Which of the expressions is an arithmetic progression. Algebraic progression
For example, the sequence \ (2 \); \(five\); \(eight\); \(eleven\); \ (14 \) ... is an arithmetic progression, because each next element differs from the previous one by three (can be obtained from the previous one by adding a triplet):
In this progression, the difference \ (d \) is positive (equal to \ (3 \)), and therefore each next term is greater than the previous one. Such progressions are called increasing.
However, \ (d \) can also be negative. For example, in arithmetic progression\(sixteen\); \(10\); \(4\); \ (- 2 \); \ (- 8 \) ... the difference of the progression \ (d \) is equal to minus six.
And in this case, each next element will be smaller than the previous one. These progressions are called decreasing.
Arithmetic progression notation
Progression is indicated by a small Latin letter.
The numbers forming the progression call it members of(or elements).
They are denoted by the same letter as the arithmetic progression, but with a numerical index equal to the number of the element in order.
For example, the arithmetic progression \ (a_n = \ left \ (2; 5; 8; 11; 14 ... \ right \) \) consists of the elements \ (a_1 = 2 \); \ (a_2 = 5 \); \ (a_3 = 8 \) and so on.
In other words, for the progression \ (a_n = \ left \ (2; 5; 8; 11; 14 ... \ right \) \)
Problem solving for arithmetic progression
In principle, the above information is already enough to solve almost any arithmetic progression problem (including those offered at the OGE).
Example (OGE).
The arithmetic progression is specified by the conditions \ (b_1 = 7; d = 4 \). Find \ (b_5 \).
Solution:
Answer: \ (b_5 = 23 \)
Example (OGE).
The first three terms of the arithmetic progression are given: \ (62; 49; 36 ... \) Find the value of the first negative term of this progression ..
Solution:
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We are given the first elements of the sequence and we know that it is an arithmetic progression. That is, each element differs from the neighboring one by the same number. Find out which one, subtracting the previous one from the next element: \ (d = 49-62 = -13 \). |
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Now we can restore our progression to the (first negative) element we need. |
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Ready. You can write an answer. |
Answer: \(-3\)
Example (OGE).
Several consecutive elements of the arithmetic progression are given: \ (… 5; x; 10; 12,5 ... \) Find the value of the element indicated by the letter \ (x \).
Solution:
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To find \ (x \), we need to know how much the next element differs from the previous one, in other words, the difference of the progression. Let's find it from two known neighboring elements: \ (d = 12.5-10 = 2.5 \). |
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And now we find the desired one without any problems: \ (x = 5 + 2.5 = 7.5 \). |
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Ready. You can write an answer. |
Answer: \(7,5\).
Example (OGE).
The arithmetic progression is specified by the following conditions: \ (a_1 = -11 \); \ (a_ (n + 1) = a_n + 5 \) Find the sum of the first six terms of this progression.
Solution:
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We need to find the sum of the first six terms of the progression. But we do not know their meanings, we are only given the first element. Therefore, first we calculate the values in turn using the given to us: \ (n = 1 \); \ (a_ (1 + 1) = a_1 + 5 = -11 + 5 = -6 \) |
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\ (S_6 = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = \) |
The amount you are looking for has been found. |
Answer: \ (S_6 = 9 \).
Example (OGE).
In arithmetic progression \ (a_ (12) = 23 \); \ (a_ (16) = 51 \). Find the difference between this progression.
Solution:
Answer: \ (d = 7 \).
Important Arithmetic Progression Formulas
As you can see, many arithmetic progression problems can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each next element in this chain is obtained by adding the same number to the previous one (the difference of the progression).
However, sometimes there are situations when it is very inconvenient to decide "head-on". For example, imagine that in the very first example we need to find not the fifth element \ (b_5 \), but the three hundred and eighty-sixth \ (b_ (386) \). What is it, we \ (385 \) times add four? Or imagine that in the penultimate example, you need to find the sum of the first seventy-three elements. You will be tortured to count ...
Therefore, in such cases, they do not solve "head-on", but use special formulas derived for the arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum \ (n \) of the first terms.
Formula \ (n \) - th member: \ (a_n = a_1 + (n-1) d \), where \ (a_1 \) is the first term of the progression;
\ (n \) - number of the element being searched for;
\ (a_n \) is a member of the progression with the number \ (n \).
This formula allows us to quickly find at least the three hundredth, even the millionth element, knowing only the first and the difference of the progression.
Example.
The arithmetic progression is specified by the conditions: \ (b_1 = -159 \); \ (d = 8.2 \). Find \ (b_ (246) \).
Solution:
Answer: \ (b_ (246) = 1850 \).
The formula for the sum of the first n terms: \ (S_n = \ frac (a_1 + a_n) (2) \ cdot n \), where
\ (a_n \) - the last summed term;
Example (OGE).
The arithmetic progression is specified by the conditions \ (a_n = 3,4n-0,6 \). Find the sum of the first \ (25 \) members of this progression.
Solution:
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\ (S_ (25) = \) \ (\ frac (a_1 + a_ (25)) (2) \) \ (\ cdot 25 \) |
To calculate the sum of the first twenty-five elements, we need to know the value of the first and twenty-fifth terms. |
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\ (n = 1; \) \ (a_1 = 3.4 1-0.6 = 2.8 \) |
Now we find the twenty-fifth term, substituting twenty-five instead of \ (n \). |
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\ (n = 25; \) \ (a_ (25) = 3.4 25-0.6 = 84.4 \) |
Well, now we can calculate the required amount without any problems. |
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\ (S_ (25) = \) \ (\ frac (a_1 + a_ (25)) (2) \) \ (\ cdot 25 = \) |
The answer is ready. |
Answer: \ (S_ (25) = 1090 \).
For the sum \ (n \) of the first terms, you can get another formula: you just need to \ (S_ (25) = \) \ (\ frac (a_1 + a_ (25)) (2) \) \ (\ cdot 25 \ ) instead of \ (a_n \) substitute the formula for it \ (a_n = a_1 + (n-1) d \). We get:
The formula for the sum of the first n terms: \ (S_n = \) \ (\ frac (2a_1 + (n-1) d) (2) \) \ (\ cdot n \), where
\ (S_n \) - the required sum \ (n \) of the first elements;
\ (a_1 \) - the first summed term;
\ (d \) - progression difference;
\ (n \) - the number of elements in the sum.
Example.
Find the sum of the first \ (33 \) - ex members of the arithmetic progression: \ (17 \); \ (15.5 \); \(fourteen\)…
Solution:
Answer: \ (S_ (33) = - 231 \).
More complex arithmetic progression problems
Now you have all the information you need to solve almost any arithmetic progression problem. We conclude the topic by considering problems in which you need not only to apply formulas, but also to think a little (in mathematics, this can be useful ☺)
Example (OGE).
Find the sum of all negative terms of the progression: \ (- 19,3 \); \(-nineteen\); \ (- 18.7 \) ...
Solution:
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\ (S_n = \) \ (\ frac (2a_1 + (n-1) d) (2) \) \ (\ cdot n \) |
The task is very similar to the previous one. We start to solve also: first we find \ (d \). |
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\ (d = a_2-a_1 = -19 - (- 19.3) = 0.3 \) |
Now we would substitute \ (d \) in the formula for the sum ... and here a small nuance emerges - we do not know \ (n \). In other words, we do not know how many terms will need to be added. How to find out? Let's think. We'll stop adding elements when we get to the first positive element. That is, you need to find out the number of this element. How? Let's write down the formula for calculating any element of the arithmetic progression: \ (a_n = a_1 + (n-1) d \) for our case. |
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\ (a_n = a_1 + (n-1) d \) |
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\ (a_n = -19.3 + (n-1) 0.3 \) |
We need \ (a_n \) to be greater than zero. Let's find out at what \ (n \) this will happen. |
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\ (- 19.3+ (n-1) 0.3> 0 \) |
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\ ((n-1) 0.3> 19.3 \) \ (|: 0.3 \) |
We divide both sides of the inequality by \ (0,3 \). |
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\ (n-1> \) \ (\ frac (19,3) (0,3) \) |
Move minus one, remembering to change signs |
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\ (n> \) \ (\ frac (19,3) (0,3) \) \ (+ 1 \) |
We calculate ... |
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\ (n> 65,333 ... \) |
... and it turns out that the first positive element will have the number \ (66 \). Accordingly, the last negative has \ (n = 65 \). Let's check it out just in case. |
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\ (n = 65; \) \ (a_ (65) = - 19.3+ (65-1) 0.3 = -0.1 \) |
Thus, we need to add the first \ (65 \) elements. |
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\ (S_ (65) = \) \ (\ frac (2 \ cdot (-19.3) + (65-1) 0.3) (2) \)\ (\ cdot 65 \) |
The answer is ready. |
Answer: \ (S_ (65) = - 630.5 \).
Example (OGE).
The arithmetic progression is specified by the conditions: \ (a_1 = -33 \); \ (a_ (n + 1) = a_n + 4 \). Find the sum from \ (26 \) th to \ (42 \) element inclusive.
Solution:
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\ (a_1 = -33; \) \ (a_ (n + 1) = a_n + 4 \) |
In this problem, you also need to find the sum of the elements, but starting not from the first, but from \ (26 \) - th. For such a case, we have no formula. How to decide? |
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For our progression \ (a_1 = -33 \), and the difference \ (d = 4 \) (after all, it is the four that we add to the previous element to find the next one). Knowing this, we find the sum of the first \ (42 \) - yh elements. |
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\ (S_ (42) = \) \ (\ frac (2 \ cdot (-33) + (42-1) 4) (2) \)\ (\ cdot 42 = \) |
Now the sum of the first \ (25 \) - ty elements. |
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\ (S_ (25) = \) \ (\ frac (2 \ cdot (-33) + (25-1) 4) (2) \)\ (\ cdot 25 = \) |
Finally, we calculate the answer. |
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\ (S = S_ (42) -S_ (25) = 2058-375 = 1683 \) |
Answer: \ (S = 1683 \).
There are several more formulas for the arithmetic progression that we did not consider in this article due to their low practical usefulness. However, you can easily find them.
I. V. Yakovlev | Mathematics Materials | MathUs.ru
Arithmetic progression
An arithmetic progression is a special kind of sequence. Therefore, before defining an arithmetic (and then geometric) progression, we need to briefly discuss the important concept of a number sequence.
Subsequence
Imagine a device on the screen of which some numbers are displayed one after another. Let's say 2; 7; 13; one; 6; 0; 3; ::: This set of numbers is just an example of a sequence.
Definition. A numerical sequence is a set of numbers in which each number can be assigned a unique number (that is, to associate a single natural number) 1. The number n is called nth member sequence.
So, in the above example, the first number is 2, this is the first member of the sequence, which can be denoted a1; number five has number 6 this is the fifth term in the sequence, which can be denoted as a5. Generally, nth term sequence is denoted by an (or bn, cn, etc.).
The situation is very convenient when the n-th term of the sequence can be specified by some formula. For example, the formula an = 2n 3 defines the sequence: 1; one; 3; five; 7; ::: The formula an = (1) n defines the sequence: 1; one; one; one; :::
Not every set of numbers is a sequence. So, a segment is not a sequence; it contains “too many” numbers to be renumbered. The set R of all real numbers is also not a sequence. These facts are proved in the course of mathematical analysis.
Arithmetic progression: basic definitions
Now we are ready to define an arithmetic progression.
Definition. An arithmetic progression is a sequence, each term of which (starting from the second) is equal to the sum of the previous term and some fixed number (called the difference of the arithmetic progression).
For example, sequence 2; five; eight; eleven; ::: is an arithmetic progression with the first term 2 and difference 3. Sequence 7; 2; 3; eight; ::: is an arithmetic progression with the first term 7 and difference 5. Sequence 3; 3; 3; ::: is an arithmetic progression with zero difference.
Equivalent definition: a sequence an is called an arithmetic progression if the difference an + 1 an is a constant value (independent of n).
An arithmetic progression is called increasing if its difference is positive, and decreasing if its difference is negative.
1 And here is a more laconic definition: a sequence is a function defined on a set natural numbers... For example, a sequence of real numbers is a function f: N! R.
By default, sequences are considered infinite, that is, containing an infinite number of numbers. But no one bothers to consider finite sequences as well; in fact, any finite set of numbers can be called a finite sequence. For example, the final sequence is 1; 2; 3; 4; 5 consists of five numbers.
Formula of the nth term of an arithmetic progression
It is easy to understand that the arithmetic progression is completely determined by two numbers: the first term and the difference. Therefore, the question arises: how, knowing the first term and the difference, to find an arbitrary member of the arithmetic progression?
It is not difficult to obtain the required formula for the nth term of an arithmetic progression. Let an
arithmetic progression with difference d. We have: |
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an + 1 = an + d (n = 1; 2;:: :): |
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In particular, we write: |
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a2 = a1 + d; |
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a3 = a2 + d = (a1 + d) + d = a1 + 2d; |
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a4 = a3 + d = (a1 + 2d) + d = a1 + 3d; |
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and now it becomes clear that the formula for an is: |
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an = a1 + (n 1) d: |
Problem 1. In arithmetic progression 2; five; eight; eleven; ::: find the formula for the nth term and calculate the hundredth term.
Solution. According to formula (1), we have:
an = 2 + 3 (n 1) = 3n 1:
a100 = 3 100 1 = 299:
Property and sign of arithmetic progression
Arithmetic progression property. In arithmetic progression an for any
In other words, each member of the arithmetic progression (starting from the second) is the arithmetic mean of the neighboring members.
Proof. We have: |
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a n 1 + a n + 1 |
(an d) + (an + d) |
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as required.
More in a general way, the arithmetic progression an satisfies the equality
a n = a n k + a n + k
for any n> 2 and any natural k< n. Попробуйте самостоятельно доказать эту формулу тем же самым приёмом, что и формулу (2 ).
It turns out that formula (2) is not only a necessary, but also a sufficient condition for a sequence to be an arithmetic progression.
A sign of an arithmetic progression. If equality (2) holds for all n> 2, then the sequence an is an arithmetic progression.
Proof. Let's rewrite formula (2) as follows:
a n a n 1 = a n + 1 a n:
This shows that the difference an + 1 an does not depend on n, and this just means that the sequence an is an arithmetic progression.
The property and feature of an arithmetic progression can be formulated as a single statement; For convenience, we will do this for three numbers (this is the situation that often occurs in problems).
Characterization of the arithmetic progression. Three numbers a, b, c form an arithmetic progression if and only if 2b = a + c.
Problem 2. (Moscow State University, Economics Faculty, 2007) Three numbers 8x, 3 x2 and 4 in the indicated order form a decreasing arithmetic progression. Find x and indicate the difference of this progression.
Solution. By the property of the arithmetic progression, we have:
2 (3 x2) = 8x 4, 2x2 + 8x 10 = 0, x2 + 4x 5 = 0, x = 1; x = 5:
If x = 1, then we get a decreasing progression 8, 2, 4 with a difference 6. If x = 5, then we get an increasing progression 40, 22, 4; this case is not good.
Answer: x = 1, the difference is 6.
Sum of the first n terms of an arithmetic progression
Legend has it that once the teacher told the children to find the sum of numbers from 1 to 100 and sat down to read the newspaper calmly. However, less than a few minutes later, one boy said that he had solved the problem. It was 9-year-old Karl Friedrich Gauss, later one of the greatest mathematicians in history.
Little Gauss's idea was this. Let be
S = 1 + 2 + 3 +::: + 98 + 99 + 100:
Let's write this amount in reverse order:
S = 100 + 99 + 98 +::: + 3 + 2 + 1;
and add these two formulas:
2S = (1 + 100) + (2 + 99) + (3 + 98) +::: + (98 + 3) + (99 + 2) + (100 + 1):
Each term in parentheses is equal to 101, and there are 100 such terms in total. Therefore,
2S = 101 100 = 10100;
We use this idea to derive the sum formula
S = a1 + a2 +::: + an + a n n: (3)
A useful modification of formula (3) is obtained by substituting the formula for the nth term an = a1 + (n 1) d into it:
2a1 + (n 1) d |
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Problem 3. Find the sum of all positive three-digit numbers divisible by 13.
Solution. Three-digit numbers divisible by 13 form an arithmetic progression with the first term 104 and the difference 13; The nth term of this progression is:
an = 104 + 13 (n 1) = 91 + 13n:
Let's find out how many members our progression contains. To do this, we solve the inequality:
an 6 999; 91 + 13n 6 999;
n 6 908 13 = 6911 13; n 6 69:
So, there are 69 members in our progression. Using formula (4), we find the required sum:
S = 2 104 + 68 13 69 = 37674: 2
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 one 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:
Number sequence
For example, for our sequence:
The assigned number is specific to only one number in the sequence. In other words, there are no three second numbers in the sequence. The second number (like the -th number) is always one.
The number with the number is called the th member of the sequence.
We usually call the entire sequence some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member:.
In our case: 
Let's say we have numerical sequence, in which the difference between adjacent numbers is the same and equal.
For example: 
etc.
This number sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius in the 6th century and was understood in a broader sense as an endless number sequence. The name "arithmetic" was carried over from the theory of continuous proportions, which was occupied by the ancient Greeks.
This is a numerical sequence, each member of which is equal to the previous one, added to the same number. This number is called the difference of the arithmetic progression and is denoted by.
Try to determine which number sequences are arithmetic progression and which are not:
a)
b)
c)
d)
Understood? Let's compare our answers:
Is an arithmetic progression - b, c.
Is not arithmetic progression - a, d.
Let's return to the given progression () and try to find the value of its th member. Exist two the way to find it.
1. Method
We can add to the previous value of the number of the progression until we get to the th term of the progression. It's good that we don't have much left to sum up - only three values:
So, the th member of the described arithmetic progression is equal to.
2. Method
What if we needed to find the value of the th term of the progression? The summation would take us more than one hour, and it is not a fact that we would not be mistaken when adding numbers.
Of course, mathematicians have come up with a way in which you do not need to add the difference of the arithmetic progression to the previous value. Take a closer look at the drawn picture ... Surely you have already noticed a certain pattern, namely:
For example, let's see how the value of the th member of this arithmetic progression is added: 
In other words:
Try to independently find the value of a member of a given arithmetic progression in this way.
Calculated? Compare your notes to the answer:
Please note that you got exactly the same number as in the previous method, when we successively added the members of the arithmetic progression to the previous value.
Let's try to "depersonalize" this formula - we will bring it into general form and get:
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Arithmetic progression equation. |
Arithmetic progressions are ascending and sometimes decreasing.
Ascending- progressions in which each subsequent value of the members is greater than the previous one.
For example:
Decreasing- progressions in which each subsequent value of the members is less than the previous one.
For example:
The derived formula is used in calculating the terms in both increasing and decreasing terms of an arithmetic progression.
Let's check it out in practice.
We are given an arithmetic progression consisting of the following numbers: Let's check what the th number of this arithmetic progression will turn out if we use our formula to calculate it:
Since, then:
Thus, we made sure that the formula operates in both decreasing and increasing arithmetic progression.
Try to find the th and th terms of this arithmetic progression on your own.
Let's compare the results obtained:
Arithmetic progression property
Let's complicate the task - we will derive the property of the arithmetic progression.
Let's say we are given the following condition:
- arithmetic progression, find the value.
Easy, you say and start counting according to the formula you already know:
Let, a, then:
Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but if we are given numbers in the condition? Admit it, there is a chance of making a mistake in the calculations.
Now think, is it possible to solve this problem in one action using any formula? Of course, yes, and it is her that we will try to withdraw now.
Let's denote the required term of the arithmetic progression as, we know the formula for finding it - this is the same formula we derived at the beginning:
, then:
- the previous member of the progression is:
- the next member of the progression is:
Let's summarize the previous and subsequent members of the progression:
It turns out that the sum of the previous and subsequent members of the progression is the doubled value of the member of the progression located between them. In other words, in order to find the value of a member of the progression with known previous and successive values, it is necessary to add them up and divide by.
That's right, we got the same number. Let's fix the material. Calculate the value for the progression yourself, because it's not difficult at all.
Well done! You know almost everything about progression! There is only one formula left to learn, which, according to legend, was easily deduced for himself by one of the greatest mathematicians of all time, the "king of mathematicians" - Karl Gauss ...
When Karl Gauss was 9 years old, a teacher engaged in checking the work of students in other grades asked the following task in the lesson: "Calculate the sum of all natural numbers from up to (according to other sources up to) inclusive." Imagine the teacher's surprise when one of his students (it was Karl Gauss) gave the correct answer to the problem in a minute, while most of the daredevil's classmates, after long calculations, received the wrong result ...
Young Karl Gauss noticed a certain pattern that you can easily notice.
Let's say we have an arithmetic progression consisting of -th members: We need to find the sum of the given members of the arithmetic progression. Of course, we can manually sum all the values, but what if in the task it is necessary to find the sum of its members, as Gauss was looking for?
Let's depict the given progression. Look closely at the highlighted numbers and try to perform various mathematical operations with them. 
Have you tried it? What have you noticed? Right! Their sums are equal
Now tell me, how many such pairs are there in the given progression? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two members of the arithmetic progression is equal, and similar equal pairs, we get that the total sum is:
.
Thus, the formula for the sum of the first terms of any arithmetic progression will be as follows:
In some problems, we do not know the th term, but we know the difference in the progression. Try to substitute in the formula for the sum, the formula of the th term.
What did you do?
Well done! Now let's return to the problem that was given to Karl Gauss: calculate yourself what is the sum of the numbers starting from the -th, and the sum of the numbers starting from the -th.
How much did you get?
Gauss found that the sum of the members is equal, and the sum of the members. Is that how you decided?
In fact, the formula for the sum of the members of an arithmetic progression was proved by the ancient Greek scientist Diophantus in the 3rd century, and throughout this time, witty people were using the properties of an arithmetic progression to the utmost.
For example, imagine Ancient Egypt and the most ambitious construction site of that time - the construction of the pyramid ... The figure shows one side of it. 
Where is the progression here you say? Look closely and find a pattern in the number of sand blocks in each row of the pyramid wall. 
Isn't it an arithmetic progression? Calculate how many blocks are needed to build one wall if block bricks are placed in the base. I hope you won't count by running your finger across the monitor, do you remember the last formula and everything we said about the arithmetic progression?
IN this case the progression looks like this:.
Difference of arithmetic progression.
The number of members of the arithmetic progression.
Let's substitute our data into the last formulas (we will count the number of blocks in 2 ways).
Method 1.
Method 2.
And now you can calculate on the monitor: compare the obtained values with the number of blocks that are in our pyramid. Did it come together? Well done, you have mastered the sum of the terms of the arithmetic progression.
Of course, you can't build a pyramid from blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:
Workout
Tasks:
- Masha is getting in shape by summer. Every day she increases the number of squats by. How many times will Masha squat in weeks, if at the first workout she did squats.
- What is the sum of all the odd numbers contained in.
- When storing logs, lumberjacks stack them in such a way that each top layer contains one log less than the previous one. How many logs are in one masonry, if logs serve as the basis of the masonry.
Answers:
- Let's define the parameters of the arithmetic progression. In this case
(weeks = days).Answer: After two weeks, Masha should squat once a day.
- First odd number, the last number.
Difference of arithmetic progression.
The number of odd numbers in is half, however, we will check this fact using the formula for finding the -th term of an arithmetic progression:The numbers do contain odd numbers.
Substitute the available data into the formula:Answer: The sum of all odd numbers contained in is equal to.
- Let's remember the pyramid problem. For our case, a, since each top layer is reduced by one log, then only in a bunch of layers, that is.
Let's substitute the data into the formula:Answer: There are logs in the masonry.
Let's sum up
- - a numerical sequence in which the difference between adjacent numbers is the same and equal. It can be ascending and decreasing.
- Finding formula-th member of the arithmetic progression is written by the formula -, where is the number of numbers in the progression.
- Property of members of an arithmetic progression- - where is the number of numbers in the progression.
- The sum of the members of an arithmetic progression can be found in two ways:
, where is the number of values.
ARITHMETIC PROGRESSION. AVERAGE LEVEL
Number sequence
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. But you can always say which one is the first, which is the second, and so on, that is, we can number them. This is an example of a number sequence.
Number sequence is a set of numbers, each of which can be assigned a unique number.
In other words, each number can be associated with a certain natural number, and the only one. And we will not assign this number to any other number from this set.
The number with the number is called the th member of the sequence.
We usually call the entire sequence some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member:.
It is very convenient if the th term of the sequence can be specified by some formula. For example, the formula
specifies the sequence:
And the formula is the following sequence:
For example, an arithmetic progression is a sequence (the first term here is equal, and the difference). Or (, difference).
Nth term formula
We call recurrent a formula in which to find out the th member, you need to know the previous or several previous ones:
To find, for example, the th term of the progression using such a formula, we will have to calculate the previous nine. For example, let. Then:
Well, what is the formula now?
In each line we add to, multiplied by some number. For what? Very simple: this is the number of the current member minus:
Much more convenient now, right? We check:
Decide for yourself:
In an arithmetic progression, find the formula for the nth term and find the hundredth term.
Solution:
The first term is equal. What is the difference? And here's what:
(it is because it is called the difference, which is equal to the difference of the successive members of the progression).
So the formula is:
Then the hundredth term is:
What is the sum of all natural numbers from to?
According to legend, the great mathematician Karl Gauss, being a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last numbers is equal, the sum of the second and the last but one is the same, the sum of the third and the third from the end is the same, and so on. How many such pairs will there be? That's right, exactly half the number of all numbers, that is. So,
The general formula for the sum of the first terms of any arithmetic progression would be:
Example:
Find the sum of all two-digit multiples.
Solution:
The first such number is. Each next is obtained by adding to the previous number. Thus, the numbers we are interested in form an arithmetic progression with the first term and the difference.
The th term formula for this progression is:
How many members are in the progression if they all have to be double digits?
Very easy: .
The last term in the progression will be equal. Then the sum:
Answer: .
Now decide for yourself:
- Every day, the athlete runs more m than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
- A cyclist drives more kilometers every day than the previous one. On the first day, he drove km. How many days does he need to travel to cover the km? How many kilometers will he travel in the last day of the journey?
- The price of a refrigerator in a store decreases by the same amount every year. Determine how much the price of the refrigerator has decreased every year, if, put up for sale for rubles, six years later it was sold for rubles.
Answers:
- The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first members of this progression:
.
Answer: - It is given here:, it is necessary to find.
Obviously, you need to use the same sum formula as in the previous problem:
.
Substitute the values:The root obviously doesn't fit, so the answer is.
Let's calculate the distance traveled for the last day using the th term formula:
(km).
Answer: - Given:. To find: .
It couldn't be easier:
(rub).
Answer:
ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN
This is a numerical sequence in which the difference between adjacent numbers is the same and equal.
Arithmetic progression can be ascending () and decreasing ().
For example:
The formula for finding the n-th term of an arithmetic progression
written by the formula, where is the number of numbers in the progression.
Property of members of an arithmetic progression
It allows you to easily find a member of the progression if its neighboring members are known - where is the number of numbers in the progression.
The sum of the members of an arithmetic progression
There are two ways to find the amount:
Where is the number of values.
Where is the number of values.
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