INSTRUCTIONS AND PROPHECIES OF THE Blessed MOTHER ALIPIA GOLOSEEVSKY, Kyiv...
When studying algebra in a secondary school (grade 9), one of the important topics is the study of numerical sequences, which include progressions - geometric and arithmetic. In this article, we will consider an arithmetic progression and examples with solutions.
What is an arithmetic progression?
To understand this, it is necessary to give a definition of the progression under consideration, as well as to give the basic formulas that will be further used in solving problems.
Arithmetic or is such a set of ordered rational numbers, each member of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of an ordered series of numbers and the difference, you can restore the entire arithmetic progression.
Let's take an example. The next sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the considered type of progression, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).
Important Formulas
We now give the basic formulas that will be needed to solve problems using an arithmetic progression. Denote by symbol a n nth member sequences where n is an integer. The difference is denoted by the Latin letter d. Then the following expressions are true:
- To determine the value of the nth term, the formula is suitable: a n \u003d (n-1) * d + a 1.
- To determine the sum of the first n terms: S n = (a n + a 1)*n/2.
To understand any examples of an arithmetic progression with a solution in grade 9, it is enough to remember these two formulas, since any problems of the type under consideration are built on their use. Also, do not forget that the progression difference is determined by the formula: d = a n - a n-1 .
Example #1: Finding an Unknown Member
We give a simple example of an arithmetic progression and the formulas that must be used to solve.
Let the sequence 10, 8, 6, 4, ... be given, it is necessary to find five terms in it.
It already follows from the conditions of the problem that the first 4 terms are known. The fifth can be defined in two ways:
- Let's calculate the difference first. We have: d = 8 - 10 = -2. The same way one could take any two other members standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d \u003d a n - a n-1, then d \u003d a 5 - a 4, from where we get: a 5 \u003d a 4 + d. Substitute known values: a 5 = 4 + (-2) = 2.
- The second method also requires knowledge of the difference of the progression in question, so you first need to determine it, as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for the n number of the sequence. We have: a n \u003d (n - 1) * d + a 1 \u003d (n - 1) * (-2) + 10 \u003d 12 - 2 * n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.
As you can see, both solutions lead to the same result. Note that in this example the difference d of the progression is negative. Such sequences are called decreasing because each successive term is less than the previous one.
Example #2: progression difference
Now let's complicate the task a bit, give an example of how to find the difference of an arithmetic progression.
It is known that in some algebraic progression the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.
Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . We substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 \u003d 6 + 6 * d. From this expression, you can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, the first part of the problem was answered.
To restore a sequence up to 7 terms, one should use the definition algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2=8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16 and 7 = 18.
Example #3: making a progression
Let's make it harder stronger condition tasks. Now you need to answer the question of how to find an arithmetic progression. We can give the following example: two numbers are given, for example, 4 and 5. It is necessary to make an algebraic progression so that three more terms fit between these.
Before starting to solve this problem, it is necessary to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 \u003d -4 and a 5 \u003d 5. Having established this, we proceed to a task that is similar to the previous one. Again, for the nth term, we use the formula, we get: a 5 \u003d a 1 + 4 * d. From: d \u003d (a 5 - a 1) / 4 \u003d (5 - (-4)) / 4 \u003d 2.25. Here we received not an integer value of the difference, but it is rational number, so the formulas for the algebraic progression remain the same.
Now let's add the found difference to a 1 and restore the missing members of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 \u003d 2.75 + 2.25 \u003d 5, which coincided with the condition of the problem.
Example #4: The first member of the progression
We continue to give examples of an arithmetic progression with a solution. In all previous problems, the first number of the algebraic progression was known. Now consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find from what number this sequence begins.
The formulas that have been used so far assume knowledge of a 1 and d. Nothing is known about these numbers in the condition of the problem. Nevertheless, let's write out the expressions for each term about which we have information: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We got two equations in which there are 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.
The specified system is easiest to solve if you express a 1 in each equation, and then compare the resulting expressions. First equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 \u003d a 43 - 42 * d \u003d 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d \u003d 37 - 42 * d, whence the difference d \u003d (37 - 50) / (42 - 14) \u003d - 0.464 (only 3 decimal places are given).
Knowing d, you can use any of the 2 expressions above for a 1 . For example, first: a 1 \u003d 50 - 14 * d \u003d 50 - 14 * (- 0.464) \u003d 56.496.
If there are doubts about the result, you can check it, for example, determine the 43rd member of the progression, which is specified in the condition. We get: a 43 \u003d a 1 + 42 * d \u003d 56.496 + 42 * (- 0.464) \u003d 37.008. A small error is due to the fact that rounding to thousandths was used in the calculations.
Example #5: Sum
Now let's look at some examples with solutions for the sum of an arithmetic progression.
Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?
Thanks to the development of computer technology, this problem can be solved, that is, sequentially add up all the numbers, which the computer will do as soon as a person presses the Enter key. However, the problem can be solved mentally if you pay attention that the presented series of numbers is an algebraic progression, and its difference is 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.
It is curious to note that this problem is called "Gaussian", since at the beginning of the 18th century the famous German, still at the age of only 10 years old, was able to solve it in his mind in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add pairs of numbers located at the edges of the sequence, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer, it is enough to multiply 50 by 101.
Example #6: sum of terms from n to m
Another typical example of the sum of an arithmetic progression is the following: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its terms from 8 to 14 will be.
The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them up sequentially. Since there are few terms, this method is not laborious enough. Nevertheless, it is proposed to solve this problem by the second method, which is more universal.
The idea is to get a formula for the sum of an algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:
- S m \u003d m * (a m + a 1) / 2.
- S n \u003d n * (a n + a 1) / 2.
Since n > m, it is obvious that the 2 sum includes the first one. The last conclusion means that if we take the difference between these sums, and add the term a m to it (in the case of taking the difference, it is subtracted from the sum S n), then we get the necessary answer to the problem. We have: S mn \u003d S n - S m + a m \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m \u003d a 1 * (n - m) / 2 + a n * n / 2 + a m * (1- m / 2). It is necessary to substitute formulas for a n and a m into this expression. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d * (3 * m - m 2 - 2) / 2.
The resulting formula is somewhat cumbersome, however, the sum S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.
As can be seen from the above solutions, all problems are based on the knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before you start solving any of these problems, it is recommended that you carefully read the condition, clearly understand what you want to find, and only then proceed with the solution.
Another tip is to strive for simplicity, that is, if you can answer the question without using complex mathematical calculations, then you need to do just that, since in this case the probability of making a mistake is less. For example, in the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn \u003d n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and split common task into separate subtasks (in this case first find the terms a n and a m).
If there are doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. How to find an arithmetic progression, found out. Once you figure it out, it's not that hard.
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. 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 members 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
Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")
An arithmetic progression is a series of numbers in which each number is greater (or less) than the previous one by the same amount.
This topic is often difficult and incomprehensible. Letter indices, the nth term of the progression, the difference of the progression - all this is somehow confusing, yes ... Let's figure out the meaning of the arithmetic progression and everything will work out right away.)
The concept of arithmetic progression.
Arithmetic progression is a very simple and clear concept. Doubt? In vain.) See for yourself.
I'll write an unfinished series of numbers:
1, 2, 3, 4, 5, ...
Can you extend this line? What numbers will go next, after the five? Everyone ... uh ..., in short, everyone will figure out that the numbers 6, 7, 8, 9, etc. will go further.
Let's complicate the task. I give an unfinished series of numbers:
2, 5, 8, 11, 14, ...
You can catch the pattern, extend the series, and name seventh row number?
If you figured out that this number is 20 - I congratulate you! You not only felt key points arithmetic progression, but also successfully used them in business! If you don't understand, read on.
Now let's translate the key points from sensations into mathematics.)
First key point.
Arithmetic progression deals with series of numbers. This is confusing at first. We are used to solving equations, building graphs and all that ... And then extend the series, find the number of the series ...
It's OK. It's just that progressions are the first acquaintance with a new branch of mathematics. The section is called "Series" and works with series of numbers and expressions. Get used to it.)
Second key point.
In an arithmetic progression, any number differs from the previous one by the same amount.
In the first example, this difference is one. Whatever number you take, it is one more than the previous one. In the second - three. Any number is three times greater than the previous one. Actually, it is this moment that gives us the opportunity to catch the pattern and calculate the subsequent numbers.
Third key point.
This moment is not striking, yes ... But very, very important. Here he is: each progression number stands in its place. There is the first number, there is the seventh, there is the forty-fifth, and so on. If you confuse them haphazardly, the pattern will disappear. The arithmetic progression will also disappear. It's just a series of numbers.
That's the whole point.
Of course, in new topic new terms and notation appear. They need to know. Otherwise, you won't understand the task. For example, you have to decide something like:
Write down the first six terms of the arithmetic progression (a n) if a 2 = 5, d = -2.5.
Does it inspire?) Letters, some indexes... And the task, by the way, couldn't be easier. You just need to understand the meaning of the terms and notation. Now we will master this matter and return to the task.
Terms and designations.
Arithmetic progression is a series of numbers in which each number is different from the previous one by the same amount.
This value is called . Let's deal with this concept in more detail.
Arithmetic progression difference.
Arithmetic progression difference is the amount by which any progression number more the previous one.
One important point. Please pay attention to the word "more". Mathematically, this means that each progression number is obtained adding the difference of an arithmetic progression to the previous number.
To calculate, let's say second numbers of the row, it is necessary to first number add this very difference of an arithmetic progression. For calculation fifth- the difference is necessary add to fourth well, etc.
Arithmetic progression difference may be positive then each number of the series will turn out to be real more than the previous one. This progression is called increasing. For example:
8; 13; 18; 23; 28; .....
Here each number is adding positive number, +5 to the previous one.
The difference can be negative then each number in the series will be less than the previous one. This progression is called (you won't believe it!) decreasing.
For example:
8; 3; -2; -7; -12; .....
Here every number is obtained too adding to the previous one, but negative number, -5.
By the way, when working with a progression, it is very useful to immediately determine its nature - whether it is increasing or decreasing. It helps a lot to find your bearings in the decision, to detect your mistakes and correct them before it's too late.
Arithmetic progression difference usually denoted by the letter d.
How to find d? Very simple. It is necessary to subtract from any number of the series previous number. Subtract. By the way, the result of subtraction is called "difference".)
Let's define, for example, d for an increasing arithmetic progression:
2, 5, 8, 11, 14, ...
We take any number of the row that we want, for example, 11. Subtract from it the previous number those. eight:
This is the correct answer. For this arithmetic progression, the difference is three.
You can just take any number of progressions, because for a specific progression d-always the same. At least somewhere at the beginning of the row, at least in the middle, at least anywhere. You can not take only the very first number. Just because the very first number no previous.)
By the way, knowing that d=3, finding the seventh number of this progression is very simple. We add 3 to the fifth number - we get the sixth, it will be 17. We add three to the sixth number, we get the seventh number - twenty.
Let's define d for a decreasing arithmetic progression:
8; 3; -2; -7; -12; .....
I remind you that, regardless of the signs, to determine d needed from any number take away the previous one. We choose any number of progression, for example -7. His previous number is -2. Then:
d = -7 - (-2) = -7 + 2 = -5
The difference of an arithmetic progression can be any number: integer, fractional, irrational, any.
Other terms and designations.
Each number in the series is called member of an arithmetic progression.
Each member of the progression has his number. The numbers are strictly in order, without any tricks. First, second, third, fourth, etc. For example, in the progression 2, 5, 8, 11, 14, ... two is the first member, five is the second, eleven is the fourth, well, you understand ...) Please clearly understand - the numbers themselves can be absolutely any, whole, fractional, negative, whatever, but numbering- strictly in order!
How to record a progression in general view? No problem! Each number in the series is written as a letter. To denote an arithmetic progression, as a rule, the letter is used a. The member number is indicated by the index at the bottom right. Members are written separated by commas (or semicolons), like this:
a 1 , a 2 , a 3 , a 4 , a 5 , .....
a 1 is the first number a 3- third, etc. Nothing tricky. You can write this series briefly like this: (a n).
There are progressions finite and infinite.
Ultimate the progression has a limited number of members. Five, thirty-eight, whatever. But it's a finite number.
Endless progression - has an infinite number of members, as you might guess.)
You can write a final progression through a series like this, all members and a dot at the end:
a 1 , a 2 , a 3 , a 4 , a 5 .
Or like this, if there are many members:
a 1 , a 2 , ... a 14 , a 15 .
In a short entry, you will have to additionally indicate the number of members. For example (for twenty members), like this:
(a n), n = 20
An infinite progression can be recognized by the ellipsis at the end of the row, as in the examples in this lesson.
Now you can already solve tasks. The tasks are simple, purely for understanding the meaning of the arithmetic progression.
Examples of tasks for arithmetic progression.
Let's take a closer look at the task above:
1. Write down the first six members of the arithmetic progression (a n), if a 2 = 5, d = -2.5.
We translate the task into understandable language. Given an infinite arithmetic progression. The second number of this progression is known: a 2 = 5. Known progression difference: d = -2.5. We need to find the first, third, fourth, fifth and sixth members of this progression.
For clarity, I will write down a series according to the condition of the problem. The first six members, where the second member is five:
a 1 , 5 , a 3 , a 4 , a 5 , a 6 ,....
a 3 = a 2 + d
We substitute in the expression a 2 = 5 and d=-2.5. Don't forget the minus!
a 3=5+(-2,5)=5 - 2,5 = 2,5
The third term is less than the second. Everything is logical. If the number is greater than the previous one negative value, so the number itself will be less than the previous one. Progression is decreasing. Okay, let's take it into account.) We consider the fourth member of our series:
a 4 = a 3 + d
a 4=2,5+(-2,5)=2,5 - 2,5 = 0
a 5 = a 4 + d
a 5=0+(-2,5)= - 2,5
a 6 = a 5 + d
a 6=-2,5+(-2,5)=-2,5 - 2,5 = -5
So, the terms from the third to the sixth have been calculated. This resulted in a series:
a 1 , 5 , 2.5 , 0 , -2.5 , -5 , ....
It remains to find the first term a 1 according to the well-known second. This is a step in the other direction, to the left.) Hence, the difference of the arithmetic progression d should not be added to a 2, a take away:
a 1 = a 2 - d
a 1=5-(-2,5)=5 + 2,5=7,5
That's all there is to it. Task response:
7,5, 5, 2,5, 0, -2,5, -5, ...
In passing, I note that we solved this task recurrent way. This terrible word means, only, the search for a member of the progression by the previous (adjacent) number. Other ways to work with progression will be discussed later.
One important conclusion can be drawn from this simple task.
Remember:
If we know at least one member and the difference of an arithmetic progression, we can find any member of this progression.
Remember? This simple derivation allows us to solve most problems school course on this topic. All tasks revolve around three main parameters: member of an arithmetic progression, difference of a progression, number of a member of a progression. Everything.
Of course, all previous algebra is not cancelled.) Inequalities, equations, and other things are attached to the progression. But according to the progression- everything revolves around three parameters.
For example, consider some popular tasks on this topic.
2. Write the final arithmetic progression as a series if n=5, d=0.4, and a 1=3.6.
Everything is simple here. Everything is already given. You need to remember how the members of an arithmetic progression are calculated, count, and write down. It is advisable not to skip the words in the task condition: "final" and " n=5". In order not to count until you are completely blue in the face.) There are only 5 (five) members in this progression:
a 2 \u003d a 1 + d \u003d 3.6 + 0.4 \u003d 4
a 3 \u003d a 2 + d \u003d 4 + 0.4 \u003d 4.4
a 4 = a 3 + d = 4.4 + 0.4 = 4.8
a 5 = a 4 + d = 4.8 + 0.4 = 5.2
It remains to write down the answer:
3,6; 4; 4,4; 4,8; 5,2.
Another task:
3. Determine if the number 7 will be a member of an arithmetic progression (a n) if a 1 \u003d 4.1; d = 1.2.
Hmm... Who knows? How to define something?
How-how ... Yes, write down the progression in the form of a series and see if there will be a seven or not! We believe:
a 2 \u003d a 1 + d \u003d 4.1 + 1.2 \u003d 5.3
a 3 \u003d a 2 + d \u003d 5.3 + 1.2 \u003d 6.5
a 4 = a 3 + d = 6.5 + 1.2 = 7.7
4,1; 5,3; 6,5; 7,7; ...
Now it is clearly seen that we are just seven slipped through between 6.5 and 7.7! The seven did not get into our series of numbers, and, therefore, the seven will not be a member of the given progression.
Answer: no.
And here is a task based on a real version of the GIA:
4. Several consecutive members of the arithmetic progression are written out:
...; fifteen; X; 9; 6; ...
Here is a series without end and beginning. No member numbers, no difference d. It's OK. To solve the problem, it is enough to understand the meaning of an arithmetic progression. Let's see and see what we can to know from this line? What are the parameters of the three main ones?
Member numbers? There is not a single number here.
But there are three numbers and - attention! - word "consecutive" in condition. This means that the numbers are strictly in order, without gaps. Are there two in this row? neighboring known numbers? Yes there is! These are 9 and 6. So we can calculate the difference of an arithmetic progression! We subtract from the six previous number, i.e. nine:
There are empty spaces left. What number will be the previous one for x? Fifteen. So x can be easily found by simple addition. To 15 add the difference of an arithmetic progression:
That's all. Answer: x=12
We solve the following problems ourselves. Note: these puzzles are not for formulas. Purely for understanding the meaning of an arithmetic progression.) We just write down a series of numbers-letters, look and think.
5. Find the first positive term of the arithmetic progression if a 5 = -3; d = 1.1.
6. It is known that the number 5.5 is a member of the arithmetic progression (a n), where a 1 = 1.6; d = 1.3. Determine the number n of this member.
7. It is known that in an arithmetic progression a 2 = 4; a 5 \u003d 15.1. Find a 3 .
8. Several consecutive members of the arithmetic progression are written out:
...; 15.6; X; 3.4; ...
Find the term of the progression, denoted by the letter x.
9. The train started moving from the station, gradually increasing its speed by 30 meters per minute. What will be the speed of the train in five minutes? Give your answer in km/h.
10. It is known that in an arithmetic progression a 2 = 5; a 6 = -5. Find a 1.
Answers (in disarray): 7.7; 7.5; 9.5; 9; 0.3; four.
Everything worked out? Wonderful! You can master the arithmetic progression for more high level, in the next lessons.
Didn't everything work out? No problem. In Special Section 555, all these puzzles are broken down piece by piece.) And, of course, a simple practical technique is described that immediately highlights the solution of such tasks clearly, clearly, as in the palm of your hand!
By the way, in the puzzle about the train there are two problems on which people often stumble. One - purely by progression, and the second - common to any tasks in mathematics, and physics too. This is a translation of dimensions from one to another. It shows how these problems should be solved.
In this lesson, we examined the elementary meaning of an arithmetic progression and its main parameters. This is enough to solve almost all problems on this topic. Add d to the numbers, write a series, everything will be decided.
The finger solution works well for very short pieces of the series, as in the examples in this lesson. If the series is longer, the calculations become more difficult. For example, if in problem 9 in the question, replace "five minutes" on the "thirty-five minutes" the problem will become much worse.)
And there are also tasks that are simple in essence, but utterly absurd in terms of calculations, for example:
Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.
And what, we will add 1/6 many, many times?! Is it possible to kill yourself!?
You can.) If you do not know a simple formula by which you can solve such tasks in a minute. This formula will be in the next lesson. And that problem is solved there. In a minute.)
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
What is the essence of the formula?
This formula allows you to find any BY HIS NUMBER" n" .
Of course, you need to know the first term a 1 and progression difference d, well, without these parameters, you can’t write down a specific progression.
It is not enough to memorize (or cheat) this formula. It is necessary to assimilate its essence and apply the formula in various problems. Yes, and do not forget at the right time, yes ...) How not forget- I do not know. But how to remember If needed, I'll give you a hint. For those who master the lesson to the end.)
So, let's deal with the formula of the n-th member of an arithmetic progression.
What is a formula in general - we imagine.) What is an arithmetic progression, a member number, a progression difference - is clearly stated in the previous lesson. Take a look if you haven't read it. Everything is simple there. It remains to figure out what nth member.
The progression in general can be written as a series of numbers:
a 1 , a 2 , a 3 , a 4 , a 5 , .....
a 1- denotes the first term of an arithmetic progression, a 3- third member a 4- fourth, and so on. If we are interested in the fifth term, let's say we are working with a 5, if one hundred and twentieth - from a 120.
How to define in general any member of an arithmetic progression, s any number? Very simple! Like this:
a n
That's what it is n-th member of an arithmetic progression. Under the letter n all the numbers of members are hidden at once: 1, 2, 3, 4, and so on.
And what does such a record give us? Just think, instead of a number, they wrote down a letter ...
This notation gives us a powerful tool for working with arithmetic progressions. Using the notation a n, we can quickly find any member any arithmetic progression. And a bunch of tasks to solve in progression. You will see further.
In the formula of the nth member of an arithmetic progression:
| a n = a 1 + (n-1)d |
a 1- the first member of the arithmetic progression;
n- member number.
The formula links the key parameters of any progression: a n ; a 1 ; d and n. Around these parameters, all the puzzles revolve in progression.
The nth term formula can also be used to write a specific progression. For example, in the problem it can be said that the progression is given by the condition:
a n = 5 + (n-1) 2.
Such a problem can even confuse ... There is no series, no difference ... But, comparing the condition with the formula, it is easy to figure out that in this progression a 1 \u003d 5, and d \u003d 2.
And it can be even angrier!) If we take the same condition: a n = 5 + (n-1) 2, yes, open the brackets and give similar ones? We get a new formula:
an = 3 + 2n.
it Only not general, but for a specific progression. This is where the pitfall lies. Some people think that the first term is a three. Although in reality the first member is a five ... A little lower we will work with such a modified formula.
In tasks for progression, there is another notation - a n+1. This is, you guessed it, the "n plus the first" term of the progression. Its meaning is simple and harmless.) This is a member of the progression, the number of which is greater than the number n by one. For example, if in some problem we take for a n fifth term, then a n+1 will be the sixth member. Etc.
Most often the designation a n+1 occurs in recursive formulas. Do not be afraid of this terrible word!) This is just a way of expressing a term of an arithmetic progression through the previous one. Suppose we are given an arithmetic progression in this form, using the recurrent formula:
a n+1 = a n +3
a 2 = a 1 + 3 = 5+3 = 8
a 3 = a 2 + 3 = 8+3 = 11
The fourth - through the third, the fifth - through the fourth, and so on. And how to count immediately, say the twentieth term, a 20? But no way!) While the 19th term is not known, the 20th cannot be counted. This is the fundamental difference between the recursive formula and the formula of the nth term. Recursive works only through previous term, and the formula of the nth term - through the first and allows straightaway find any member by its number. Not counting the whole series of numbers in order.
In an arithmetic progression, a recursive formula can easily be turned into a regular one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write the formula in the usual form, and work with it. In the GIA, such tasks are often found.
Application of the formula of the n-th member of an arithmetic progression.
To begin, consider direct application formulas. At the end of the previous lesson there was a problem:
Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.
This problem can be solved without any formulas, simply based on the meaning of the arithmetic progression. Add, yes add ... An hour or two.)
And according to the formula, the solution will take less than a minute. You can time it.) We decide.
The conditions provide all the data for using the formula: a 1 \u003d 3, d \u003d 1/6. It remains to be seen what n. No problem! We need to find a 121. Here we write:
Please pay attention! Instead of an index n a specific number appeared: 121. Which is quite logical.) We are interested in the member of the arithmetic progression number one hundred twenty one. This will be our n. It is this meaning n= 121 we will substitute further into the formula, in brackets. Substitute all the numbers in the formula and calculate:
a 121 = 3 + (121-1) 1/6 = 3+20 = 23
That's all there is to it. Just as quickly one could find the five hundred and tenth member, and the thousand and third, any. We put instead n the desired number in the index of the letter " a" and in brackets, and we consider.
Let me remind you the essence: this formula allows you to find any term of an arithmetic progression BY HIS NUMBER" n" .
Let's solve the problem smarter. Let's say we have the following problem:
Find the first term of the arithmetic progression (a n) if a 17 =-2; d=-0.5.
If you have any difficulties, I will suggest the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Hand write, right in your notebook:
| a n = a 1 + (n-1)d |
And now, looking at the letters of the formula, we understand what data we have and what is missing? Available d=-0.5, there is a seventeenth member ... Everything? If you think that's all, then you can't solve the problem, yes ...
We also have a number n! In the condition a 17 =-2 hidden two options. This is both the value of the seventeenth member (-2) and its number (17). Those. n=17. This "little thing" often slips past the head, and without it, (without the "little thing", not the head!) The problem cannot be solved. Although ... and without a head too.)
Now we can just stupidly substitute our data into the formula:
a 17 \u003d a 1 + (17-1) (-0.5)
Oh yes, a 17 we know it's -2. Okay, let's put it in:
-2 \u003d a 1 + (17-1) (-0.5)
That, in essence, is all. It remains to express the first term of the arithmetic progression from the formula, and calculate. You get the answer: a 1 = 6.
Such a technique - writing a formula and simply substituting known data - helps a lot in simple tasks. Well, you must, of course, be able to express a variable from a formula, but what to do!? Without this skill, mathematics can not be studied at all ...
Another popular problem:
Find the difference of the arithmetic progression (a n) if a 1 =2; a 15 =12.
What are we doing? You will be surprised, we write the formula!)
| a n = a 1 + (n-1)d |
Consider what we know: a 1 =2; a 15 =12; and (special highlight!) n=15. Feel free to substitute in the formula:
12=2 + (15-1)d
Let's do the arithmetic.)
12=2 + 14d
d=10/14 = 5/7
This is the correct answer.
So, tasks a n , a 1 and d decided. It remains to learn how to find the number:
The number 99 is a member of an arithmetic progression (a n), where a 1 =12; d=3. Find the number of this member.
We substitute the known quantities into the formula of the nth term:
a n = 12 + (n-1) 3
At first glance, there are two unknown quantities here: a n and n. But a n is some member of the progression with the number n... And this member of the progression we know! It's 99. We don't know his number. n, so this number also needs to be found. Substitute the progression term 99 into the formula:
99 = 12 + (n-1) 3
We express from the formula n, we think. We get the answer: n=30.
And now a problem on the same topic, but more creative):
Determine if the number 117 will be a member of an arithmetic progression (a n):
-3,6; -2,4; -1,2 ...
Let's write the formula again. What, there are no options? Hm... Why do we need eyes?) Do we see the first member of the progression? We see. This is -3.6. You can safely write: a 1 \u003d -3.6. Difference d can be determined from the series? It's easy if you know what the difference of an arithmetic progression is:
d = -2.4 - (-3.6) = 1.2
Yes, we did the simplest thing. It remains to deal with an unknown number n and an incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know that ... How to be!? Well, how to be, how to be... Turn on Creative skills!)
We suppose that 117 is, after all, a member of our progression. With an unknown number n. And, just like in the previous problem, let's try to find this number. Those. we write the formula (yes-yes!)) and substitute our numbers:
117 = -3.6 + (n-1) 1.2
Again we express from the formulan, we count and get:
Oops! The number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion do we draw? Yes! Number 117 is not member of our progression. It is somewhere between the 101st and 102nd members. If the number turned out to be natural, i.e. positive integer, then the number would be a member of the progression with the found number. And in our case, the answer to the problem will be: no.
Task based on a real version of the GIA:
The arithmetic progression is given by the condition:
a n \u003d -4 + 6.8n
Find the first and tenth terms of the progression.
Here the progression is set in an unusual way. Some kind of formula ... It happens.) However, this formula (as I wrote above) - also the formula of the n-th member of an arithmetic progression! She also allows find any member of the progression by its number.
We are looking for the first member. The one who thinks. that the first term is minus four, is fatally mistaken!) Because the formula in the problem is modified. The first term of an arithmetic progression in it hidden. Nothing, we'll find it now.)
Just as in the previous tasks, we substitute n=1 into this formula:
a 1 \u003d -4 + 6.8 1 \u003d 2.8
Here! The first term is 2.8, not -4!
Similarly, we are looking for the tenth term:
a 10 \u003d -4 + 6.8 10 \u003d 64
That's all there is to it.
And now, for those who have read up to these lines, the promised bonus.)
Suppose, in a difficult combat situation of the GIA or the Unified State Exam, you forgot the useful formula of the n-th member of an arithmetic progression. Something comes to mind, but somehow uncertainly ... Whether n there, or n+1, or n-1... How to be!?
Calm! This formula is easy to derive. Not very strict, but to be sure and right decision that's enough!) For the conclusion, it is enough to remember the elementary meaning of the arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.
We draw a numerical axis and mark the first one on it. second, third, etc. members. And note the difference d between members. Like this:
We look at the picture and think: what is the second term equal to? Second one d:
a 2 =a 1 + 1 d
What is the third term? Third term equals first term plus two d.
a 3 =a 1 + 2 d
Do you get it? I don't put some words in bold for nothing. Okay, one more step.)
What is the fourth term? Fourth term equals first term plus three d.
a 4 =a 1 + 3 d
It's time to realize that the number of gaps, i.e. d, always one less than the number of the member you are looking for n. That is, up to the number n, number of gaps will be n-1. So, the formula will be (no options!):
| a n = a 1 + (n-1)d |
In general, visual pictures are very helpful in solving many problems in mathematics. Don't neglect the pictures. But if it's difficult to draw a picture, then ... only a formula!) In addition, the formula of the nth term allows you to connect the entire powerful arsenal of mathematics to the solution - equations, inequalities, systems, etc. You can't put a picture in an equation...
Tasks for independent decision.
For warm-up:
1. In arithmetic progression (a n) a 2 =3; a 5 \u003d 5.1. Find a 3 .
Hint: according to the picture, the problem is solved in 20 seconds ... According to the formula, it turns out more difficult. But for mastering the formula, it is more useful.) In Section 555, this problem is solved both by the picture and by the formula. Feel the difference!)
And this is no longer a warm-up.)
2. In arithmetic progression (a n) a 85 \u003d 19.1; a 236 =49, 3. Find a 3 .
What, reluctance to draw a picture?) Still! It's better in the formula, yes ...
3. Arithmetic progression is given by the condition:a 1 \u003d -5.5; a n+1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.
In this task, the progression is given in a recurrent way. But counting up to the one hundred and twenty-fifth term... Not everyone can do such a feat.) But the formula of the nth term is within the power of everyone!
4. Given an arithmetic progression (a n):
-148; -143,8; -139,6; -135,4, .....
Find the number of the smallest positive term of the progression.
5. According to the condition of task 4, find the sum of the smallest positive and largest negative members of the progression.
6. The product of the fifth and twelfth terms of an increasing arithmetic progression is -2.5, and the sum of the third and eleventh terms is zero. Find a 14 .
Not the easiest task, yes ...) Here the method "on the fingers" will not work. You have to write formulas and solve equations.
Answers (in disarray):
3,7; 3,5; 2,2; 37; 2,7; 56,5
Happened? It's nice!)
Not everything works out? It happens. By the way, in the last task there is one subtle point. Attentiveness when reading the problem will be required. And logic.
The solution to all these problems is discussed in detail in Section 555. And the fantasy element for the fourth, and the subtle moment for the sixth, and general approaches for solving any problems for the formula of the nth term - everything is painted. I recommend.
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
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,
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 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 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
| 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, 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:
| 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 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
| 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 . ◄
