b 5 = b 3 · q 2,
b 5 = b 4 · q. ◄
b n = b k · qn - k,
b n = b n - k · q k,
then obviously
b n 2 = b n - k· b n + k
the square of any term of a geometric progression, starting from the second, is equal to the product of the terms of this progression equidistant from it.
In addition, for any geometric progression the equality is true:
b m· b n= b k· b l,
m+ n= k+ l.
For example,
in geometric progression
1) b 6 2 = 32 2 = 1024 = 16 · 64 = b 5 · b 7 ;
2) 1024 = b 11 = b 6 · q 5 = 32 · 2 5 = 1024;
3) b 6 2 = 32 2 = 1024 = 8 · 128 = b 4 · b 8 ;
4) b 2 · b 7 = b 4 · b 5 , because
b 2 · b 7 = 2 · 64 = 128,
b 4 · b 5 = 8 · 16 = 128. ◄
S n= b 1 + b 2 + b 3 + . . . + b n
first n members of a geometric progression with denominator q ≠ 0 calculated by the formula:
And when q = 1 - according to the formula
S n= nb 1
Note that if you need to sum the terms
b k, b k +1 , . . . , b n,
then the formula is used:
| S n- S k -1 = b k + b k +1 + . . . + b n = b k · | 1 - qn -
k +1
| . |
| 1 - q
|
For example,
in geometric progression 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, . . .
S 10 = 1 + 2 + . . . + 512 = 1 · (1 - 2 10) / (1 - 2) = 1023;
64 + 128 + 256 + 512 = S 10 - S 6 = 64 · (1 - 2 10-7+1) / (1 - 2) = 960. ◄
If a geometric progression is given, then the quantities b 1 , b n, q, n And S n connected by two formulas:
Therefore, if the values of any three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas, combined into a system of two equations with two unknowns.
For a geometric progression with the first term b 1 and denominator q the following take place properties of monotonicity :
- progression is increasing if one of the following conditions is met:
b 1 > 0 And q> 1;
b 1 < 0 And 0 < q< 1;
- The progression is decreasing if one of the following conditions is met:
b 1 > 0 And 0 < q< 1;
b 1 < 0 And q> 1.
If q< 0 , then the geometric progression is alternating: its terms with odd numbers have the same sign as its first term, and terms with even numbers have the opposite sign. It is clear that an alternating geometric progression is not monotonic.
Product of the first n terms of a geometric progression can be calculated using the formula:
Pn= b 1 · b 2 · b 3 · . . . · b n = (b 1 · b n) n / 2 .
For example,
1 · 2 · 4 · 8 · 16 · 32 · 64 · 128 = (1 · 128) 8/2 = 128 4 = 268 435 456;
3 · 6 · 12 · 24 · 48 = (3 · 48) 5/2 = (144 1/2) 5 = 12 5 = 248 832.◄
Infinitely decreasing geometric progression
Infinitely decreasing geometric progression called an infinite geometric progression whose denominator modulus is less 1 , that is
|q| < 1 .
Note that an infinitely decreasing geometric progression may not be a decreasing sequence. It fits the occasion
1 < q< 0 .
With such a denominator, the sequence is alternating. For example,
1, - 1 / 2 , 1 / 4 , - 1 / 8 , . . . .
The sum of an infinitely decreasing geometric progression name the number to which the sum of the first ones approaches without limit n members of a progression with an unlimited increase in the number n . This number is always finite and is expressed by the formula
| S= b 1 + b 2 + b 3 + . . . = | b 1
| . |
| 1 - q
|
For example,
10 + 1 + 0,1 + 0,01 + . . . = 10 / (1 - 0,1) = 11 1 / 9 ,
10 - 1 + 0,1 - 0,01 + . . . = 10 / (1 + 0,1) = 9 1 / 11 . ◄
Relationship between arithmetic and geometric progressions
Arithmetic and geometric progressions are closely related. Let's look at just two examples.
a 1 , a 2 , a 3 , . . . d , That
b a 1 , b a 2 , b a 3 , . . . b d .
For example,
1, 3, 5, . . . - arithmetic progression with difference 2 And
7 1 , 7 3 , 7 5 , . . . - geometric progression with denominator 7 2 . ◄
b 1 , b 2 , b 3 , . . . - geometric progression with denominator q , That
log a b 1, log a b 2, log a b 3, . . . - arithmetic progression with difference log aq .
For example,
2, 12, 72, . . . - geometric progression with denominator 6 And
lg 2, lg 12, lg 72, . . . - arithmetic progression with difference lg 6 . ◄
Some problems in physics and mathematics can be solved using the properties number series. The two simplest number sequences taught in schools are algebraic and geometric. In this article we will take a closer look at the question of how to find the sum of finite progression geometric decreasing.
Progression geometric
These words mean the following series real numbers, whose elements a i satisfy the expression:
Here i is the number of the element in the row, r is constant number, which is called the denominator.
This definition shows that, knowing any member of the progression and its denominator, you can restore the entire series of numbers. For example, if the 10th element is known, then dividing it by r will get the 9th element, then dividing it again will get the 8th and so on. These simple reasoning allow us to write an expression that is valid for the series of numbers under consideration:
An example of a progression with a denominator of 2 would be the following series:
1, 2, 4, 8, 16, 32, ...
If the denominator is equal to -2, then a completely different series is obtained:
1, -2, 4, -8, 16, -32, ...
Geometric progression is much faster than algebraic progression, that is, its terms increase quickly and decrease quickly.
Sum of i terms of progression
For solutions practical problems Often you have to calculate the sum of several elements of the numerical sequence in question. For this case the following formula is valid:
S i = a 1 *(r i -1)/(r-1)
It can be seen that to calculate the sum of i terms, you need to know only two numbers: a 1 and r, which is logical, since they uniquely determine the entire sequence.
Decreasing sequence and the sum of its terms
Now let's consider special case. We will assume that the modulus of the denominator r does not exceed one, that is -1 A decreasing geometric progression is interesting to consider because the infinite sum of its terms tends to a finite real number. Let's get the formula for the sum. This is easy to do if you write out the expression for S i given in the previous paragraph. We have: S i = a 1 *(r i -1)/(r-1) Let's consider the case when i->∞. Since the modulus of the denominator is less than 1, raising it to an infinite power will give zero. This can be checked using the example of r=0.5: 0,5 2 = 0,25; 0,5 3 = 0,125; ...., 0,5 20 = 0,0000009. As a result, the sum of the terms of an infinite decreasing geometric progression will take the form: This formula is often used in practice, for example, to calculate the areas of figures. It is also used to solve the paradox of Zeno of Elea with the tortoise and Achilles. Obviously, considering the amount endless progression geometric increasing (r>1), will lead to the result S ∞ = +∞. Let us show how to apply the above formulas using an example of solving a problem. It is known that the sum of an infinite geometric progression is 11. Moreover, its 7th term is 6 times less than the third term. What is the first element for this number series? First, let's write out two expressions to determine the 7th and 3rd elements. We get: Dividing the first expression by the second and expressing the denominator, we have: a 7 /a 3 = r 4 => r = 4 √(a 7 /a 3) Since the ratio of the seventh and third terms is given in the problem statement, you can substitute it and find r: r = 4 √(a 7 /a 3) = 4 √(1/6) ≈ 0.63894 We calculated r to five decimal places. Since the resulting value is less than one, the progression is decreasing, which justifies the use of the formula for its infinite sum. Let's write the expression for the first term through the sum S ∞: We substitute known values into this formula and get the answer: a 1 = 11*(1-0.63894) = 3.97166. Zeno of Elea is a famous Greek philosopher who lived in the 5th century BC. e. A number of its apogees or paradoxes have reached the present day, in which the problem of the infinitely large and the infinitely small in mathematics is formulated. One of Zeno's famous paradoxes is the competition between Achilles and the tortoise. Zeno believed that if Achilles gave the tortoise some advantage in distance, he would never be able to catch up with it. For example, let Achilles run 10 times faster than an animal crawling, which, for example, is 100 meters in front of him. When the warrior runs 100 meters, the turtle crawls away 10 meters. Having run 10 meters again, Achilles sees that the turtle crawls another 1 meter. You can argue this way ad infinitum, the distance between the competitors will indeed decrease, but the turtle will always be in front. Led Zeno to the conclusion that movement does not exist, and all surrounding movements of objects are an illusion. Of course, the ancient Greek philosopher was wrong. The solution to the paradox lies in the fact that an infinite sum of constantly decreasing segments tends to a finite number. In the above case, for the distance that Achilles ran, we get: 100 + 10 + 1 + 0,1 + 0,01 + ... Applying the formula for the sum of an infinite geometric progression, we obtain: S ∞ = 100 /(1-0.1) ≈ 111.111 meters This result shows that Achilles will catch up with the tortoise when it crawls only 11.111 meters. The ancient Greeks did not know how to work with infinite quantities in mathematics. However, this paradox can be resolved if we pay attention not to the infinite number of gaps that Achilles must overcome, but to the finite number of steps the runner needs to reach his goal. Additional materials Educational aids and simulators in the Integral online store for grade 9
Guys, today we will get acquainted with another type of progression. Example. 1,2,4,8,16... A geometric progression in which the first term is equal to one, and $q=2$. Example. 8,8,8,8... A geometric progression in which the first term is equal to eight, Example. 3,-3,3,-3,3... Geometric progression in which the first term is equal to three, Geometric progression has the properties of monotony. Just like in an arithmetic progression, if in a geometric progression the number of elements is finite, then the progression is called a finite geometric progression. $b_(1), b_(2), b_(3), ..., b_(n-2), b_(n-1), b_(n)$. Let's return to our examples. Example. 1,2,4,8,16... Geometric progression in which the first term is equal to one, Example. 16,8,4,2,1,1/2… A geometric progression in which the first term is equal to sixteen, and $q=\frac(1)(2)$. Example. 8,8,8,8... A geometric progression in which the first term is equal to eight, and $q=1$. Example. 3,-3,3,-3,3... A geometric progression in which the first term is equal to three, and $q=-1$. Example. Given a geometric progression $b_(1), b_(2), …, b_(n), … $. Solution. Example. The difference between the seventh and fifth terms of the geometric progression is 192, the sum of the fifth and sixth terms of the progression is 192. Find the tenth term of this progression. Solution. Let a finite geometric progression be given: $b_(1),b_(2),…,b_(n-1),b_(n)$. $S_(n)*q-S_(n)=(b_(2)+⋯+b_(n-1)+b_(n))+b_(n)*q-b_(1)-(b_(2 )+⋯+b_(n-1)+b_(n))=b_(n)*q-b_(1)$. $S_(n)(q-1)=b_(n)*q-b_(1)$. $S_(n)=\frac(b_(n)*q-b_(1))(q-1)=\frac(b_(1)*q^(n-1)*q-b_(1)) (q-1)=\frac(b_(1)(q^(n)-1))(q-1)$. $S_(n)=\frac(b_(1)(q^(n)-1))(q-1)$. We have obtained the formula for the sum of a finite geometric progression. Solution. Example. Solution. $S_(n)=\frac(-3*(q^(n)-1))(q-1)=-4095$. A number sequence is a geometric progression only when the square of each member is equal to the product of the two adjacent members of the progression. Do not forget that for a finite progression this condition is not satisfied for the first and last terms. The modulus of any term of a geometric progression is equal to the geometric mean of its two neighboring terms. Solution. Lesson on the topic “Infinitely decreasing geometric progression” (algebra, 10th grade)
The purpose of the lesson: introducing students to a new type of sequence - an infinitely decreasing geometric progression. Equipment: projector, screen. Lesson type: lesson - learning new topic. During the classes I
. Org. moment. State the topic and purpose of the lesson. II
. Updating students' knowledge. In 9th grade you studied arithmetic and geometric progressions. Questions 1. Definition of arithmetic progression. (An arithmetic progression is a sequence in which each member, starting from the second, is equal to the previous member added to the same number). 2. Formula n th term of the arithmetic progression ( 3. Formula for the sum of the first n terms of an arithmetic progression. ( 4. Definition of geometric progression. (A geometric progression is a sequence of non-zero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number). 5. Formula n th term of the geometric progression ( 6. Formula for the sum of the first n members of a geometric progression. ( 7. What other formulas do you know? ( 5. For geometric progression 6. For geometric progression 7. Exponentially b
3
= 8
And b
5
= 2
. Find b
4
. (4) 8. Exponentially b
3
= 8
And b
5
= 2
. Find b
1
And
q
. 9. Exponentially b
3
= 8
And b
5
= 2
. Find S
5
. (62) III
. Learning a new topic(demonstration of presentation). Consider a square with a side equal to 1. Let's draw another square whose side is half the size of the first square, then another one whose side is half the second, then the next one, etc. Each time the side of the new square is equal to half of the previous one. As a result, we received a sequence of sides of squares And, what is very important, the more we build such squares, the smaller the side of the square will be. For example, Those. As the number n increases, the terms of the progression approach zero. Using this figure, you can consider another sequence. For example, the sequence of areas of squares: Let's look at another example. Equilateral triangle with a side equal to 1 cm. Let's construct the next triangle with vertices at the midpoints of the sides of the 1st triangle, according to the theorem about midline triangle - the side of the 2nd is equal to half the side of the first, the side of the 3rd is equal to half the side of the 2nd, etc. Again we obtain a sequence of lengths of the sides of triangles. If we consider a geometric progression with a negative denominator. Then, again, with increasing numbers n terms of the progression approach zero. Let's pay attention to the denominators of these sequences. Everywhere the denominators were less than 1 in absolute value. We can conclude: a geometric progression will be infinitely decreasing if the modulus of its denominator is less than 1. Definition:
A geometric progression is said to be infinitely decreasing if the modulus of its denominator is less than one. Using the definition, you can decide whether a geometric progression is infinitely decreasing or not. Task Is the sequence an infinitely decreasing geometric progression if it is given by the formula: Solution: this geometric progression is infinitely decreasing. b) this sequence is not an infinitely decreasing geometric progression. Consider a square with a side equal to 1. Divide it in half, one of the halves in half, etc. The areas of all the resulting rectangles form an infinitely decreasing geometric progression: The sum of the areas of all rectangles obtained in this way will be equal to the area of the 1st square and equal to 1.
The task of finding the first term of a progression
Zeno's famous paradox with the fast Achilles and the slow tortoise
Lesson and presentation on the topic: "Number sequences. Geometric progression"
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Powers and roots Functions and graphs
The topic of today's lesson is geometric progression.Geometric progression
Definition. A numerical sequence in which each term, starting from the second, is equal to the product of the previous one and some fixed number is called a geometric progression.
Let's define our sequence recursively: $b_(1)=b$, $b_(n)=b_(n-1)*q$,
where b and q are certain given numbers. The number q is called the denominator of the progression.
and $q=1$.
and $q=-1$.
If $b_(1)>0$, $q>1$,
then the sequence is increasing.
If $b_(1)>0$, $0 The sequence is usually denoted in the form: $b_(1), b_(2), b_(3), ..., b_(n), ...$.
Note that if a sequence is a geometric progression, then the sequence of squares of terms is also a geometric progression. In the second sequence, the first term is equal to $b_(1)^2$, and the denominator is equal to $q^2$.Formula for the nth term of a geometric progression
Geometric progression can also be specified in analytical form. Let's see how to do this:
$b_(1)=b_(1)$.
$b_(2)=b_(1)*q$.
$b_(3)=b_(2)*q=b_(1)*q*q=b_(1)*q^2$.
$b_(4)=b_(3)*q=b_(1)*q^3$.
$b_(5)=b_(4)*q=b_(1)*q^4$.
We easily notice the pattern: $b_(n)=b_(1)*q^(n-1)$.
Our formula is called the "formula of the nth term of a geometric progression."
and $q=2$.
$b_(n)=1*2^(n)=2^(n-1)$.
$b_(n)=16*(\frac(1)(2))^(n-1)$.
$b_(n)=8*1^(n-1)=8$.
$b_(n)=3*(-1)^(n-1)$.
a) It is known that $b_(1)=6, q=3$. Find $b_(5)$.
b) It is known that $b_(1)=6, q=2, b_(n)=768$. Find n.
c) It is known that $q=-2, b_(6)=96$. Find $b_(1)$.
d) It is known that $b_(1)=-2, b_(12)=4096$. Find q.
a) $b_(5)=b_(1)*q^4=6*3^4=486$.
b) $b_n=b_1*q^(n-1)=6*2^(n-1)=768$.
$2^(n-1)=\frac(768)(6)=128$, since $2^7=128 => n-1=7; n=8$.
c) $b_(6)=b_(1)*q^5=b_(1)*(-2)^5=-32*b_(1)=96 => b_(1)=-3$.
d) $b_(12)=b_(1)*q^(11)=-2*q^(11)=4096 => q^(11)=-2048 => q=-2$.
We know that: $b_(7)-b_(5)=192$ and $b_(5)+b_(6)=192$.
We also know: $b_(5)=b_(1)*q^4$; $b_(6)=b_(1)*q^5$; $b_(7)=b_(1)*q^6$.
Then:
$b_(1)*q^6-b_(1)*q^4=192$.
$b_(1)*q^4+b_(1)*q^5=192$.
We received a system of equations:
$\begin(cases)b_(1)*q^4(q^2-1)=192\\b_(1)*q^4(1+q)=192\end(cases)$.
Equating our equations we get:
$b_(1)*q^4(q^2-1)=b_(1)*q^4(1+q)$.
$q^2-1=q+1$.
$q^2-q-2=0$.
We got two solutions q: $q_(1)=2, q_(2)=-1$.
Substitute sequentially into the second equation:
$b_(1)*2^4*3=192 => b_(1)=4$.
$b_(1)*(-1)^4*0=192 =>$ no solutions.
We got that: $b_(1)=4, q=2$.
Let's find the tenth term: $b_(10)=b_(1)*q^9=4*2^9=2048$.Sum of a finite geometric progression
Let us have a finite geometric progression. Let's, just like for an arithmetic progression, calculate the sum of its terms.
Let us introduce the designation for the sum of its terms: $S_(n)=b_(1)+b_(2)+⋯+b_(n-1)+b_(n)$.
In the case when $q=1$. All terms of the geometric progression are equal to the first term, then it is obvious that $S_(n)=n*b_(1)$.
Let us now consider the case $q≠1$.
Let's multiply the above amount by q.
$S_(n)*q=(b_(1)+b_(2)+⋯+b_(n-1)+b_(n))*q=b_(1)*q+b_(2)*q+⋯ +b_(n-1)*q+b_(n)*q=b_(2)+b_(3)+⋯+b_(n)+b_(n)*q$.
Note:
$S_(n)=b_(1)+(b_(2)+⋯+b_(n-1)+b_(n))$.
$S_(n)*q=(b_(2)+⋯+b_(n-1)+b_(n))+b_(n)*q$.
Example.
Find the sum of the first seven terms of a geometric progression whose first term is 4 and the denominator is 3.
$S_(7)=\frac(4*(3^(7)-1))(3-1)=2*(3^(7)-1)=4372$.
Find the fifth term of the geometric progression that is known: $b_(1)=-3$; $b_(n)=-3072$; $S_(n)=-4095$.
$b_(n)=(-3)*q^(n-1)=-3072$.
$q^(n-1)=1024$.
$q^(n)=1024q$.
$-4095(q-1)=-3*(q^(n)-1)$.
$-4095(q-1)=-3*(1024q-1)$.
$1365q-1365=1024q-1$.
$341q=$1364.
$q=4$.
$b_5=b_1*q^4=-3*4^4=-3*256=-768$.Characteristic property of geometric progression
Guys, a geometric progression is given. Let's look at its three consecutive members: $b_(n-1),b_(n),b_(n+1)$.
We know that:
$\frac(b_(n))(q)=b_(n-1)$.
$b_(n)*q=b_(n+1)$.
Then:
$\frac(b_(n))(q)*b_(n)*q=b_(n)^(2)=b_(n-1)*b_(n+1)$.
$b_(n)^(2)=b_(n-1)*b_(n+1)$.
If the progression is finite, then this equality holds for all terms except the first and last.
If it is not known in advance what form the sequence has, but it is known that: $b_(n)^(2)=b_(n-1)*b_(n+1)$.
Then we can safely say that this is a geometric progression.
Let's look at this identity: $\sqrt(b_(n)^(2))=\sqrt(b_(n-1)*b_(n+1))$.
$|b_(n)|=\sqrt(b_(n-1)*b_(n+1))$.
$\sqrt(a*b)$ is called the average geometric numbers a and b.
Example.
Find x such that $x+2; 2x+2; 3x+3$ were three consecutive terms of a geometric progression.
Let's use the characteristic property:
$(2x+2)^2=(x+2)(3x+3)$.
$4x^2+8x+4=3x^2+3x+6x+6$.
$x^2-x-2=0$.
$x_(1)=2$ and $x_(2)=-1$.
Let us sequentially substitute our solutions into the original expression:
With $x=2$, we got the sequence: 4;6;9 – a geometric progression with $q=1.5$.
For $x=-1$, we get the sequence: 1;0;0.
Answer: $x=2.$Problems to solve independently
1. Find the eighth first term of the geometric progression 16;-8;4;-2….
2. Find the tenth term of the geometric progression 11,22,44….
3. It is known that $b_(1)=5, q=3$. Find $b_(7)$.
4. It is known that $b_(1)=8, q=-2, b_(n)=512$. Find n.
5. Find the sum of the first 11 terms of the geometric progression 3;12;48….
6. Find x such that $3x+4; 2x+4; x+5$ are three consecutive terms of a geometric progression. 
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