How to find the sum of a geometric progression. Geometric progression. Comprehensive guide with examples (2019)

If every natural number n match a real number a n , then they say that given number sequence :

a 1 , a 2 , a 3 , . . . , a n , . . . .

So, a numerical sequence is a function of a natural argument.

Number a 1 called the first member of the sequence , number a 2 — the second member of the sequence , number a 3 — third and so on. Number a n called nth member sequences , a natural number n — his number .

From two neighboring members a n and a n +1 member sequences a n +1 called subsequent (towards a n ), a a n — previous (towards a n +1 ).

To specify a sequence, you must specify a method that allows you to find a sequence member with any number.

Often the sequence is given with nth term formulas , that is, a formula that allows you to determine a sequence member 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 of the numerical sequence are set 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 it has a finite number of 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.

Prime number sequence:

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 waning , if each of its members, starting from the second, is less than the previous one.

For example,

2, 4, 6, 8, . . . , 2n, . . . is an ascending sequence;

1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . is a descending sequence. ◄

A sequence whose elements do not decrease with increasing number, or, conversely, do not increase, is called monotonous sequence .

Monotonic sequences, in particular, are increasing sequences and decreasing sequences.

Arithmetic progression

Arithmetic progression a sequence is called, each member of which, 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 condition is met:

a n +1 = a n + d,

where d - some number.

Thus, the difference between the next and the previous members of a given arithmetic progression always constant:

a 2 - a 1 = a 3 - a 2 = . . . = a n +1 - a n = d.

Number d called the difference of an arithmetic progression.

To set an arithmetic progression, it is enough to specify its first term and difference.

For example,

if a 1 = 3, d = 4 , then the first five terms of the sequence are found 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 difference d her n

a n = a 1 + (n- 1)d.

For example,

find the thirtieth term of an 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 the arithmetic progression, starting from the second, is equal to the arithmetic mean of the previous and subsequent members.

numbers a, b and c are consecutive members 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 statement above. We have:

a n = 2n- 7,

a n-1 = 2(n- 1) - 7 = 2n- 9,

a n+1 = 2(n+ 1) - 7 = 2n- 5.

Consequently,

a n+1 + a n-1	=	2n- 5 + 2n- 9	= 2n- 7 = a n,
2		2

◄

Note that n -th member 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

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 members of this arithmetic progression equally spaced from it.

In addition, for any arithmetic progression, the equality is true:

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, as

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 members of an arithmetic progression is equal to the product of half the sum of the extreme terms by the number of terms:

From this, in particular, it follows that if it is necessary 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 linked by two formulas:

Therefore, if the values ​​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 a sequence is called, each term of which, 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 condition is met:

b n +1 = b n · q,

where q ≠ 0 - some number.

Thus, the ratio of the next term of this 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 a geometric progression.

To set a geometric progression, it is enough to specify its first term and denominator.

For example,

if b 1 = 1, q = -3 , then the first five terms of the sequence are found 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 -th term can be found by the formula:

b n = b 1 · q n -1 .

For example,

find the seventh term of a geometric progression 1, 2, 4, . . .

b 1 = 1, q = 2,

b 7 = b 1 · q 6 = 1 2 6 = 64. ◄

bn-1 = b 1 · q n -2 ,

b n = b 1 · q n -1 ,

b n +1 = b 1 · q n,

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 previous and subsequent members.

Since the converse is also true, the following assertion holds:

numbers a, b and c are consecutive members of some geometric progression if and only if the square of one of them is equal to the product of 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 statement above. We have:

b n= -3 2 n,

b n -1 = -3 2 n -1 ,

b n +1 = -3 2 n +1 .

Consequently,

b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) (-3 2 n +1 ) = b n -1 · b n +1 ,

which proves the required assertion. ◄

Note that n th term of a geometric progression can be found not only through b 1 , but also any previous term b k , for which it suffices to use the formula

b n = b k · q n - k.

For example,

for b 5 can be written

b 5 = b 1 · q 4 ,

b 5 = b 2 · q 3,

b 5 = b 3 · q2,

b 5 = b 4 · q. ◄

b n = b k · q n - k,

b n = b n - k · q k,

then obviously

b n 2 = b n - k· b n + k

the square of any member of a geometric progression, starting from the second, is equal to the product of the members 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,

exponentially

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 , as

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 a denominator q ≠ 0 calculated by the formula:

And when q = 1 - according to the formula

S n= n.b. 1

Note that if we 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 - q n - k +1	.
	1 - q

For example,

exponentially 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 linked 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 monotonicity properties :

• the progression is increasing if one of the following conditions is met:

b 1 > 0 and q> 1;

b 1 < 0 and 0 < q< 1;

• A 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 sign-alternating: its odd-numbered terms have the same sign as its first term, and even-numbered terms 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 by the formula:

P n= 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 is called an infinite geometric progression whose denominator modulus is less than 1 , that is

|q| < 1 .

Note that an infinitely decreasing geometric progression may not be a decreasing sequence. This fits the case

1 < q< 0 .

With such a denominator, the sequence is sign-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 n terms of the 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 consider just two examples.

a 1 , a 2 , a 3 , . . . d , then

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 , . . . is a geometric progression with a denominator 7 2 . ◄

b 1 , b 2 , b 3 , . . . is a geometric progression with a denominator q , then

log a b 1, log a b 2, log a b 3, . . . — arithmetic progression with difference log aq .

For example,

2, 12, 72, . . . is a geometric progression with a denominator 6 and

lg 2, lg 12, lg 72, . . . — arithmetic progression with difference lg 6 . ◄

>>Math: Geometric progression

For the convenience of the reader, this section follows exactly the same plan as we followed in the previous section.

1. Basic concepts.

Definition. A numerical sequence, all members of which are different from 0 and each member of which, starting from the second, is obtained from the previous member by multiplying it by the same number is called a geometric progression. In this case, the number 5 is called the denominator of a geometric progression.

Thus, a geometric progression is a numerical sequence (b n) given recursively by the relations

Is it possible, by looking at a number sequence, to determine whether it is a geometric progression? Can. If you are convinced that the ratio of any member of the sequence to the previous member is constant, then you have a geometric progression.
Example 1

1, 3, 9, 27, 81,... .
b 1 = 1, q = 3.

Example 2

This is a geometric progression that
Example 3

This is a geometric progression that
Example 4

8, 8, 8, 8, 8, 8,....

This is a geometric progression where b 1 - 8, q = 1.

Note that this sequence is also an arithmetic progression (see Example 3 from § 15).

Example 5

2,-2,2,-2,2,-2.....

This is a geometric progression, in which b 1 \u003d 2, q \u003d -1.

Obviously, a geometric progression is an increasing sequence if b 1 > 0, q > 1 (see Example 1), and a decreasing sequence if b 1 > 0, 0< q < 1 (см. пример 2).

To indicate that the sequence (b n) is a geometric progression, the following notation is sometimes convenient:

The icon replaces the phrase "geometric progression".
We note one curious and at the same time quite obvious property of a geometric progression:
If the sequence is a geometric progression, then the sequence of squares, i.e. is a geometric progression.
In the second geometric progression, the first term is equal to a equal to q 2.
If we discard all the terms following b n exponentially, then we get a finite geometric progression
In the following paragraphs of this section, we will consider the most important properties geometric progression.

2. Formula of the n-th term of a geometric progression.

Consider a geometric progression denominator q. We have:

It is not difficult to guess that for any number n the equality

This is the formula for the nth term of a geometric progression.

Comment.

If you have read the important remark from the previous paragraph and understood it, then try to prove formula (1) by mathematical induction, just as it was done for the formula of the nth term of an arithmetic progression.

Let's rewrite the formula of the nth term of the geometric progression

and introduce the notation: We get y \u003d mq 2, or, in more detail,
The argument x is contained in the exponent, so such a function is called an exponential function. This means that a geometric progression can be considered as an exponential function given on the set N of natural numbers. On fig. 96a shows a graph of the function of Fig. 966 - function graph In both cases, we have isolated points (with abscissas x = 1, x = 2, x = 3, etc.) lying on some curve (both figures show the same curve, only differently located and depicted in different scales). This curve is called the exponent. More about the exponential function and its graph will be discussed in the 11th grade algebra course.

Let's return to examples 1-5 from the previous paragraph.

1) 1, 3, 9, 27, 81,... . This is a geometric progression, in which b 1 \u003d 1, q \u003d 3. Let's make a formula for the nth term
2) This is a geometric progression, in which Let's formulate the n-th term

This is a geometric progression that Compose the formula for the nth term
4) 8, 8, 8, ..., 8, ... . This is a geometric progression, in which b 1 \u003d 8, q \u003d 1. Let's make a formula for the nth term
5) 2, -2, 2, -2, 2, -2,.... This is a geometric progression, in which b 1 = 2, q = -1. Compose the formula for the nth term

Example 6

Given a geometric progression

In all cases, the solution is based on the formula of the nth member of a geometric progression

a) Putting n = 6 in the formula of the nth term of the geometric progression, we get

b) We have

Since 512 \u003d 2 9, we get n - 1 \u003d 9, n \u003d 10.

d) We have

Example 7

The difference between the seventh and fifth members of the geometric progression is 48, the sum of the fifth and sixth members of the progression is also 48. Find the twelfth member of this progression.

First step. Drawing up a mathematical model.

The conditions of the task can be briefly written as follows:

Using the formula of the n-th member of a geometric progression, we get:
Then the second condition of the problem (b 7 - b 5 = 48) can be written as

The third condition of the problem (b 5 +b 6 = 48) can be written as

As a result, we obtain a system of two equations with two variables b 1 and q:

which, in combination with condition 1) written above, is the mathematical model of the problem.

Second phase.

Working with the compiled model. Equating the left parts of both equations of the system, we get:

(we have divided both sides of the equation into the expression b 1 q 4 , which is different from zero).

From the equation q 2 - q - 2 = 0 we find q 1 = 2, q 2 = -1. Substituting the value q = 2 into the second equation of the system, we obtain
Substituting the value q = -1 into the second equation of the system, we get b 1 1 0 = 48; this equation has no solutions.

So, b 1 \u003d 1, q \u003d 2 - this pair is the solution to the compiled system of equations.

Now we can write down a geometric progression, about which in question in the problem: 1, 2, 4, 8, 16, 32, ... .

Third stage.

The answer to the problem question. It is required to calculate b 12 . We have

Answer: b 12 = 2048.

3. The formula for the sum of members of a finite geometric progression.

Let there be a finite geometric progression

Denote by S n the sum of its terms, i.e.

Let's derive a formula for finding this sum.

Let's start with the simplest case, when q = 1. Then the geometric progression b 1 ,b 2 , b 3 ,..., bn consists of n numbers equal to b 1 , i.e. the progression is b 1 , b 2 , b 3 , ..., b 4 . The sum of these numbers is nb 1 .

Let now q = 1 To find S n we use an artificial method: let's perform some transformations of the expression S n q. We have:

Performing transformations, we, firstly, used the definition of a geometric progression, according to which (see the third line of reasoning); secondly, they added and subtracted why the meaning of the expression, of course, did not change (see the fourth line of reasoning); thirdly, we used the formula of the n-th member of a geometric progression:

From formula (1) we find:

This is the formula for the sum of n members of a geometric progression (for the case when q = 1).

Example 8

Given a finite geometric progression

a) the sum of the members of the progression; b) the sum of the squares of its terms.

b) Above (see p. 132) we have already noted that if all members of a geometric progression are squared, then a geometric progression with the first member b 2 and the denominator q 2 will be obtained. Then the sum of the six terms of the new progression will be calculated by

Example 9

Find the 8th term of a geometric progression for which

In fact, we have proved the following theorem.

A numerical sequence is a geometric progression if and only if the square of each of its terms, except for the first one (and the last one, in the case of a finite sequence), is equal to the product of the previous and subsequent terms (a characteristic property of a geometric progression).

Lesson and presentation on the topic: "Number sequences. Geometric progression"

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Guys, today we will get acquainted with another type of progression.
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.

Example. 1,2,4,8,16… Geometric progression, in which the first member is equal to one, and $q=2$.

Example. 8,8,8,8… A geometric progression whose first term is eight,
and $q=1$.

Example. 3,-3,3,-3,3... A geometric progression whose first term is three,
and $q=-1$.

The geometric progression has the properties of monotonicity.
If $b_(1)>0$, $q>1$,
then the sequence is increasing.
If $b_(1)>0$, $0 The sequence is usually denoted as: $b_(1), b_(2), b_(3), ..., b_(n), ...$.

Just like in an arithmetic progression, if the number of elements in a geometric progression is finite, then the progression is called a finite geometric progression.

$b_(1), b_(2), b_(3), ..., b_(n-2), b_(n-1), b_(n)$.
Note that if the sequence is a geometric progression, then the sequence of squared terms is also a geometric progression. The second sequence has the first term $b_(1)^2$ and the denominator $q^2$.

Formula of the nth member of a geometric progression

Geometric progression can also be specified in analytical form. Let's see how to do it:
$b_(1)=b_(1)$.
$b_(2)=b_(1)*q$.
$b_(3)=b_(2)*q=b_(1)*q*q=b_(1)*q^2$.
$b_(4)=b_(3)*q=b_(1)*q^3$.
$b_(5)=b_(4)*q=b_(1)*q^4$.
We can easily see the pattern: $b_(n)=b_(1)*q^(n-1)$.
Our formula is called "formula of the n-th member of a geometric progression".

Let's go back to our examples.

Example. 1,2,4,8,16… A geometric progression whose first term is equal to one,
and $q=2$.
$b_(n)=1*2^(n)=2^(n-1)$.

Example. 16,8,4,2,1,1/2… A geometric progression whose first term is sixteen and $q=\frac(1)(2)$.
$b_(n)=16*(\frac(1)(2))^(n-1)$.

Example. 8,8,8,8… A geometric progression where the first term is eight and $q=1$.
$b_(n)=8*1^(n-1)=8$.

Example. 3,-3,3,-3,3… A geometric progression whose first term is three and $q=-1$.
$b_(n)=3*(-1)^(n-1)$.

Example. Given a geometric progression $b_(1), b_(2), …, b_(n), … $.
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.

Decision.
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) $b_(12)=b_(1)*q^(11)=-2*q^(11)=4096 => q^(11)=-2048 => q=-2$.

Example. The difference between the seventh and fifth members of the geometric progression is 192, the sum of the fifth and sixth members of the progression is 192. Find the tenth member of this progression.

Decision.
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 got 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 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 successively 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: $b_(10)=b_(1)*q^9=4*2^9=2048$.

The sum of a finite geometric progression

Suppose we have a finite geometric progression. Let's, as well as for an arithmetic progression, calculate the sum of its members.

Let a finite geometric progression be given: $b_(1),b_(2),…,b_(n-1),b_(n)$.
Let's introduce the notation for the sum of its terms: $S_(n)=b_(1)+b_(2)+⋯+b_(n-1)+b_(n)$.
In the case when $q=1$. All members of the geometric progression are equal to the first member, then it is obvious that $S_(n)=n*b_(1)$.
Consider now the case $q≠1$.
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:
$S_(n)=b_(1)+(b_(2)+⋯+b_(n-1)+b_(n))$.
$S_(n)*q=(b_(2)+⋯+b_(n-1)+b_(n))+b_(n)*q$.

$S_(n)*q-S_(n)=(b_(2)+⋯+b_(n-1)+b_(n))+b_(n)*q-b_(1)-(b_(2 )+⋯+b_(n-1)+b_(n))=b_(n)*q-b_(1)$.

$S_(n)(q-1)=b_(n)*q-b_(1)$.

$S_(n)=\frac(b_(n)*q-b_(1))(q-1)=\frac(b_(1)*q^(n-1)*q-b_(1)) (q-1)=\frac(b_(1)(q^(n)-1))(q-1)$.

$S_(n)=\frac(b_(1)(q^(n)-1))(q-1)$.

We have obtained the formula for the sum of a finite geometric progression.

Example.
Find the sum of the first seven terms of a geometric progression whose first term is 4 and the denominator is 3.

Decision.
$S_(7)=\frac(4*(3^(7)-1))(3-1)=2*(3^(7)-1)=4372$.

Example.
Find the fifth member of the geometric progression, which is known: $b_(1)=-3$; $b_(n)=-3072$; $S_(n)=-4095$.

Decision.
$b_(n)=(-3)*q^(n-1)=-3072$.
$q^(n-1)=1024$.
$q^(n)=1024q$.

$S_(n)=\frac(-3*(q^(n)-1))(q-1)=-4095$.
$-4095(q-1)=-3*(q^(n)-1)$.
$-4095(q-1)=-3*(1024q-1)$.
$1365q-1365=1024q-1$.
$341q=1364$.
$q=4$.
$b_5=b_1*q^4=-3*4^4=-3*256=-768$.

Characteristic property of a geometric progression

Guys, given a geometric progression. Let's consider 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 kind of sequence 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.

A number sequence is a geometric progression only when the square of each of its terms is equal to the product of its two neighboring terms of the progression. Let's not forget that for finite progression this condition is not satisfied for the first and last member.

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 geometric mean of a and b.

The modulus of any member of a geometric progression is equal to the geometric mean of the two members adjacent to it.

Example.
Find x such that $x+2; 2x+2; 3x+3$ were three consecutive members of a geometric progression.

Decision.
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 $x_(2)=-1$.
Substitute sequentially in the original expression, our solutions:
With $x=2$, we got the sequence: 4;6;9 is a geometric progression with $q=1.5$.
With $x=-1$, we got the sequence: 1;0;0.
Answer: $x=2.$

Tasks for independent solution

1. Find the eighth first member of the geometric progression 16; -8; 4; -2 ....
2. Find the tenth member 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 members of the geometric progression 3;12;48….
6. Find x such that $3x+4; 2x+4; x+5$ are three consecutive members of a geometric progression.

Geometric progression no less important in mathematics than in arithmetic. A geometric progression is such a sequence of numbers b1, b2,..., b[n] each next member of which is obtained by multiplying the previous one by a constant number. This number, which also characterizes the rate of growth or decrease of the progression, is called denominator of a geometric progression and denote

For complete task geometric progression, in addition to the denominator, it is necessary to know or determine its first term. For a positive value of the denominator, the progression is a monotone sequence, and if this sequence of numbers is monotonically decreasing and monotonically increasing when. The case when the denominator is equal to one is not considered in practice, since we have a sequence same numbers, and their summation is of no practical interest

General term of a geometric progression calculated according to the formula

The sum of the first n terms of a geometric progression determined by the formula

Let us consider solutions of classical geometric progression problems. Let's start with the simplest to understand.

Example 1. The first term of a geometric progression is 27, and its denominator is 1/3. Find the first six terms of a geometric progression.

Solution: We write the condition of the problem in the form

For calculations, we use the formula for the nth member of a geometric progression

Based on it, we find unknown members of the progression

As you can see, calculating the terms of a geometric progression is not difficult. The progression itself will look like this

Example 2. The first three members of a geometric progression are given: 6; -12; 24. Find the denominator and the seventh term.

Solution: We calculate the denominator of the geometric progression based on its definition

We got an alternating geometric progression whose denominator is -2. The seventh term is calculated by the formula

On this task is solved.

Example 3. A geometric progression is given by two of its members . Find the tenth term of the progression.

Decision:

Let's write the given values ​​​​through the formulas

According to the rules, one would have to find the denominator, and then look for desired value, but for the tenth term we have

The same formula can be obtained on the basis of simple manipulations with the input data. We divide the sixth term of the series by another, as a result we get

If the resulting value is multiplied by the sixth term, we get the tenth

Thus, for such problems, with the help of simple transformations into fast way you can find the right solution.

Example 4. Geometric progression is given by recurrent formulas

Find the denominator of the geometric progression and the sum of the first six terms.

Decision:

We write the given data in the form of a system of equations

Express the denominator by dividing the second equation by the first

Find the first term of the progression from the first equation

Compute the following five terms to find the sum of the geometric progression

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 the inequality zero of the first term and denominator of the progression, 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 coincides with the first term and is equal to the number a. The value of the function y can be understood as nth term progressions, if the argument x is taken as a natural number n (counter).

Exists for the sum of the first n members of a geometric progression: S(n)=b1 (1-q^n)/(1-q). This formula valid for q≠1. If q=1, then the sum of the first n terms is calculated by the formula S(n)=n b1. By the way, the progression will be called increasing for q greater than one and positive b1. When the denominator of the progression, modulo not exceeding one, the progression will be called decreasing.

special case geometric progression - an infinitely decreasing geometric progression (b.u.g.p.). The fact is that the members of a decreasing geometric progression will decrease over and over again, but will never reach zero. Despite this, it is possible to find the sum of all terms of such a progression. It is determined by the formula S=b1/(1-q). Total n members are infinite.

To visualize how you can add an infinite number of numbers and not get infinity, bake a cake. Cut off half of it. Then cut 1/2 off the half, and so on. The pieces that you will get are nothing more than members of an infinitely decreasing geometric progression with a denominator of 1/2. If you put all these pieces together, you get the original cake.

Geometry problems are a special kind of exercise that requires spatial thinking. If you can't solve the geometric task try to follow the rules below.

Instruction

Read the condition of the problem very carefully, if you don’t remember or don’t understand something, re-read it again.

Try to determine what type of geometric problems it is, for example: computational, when you need to find out some value, tasks for, requiring logical chain reasoning, construction problems with the help of a compass and ruler. More tasks mixed type. Once you've figured out the type of problem, try to think logically.

Apply the necessary theorem for this problem, if there are doubts or there are no options at all, then try to remember the theory that you studied on the relevant topic.

Make a draft of the problem as well. Try to apply known ways checking the correctness of your solution.

Complete the solution of the problem neatly in a notebook, without blots and strikethroughs, and most importantly -. Perhaps it will take time and effort to solve the first geometric problems. However, once you get the hang of this process, you'll start clicking tasks like nuts and having fun doing it!

A geometric progression is a sequence of numbers b1, b2, b3, ... , b(n-1), b(n) such that b2=b1*q, b3=b2*q, ... , b(n) =b(n-1)*q, b1≠0, q≠0. In other words, each member of the progression is obtained from the previous one by multiplying it by some non-zero denominator of the progression q.

Instruction

Problems on a progression are most often solved by compiling and following a system with respect to the first term of the progression b1 and the denominator of the progression q. To write equations, it is useful to remember some formulas.

How to express the n-th member of the progression through the first member of the progression and the denominator of the progression: b(n)=b1*q^(n-1).

Consider separately the case |q|<1. Если знаменатель прогрессии по модулю меньше единицы, имеем бесконечно убывающую геометрическую . Сумма первых n членов бесконечно убывающей геометрической прогрессии ищется так же, как и для неубывающей геометрической прогрессии. Однако в случае бесконечно убывающей геометрической прогрессии можно найти также сумму всех членов этой прогрессии, поскольку при бесконечном n будет бесконечно уменьшаться значение b(n), и сумма всех членов будет стремиться к определенному пределу. Итак, сумма всех членов бесконечно убывающей геометрической прогрессии