Mathematics has its own beauty, just like painting and poetry.
Russian scientist, mechanic N.E. Zhukovsky
Very common problems in entrance examinations in mathematics are problems related to the concept of arithmetic progression. To successfully solve such problems, you must have a good knowledge of the properties of arithmetic progression and have certain skills in their application.
Let us first recall the basic properties of an arithmetic progression and present the most important formulas, associated with this concept.
Definition. Number sequence, in which each subsequent term differs from the previous one by the same number, called an arithmetic progression. In this case, the numbercalled the progression difference.
For an arithmetic progression, the following formulas are valid:
, (1)
Where . Formula (1) is called the formula of the general term of an arithmetic progression, and formula (2) represents the main property of an arithmetic progression: each term of the progression coincides with the arithmetic mean of its neighboring terms and .
Note that it is precisely because of this property that the progression under consideration is called “arithmetic”.
The above formulas (1) and (2) are generalized as follows:
(3)
To calculate the amount first terms of an arithmetic progressionthe formula is usually used
(5) where and .
If we take into account the formula (1), then from formula (5) it follows
If we denote , then
Where . Since , formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).
In particular , from formula (5) it follows, What
Little known to most students is the property of arithmetic progression, formulated through the following theorem.
Theorem. If , then
Proof. If , then
The theorem has been proven.
For example , using the theorem, it can be shown that
Let's move on to consider typical examples of solving problems on the topic " Arithmetic progression».
Example 1. Let it be. Find .
Solution. Applying formula (6), we obtain . Since and , then or .
Example 2. Let it be three times greater, and when divided by the quotient, the result is 2 and the remainder is 8. Determine and .
Solution. From the conditions of the example, the system of equations follows
Since , , and , then from the system of equations (10) we obtain
The solution to this system of equations is and .
Example 3. Find if and .
Solution. According to formula (5) we have or . However, using property (9), we obtain .
Since and , then from the equality the equation follows or .
Example 4. Find if .
Solution.According to formula (5) we have
However, using the theorem, we can write
From here and from formula (11) we obtain .
Example 5. Given: . Find .
Solution. Since, then. However, therefore.
Example 6. Let , and . Find .
Solution. Using formula (9), we obtain . Therefore, if , then or .
Since and then here we have a system of equations
Solving which, we get and .
Natural root equations is .
Example 7. Find if and .
Solution. Since according to formula (3) we have that , then the system of equations follows from the problem conditions
If we substitute the expressioninto the second equation of the system, then we get or .
Roots quadratic equation are And .
Let's consider two cases.
1. Let , then . Since and , then .
In this case, according to formula (6), we have
2. If , then , and
Answer: and.
Example 8. It is known that and. Find .
Solution. Taking into account formula (5) and the condition of the example, we write and .
This implies the system of equations
If we multiply the first equation of the system by 2 and then add it to the second equation, we get
According to formula (9) we have. In this regard, it follows from (12) or .
Since and , then .
Answer: .
Example 9. Find if and .
Solution. Since , and by condition , then or .
From formula (5) it is known, What . Since, then.
Hence , here we have a system of linear equations
From here we get and . Taking into account formula (8), we write .
Example 10. Solve the equation.
Solution. From the given equation it follows that . Let us assume that , , and . In this case .
According to formula (1), we can write or .
Since , then equation (13) has the only suitable root .
Example 11. Find the maximum value provided that and .
Solution. Since , then the arithmetic progression under consideration is decreasing. In this regard, the expression takes on its maximum value when it is the number of the minimum positive term of the progression.
Let us use formula (1) and the fact, that and . Then we get that or .
Since , then or . However, in this inequalitylargest natural number, That's why .
If the values of , and are substituted into formula (6), we get .
Answer: .
Example 12. Determine the sum of all two-digit natural numbers, which when divided by 6 leaves a remainder of 5.
Solution. Let us denote by the set of all two-digit natural numbers, i.e. . Next, we will construct a subset consisting of those elements (numbers) of the set that, when divided by the number 6, give a remainder of 5.
Easy to install, What . Obviously , that the elements of the setform an arithmetic progression, in which and .
To establish the cardinality (number of elements) of the set, we assume that . Since and , it follows from formula (1) or . Taking into account formula (5), we obtain .
The above examples of problem solving can by no means claim to be exhaustive. This article is written based on the analysis modern methods solving typical problems on a given topic. For a more in-depth study of methods for solving problems related to arithmetic progression, it is advisable to refer to the list of recommended literature.
1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace and Education, 2013. – 608 p.
2. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. – 216 p.
3. Medynsky M.M. A complete course of elementary mathematics in problems and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. – 208 p.
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For example, the sequence \(2\); \(5\); \(8\); \(eleven\); \(14\)... is an arithmetic progression, because each subsequent element differs from the previous one by three (can be obtained from the previous one by adding three):
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 number. For example, in arithmetic progression \(16\); \(10\); \(4\); \(-2\); \(-8\)... the progression difference \(d\) 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.
Numbers that form a progression are called members(or elements).
They are denoted by the same letter as an 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\(2; 5; 8; 11; 14…\right\)\)
Solving arithmetic progression problems
In principle, the information presented above 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 an 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 know that it is an arithmetic progression. That is, each element differs from its neighbor by the same number. Let's find out which one by 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).
Given several consecutive elements of an arithmetic progression: \(…5; x; 10; 12.5...\) Find the value of the element designated 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 progression difference. Let's find it from two known neighboring elements: \(d=12.5-10=2.5\). |
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And now we can easily find what we are looking for: \(x=5+2.5=7.5\). |
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Ready. You can write an answer. |
Answer: \(7,5\).
Example (OGE).
Arithmetic progression is given 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 given only the first element. Therefore, we first calculate the values one by one, using what is 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 required amount has been found. |
Answer: \(S_6=9\).
Example (OGE).
In arithmetic progression \(a_(12)=23\); \(a_(16)=51\). Find the difference of this progression.
Solution:
Answer: \(d=7\).
Important formulas for arithmetic progression
As you can see, many problems on arithmetic progression can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each subsequent 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 \(b_(386)\). Do we need to add four \(385\) times? Or imagine that in the penultimate example you need to find the sum of the first seventy-three elements. You'll be tired of counting...
Therefore, in such cases they do not solve things “head-on”, but use special formulas derived for arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum of \(n\) first terms.
Formula of the \(n\)th term: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first term of the progression;
\(n\) – number of the required element;
\(a_n\) – term of the progression with number \(n\).
This formula allows us to quickly find even the three-hundredth or 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: \(b_(246)=1850\).
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 \(25\) terms 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 terms, 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 let's find the twenty-fifth term by 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 easily calculate the required amount. |
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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:
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 of \(n\) first elements;
\(a_1\) – the first summed term;
\(d\) – progression difference;
\(n\) – number of elements in total.
Example.
Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15.5\); \(14\)…
Solution:
Answer: \(S_(33)=-231\).
More complex arithmetic progression problems
Now you have everything necessary information for solving almost any arithmetic progression problem. Let’s finish the topic by considering problems in which you not only need to apply formulas, but also think a little (in mathematics this can be useful ☺)
Example (OGE).
Find the sum of all negative terms of the progression: \(-19.3\); \(-19\); \(-18.7\)…
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 begin to solve the same thing: first we find \(d\). |
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\(d=a_2-a_1=-19-(-19.3)=0.3\) |
Now we would like to substitute \(d\) into the formula for the sum... and here a small nuance emerges - we do not know \(n\). In other words, we don’t know how many terms will need to be added. How to find out? Let's think. We will stop adding elements when we reach 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 an 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 become 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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\(n-1>\)\(\frac(19.3)(0.3)\) |
We transfer minus one, not forgetting to change the signs |
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\(n>\)\(\frac(19.3)(0.3)\) \(+1\) |
Let's 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 one has \(n=65\). Just in case, let's check this. |
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\(n=65;\) \(a_(65)=-19.3+(65-1)·0.3=-0.1\) |
So 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 the \(26\)th to the \(42\) 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 elements, but starting not from the first, but from the \(26\)th. For such a case we do not have a 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 \(42\)-y elements. |
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\(S_(42)=\) \(\frac(2 \cdot (-33)+(42-1)4)(2)\)\(\cdot 42=\) |
Now the sum of the first \(25\) elements. |
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\(S_(25)=\) \(\frac(2 \cdot (-33)+(25-1)4)(2)\)\(\cdot 25=\) |
And finally, we calculate the answer. |
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\(S=S_(42)-S_(25)=2058-375=1683\) |
Answer: \(S=1683\).
For arithmetic progression, there are several more formulas that we did not consider in this article due to their low practical usefulness. However, you can easily find them.
What is the main essence of the formula?
This formula allows you to find any BY HIS NUMBER " n" .
Of course, you also need to know the first term a 1 and progression difference d, well, without these parameters you can’t write down a specific progression.
Memorizing (or cribing) this formula is not enough. You need to understand its essence and apply the formula in various problems. And don’t forget in right moment, but how not forget- I don't know. And here how to remember If necessary, I will definitely advise you. For those who complete the lesson to the end.)
So, let's look at the formula for the nth term of an arithmetic progression.
What is a formula in general? By the way, take a look if you haven’t read it. Everything is simple there. It remains to figure out what it is nth term.
Progression in general view 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- the 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 - s a 120.
How can we define it in general terms? any term of an arithmetic progression, with any number? Very simple! Like this:
a n
That's what it is nth term of an arithmetic progression. The letter n hides all the member numbers 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 progression. Using the notation a n, we can quickly find any member any arithmetic progression. And solve a bunch of other progression problems. You'll see for yourself further.
In the formula for the nth term of an arithmetic progression:
| a n = a 1 + (n-1)d |
a 1- the first term of an arithmetic progression;
n- member number.
The formula connects the key parameters of any progression: a n ; a 1 ; d And n. All progression problems revolve around these parameters.
The nth term formula can also be used to write a specific progression. For example, the problem may say that the progression is specified by the condition:
a n = 5 + (n-1) 2.
Such a problem can be a dead end... There is neither a series nor a difference... But, comparing the condition with the formula, it is easy to understand that in this progression a 1 =5, and d=2.
And it can be even worse!) If we take the same condition: a n = 5 + (n-1) 2, Yes, open the parentheses and bring similar ones? We get a new formula:
a n = 3 + 2n.
This Just not general, but for a specific progression. This is where the pitfall lurks. Some people think that the first term is a three. Although in reality the first term is five... A little lower we will work with such a modified formula.
In progression problems there is another notation - a n+1. This is, as you guessed, the “n plus first” term of the progression. Its meaning is simple and harmless.) This is a member of the progression whose number is greater than number n by one. For example, if in some problem we take a n fifth term then a n+1 will be the sixth member. Etc.
Most often the designation a n+1 found in recurrence formulas. Don't be afraid of this scary word!) This is just a way of expressing a member of an arithmetic progression through the previous one. Let's say we are given an arithmetic progression in this form, using a 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. How can we immediately count, say, the twentieth term? a 20? But there’s no way!) Until we find out the 19th term, we can’t count the 20th. This is it fundamental difference recurrent formula from the formula of the nth term. Recurrent works only through previous term, and the formula of the nth term is through first and allows straightaway find any member by its number. Without calculating the entire series of numbers in order.
In an arithmetic progression, it is easy to turn a recurrent formula 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 its usual form, and work with it. In the State Academy of Sciences, such tasks are often encountered.
Application of the formula for the nth term of an arithmetic progression.
First, let's look at direct application formulas. At the end of the previous lesson there was a problem:
An arithmetic progression (a n) is given. 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 an arithmetic progression. Add and add... An hour or two.)
And according to the formula, the solution will take less than a minute. You can time it.) Let's decide.
The conditions provide all the data for using the formula: a 1 =3, d=1/6. It remains to figure out what is equal n. No problem! We need to find a 121. So 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 ours n. This is the meaning n= 121 we will substitute further into the formula, in brackets. We substitute all the numbers into the formula and calculate:
a 121 = 3 + (121-1) 1/6 = 3+20 = 23
That's it. Just as quickly one could find the five hundred and tenth term, and the thousand and third, any one. We put instead n the desired number in the index of the letter " a" and in brackets, and we count.
Let me remind you the point: this formula allows you to find any arithmetic progression term BY HIS NUMBER " n" .
Let's solve the problem in a more cunning way. Let us come across 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 tell you the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Write down with your hands, 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... Is that it? If you think that’s it, then you won’t solve the problem, yes...
We still have a number n! In condition a 17 =-2 hidden two parameters. This is both the value of the seventeenth term (-2) and its number (17). Those. n=17. This “trifle” often slips past the head, and without it, (without the “trifle”, not the head!) the problem cannot be solved. Although... and without a head too.)
Now we can simply stupidly substitute our data into the formula:
a 17 = a 1 + (17-1)·(-0.5)
Oh yes, a 17 we know it's -2. Okay, let's substitute:
-2 = a 1 + (17-1)·(-0.5)
That's basically all. It remains to express the first term of the arithmetic progression from the formula and calculate it. The answer will be: a 1 = 6.
This technique - writing down a formula and simply substituting known data - is a great help in simple tasks. Well, of course, you must be able to express a variable from a formula, but what to do!? Without this skill, mathematics may not be studied at all...
Another popular puzzle:
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 are writing the formula!)
| a n = a 1 + (n-1)d |
Let's consider what we know: a 1 =2; a 15 =12; and (I’ll especially highlight!) n=15. Feel free to substitute this into the formula:
12=2 + (15-1)d
We do the arithmetic.)
12=2 + 14d
d=10/14 = 5/7
This is the correct answer.
So, the tasks for a n, a 1 And d decided. All that remains is to learn how to find the number:
The number 99 is a member of the arithmetic progression (a n), where a 1 =12; d=3. Find this member's number.
We substitute the quantities known to us 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- this is some member of the progression with a number n...And we know this member of the progression! It's 99. We don't know its number. n, So this number is what you need to find. We substitute the term of the progression 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 whether the number 117 is a member of the arithmetic progression (a n):
-3,6; -2,4; -1,2 ...
Let's write the formula again. What, there are no parameters? Hm... Why are we given eyes?) Do we see the first term of the progression? We see. This is -3.6. You can safely write: a 1 = -3.6. Difference d Can you tell from the series? It’s easy if you know what the difference of an arithmetic progression is:
d = -2.4 - (-3.6) = 1.2
So, we did the simplest thing. It remains to deal with the unknown number n and the 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... What to do!? Well, how to be, how to be... Turn on your creative abilities!)
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 can we draw? Yes! Number 117 is not member of our progression. It is somewhere between the one hundred and first and one hundred and second terms. If the number turned out natural, i.e. is a positive integer, then the number would be a member of the progression with the number found. And in our case, the answer to the problem will be: No.
Task based real option GIA:
The arithmetic progression is given by the condition:
a n = -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 for the nth term 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 the arithmetic progression in it hidden. It’s okay, we’ll find it now.)
Just as in previous problems, we substitute n=1 into this formula:
a 1 = -4 + 6.8 1 = 2.8
Here! The first term is 2.8, not -4!
We look for the tenth term in the same way:
a 10 = -4 + 6.8 10 = 64
That's it.
And now, for those who have read to these lines, the promised bonus.)
Suppose, in a difficult combat situation of the State Examination or Unified State Examination, you have forgotten the useful formula for the nth term of an arithmetic progression. I remember something, but somehow uncertainly... Or n there, or n+1, or n-1... How to be!?
Calm! This formula is easy to derive. It’s not very strict, but it’s definitely enough for confidence and the right decision!) To make a conclusion, it’s enough to remember the elementary meaning of an arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.
Draw a number line and mark the first one on it. second, third, etc. members. And we note the difference d between members. Like this:
We look at the picture and think: what does the second term equal? 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? It’s not for nothing that I highlight some words in bold. 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, to the number n, number of spaces will n-1. Therefore, the formula will be (without variations!):
| 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 insert a picture into the equation...
Tasks for independent solution.
To warm up:
1. In arithmetic progression (a n) a 2 =3; a 5 =5.1. Find a 3 .
Hint: according to the picture, the problem can be 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 using both the picture and the formula. Feel the difference!)
And this is no longer a warm-up.)
2. In arithmetic progression (a n) a 85 =19.1; a 236 =49, 3. Find a 3 .
What, you don’t want to draw a picture?) Of course! Better according to the formula, yes...
3. The arithmetic progression is given by the condition:a 1 = -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 specified in a recurrent manner. But counting to the one hundred and twenty-fifth term... Not everyone can do such a feat.) But the formula for 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 conditions of task 4, find the sum of the smallest positive and largest negative terms 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...) The “fingertip” method won’t work here. You will 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? Happens. By the way, there is one subtle point in the last task. Care will be required when reading the problem. And logic.
The solution to all these problems is discussed in detail in Section 555. And the element of fantasy for the fourth, and the subtle point for the sixth, and general approaches for solving any problems involving the formula of the nth term - everything is described. I recommend.
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You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
Lesson type: learning new material.
Lesson objectives:
- expanding and deepening students’ understanding of problems solved using arithmetic progression; organizing students' search activities when deriving the formula for the sum of the first n terms of an arithmetic progression;
- developing the ability to independently acquire new knowledge and use already acquired knowledge to achieve a given task;
- developing the desire and need to generalize the facts obtained, developing independence.
Tasks:
- summarize and systematize existing knowledge on the topic “Arithmetic progression”;
- derive formulas for calculating the sum of the first n terms of an arithmetic progression;
- teach how to apply the obtained formulas when solving various problems;
- draw students' attention to the procedure for finding the value of a numerical expression.
Equipment:
- cards with tasks for working in groups and pairs;
- evaluation paper;
- presentation“Arithmetic progression.”
I. Updating of basic knowledge.
1. Independent work in pairs.
1st option:
Define arithmetic progression. Write down a recurrence formula that defines an arithmetic progression. Please provide an example of an arithmetic progression and indicate its difference.
2nd option:
Write down the formula for the nth term of an arithmetic progression. Find the 100th term of the arithmetic progression ( a n}: 2, 5, 8 …
At this time, two students back side boards are preparing answers to these same questions.
Students evaluate their partner's work by checking them on the board. (Sheets with answers are handed in.)
2. Game moment.
Exercise 1.
Teacher. I thought of some arithmetic progression. Ask me only two questions so that after the answers you can quickly name the 7th term of this progression. (1, 3, 5, 7, 9, 11, 13, 15…)
Questions from students.
- What is the sixth term of the progression and what is the difference?
- What is the eighth term of the progression and what is the difference?
If there are no more questions, then the teacher can stimulate them - a “ban” on d (difference), that is, it is not allowed to ask what the difference is equal to. You can ask questions: what is the 6th term of the progression equal to and what is the 8th term of the progression equal to?
Task 2.
There are 20 numbers written on the board: 1, 4, 7 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58.
The teacher stands with his back to the board. Students call out the number, and the teacher instantly calls out the number itself. Explain how I can do this?
The teacher remembers the formula for the nth term a n = 3n – 2 and, substituting the specified values n, finds the corresponding values a n.
II. Setting a learning task.
I propose to solve an ancient problem dating back to the 2nd millennium BC, found in Egyptian papyri.
Task:“Let it be said to you: divide 10 measures of barley among 10 people, the difference between each person and his neighbor is 1/8 of the measure.”
- How is this problem related to the topic arithmetic progression? (Each next person receives 1/8 of the measure more, which means the difference is d=1/8, 10 people, which means n=10.)
- What do you think the number 10 measures means? (Sum of all terms of the progression.)
- What else do you need to know to make it easy and simple to divide the barley according to the conditions of the problem? (First term of progression.)
Lesson Objective– obtaining the dependence of the sum of the terms of the progression on their number, the first term and the difference, and checking whether the problem was solved correctly in ancient times.
Before we deduce the formula, let's look at how the ancient Egyptians solved the problem.
And they solved it as follows:
1) 10 measures: 10 = 1 measure – average share;
2) 1 measure ∙ = 2 measures – doubled average share.
Doubled average share is the sum of the shares of the 5th and 6th person.
3) 2 measures – 1/8 measures = 1 7/8 measures – double the share of the fifth person.
4) 1 7/8: 2 = 5/16 – fraction of a fifth; and so on, you can find the share of each previous and subsequent person.
We get the sequence:
III. Solving the problem.
1. Work in groups
Group I: Find the sum of 20 consecutive natural numbers: S 20 =(20+1)∙10 =210.
In general 
II group: Find the sum of natural numbers from 1 to 100 (The Legend of Little Gauss).
S 100 = (1+100)∙50 = 5050
Conclusion:
III group: Find the sum of natural numbers from 1 to 21.
Solution: 1+21=2+20=3+19=4+18…
Conclusion:
IV group: Find the sum of natural numbers from 1 to 101.
Conclusion:
This method of solving the problems considered is called the “Gauss Method”.
2. Each group presents the solution to the problem on the board.
3. Generalization of the proposed solutions for an arbitrary arithmetic progression:
a 1, a 2, a 3,…, a n-2, a n-1, a n.
S n =a 1 + a 2 + a 3 + a 4 +…+ a n-3 + a n-2 + a n-1 + a n.
Let's find this sum using similar reasoning:
4. Have we solved the problem?(Yes.)
IV. Primary understanding and application of the obtained formulas when solving problems.
1. Checking the solution to an ancient problem using the formula.
2. Application of the formula in solving various problems.
3. Exercises to develop the ability to apply formulas when solving problems.
A) No. 613
Given: ( a n) – arithmetic progression;
(a n): 1, 2, 3, …, 1500
Find: S 1500
Solution: , a 1 = 1, and 1500 = 1500,
B) Given: ( a n) – arithmetic progression;
(a n): 1, 2, 3, …
S n = 210
Find: n
Solution:
V. Independent work with mutual verification.
Denis started working as a courier. In the first month his salary was 200 rubles, in each subsequent month it increased by 30 rubles. How much did he earn in total in a year?
Given: ( a n) – arithmetic progression;
a 1 = 200, d=30, n=12
Find: S 12
Solution:
Answer: Denis received 4380 rubles for the year.
VI. Homework instruction.
- Section 4.3 – learn the derivation of the formula.
- №№ 585, 623 .
- Create a problem that can be solved using the formula for the sum of the first n terms of an arithmetic progression.
VII. Summing up the lesson.
1. Score sheet
2. Continue the sentences
- Today in class I learned...
- Formulas learned...
- I believe that …
3. Can you find the sum of numbers from 1 to 500? What method will you use to solve this problem?
Bibliography.
1. Algebra, 9th grade. Textbook for general education institutions. Ed. G.V. Dorofeeva. M.: “Enlightenment”, 2009.