How to find x exponentially. Geometric progression

Geometric progression, along with arithmetic, is an important number series, which is studied in school course algebra in 9th grade. In this article, we will consider the denominator of a geometric progression, and how its value affects its properties.

Definition of geometric progression

To begin with, we give the definition of this number series. A geometric progression is a series rational numbers, which is formed by sequentially multiplying its first element by a constant number called the denominator.

For example, the numbers in the series 3, 6, 12, 24, ... are a geometric progression, because if we multiply 3 (the first element) by 2, we get 6. If we multiply 6 by 2, we get 12, and so on.

The members of the sequence under consideration are usually denoted by the symbol ai, where i is an integer indicating the number of the element in the series.

The above definition of a progression can be written in the language of mathematics as follows: an = bn-1 * a1, where b is the denominator. It is easy to check this formula: if n = 1, then b1-1 = 1, and we get a1 = a1. If n = 2, then an = b * a1, and we again come to the definition of the series of numbers under consideration. Similar reasoning can be continued for large values ​​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 in order to find the sum of an arbitrary number of terms.

Infinitely decreasing sequence

Above was an explanation of what it is. Now, knowing the formula for Sn, let's apply it to this number series. Since any number whose modulus does not exceed 1 tends to zero when raised to large powers, that is, b∞ => 0 if -1

Since the difference (1 - b) will always be positive, regardless of the value of the denominator, the sign of the sum of an infinitely decreasing geometric progression S∞ is uniquely determined by the sign of its first element a1.

Now we will consider several problems, where we will show how to apply the acquired knowledge to specific numbers.

Task number 1. Calculation of unknown elements of the progression and the sum

Given a geometric progression, the denominator of the progression is 2, and its first element is 3. What will be its 7th and 10th terms, and what is the sum of its seven initial elements?

The condition of the problem is quite simple and involves the direct use of the above formulas. So, to calculate the element with number n, we use the expression an = bn-1 * a1. For the 7th element we have: a7 = b6 * a1, substituting the known data, we get: a7 = 26 * 3 = 192. We do the same for the 10th member: a10 = 29 * 3 = 1536.

We use the well-known formula for the sum and determine this value for the first 7 elements of the series. We have: S7 = (27 - 1) * 3 / (2 - 1) = 381.

Task number 2. Determining the sum of arbitrary elements of the progression

Let -2 be the denominator of the exponential progression bn-1 * 4, where n is an integer. It is necessary to determine the sum from the 5th to the 10th element of this series, inclusive.

The problem posed cannot be solved directly using known formulas. You can solve it with 2 various methods. For the sake of completeness, we present both.

Method 1. Its idea is simple: you need to calculate the two corresponding sums of the first terms, and then subtract the other from one. Calculate the smaller sum: S10 = ((-2)10 - 1) * 4 / (-2 - 1) = -1364. Now we calculate a large amount: 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 known values and get the required number: b = 1 - 2 / 3 = -1 / 3 or -0.333(3). We can check this result qualitatively if we remember that for this type of sequence, the modulus b must not go beyond 1. As you can see, |-1 / 3|

Task number 4. Restoring a series of numbers

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

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

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 , and the 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,

the 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 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 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 is 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

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,

◄

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

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, 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 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 , 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 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:

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 a 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

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 . ◄

Let's consider a series.

7 28 112 448 1792...

It is absolutely clear that the value of any of its elements is exactly four times greater than the previous one. So this series is a progression.

A geometric progression is an infinite sequence of numbers main feature which is that the next number is obtained from the previous one by multiplying by some specific number. This is expressed by the following formula.

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

Accordingly, z ∈ N.

The period when 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.

• Terms:a 1 = 3, a 3 = 48. Findq.

Solution: each subsequent element is greater than the previous one inq once.It is necessary to express some elements through others using a denominator.

Consequently,a 3 = q 2 · a 1

When substitutingq= 4

• Terms: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 , Consequently,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.

First level

Geometric progression. Comprehensive guide with examples (2019)

Numeric sequence

So let's sit down and start writing some numbers. For example:

You can write any numbers, and there can be as many as you like (in our case, them). No matter how many numbers we write, we can always say which of them is the first, which is the second, and so on to the last, that is, we can number them. This is an example of a number sequence:

Numeric sequence is a set of numbers, each of which can be assigned a unique number.

For example, for our sequence:

The assigned number is specific to only one sequence number. In other words, there are no three second numbers in the sequence. The second number (like the -th number) is always the same.

The number with the number is called the -th member of the sequence.

We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .

In our case:

The most common types of progression are arithmetic and geometric. In this topic, we will talk about the second kind - geometric progression.

Why do we need a geometric progression and its history.

Even in ancient times, the Italian mathematician, the monk Leonardo of Pisa (better known as Fibonacci), dealt with the practical needs of trade. The monk was faced with the task of determining what is the smallest number of weights that can be used to weigh the goods? In his writings, Fibonacci proves that such a system of weights is optimal: This is one of the first situations in which people had to deal with a geometric progression, which you have probably heard about and have at least a general idea of. Once you fully understand the topic, think about why such a system is optimal?

At present, in life practice, a geometric progression manifests itself when investing money in a bank, when the amount of interest is charged on the amount accumulated in the account for the previous period. In other words, if you put money on a term deposit in a savings bank, then in a year the deposit will increase by from the original amount, i.e. the new amount will be equal to the contribution multiplied by. In another year, this amount will increase by, i.е. the amount obtained at that time is again multiplied by and so on. A similar situation is described in the problems of computing the so-called compound interest- the percentage is taken each time from the amount that is on the account, taking into account the previous interest. We will talk about these tasks a little later.

There are many more simple cases where a geometric progression is applied. For example, the spread of influenza: one person infected a person, they, in turn, infected another person, and thus the second wave of infection - a person, and they, in turn, infected another ... and so on ...

By the way, a financial pyramid, the same MMM, is a simple and dry calculation according to the properties of a geometric progression. Interesting? Let's figure it out.

Geometric progression.

Let's say we have a number sequence:

You will immediately answer that it is easy and the name of such a sequence is an arithmetic progression with the difference of its members. How about something like this:

If you subtract the previous number from the next number, then you will see that each time you get a new difference (and so on), but the sequence definitely exists and is easy to notice - each next number is times larger than the previous one!

This type of sequence is called geometric progression and is marked.

A geometric progression ( ) is a numerical sequence, the first term of which is different from zero, and each term, starting from the second, is equal to the previous one, multiplied by the same number. This number is called the denominator of a geometric progression.

The constraints that the first term ( ) is not equal and are not random. Let's say that there are none, and the first term is still equal, and q is, hmm .. let, then it turns out:

Agree that this is no progression.

As you understand, we will get the same results if it is any number other than zero, but. In these cases, there will simply be no progression, since the entire number series will be either all zeros, or one number, and all the rest zeros.

Now let's talk in more detail about the denominator of a geometric progression, that is, about.

Let's repeat: - this is a number, how many times does each subsequent term change geometric progression.

What do you think it could be? That's right, positive and negative, but not zero (we talked about this a little higher).

Let's say we have a positive. Let in our case, a. What is the second term and? You can easily answer that:

All right. Accordingly, if, then all subsequent members of the progression have the same sign - they positive.

What if it's negative? For example, a. What is the second term and?

It's a completely different story

Try to count the term of this progression. How much did you get? I have. Thus, if, then the signs of the terms of the geometric progression alternate. That is, if you see a progression with alternating signs in its members, then its denominator is negative. This knowledge can help you test yourself when solving problems on this topic.

Now let's practice a little: try to determine which numerical sequences are a geometric progression, and which are an arithmetic one:

Got it? Compare our answers:

• Geometric progression - 3, 6.
• Arithmetic progression - 2, 4.
• It is neither an arithmetic nor a geometric progression - 1, 5, 7.

Let's return to our last progression, and let's try to find its term in the same way as in arithmetic. As you may have guessed, there are two ways to find it.

We successively multiply each term by.

So, the -th member of the described geometric progression is equal to.

As you already guess, now you yourself will derive a formula that will help you find any member of a geometric progression. Or have you already brought it out for yourself, describing how to find the th member in stages? If so, then check the correctness of your reasoning.

Let's illustrate this by the example of finding the -th member of this progression:

In other words:

Find yourself the value of a member of a given geometric progression.

Happened? Compare our answers:

Pay attention that you got exactly the same number as in the previous method, when we successively multiplied by each previous member of the geometric progression.
Let's try to "depersonalize" this formula - we bring it into a general form and get:

The derived formula is true for all values ​​- both positive and negative. Check it yourself by calculating the terms of a geometric progression with the following conditions: , a.

Did you count? Let's compare the results:

Agree that it would be possible to find a member of the progression in the same way as a member, however, there is a possibility of miscalculating. And if we have already found the th term of a geometric progression, a, then what could be easier than using the “truncated” part of the formula.

An infinitely decreasing geometric progression.

More recently, we talked about what can be either greater or less than zero, however, there are special values ​​for which the geometric progression is called infinitely decreasing.

Why do you think it has such a name?
To begin with, let's write down some geometric progression consisting of members.
Let's say, then:

We see that each subsequent term is less than the previous one in times, but will there be any number? You immediately answer - "no". That is why the infinitely decreasing - decreases, decreases, but never becomes zero.

To clearly understand what this looks like visually, let's try to draw a graph of our progression. So, for our case, the formula takes the following form:

On the charts, we are accustomed to build dependence on, therefore:

The essence of the expression has not changed: in the first entry, we showed the dependence of the value of a geometric progression member on its ordinal number, and in the second entry, we simply took the value of a geometric progression member for, and the ordinal number was designated not as, but as. All that's left to do is plot the graph.
Let's see what you got. Here's the chart I got:

See? The function decreases, tends to zero, but never crosses it, so it is infinitely decreasing. Let's mark our points on the graph, and at the same time what the coordinate and means:

Try to schematically depict a graph of a geometric progression if its first term is also equal. Analyze what is the difference with our previous chart?

Did you manage? Here's the chart I got:

Now that you have fully understood the basics of the geometric progression topic: you know what it is, you know how to find its term, and you also know what an infinitely decreasing geometric progression is, let's move on to its main property.

property of a geometric progression.

Do you remember the property of the members of an arithmetic progression? Yes, yes, how to find the value of a certain number of a progression when there are previous and subsequent values ​​​​of the members of this progression. Remembered? This:

Now we are faced with exactly the same question for the terms of a geometric progression. To derive such a formula, let's start drawing and reasoning. You'll see, it's very easy, and if you forget, you can bring it out yourself.

Let's take another simple geometric progression, in which we know and. How to find? With an arithmetic progression, this is easy and simple, but how is it here? In fact, there is nothing complicated in geometry either - you just need to paint each value given to us according to the formula.

You ask, and now what do we do with it? Yes, very simple. To begin with, let's depict these formulas in the figure, and try to do various manipulations with them in order to come to a value.

We abstract from the numbers that we are given, we will focus only on their expression through a formula. We need to find the value highlighted in orange, knowing the terms adjacent to it. Let's try to perform various actions with them, as a result of which we can get.

Addition.
Let's try to add two expressions and we get:

From this expression, as you can see, we will not be able to express in any way, therefore, we will try another option - subtraction.

Subtraction.

As you can see, we cannot express from this either, therefore, we will try to multiply these expressions by each other.

Multiplication.

Now look carefully at what we have, multiplying the terms of a geometric progression given to us in comparison with what needs to be found:

Guess what I'm talking about? Correctly, in order to find it, we need to take the square root of the geometric progression numbers adjacent to the desired number multiplied by each other:

Here you go. You yourself deduced the property of a geometric progression. Try to write this formula in general form. Happened?

Forgot condition when? Think about why it is important, for example, try to calculate it yourself, at. What happens in this case? That's right, complete nonsense, since the formula looks like this:

Accordingly, do not forget this limitation.

Now let's calculate what is

Correct answer - ! If you didn’t forget the second possible value when calculating, then you are a great fellow and you can immediately proceed to training, and if you forgot, read what is analyzed below and pay attention to why both roots must be written in the answer.

Let's draw both of our geometric progressions - one with a value, and the other with a value, and check if both of them have the right to exist:

In order to check whether such a geometric progression exists or not, it is necessary to see if it is the same between all its given members? Calculate q for the first and second cases.

See why we have to write two answers? Because the sign of the required term depends on whether it is positive or negative! And since we do not know what it is, we need to write both answers with a plus and a minus.

Now that you have mastered the main points and deduced the formula for the property of a geometric progression, find, knowing and

Compare your answers with the correct ones:

What do you think, what if we were given not the values ​​of the members of the geometric progression adjacent to the desired number, but equidistant from it. For example, we need to find, and given and. Can we use the formula we derived in this case? Try to confirm or refute this possibility in the same way, describing what each value consists of, as you did when deriving the formula from the beginning, with.
What did you get?

Now look carefully again.
and correspondingly:

From this we can conclude that the formula works not only with neighboring with the desired terms of a geometric progression, but also with equidistant from what the members are looking for.

Thus, our original formula becomes:

That is, if in the first case we said that, now we say that it can be equal to any natural number that is less. The main thing is to be the same for both given numbers.

Practice on specific examples, just be extremely careful!

1. , . Find.
2. , . Find.
3. , . Find.

I decided? I hope you were extremely attentive and noticed a small catch.

We compare the results.

In the first two cases, we calmly apply the above formula and get the following values:

In the third case, upon careful consideration of the serial numbers of the numbers given to us, we understand that they are not equidistant from the number we are looking for: it is the previous number, but removed in position, so it is not possible to apply the formula.

How to solve it? It's actually not as difficult as it seems! Let's write down with you what each number given to us and the desired number consists of.

So we have and. Let's see what we can do with them. I suggest splitting. We get:

We substitute our data into the formula:

The next step we can find - for this we need to take the cube root of the resulting number.

Now let's look again at what we have. We have, but we need to find, and it, in turn, is equal to:

We found all the necessary data for the calculation. Substitute in the formula:

Our answer: .

Try to solve another same problem yourself:
Given: ,
Find:

How much did you get? I have - .

As you can see, in fact, you need remember only one formula- . All the rest you can withdraw without any difficulty yourself at any time. To do this, simply write the simplest geometric progression on a piece of paper and write down what, according to the above formula, each of its numbers is equal to.

The sum of the terms of a geometric progression.

Now consider the formulas that allow us to quickly calculate the sum of the terms of a geometric progression in a given interval:

To derive the formula for the sum of terms of a finite geometric progression, we multiply all parts of the above equation by. We get:

Look closely: what do the last two formulas have in common? That's right, common members, for example and so on, except for the first and last member. Let's try to subtract the 1st equation from the 2nd equation. What did you get?

Now express through the formula of a member of a geometric progression and substitute the resulting expression in our last formula:

Group the expression. You should get:

All that's left to do is express:

Accordingly, in this case.

What if? What formula works then? Imagine a geometric progression at. What is she like? Correctly a series of identical numbers, respectively, the formula will look like this:

As with arithmetic and geometric progression, there are many legends. One of them is the legend of Seth, the creator of chess.

Many people know that the game of chess was invented in India. When the Hindu king met her, he was delighted with her wit and the variety of positions possible in her. Upon learning that it was invented by one of his subjects, the king decided to personally reward him. He called the inventor to him and ordered to ask him for whatever he wanted, promising to fulfill even the most skillful desire.

Seta asked for time to think, and when the next day Seta appeared before the king, he surprised the king with the unparalleled modesty of his request. He asked for a grain of wheat for the first square of the chessboard, wheat for the second, for the third, for the fourth, and so on.

The king was angry and drove Seth away, saying that the servant's request was unworthy of royal generosity, but promised that the servant would receive his grains for all the cells of the board.

And now the question is: using the formula for the sum of members of a geometric progression, calculate how many grains Seth should receive?

Let's start discussing. Since, according to the condition, Seth asked for a grain of wheat for the first cell of the chessboard, for the second, for the third, for the fourth, etc., we see that the problem is about a geometric progression. What is equal in this case?
Correctly.

Total cells of the chessboard. Respectively, . We have all the data, it remains only to substitute into the formula and calculate.

To represent at least approximately the "scales" of a given number, we transform using the properties of the degree:

Of course, if you want, you can take a calculator and calculate what kind of number you end up with, and if not, you'll have to take my word for it: the final value of the expression will be.
That is:

quintillion quadrillion trillion billion million thousand.

Fuh) If you want to imagine the enormity of this number, then estimate what size barn would be required to accommodate the entire amount of grain.
With a barn height of m and a width of m, its length would have to extend to km, i.e. twice as far as from the Earth to the Sun.

If the king were strong in mathematics, he could offer the scientist himself to count the grains, because in order to count a million grains, he would need at least a day of tireless counting, and given that it is necessary to count the quintillions, the grains would have to be counted all his life.

And now we will solve a simple problem on the sum of terms of a geometric progression.
Vasya, a 5th grade student, fell ill with the flu, but continues to go to school. Every day, Vasya infects two people who, in turn, infect two more people, and so on. Just one person in the class. In how many days will the whole class get the flu?

So, the first member of a geometric progression is Vasya, that is, a person. th member of the geometric progression, these are the two people whom he infected on the first day of his arrival. The total sum of the members of the progression is equal to the number of students 5A. Accordingly, we are talking about a progression in which:

Let's substitute our data into the formula for the sum of the terms of a geometric progression:

The whole class will get sick within days. Don't believe in formulas and numbers? Try to portray the "infection" of the students yourself. Happened? See what it looks like for me:

Calculate for yourself how many days the students would get the flu if everyone would infect a person, and there was a person in the class.

What value did you get? It turned out that everyone started to get sick after a day.

As you can see, such a task and the drawing for it resembles a pyramid, in which each subsequent “brings” new people. However, sooner or later a moment comes when the latter cannot attract anyone. In our case, if we imagine that the class is isolated, the person from closes the chain (). Thus, if a person were involved in a financial pyramid in which money was given if you brought two other participants, then the person (or in the general case) would not bring anyone, respectively, would lose everything that they invested in this financial scam.

Everything that was said above refers to a decreasing or increasing geometric progression, but, as you remember, we have a special kind - an infinitely decreasing geometric progression. How to calculate the sum of its members? And why does this type of progression have certain features? Let's figure it out together.

So, for starters, let's look again at this picture of an infinitely decreasing geometric progression from our example:

And now let's look at the formula for the sum of a geometric progression, derived a little earlier:
or

What are we striving for? That's right, the graph shows that it tends to zero. That is, when, it will be almost equal, respectively, when calculating the expression, we will get almost. In this regard, we believe that when calculating the sum of an infinitely decreasing geometric progression, this bracket can be neglected, since it will be equal.

- the formula is the sum of the terms of an infinitely decreasing geometric progression.

IMPORTANT! We use the formula for the sum of terms of an infinitely decreasing geometric progression only if the condition explicitly states that we need to find the sum endless the number of members.

If a specific number n is indicated, then we use the formula for the sum of n terms, even if or.

And now let's practice.

1. Find the sum of the first terms of a geometric progression with and.
2. Find the sum of the terms of an infinitely decreasing geometric progression with and.

I hope you were very careful. Compare our answers:

Now you know everything about geometric progression, and it's time to move from theory to practice. The most common exponential problems found on the exam are compound interest problems. It is about them that we will talk.

Problems for calculating compound interest.

You must have heard of the so-called compound interest formula. Do you understand what she means? If not, let's figure it out, because having realized the process itself, you will immediately understand what the geometric progression has to do with it.

We all go to the bank and know that there are different conditions for deposits: this is the term, and additional maintenance, and interest with two different ways of calculating it - simple and complex.

FROM simple interest everything is more or less clear: interest is charged once at the end of the deposit term. That is, if we are talking about putting 100 rubles a year under, then they will be credited only at the end of the year. Accordingly, by the end of the deposit, we will receive rubles.

Compound interest is an option in which interest capitalization, i.e. their addition to the amount of the deposit and the subsequent calculation of income not from the initial, but from the accumulated amount of the deposit. Capitalization does not occur constantly, but with some periodicity. As a rule, such periods are equal and most often banks use a month, a quarter or a year.

Let's say that we put all the same rubles per annum, but with a monthly capitalization of the deposit. What do we get?

Do you understand everything here? If not, let's take it step by step.

We brought rubles to the bank. By the end of the month, we should have an amount in our account consisting of our rubles plus interest on them, that is:

I agree?

We can take it out of the bracket and then we get:

Agree, this formula is already more similar to the one we wrote at the beginning. It remains to deal with percentages

In the condition of the problem, we are told about the annual. As you know, we do not multiply by - we convert percentages to decimals, that is:

Right? Now you ask, where did the number come from? Very simple!
I repeat: the condition of the problem says about ANNUAL interest accrued MONTHLY. As you know, in a year of months, respectively, the bank will charge us a part of the annual interest per month:

Realized? Now try to write what this part of the formula would look like if I said that interest is calculated daily.
Did you manage? Let's compare the results:

Well done! Let's return to our task: write down how much will be credited to our account for the second month, taking into account that interest is charged on the accumulated deposit amount.
Here's what happened to me:

Or, in other words:

I think that you have already noticed a pattern and saw a geometric progression in all this. Write what its member will be equal to, or, in other words, how much money we will receive at the end of the month.
Did? Checking!

As you can see, if you put money in a bank for a year at a simple interest, then you will receive rubles, and if you put it at a compound rate, you will receive rubles. The benefit is small, but this happens only during the th year, but for a longer period, capitalization is much more profitable:

Consider another type of compound interest problem. After what you figured out, it will be elementary for you. So the task is:

Zvezda started investing in the industry in 2000 with a dollar capital. Every year since 2001, it has made a profit that is equal to the previous year's capital. How much profit will the Zvezda company receive at the end of 2003, if the profit was not withdrawn from circulation?

The capital of the Zvezda company in 2000.
- the capital of the Zvezda company in 2001.
- the capital of the Zvezda company in 2002.
- the capital of the Zvezda company in 2003.

Or we can write briefly:

For our case:

2000, 2001, 2002 and 2003.

Respectively:
rubles
Note that in this problem we do not have a division either by or by, since the percentage is given ANNUALLY and it is calculated ANNUALLY. That is, when reading the problem for compound interest, pay attention to what percentage is given, and in what period it is charged, and only then proceed to the calculations.
Now you know everything about geometric progression.

Workout.

1. Find a term of a geometric progression if it is known that, and
2. Find the sum of the first terms of a geometric progression, if it is known that, and
3. MDM Capital started investing in the industry in 2003 with a dollar capital. Every year since 2004, she has made a profit that is equal to the previous year's capital. The company "MSK Cash Flows" began to invest in the industry in 2005 in the amount of $10,000, starting to make a profit in 2006 in the amount of. By how many dollars does the capital of one company exceed that of another at the end of 2007, if profits were not withdrawn from circulation?

Answers:

1. Since the condition of the problem does not say that the progression is infinite and it is required to find the sum of a specific number of its members, the calculation is carried out according to the formula:
2. 
   Company "MDM Capital":

   2003, 2004, 2005, 2006, 2007.
   - increases by 100%, that is, 2 times.
   Respectively:
   rubles
   MSK Cash Flows:

   2005, 2006, 2007.
   - increases by, that is, times.
   Respectively:
   rubles
   rubles

Let's summarize.

1) A geometric progression ( ) is a numerical sequence, the first term of which is different from zero, and each term, starting from the second, is equal to the previous one, multiplied by the same number. This number is called the denominator of a geometric progression.

2) The equation of the members of a geometric progression -.

3) can take any value, except for and.

• if, then all subsequent members of the progression have the same sign - they positive;
• if, then all subsequent members of the progression alternate signs;
• when - the progression is called infinitely decreasing.

4) , at - property of a geometric progression (neighboring members)

or
, at (equidistant terms)

When you find it, do not forget that there should be two answers..

For example,

5) The sum of the members of a geometric progression is calculated by the formula:
or

If the progression is infinitely decreasing, then:
or

IMPORTANT! We use the formula for the sum of terms of an infinitely decreasing geometric progression only if the condition explicitly states that it is necessary to find the sum of an infinite number of terms.

6) Tasks for compound interest are also calculated according to the formula of the th member of a geometric progression, provided that the funds were not withdrawn from circulation:

GEOMETRIC PROGRESSION. BRIEFLY ABOUT THE MAIN

Geometric progression( ) is a numerical sequence, the first term of which is different from zero, and each term, starting from the second, is equal to the previous one, multiplied by the same number. This number is called the denominator of a geometric progression.

Denominator of a geometric progression can take any value except for and.

• If, then all subsequent members of the progression have the same sign - they are positive;
• if, then all subsequent members of the progression alternate signs;
• when - the progression is called infinitely decreasing.

Equation of members of a geometric progression - .

The sum of the terms of a geometric progression calculated by the formula:
or

>>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 of a 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 the method 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 dots(with abscissas x = 1, x = 2, x = 3, etc.) lying on some curve (both figures show the same curve, only differently located and depicted at 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 stage. 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 mathematical model tasks.

Second phase.

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

(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 the geometric progression in question: 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 trick: 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 members.

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).