Arithmetic progression problems have existed since ancient times. They appeared and demanded a solution, because they had a practical need.
So, in one of the papyri of Ancient Egypt, which has a mathematical content - the Rhind papyrus (XIX century BC) - contains the following task: divide ten measures of bread into ten people, provided that the difference between each of them is one eighth of a measure.
And in the mathematical works of the ancient Greeks there are elegant theorems related to arithmetic progression. So, Hypsicles of Alexandria (2nd century, who compiled many interesting problems and added the fourteenth book to Euclid's "Elements", formulated the idea: "In an arithmetic progression with an even number of members, the sum of the members of the 2nd half is greater than the sum of the members of the 1st by the square 1 / 2 members.
The sequence an is denoted. The numbers of the sequence are called its members and are usually denoted by letters with indices that indicate the serial number of this member (a1, a2, a3 ... it reads: “a 1st”, “a 2nd”, “a 3rd” and so on ).
The sequence can be infinite or finite.
What is an arithmetic progression? It is understood as obtained by adding the previous term (n) with the same number d, which is the difference of the progression.
If d<0, то мы имеем убывающую прогрессию. Если d>0, then such a progression is considered to be increasing.
An arithmetic progression is said to be finite if only a few of its first terms are taken into account. With a very large number of members, this is already an infinite progression.
Any arithmetic progression is given by the following formula:
an =kn+b, while b and k are some numbers.
The statement, which is the opposite, is absolutely true: if the sequence is given by a similar formula, then this is exactly an arithmetic progression, which has the properties:
- Each member of the progression is the arithmetic mean of the previous member and the next one.
- The opposite: if, starting from the 2nd, each term is the arithmetic mean of the previous term and the next, i.e. if the condition is met, then the given sequence is an arithmetic progression. This equality is at the same time a sign of progression, so it is usually called a characteristic property of progression.
In the same way, the theorem that reflects this property is true: a sequence is an arithmetic progression only if this equality is true for any of the members of the sequence, starting from the 2nd.
The characteristic property for any four numbers of an arithmetic progression can be expressed by the formula an + am = ak + al if n + m = k + l (m, n, k are the numbers of the progression).
In an arithmetic progression, any necessary (Nth) term can be found by applying the following formula:
For example: the first term (a1) in an arithmetic progression is given and equals three, and the difference (d) equals four. You need to find the forty-fifth term of this progression. a45 = 1+4(45-1)=177
The formula an = ak + d(n - k) allows you to determine the n-th member of an arithmetic progression through any of its k-th member, provided that it is known.
The sum of the members of an arithmetic progression (assuming the 1st n members of the final progression) is calculated as follows:
Sn = (a1+an) n/2.
If the 1st term is also known, then another formula is convenient for calculation:
Sn = ((2a1+d(n-1))/2)*n.
The sum of an arithmetic progression that contains n terms is calculated as follows:
The choice of formulas for calculations depends on the conditions of the tasks and the initial data.
The natural series of any numbers such as 1,2,3,...,n,... is the simplest example of an arithmetic progression.
In addition to the arithmetic progression, there is also a geometric one, which has its own properties and characteristics.
Arithmetic progression name a sequence of numbers (members of a progression)
In which each subsequent term differs from the previous one by a steel term, which is also called step or progression difference.
Thus, by setting the step of the progression and its first term, you can find any of its elements using the formula
Properties of an arithmetic progression
1) Each member of the arithmetic progression, starting from the second number, is the arithmetic mean of the previous and next member of the progression
The converse is also true. If the arithmetic mean of neighboring odd (even) members of the progression is equal to the member that stands between them, then this sequence of numbers is an arithmetic progression. By this assertion it is very easy to check any sequence.
Also by the property of arithmetic progression, the above formula can be generalized to the following
This is easy to verify if we write the terms to the right of the equal sign
It is often used in practice to simplify calculations in problems.
2) The sum of the first n terms of an arithmetic progression is calculated by the formula
Remember well the formula for the sum of an arithmetic progression, it is indispensable in calculations and is quite common in simple life situations.
3) If you need to find not the entire sum, but a part of the sequence starting from its k -th member, then the following sum formula will come in handy in you
4) It is of practical interest to find the sum of n members of an arithmetic progression starting from the kth number. To do this, use the formula
This is where the theoretical material ends and we move on to solving problems that are common in practice.
Example 1. Find the fortieth term of the arithmetic progression 4;7;...
Solution:
According to the condition, we have
Define the progression step
According to the well-known formula, we find the fortieth term of the progression
Example2. The arithmetic progression is given by its third and seventh members. Find the first term of the progression and the sum of ten.
Solution:
We write the given elements of the progression according to the formulas
We subtract the first equation from the second equation, as a result we find the progression step
The found value is substituted into any of the equations to find the first term of the arithmetic progression
Calculate the sum of the first ten terms of the progression
Without applying complex calculations, we found all the required values.
Example 3. An arithmetic progression is given by the denominator and one of its members. Find the first term of the progression, the sum of its 50 terms starting from 50, and the sum of the first 100.
Solution:
Let's write the formula for the hundredth element of the progression
and find the first
Based on the first, we find the 50th term of the progression
Finding the sum of the part of the progression
and the sum of the first 100
The sum of the progression is 250.
Example 4
Find the number of members of an arithmetic progression if:
a3-a1=8, a2+a4=14, Sn=111.
Solution:
We write the equations in terms of the first term and the step of the progression and define them
We substitute the obtained values into the sum formula to determine the number of terms in the sum
Making simplifications
and solve the quadratic equation
Of the two values found, only the number 8 is suitable for the condition of the problem. Thus the sum of the first eight terms of the progression is 111.
Example 5
solve the equation
1+3+5+...+x=307.
Solution: This equation is the sum of an arithmetic progression. We write out its first term and find the difference of the progression
Examples for arithmetic and geometric progression taken from the "Collection of problems for applicants. Mathematics" published by Volyn State University named after Lesya Ukrainka in 2001. Read the answers carefully and choose the most necessary for yourself.
Group A (level 1)
Example 1. Calculate the sixth term of the arithmetic progression 21.3; 22.4; … ,
Solution: Find the difference (step) of the progression
d \u003d a 2 -a 1 \u003d 22.4-21.3 \u003d 1.1.
Next, we calculate the sixth term of the arithmetic progression
a 6 \u003d a 1 + (6-1) d \u003d 21.3 + 5 * 1.1 \u003d 26.8.
Example 2. Calculate the sixth term of the geometric progression 5; ten; twenty; ...
Solution: Find the denominator of a geometric progression
q \u003d b 2 / b 1 \u003d 10/5 \u003d 2.
We calculate the sixth term of a geometric progression
b 6 \u003d b 1 q 6-1 \u003d 5 * 25 \u003d 5 * 32 \u003d 160.
Example 3. In an arithmetic progression, a 1 \u003d 2.1 a 10 \u003d 12.9. Calculate progression difference.
Solution: Let's represent the tenth term of the progression as a formula
a 10 \u003d a 1 + (10-1) d \u003d a 1 + 9d.
Substitute the known values and solve
12.9=2.1+9d;
9d=12.9-2.1=10.8;
d=10.8/9=1.2.
Answer: progression difference d=1.2.
Example 4. In geometric progression b 1 =2.56; b 4 \u003d 4.42368. Calculate the denominator of the progression.
Solution: Find the denominator of the progression
q \u003d b 2 / b 1 \u003d 4.42368 / 2.56 \u003d 1.728.
You can't do without a calculator here.
Answer: the denominator of the progression is q=1.728.
Example 5. In an arithmetic progression, a 1 \u003d 20.1, d \u003d 1.3. Calculate the sum of the first eight terms of the progression.
Solution: The sum of the arithmetic progression is found by the formula
Performing Calculations
S 8 \u003d (2 * 20.1 + (8-1) * 1.3) * 8 / 2 \u003d 197.2.
Answer: S 8 \u003d 197.2.
Example 6 . In a geometric progression b 1 =1.5; q=1.2. Calculate the sum of the first four terms of the progression.
Solution: The sum of the geometric progression is calculated by the formula
Finding the sum of the progression 
Answer: S 8 \u003d 8.052.
Example 7 . In arithmetic progression a 1 \u003d 1.35 d \u003d -2.4. Calculate the number of the progression term, equal to -25.05.
Solution: A member of an arithmetic progression is found by the formula
a n \u003d a 1 + (n-1) d.
By condition, everything except the ordinal number is known, we will find it
-25.05=1.35+(n-1)(-2.4) ; 
Answer: n=12.
Example 8. Calculate the seventh term of the progression 23.5; 24.82; 26.14; ...
Solution: Since the condition does not specify which progression is set, you first need to set it. Get that arithmetic
d=a 2 -a 1 = 24.82-23.5=1.32;
d \u003d a 3 -a 2 \u003d 26.14-24.82 \u003d 1.32.
Finding the seventh term of the progression
a 7 \u003d a 1 + (7-1) d \u003d 23.5 + 6 * 1.32 \u003d 31.42.
Answer: a 7 \u003d 31.42.
Example 9. Calculate the number of the progression member 2.1; 3.3; 4.5; ... , equal to 11.7 .
Solution: It is easy to verify that an arithmetic progression is given. Finding the progression difference
d \u003d a 2 -a 1 \u003d 3.3-2.1 \u003d 1.2.
According to the progression term formula
a n \u003d a 1 + (n-1) d
find the number
11.7=2.1+(n-1)*1.2; 
Answer: n= 9 .
Example 10. Calculate the fourth term of the progression 1.5; 1.8; 2.16; ... .
Solution: Without checking, we can say that the progression is geometric. Find its denominator
q \u003d b 2 / b 1 \u003d 1, 8 / 1.5 \u003d 1.2.
Calculate the 4th member of the geometric progression using the formula
b 4 \u003d b 1 q 3 \u003d 1.5 * 1.2 3 \u003d 2.592.
Answer: b 4 \u003d 2.592.
Example 11. Calculate the number of the progression member 1,2; 1.8; 2.16; ... equal to 4.05.
Solution: We have a geometric progression. Find the denominator of the progression
q \u003d b 2 / b 1 \u003d 1, 8 / 1.2 \u003d 1.5.
Find the progression number from the dependence
b n = b 1 q n-1 .
4.05=1.2*1.5n-1;
1.5 n-1 \u003d 4.05 / 1.2 \u003d 3.375 \u003d 1.5 3;
n-1=3; n=4.
Answer: n=4.
Example 12. In an arithmetic progression, a 5 \u003d 14.91 a 9 \u003d 20.11. Calculate a 1 .
Solution: We express the 9th term of the progression through 5
a 9 \u003d a 5 + (9-5) d
and find the progression step
20.11=14.91+4d;
4d=5.2; d=5.2/4=1.3.
We express the 5th term of the progression in terms of 1 and calculate the first
a 5 = a 1 +4d;
14.91 \u003d a 1 +5.2;
a 1 \u003d 14.91-5.2 \u003d 9.71.
Answer: a 1 \u003d 9.71.
Example 13 . In arithmetic progression, a 7 \u003d 12.01; a 11 \u003d 17.61. Calculate progression difference.
Solution: We express 11 terms of the progression through 7
a 11 \u003d a 7 + (11-7) d.
From here we calculate the progression step
17.61=12.01+4d;
4d=5.6; d=5.6/4=1.4.
Answer: d=1.4.
Example 14. In geometric progression b 5 =64; b 8 =1. Calculate b 3 .
Solution: We express the 8th term of the progression in terms of 5
b 8 \u003d b 5 q 8-5.
From here we find the denominator of the progression
1=64 q 3 ;
q 3 \u003d 1/64 \u003d (1/4) 3;
q=1/4.
Similarly, we find b 3 through b 5
b 3 \u003d b 5 / q 2 \u003d 64 * 4 2 \u003d 1024.
Answer: b 3 \u003d 1024.
Example 15. In an arithmetic progression, a 9 + a 15 \u003d 14.8. Calculate a 12
Solution: In this example, it should be noted that the 12th member of the progression is in the middle between its number 9 and 15. Therefore, the neighboring terms of the progression (9, 15 ) can be expressed in terms of 12 as follows
a 9 \u003d a 12 - (12-9) d;
a 15 \u003d a 12 + (15-9) d;
a 9 \u003d a 12 -3d;
a 15 = a 12 + 3d.
Let us sum the extreme terms of the progression
a 9 + a 15 = a 12 -3d+ a 12 + 3d=2a 12.
From here we find the 12th term of the progression
a 12 \u003d (a 9 + a 15) / 2 \u003d 14.8 / 2 \u003d 7.4.
Answer: a 12 \u003d 7.4.
Example 16. Exponentially b 10 *b 14 =289. Calculate module 12 of the progression term | b 12 |.
Solution: The algorithm for solving the problem is contained in the previous example. It is necessary to express 10 and 14 members of a geometric progression through 12. By the properties of a geometric progression, we get
b 10 \u003d b 12 / q 2; b 14 = b 12 * q 2 .
It is easy to see that when they work, the sign of the progression disappears.
b 10 * b 14 \u003d (b 12) 2 \u003d 289 \u003d 17 2.
From here we find the module | b 12 |
(b 12) 2 =289=17 2 -> | b 12 |=17.
Answer: | b 12 |=17.
Example 17. Exponentially b 8 =1.3. Calculate b 6 *b 10 .
Solution: The calculation scheme is similar to the previous example - we express 6 and 10 members of the progression through 8.
b 6 \u003d b 8 / q 2; b 10 = b 8 * q 2 .
When they are multiplied, the denominators are reduced and we get the square of the known term of the progression
b 6 *b 10 \u003d (b 8) 2 \u003d 1.3 2 \u003d 1.69.
Answer: b 6 * b 10 \u003d 1.69.
Example 18. In an arithmetic progression, a 10 \u003d 3.6: a 12 \u003d 8. Calculate a 8
Solution: Let's write the members of the progression in a series a 8 , a 10 , a 12 . Between them the same step, let's find it
a 12 = a 10 +2d;
2d \u003d a 12 - a 10 \u003d 8-3.6 \u003d 4.4.
In the same way we find a 8
a 10 = a 8 +2d;
a 8 \u003d a 10 -2d \u003d 3.6-4.4 \u003d -0.8.
Here are some simple calculations.
Answer: a 8 \u003d -0.8.
Example 19. Exponentially b 14 =8; b 16 =2. Calculate b 12 .
Solution: Omitting detailed explanations, we write down the product of the 14th and 16th terms of the progression
b 14 *b 16 =(b 12) 2 .
This is equivalent to the geometric mean. Finding the root of the product of terms, we get the desired value
(b 12) 2 \u003d 8 * 2 \u003d 16; b 12 =4.
Answer: b 12 \u003d 4.
Example 20. In an arithmetic progression, a 5 \u003d 3.4; a 11 \u003d 6.9. Calculate a 17 .
Solution: Between 5,11 and 17 members of the progression is the same step and it is equal to 6d. Therefore, the final solution can be written as
a 17 \u003d a 11 + 6d \u003d a 11 + (a 11 - a 5) \u003d 2 * 6.9-3.4 \u003d 10.4.
I think you understand why such a record. If not - try to paint 11 terms of the progression through 5 and turn 6d .
Answer: a 17 \u003d 10.4.
Example 21. Calculate the 6th member of the geometric progression 3; 12;... .
Solution: Find the denominator of the progression
q \u003d b 2 / b 1 \u003d 12/3 \u003d 4.
Let's use the general formula of the term of a geometric progression
b n = b 1 *q n-1 .
From here we get
b 6 \u003d b 1 * q 5 \u003d b 2 * q 4.
As you can see, the main thing in the record is that the sum of the index (2) and the degree (4) correspond to the ordinal number of the progression member (6). Performing Calculations
b 6 \u003d 12 * 4 4 \u003d 12 * 256 \u003d 3072.
We got a large number, but the geometric progression is different in that its members either grow rapidly or descend.
Answer: b 6 \u003d 3072.
Example 22. In an arithmetic progression, a 3 \u003d 48; a 5 =42. Calculate a 7 .
Solution: Since the difference between the progression between the given members and the desired one has become and is equal to 2d, then the formula for the 7th member of the progression will look like
a 7 \u003d a 5 + 2d \u003d a 5 + (a 5 - a 3);
and 7 \u003d 2 * 42-48 \u003d 36.
Answer: a 7 \u003d 36.
Arithmetic and geometric progressions
Theoretical information
Theoretical information
Arithmetic progression |
Geometric progression |
|
Definition |
Arithmetic progression a n a sequence is called, each member of which, starting from the second, is equal to the previous member, added with the same number d (d- progression difference) |
geometric progression b n a sequence of non-zero numbers is called, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression) |
Recurrent formula |
For any natural n |
For any natural n |
nth term formula |
a n = a 1 + d (n - 1) |
b n \u003d b 1 ∙ q n - 1, b n ≠ 0 |
| characteristic property |  |
|
| Sum of the first n terms |  |
 |
Examples of tasks with comments
Exercise 1
In arithmetic progression ( a n) a 1 = -6, a 2
According to the formula of the nth term:
a 22 = a 1+ d (22 - 1) = a 1+ 21d
By condition:
a 1= -6, so a 22= -6 + 21d.
It is necessary to find the difference of progressions:
d= a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = - 48.
Answer : a 22 = -48.
Task 2
Find the fifth term of the geometric progression: -3; 6;....
1st way (using n-term formula)
According to the formula of the n-th member of a geometric progression:
b 5 \u003d b 1 ∙ q 5 - 1 = b 1 ∙ q 4.
Because b 1 = -3,
2nd way (using recursive formula)
Since the denominator of the progression is -2 (q = -2), then:
b 3 = 6 ∙ (-2) = -12;
b 4 = -12 ∙ (-2) = 24;
b 5 = 24 ∙ (-2) = -48.
Answer : b 5 = -48.
Task 3
In arithmetic progression ( a n) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.
For an arithmetic progression, the characteristic property has the form 
.
Therefore:
.
Substitute the data in the formula:
Answer: 95.
Task 4
In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.
To find the sum of the first n terms of an arithmetic progression, two formulas are used:
.
Which of them is more convenient to apply in this case?
By condition, the formula of the nth member of the original progression is known ( a n) a n= 3n - 4. Can be found immediately and a 1, and a 16 without finding d . Therefore, we use the first formula.
Answer: 368.
Task 5
In arithmetic progression a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.
According to the formula of the nth term:
a 22 = a 1 + d (22 – 1) = a 1+ 21d.
By condition, if a 1= -6, then a 22= -6 + 21d. It is necessary to find the difference of progressions:
d= a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = -48.
Answer : a 22 = -48.
Task 6
Several consecutive terms of a geometric progression are recorded:
Find the term of the progression, denoted by the letter x .
When solving, we use the formula for the nth term b n \u003d b 1 ∙ q n - 1 for geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of these terms of the progression and divide by the previous one. In our example, you can take and divide by. We get that q \u003d 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.
Substituting the found values into the formula, we get:
.
Answer : .
Task 7
From the arithmetic progressions given by the formula of the nth term, choose the one for which the condition is satisfied a 27 > 9:
Since the specified condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:
.
Answer: 4.
Task 8
In arithmetic progression a 1= 3, d = -1.5. Specify the largest value of n for which the inequality holds a n > -6.
Someone treats the word "progression" with caution, as a very complex term from the sections of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi counter (where they still remain). And to understand the essence (and in mathematics there is nothing more important than “to understand the essence”) of an arithmetic sequence is not so difficult, having analyzed a few elementary concepts.
Mathematical number sequence
It is customary to call a numerical sequence a series of numbers, each of which has its own number.
and 1 is the first member of the sequence;
and 2 is the second member of the sequence;
and 7 is the seventh member of the sequence;
and n is the nth member of the sequence;
However, not any arbitrary set of figures and numbers interests us. We will focus our attention on a numerical sequence in which the value of the nth member is related to its ordinal number by a dependence that can be clearly formulated mathematically. In other words: the numerical value of the nth number is some function of n.
a - value of a member of the numerical sequence;
n is its serial number;
f(n) is a function where the ordinal in the numeric sequence n is the argument.
Definition
An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth member of an arithmetic sequence is as follows:
a n - the value of the current member of the arithmetic progression;
a n+1 - the formula of the next number;
d - difference (a certain number).
It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one, and such an arithmetic progression will be increasing.
In the graph below, it is easy to see why the number sequence is called "increasing".
In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.
The value of the specified member
Sometimes it is necessary to determine the value of some arbitrary term a n of an arithmetic progression. You can do this by calculating successively the values of all members of the arithmetic progression, from the first to the desired one. However, this way is not always acceptable if, for example, it is necessary to find the value of the five thousandth or eight millionth term. The traditional calculation will take a long time. However, a specific arithmetic progression can be investigated using certain formulas. There is also a formula for the nth term: the value of any member of an arithmetic progression can be determined as the sum of the first member of the progression with the difference of the progression, multiplied by the number of the desired member, minus one.
The formula is universal for increasing and decreasing progression.
An example of calculating the value of a given member
Let's solve the following problem of finding the value of the n-th member of an arithmetic progression.
Condition: there is an arithmetic progression with parameters:
The first member of the sequence is 3;
The difference in the number series is 1.2.
Task: it is necessary to find the value of 214 terms
Solution: to determine the value of a given member, we use the formula:
a(n) = a1 + d(n-1)
Substituting the data from the problem statement into the expression, we have:
a(214) = a1 + d(n-1)
a(214) = 3 + 1.2 (214-1) = 258.6
Answer: The 214th member of the sequence is equal to 258.6.
The advantages of this calculation method are obvious - the entire solution takes no more than 2 lines.
Sum of a given number of terms
Very often, in a given arithmetic series, it is required to determine the sum of the values of some of its segments. It also doesn't need to calculate the values of each term and then sum them up. This method is applicable if the number of terms whose sum must be found is small. In other cases, it is more convenient to use the following formula.
The sum of the members of an arithmetic progression from 1 to n is equal to the sum of the first and nth members, multiplied by the member number n and divided by two. If in the formula the value of the n-th member is replaced by the expression from the previous paragraph of the article, we get:
Calculation example
For example, let's solve a problem with the following conditions:
The first term of the sequence is zero;
The difference is 0.5.
In the problem, it is required to determine the sum of the terms of the series from 56 to 101.
Solution. Let's use the formula for determining the sum of the progression:
s(n) = (2∙a1 + d∙(n-1))∙n/2
First, we determine the sum of the values of 101 members of the progression by substituting the given conditions of our problem into the formula:
s 101 = (2∙0 + 0.5∙(101-1))∙101/2 = 2 525
Obviously, in order to find out the sum of the terms of the progression from the 56th to the 101st, it is necessary to subtract S 55 from S 101.
s 55 = (2∙0 + 0.5∙(55-1))∙55/2 = 742.5
So the sum of the arithmetic progression for this example is:
s 101 - s 55 \u003d 2,525 - 742.5 \u003d 1,782.5
Example of practical application of arithmetic progression
At the end of the article, let's return to the example of the arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider such an example.
Getting into a taxi (which includes 3 km) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles / km. Travel distance 30 km. Calculate the cost of the trip.
1. Let's discard the first 3 km, the price of which is included in the landing cost.
30 - 3 = 27 km.
2. Further calculation is nothing more than parsing an arithmetic number series.
The member number is the number of kilometers traveled (minus the first three).
The value of the member is the sum.
The first term in this problem will be equal to a 1 = 50 rubles.
Progression difference d = 22 p.
the number of interest to us - the value of the (27 + 1)th member of the arithmetic progression - the meter reading at the end of the 27th kilometer - 27.999 ... = 28 km.
a 28 \u003d 50 + 22 ∙ (28 - 1) \u003d 644
Calculations of calendar data for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of the celestial body to the luminary. In addition, various numerical series are successfully used in statistics and other applied branches of mathematics.
Another kind of number sequence is geometric
A geometric progression is characterized by a large, compared with an arithmetic, rate of change. It is no coincidence that in politics, sociology, medicine, often, in order to show the high speed of the spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops exponentially.
The N-th member of the geometric number series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first member is 1, the denominator is 2, respectively, then:
n=1: 1 ∙ 2 = 2
n=2: 2 ∙ 2 = 4
n=3: 4 ∙ 2 = 8
n=4: 8 ∙ 2 = 16
n=5: 16 ∙ 2 = 32,
b n - the value of the current member of the geometric progression;
b n+1 - the formula of the next member of the geometric progression;
q is the denominator of a geometric progression (constant number).
If the graph of an arithmetic progression is a straight line, then the geometric one draws a slightly different picture:
As in the case of arithmetic, a geometric progression has a formula for the value of an arbitrary member. Any nth term of a geometric progression is equal to the product of the first term and the denominator of the progression to the power of n reduced by one:
Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Find the 5th term of the progression
b 5 \u003d b 1 ∙ q (5-1) \u003d 3 ∙ 1.5 4 \u003d 15.1875
The sum of a given number of members is also calculated using a special formula. The sum of the first n members of a geometric progression is equal to the difference between the product of the nth member of the progression and its denominator and the first member of the progression, divided by the denominator reduced by one:
If b n is replaced using the formula discussed above, the value of the sum of the first n members of the considered number series will take the form:
Example. The geometric progression starts with the first term equal to 1. The denominator is set equal to 3. Let's find the sum of the first eight terms.
s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280
