Parallelogram definition drawing the main elements of the property. Definition of a parallelogram and its properties
Yes, yes: the arithmetic progression is not a toy for you :) Well, friends, if you are reading this text, then the internal cap-obviousness tells me that you do not yet know what an arithmetic progression is, but you really (no, like this: SOOOOO!) Want to know. Therefore, I will not torment you with long introductions and get straight to the point.
Let's start with a couple of examples. Consider several sets of numbers:
- 1; 2; 3; 4; ...
- 15; 20; 25; 30; ...
- $ \ sqrt (2); \ 2 \ sqrt (2); \ 3 \ sqrt (2); ... $
What do all these sets have in common? At first glance, nothing. But actually there is something. Namely: each next element differs from the previous one by the same number.
Judge for yourself. The first set is simply consecutive numbers, each next one more than the previous one. In the second case, the difference between the adjacent numbers is already equal to five, but this difference is still constant. In the third case, roots in general. However, $ 2 \ sqrt (2) = \ sqrt (2) + \ sqrt (2) $, and $ 3 \ sqrt (2) = 2 \ sqrt (2) + \ sqrt (2) $, i.e. and in this case, each next element simply increases by $ \ sqrt (2) $ (and don't be afraid that this number is irrational).
So: all such sequences are called arithmetic progressions. Let's give a strict definition:
Definition. A sequence of numbers in which each next differs from the previous by exactly the same amount is called arithmetic progression... The very amount by which the numbers differ is called the difference of the progression and is most often denoted by the letter $ d $.
Designation: $ \ left (((a) _ (n)) \ right) $ - the progression itself, $ d $ - its difference.
And just a couple of important remarks. First, only orderly sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. You cannot rearrange or swap numbers.
Secondly, the sequence itself can be either finite or infinite. For example, the set (1; 2; 3) is obviously a finite arithmetic progression. But if you write something in the spirit (1; 2; 3; 4; ...) - this is already an endless progression. The ellipsis after the four, as it were, hints that there are still quite a few numbers going on. Infinitely many, for example. :)
I would also like to note that progressions are increasing and decreasing. We have already seen the increasing ones - the same set (1; 2; 3; 4; ...). And here are examples of decreasing progressions:
- 49; 41; 33; 25; 17; ...
- 17,5; 12; 6,5; 1; −4,5; −10; ...
- $ \ sqrt (5); \ \ sqrt (5) -1; \ \ sqrt (5) -2; \ \ sqrt (5) -3; ... $
OK OK: last example may seem overly complicated. But the rest, I think, is clear to you. Therefore, we will introduce new definitions:
Definition. An arithmetic progression is called:
- increasing if each next element is larger than the previous one;
- decreasing if, on the contrary, each subsequent element is less than the previous one.
In addition, there are so-called "stationary" sequences - they consist of the same repeating number. For example, (3; 3; 3; ...).
There remains only one question: how to distinguish an increasing progression from a decreasing one? Fortunately, it all depends on the sign of the number $ d $, i.e. difference progression:
- If $ d \ gt 0 $, then the progression is increasing;
- If $ d \ lt 0 $, then the progression is obviously decreasing;
- Finally, there is the case $ d = 0 $ - in this case the whole progression is reduced to a stationary sequence of identical numbers: (1; 1; 1; 1; ...), etc.
Let's try to calculate the difference $ d $ for the three decreasing progressions given above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract the number on the left from the number on the right. It will look like this:
- 41−49=−8;
- 12−17,5=−5,5;
- $ \ sqrt (5) -1- \ sqrt (5) = - 1 $.
As you can see, in all three cases the difference really turned out to be negative. And now that we have more or less figured out the definitions, it's time to figure out how progressions are described and what their properties are.
Progression members and recurrent formula
Since the elements of our sequences cannot be swapped, they can be numbered:
\ [\ left (((a) _ (n)) \ right) = \ left \ (((a) _ (1)), \ ((a) _ (2)), ((a) _ (3 )), ... \ right \) \]
The individual elements of this set are called members of the progression. They are indicated by a number: the first term, the second term, etc.
In addition, as we already know, the neighboring members of the progression are related by the formula:
\ [((a) _ (n)) - ((a) _ (n-1)) = d \ Rightarrow ((a) _ (n)) = ((a) _ (n-1)) + d \]
In short, to find the $ n $ th term in the progression, you need to know the $ n-1 $ th term and the $ d $ difference. Such a formula is called recurrent, because with its help you can find any number, only knowing the previous one (and in fact - all the previous ones). This is very inconvenient, so there is a more tricky formula that reduces any calculations to the first term and the difference:
\ [((a) _ (n)) = ((a) _ (1)) + \ left (n-1 \ right) d \]
Surely you have already met this formula. They love to give it in all sorts of reference books and reshebniks. And in any sensible textbook on mathematics, she goes one of the first.
However, I suggest we practice a little.
Problem number 1. Write out the first three terms of the arithmetic progression $ \ left (((a) _ (n)) \ right) $, if $ ((a) _ (1)) = 8, d = -5 $.
Solution. So, we know the first term $ ((a) _ (1)) = 8 $ and the difference of the progression $ d = -5 $. Let's use the formula just given and substitute $ n = 1 $, $ n = 2 $ and $ n = 3 $:
\ [\ begin (align) & ((a) _ (n)) = ((a) _ (1)) + \ left (n-1 \ right) d; \\ & ((a) _ (1)) = ((a) _ (1)) + \ left (1-1 \ right) d = ((a) _ (1)) = 8; \\ & ((a) _ (2)) = ((a) _ (1)) + \ left (2-1 \ right) d = ((a) _ (1)) + d = 8-5 = 3; \\ & ((a) _ (3)) = ((a) _ (1)) + \ left (3-1 \ right) d = ((a) _ (1)) + 2d = 8-10 = -2. \\ \ end (align) \]
Answer: (8; 3; −2)
That's all! Please note: our progression is decreasing.
Of course, $ n = 1 $ could not have been substituted - the first term is already known to us. However, substituting one, we made sure that our formula works even for the first term. In other cases, it all boiled down to trivial arithmetic.
Problem number 2. Write down the first three terms of the arithmetic progression if its seventh term is −40 and the seventeenth term is −50.
Solution. Let's write down the condition of the problem in the usual terms:
\ [((a) _ (7)) = - 40; \ quad ((a) _ (17)) = - 50. \]
\ [\ left \ (\ begin (align) & ((a) _ (7)) = ((a) _ (1)) + 6d \\ & ((a) _ (17)) = ((a) _ (1)) + 16d \\ \ end (align) \ right. \]
\ [\ left \ (\ begin (align) & ((a) _ (1)) + 6d = -40 \\ & ((a) _ (1)) + 16d = -50 \\ \ end (align) \ right. \]
I put the sign of the system because these requirements must be fulfilled simultaneously. And now note that if we subtract the first from the second equation (we have the right to do this, since we have a system), we get this:
\ [\ begin (align) & ((a) _ (1)) + 16d- \ left (((a) _ (1)) + 6d \ right) = - 50- \ left (-40 \ right); \\ & ((a) _ (1)) + 16d - ((a) _ (1)) - 6d = -50 + 40; \\ & 10d = -10; \\ & d = -1. \\ \ end (align) \]
That's how easy we found the difference in the progression! It remains to substitute the found number into any of the equations of the system. For example, in the first:
\ [\ begin (matrix) ((a) _ (1)) + 6d = -40; \ quad d = -1 \\ \ Downarrow \\ ((a) _ (1)) - 6 = -40; \\ ((a) _ (1)) = - 40 + 6 = -34. \\ \ end (matrix) \]
Now, knowing the first term and the difference, it remains to find the second and third terms:
\ [\ begin (align) & ((a) _ (2)) = ((a) _ (1)) + d = -34-1 = -35; \\ & ((a) _ (3)) = ((a) _ (1)) + 2d = -34-2 = -36. \\ \ end (align) \]
Ready! The problem has been solved.
Answer: (-34; -35; -36)
Pay attention to an interesting property of the progression that we discovered: if we take the $ n $ th and $ m $ th terms and subtract them from each other, then we get the difference of the progression multiplied by the number $ n-m $:
\ [((a) _ (n)) - ((a) _ (m)) = d \ cdot \ left (n-m \ right) \]
Simple but very useful property, which you definitely need to know - with its help, you can significantly speed up the solution of many problems in progressions. Here's a prime example:
Problem number 3. The fifth term of the arithmetic progression is 8.4, and its tenth term is 14.4. Find the fifteenth term of this progression.
Solution. Since $ ((a) _ (5)) = 8.4 $, $ ((a) _ (10)) = 14.4 $, and you need to find $ ((a) _ (15)) $, then we note following:
\ [\ begin (align) & ((a) _ (15)) - ((a) _ (10)) = 5d; \\ & ((a) _ (10)) - ((a) _ (5)) = 5d. \\ \ end (align) \]
But by condition $ ((a) _ (10)) - ((a) _ (5)) = 14.4-8.4 = $ 6, therefore $ 5d = $ 6, whence we have:
\ [\ begin (align) & ((a) _ (15)) - 14.4 = 6; \\ & ((a) _ (15)) = 6 + 14.4 = 20.4. \\ \ end (align) \]
Answer: 20.4
That's all! We did not need to compose some systems of equations and calculate the first term and the difference - everything was solved in just a couple of lines.
Now let's consider another type of tasks - to find negative and positive members of the progression. It is no secret that if the progression increases, while the first term is negative, then sooner or later positive terms will appear in it. And on the contrary: the members of the decreasing progression will sooner or later become negative.
At the same time, it is far from always possible to grope this moment "head-on", sequentially going through the elements. Often, problems are designed in such a way that without knowing the formulas, the calculations would take several sheets - we would just fall asleep while we found the answer. Therefore, we will try to solve these problems in a faster way.
Problem number 4. How many negative terms are in the arithmetic progression -38.5; −35.8; ...?
Solution. So, $ ((a) _ (1)) = - 38.5 $, $ ((a) _ (2)) = - 35.8 $, from where we immediately find the difference:
Note that the difference is positive, so the progression increases. The first term is negative, so at some point we really will stumble upon positive numbers. The only question is when it will happen.
Let's try to find out: how long (i.e. up to what natural number $ n $) the negativity of the terms is preserved:
\ [\ begin (align) & ((a) _ (n)) \ lt 0 \ Rightarrow ((a) _ (1)) + \ left (n-1 \ right) d \ lt 0; \\ & -38.5+ \ left (n-1 \ right) \ cdot 2.7 \ lt 0; \ quad \ left | \ cdot 10 \ right. \\ & -385 + 27 \ cdot \ left (n-1 \ right) \ lt 0; \\ & -385 + 27n-27 \ lt 0; \\ & 27n \ lt 412; \\ & n \ lt 15 \ frac (7) (27) \ Rightarrow ((n) _ (\ max)) = 15. \\ \ end (align) \]
The last line requires clarification. So, we know that $ n \ lt 15 \ frac (7) (27) $. On the other hand, we will be satisfied with only integer values of the number (moreover: $ n \ in \ mathbb (N) $), so the largest allowed number is exactly $ n = 15 $, and in no case is 16.
Problem number 5. In arithmetic progression $ (() _ (5)) = - 150, (() _ (6)) = - 147 $. Find the number of the first positive term of this progression.
It would be exactly the same problem as the previous one, but we don't know $ ((a) _ (1)) $. But the neighboring terms are known: $ ((a) _ (5)) $ and $ ((a) _ (6)) $, so we can easily find the difference of the progression:
In addition, we will try to express the fifth term in terms of the first and the difference according to the standard formula:
\ [\ begin (align) & ((a) _ (n)) = ((a) _ (1)) + \ left (n-1 \ right) \ cdot d; \\ & ((a) _ (5)) = ((a) _ (1)) + 4d; \\ & -150 = ((a) _ (1)) + 4 \ cdot 3; \\ & ((a) _ (1)) = - 150-12 = -162. \\ \ end (align) \]
Now we proceed by analogy with the previous task. We find out at what point in our sequence there will be positive numbers:
\ [\ begin (align) & ((a) _ (n)) = - 162+ \ left (n-1 \ right) \ cdot 3 \ gt 0; \\ & -162 + 3n-3 \ gt 0; \\ & 3n \ gt 165; \\ & n \ gt 55 \ Rightarrow ((n) _ (\ min)) = 56. \\ \ end (align) \]
The smallest integer solution to this inequality is 56.
Please note: in the last task, everything was reduced to a strict inequality, so the $ n = 55 $ option will not suit us.
Now that we have learned how to solve simple problems, let's move on to more complex ones. But first, let's study another very useful property of arithmetic progressions, which will save us a lot of time and unequal cells in the future. :)
Arithmetic mean and equal indents
Consider several consecutive members of the increasing arithmetic progression $ \ left (((a) _ (n)) \ right) $. Let's try to mark them on the number line:
Members of an arithmetic progression on a number lineI specifically noted arbitrary terms $ ((a) _ (n-3)), ..., ((a) _ (n + 3)) $, not any $ ((a) _ (1)) , \ ((a) _ (2)), \ ((a) _ (3)) $, etc. Because the rule, which I will now talk about, works the same for any "segments".
And the rule is very simple. Let's remember the recurrence formula and write it down for all marked members:
\ [\ begin (align) & ((a) _ (n-2)) = ((a) _ (n-3)) + d; \\ & ((a) _ (n-1)) = ((a) _ (n-2)) + d; \\ & ((a) _ (n)) = ((a) _ (n-1)) + d; \\ & ((a) _ (n + 1)) = ((a) _ (n)) + d; \\ & ((a) _ (n + 2)) = ((a) _ (n + 1)) + d; \\ \ end (align) \]
However, these equalities can be rewritten differently:
\ [\ begin (align) & ((a) _ (n-1)) = ((a) _ (n)) - d; \\ & ((a) _ (n-2)) = ((a) _ (n)) - 2d; \\ & ((a) _ (n-3)) = ((a) _ (n)) - 3d; \\ & ((a) _ (n + 1)) = ((a) _ (n)) + d; \\ & ((a) _ (n + 2)) = ((a) _ (n)) + 2d; \\ & ((a) _ (n + 3)) = ((a) _ (n)) + 3d; \\ \ end (align) \]
Well, so what? And the fact that the terms $ ((a) _ (n-1)) $ and $ ((a) _ (n + 1)) $ lie at the same distance from $ ((a) _ (n)) $. And this distance is equal to $ d $. The same can be said about the members $ ((a) _ (n-2)) $ and $ ((a) _ (n + 2)) $ - they are also removed from $ ((a) _ (n)) $ the same distance equal to $ 2d $. You can continue indefinitely, but the meaning is well illustrated by the picture.
The members of the progression lie at the same distance from the center What does this mean for us? This means that you can find $ ((a) _ (n)) $ if the neighboring numbers are known:
\ [((a) _ (n)) = \ frac (((a) _ (n-1)) + ((a) _ (n + 1))) (2) \]
We came up with an excellent statement: every member of the arithmetic progression is equal to the arithmetic mean of the neighboring terms! Moreover: we can deviate from our $ ((a) _ (n)) $ left and right not one step, but $ k $ steps - and still the formula will be correct:
\ [((a) _ (n)) = \ frac (((a) _ (n-k)) + ((a) _ (n + k))) (2) \]
Those. we can easily find some $ ((a) _ (150)) $ if we know $ ((a) _ (100)) $ and $ ((a) _ (200)) $, because $ (( a) _ (150)) = \ frac (((a) _ (100)) + ((a) _ (200))) (2) $. At first glance, it may seem that this fact does not give us anything useful. However, in practice, many problems are specially "sharpened" for the use of the arithmetic mean. Take a look:
Problem number 6. Find all values of $ x $ for which the numbers $ -6 ((x) ^ (2)) $, $ x + 1 $ and $ 14 + 4 ((x) ^ (2)) $ are consecutive members of the arithmetic progression (in order).
Solution. Since the indicated numbers are members of the progression, the condition of the arithmetic mean is satisfied for them: the central element $ x + 1 $ can be expressed in terms of adjacent elements:
\ [\ begin (align) & x + 1 = \ frac (-6 ((x) ^ (2)) + 14 + 4 ((x) ^ (2))) (2); \\ & x + 1 = \ frac (14-2 ((x) ^ (2))) (2); \\ & x + 1 = 7 - ((x) ^ (2)); \\ & ((x) ^ (2)) + x-6 = 0. \\ \ end (align) \]
The result is a classic quadratic equation. Its roots: $ x = 2 $ and $ x = -3 $ - these are the answers.
Answer: −3; 2.
Problem number 7. Find the $$ values for which the numbers $ -1; 4-3; (() ^ (2)) + 1 $ make an arithmetic progression (in that order).
Solution. Again, we express the middle term in terms of the arithmetic mean of the neighboring terms:
\ [\ begin (align) & 4x-3 = \ frac (x-1 + ((x) ^ (2)) + 1) (2); \\ & 4x-3 = \ frac (((x) ^ (2)) + x) (2); \ quad \ left | \ cdot 2 \ right .; \\ & 8x-6 = ((x) ^ (2)) + x; \\ & ((x) ^ (2)) - 7x + 6 = 0. \\ \ end (align) \]
Again the quadratic equation. And again there are two roots: $ x = 6 $ and $ x = 1 $.
Answer: 1; 6.
If in the process of solving a problem you get out some brutal numbers, or you are not completely sure of the correctness of the answers found, then there is a wonderful technique that allows you to check: did we solve the problem correctly?
For example, in problem no. 6 we received answers -3 and 2. How to check that these answers are correct? Let's just plug them in and see what happens. Let me remind you that we have three numbers ($ -6 (() ^ (2)) $, $ + 1 $ and $ 14 + 4 (() ^ (2)) $), which must form an arithmetic progression. Substitute $ x = -3 $:
\ [\ begin (align) & x = -3 \ Rightarrow \\ & -6 ((x) ^ (2)) = - 54; \\ & x + 1 = -2; \\ & 14 + 4 ((x) ^ (2)) = 50. \ end (align) \]
Received numbers -54; −2; 50, which differ by 52, is undoubtedly an arithmetic progression. The same thing happens for $ x = 2 $:
\ [\ begin (align) & x = 2 \ Rightarrow \\ & -6 ((x) ^ (2)) = - 24; \\ & x + 1 = 3; \\ & 14 + 4 ((x) ^ (2)) = 30. \ end (align) \]
Again a progression, but with a difference of 27. Thus, the problem is solved correctly. Those interested can check the second problem on their own, but I'll say right away: everything is correct there too.
In general, while solving the last problems, we stumbled upon one more interesting fact, which also needs to be remembered:
If three numbers are such that the second is the arithmetic mean of the first and the last, then these numbers form an arithmetic progression.
In the future, understanding this statement will allow us to literally "construct" the necessary progressions, based on the condition of the problem. But before we get down to such "construction", we should pay attention to one more fact, which directly follows from what has already been considered.
Grouping and sum of elements
Let's go back to the number axis again. Let us note there several members of the progression, between which, perhaps. there are a lot of other members:
The number line has 6 elements markedLet's try to express "left tail" in terms of $ ((a) _ (n)) $ and $ d $, and "right tail" in terms of $ ((a) _ (k)) $ and $ d $. It's very simple:
\ [\ begin (align) & ((a) _ (n + 1)) = ((a) _ (n)) + d; \\ & ((a) _ (n + 2)) = ((a) _ (n)) + 2d; \\ & ((a) _ (k-1)) = ((a) _ (k)) - d; \\ & ((a) _ (k-2)) = ((a) _ (k)) - 2d. \\ \ end (align) \]
Now, note that the following sums are equal:
\ [\ begin (align) & ((a) _ (n)) + ((a) _ (k)) = S; \\ & ((a) _ (n + 1)) + ((a) _ (k-1)) = ((a) _ (n)) + d + ((a) _ (k)) - d = S; \\ & ((a) _ (n + 2)) + ((a) _ (k-2)) = ((a) _ (n)) + 2d + ((a) _ (k)) - 2d = S. \ end (align) \]
Simply put, if we consider as a start two elements of the progression, which in total are equal to some $ S $ number, and then we begin to walk from these elements in opposite directions (towards each other or vice versa to move away), then the sums of the elements that we will stumble upon will also be equal$ S $. This can be most clearly represented graphically:
Equal indentation gives equal amounts Understanding this fact will allow us to solve problems fundamentally more high level difficulties than those that we considered above. For example, such:
Problem number 8. Determine the difference of the arithmetic progression in which the first term is 66, and the product of the second and twelfth terms is the smallest possible.
Solution. Let's write down everything we know:
\ [\ begin (align) & ((a) _ (1)) = 66; \\ & d =? \\ & ((a) _ (2)) \ cdot ((a) _ (12)) = \ min. \ end (align) \]
So, we do not know the difference of the progression $ d $. Actually, the whole solution will be built around the difference, since the product $ ((a) _ (2)) \ cdot ((a) _ (12)) $ can be rewritten as follows:
\ [\ begin (align) & ((a) _ (2)) = ((a) _ (1)) + d = 66 + d; \\ & ((a) _ (12)) = ((a) _ (1)) + 11d = 66 + 11d; \\ & ((a) _ (2)) \ cdot ((a) _ (12)) = \ left (66 + d \ right) \ cdot \ left (66 + 11d \ right) = \\ & = 11 \ cdot \ left (d + 66 \ right) \ cdot \ left (d + 6 \ right). \ end (align) \]
For those in the tank: I took out the common factor of 11 from the second parenthesis. Thus, the sought product is a quadratic function with respect to the variable $ d $. Therefore, consider the function $ f \ left (d \ right) = 11 \ left (d + 66 \ right) \ left (d + 6 \ right) $ - its graph will be a parabola with branches up, since if we expand the brackets, then we get:
\ [\ begin (align) & f \ left (d \ right) = 11 \ left (((d) ^ (2)) + 66d + 6d + 66 \ cdot 6 \ right) = \\ & = 11 (( d) ^ (2)) + 11 \ cdot 72d + 11 \ cdot 66 \ cdot 6 \ end (align) \]
As you can see, the coefficient at the leading term is 11 - this is a positive number, so we really are dealing with a parabola with branches up:
schedule quadratic function- parabola
Please note: this parabola takes its minimum value at its vertex with the abscissa $ ((d) _ (0)) $. Of course, we can calculate this abscissa according to the standard scheme (there is also the formula $ ((d) _ (0)) = (- b) / (2a) \; $), but it would be much more reasonable to notice that the desired vertex lies on the axis symmetry of the parabola, so the point $ ((d) _ (0)) $ is equidistant from the roots of the equation $ f \ left (d \ right) = 0 $:
\ [\ begin (align) & f \ left (d \ right) = 0; \\ & 11 \ cdot \ left (d + 66 \ right) \ cdot \ left (d + 6 \ right) = 0; \\ & ((d) _ (1)) = - 66; \ quad ((d) _ (2)) = - 6. \\ \ end (align) \]
That is why I was in no hurry to open the brackets: in the original form, the roots were very, very easy to find. Therefore, the abscissa is equal to the mean arithmetic numbers−66 and −6:
\ [((d) _ (0)) = \ frac (-66-6) (2) = - 36 \]
What does the discovered number give us? With it, the required product takes on the smallest value (by the way, we haven't counted $ ((y) _ (\ min)) $ - we don't need this). At the same time, this number is the difference between the initial progression, i.e. we found the answer. :)
Answer: −36
Problem number 9. Insert three numbers between the numbers $ - \ frac (1) (2) $ and $ - \ frac (1) (6) $ so that they together with the given numbers form an arithmetic progression.
Solution. Basically, we need to make a sequence of five numbers, with the first and last numbers already known. Let's denote the missing numbers by the variables $ x $, $ y $ and $ z $:
\ [\ left (((a) _ (n)) \ right) = \ left \ (- \ frac (1) (2); x; y; z; - \ frac (1) (6) \ right \ ) \]
Note that the number $ y $ is the "middle" of our sequence - it is equidistant from both the numbers $ x $ and $ z $, and from the numbers $ - \ frac (1) (2) $ and $ - \ frac (1) ( 6) $. And if from the numbers $ x $ and $ z $ we are in this moment cannot get $ y $, then the situation is different with the ends of the progression. Remembering the arithmetic mean:
Now, knowing $ y $, we will find the remaining numbers. Note that $ x $ lies between the numbers $ - \ frac (1) (2) $ and the $ y = - \ frac (1) (3) $ just found. That's why
Reasoning similarly, we find the remaining number:
Ready! We found all three numbers. Let's write them down in the answer in the order in which they should be inserted between the original numbers.
Answer: $ - \ frac (5) (12); \ - \ frac (1) (3); \ - \ frac (1) (4) $
Problem number 10. Insert several numbers between the numbers 2 and 42, which together with these numbers form an arithmetic progression, if you know that the sum of the first, second and last of the inserted numbers is 56.
Solution. An even more difficult task, which, however, is solved according to the same scheme as the previous ones - through the arithmetic mean. The problem is that we don't know exactly how many numbers to insert. Therefore, for definiteness, let us assume that after inserting everything there will be exactly $ n $ numbers, and the first of them is 2, and the last is 42. In this case, the sought arithmetic progression can be represented as:
\ [\ left (((a) _ (n)) \ right) = \ left \ (2; ((a) _ (2)); ((a) _ (3)); ...; (( a) _ (n-1)); 42 \ right \) \]
\ [((a) _ (2)) + ((a) _ (3)) + ((a) _ (n-1)) = 56 \]
Note, however, that the numbers $ ((a) _ (2)) $ and $ ((a) _ (n-1)) $ are obtained from the numbers 2 and 42 at the edges by one step towards each other, i.e. ... to the center of the sequence. This means that
\ [((a) _ (2)) + ((a) _ (n-1)) = 2 + 42 = 44 \]
But then the expression written above can be rewritten as follows:
\ [\ begin (align) & ((a) _ (2)) + ((a) _ (3)) + ((a) _ (n-1)) = 56; \\ & \ left (((a) _ (2)) + ((a) _ (n-1)) \ right) + ((a) _ (3)) = 56; \\ & 44 + ((a) _ (3)) = 56; \\ & ((a) _ (3)) = 56-44 = 12. \\ \ end (align) \]
Knowing $ ((a) _ (3)) $ and $ ((a) _ (1)) $, we can easily find the difference of the progression:
\ [\ begin (align) & ((a) _ (3)) - ((a) _ (1)) = 12 - 2 = 10; \\ & ((a) _ (3)) - ((a) _ (1)) = \ left (3-1 \ right) \ cdot d = 2d; \\ & 2d = 10 \ Rightarrow d = 5. \\ \ end (align) \]
It remains only to find the rest of the members:
\ [\ begin (align) & ((a) _ (1)) = 2; \\ & ((a) _ (2)) = 2 + 5 = 7; \\ & ((a) _ (3)) = 12; \\ & ((a) _ (4)) = 2 + 3 \ cdot 5 = 17; \\ & ((a) _ (5)) = 2 + 4 \ cdot 5 = 22; \\ & ((a) _ (6)) = 2 + 5 \ cdot 5 = 27; \\ & ((a) _ (7)) = 2 + 6 \ cdot 5 = 32; \\ & ((a) _ (8)) = 2 + 7 \ cdot 5 = 37; \\ & ((a) _ (9)) = 2 + 8 \ cdot 5 = 42; \\ \ end (align) \]
Thus, already at the 9th step we will come to the left end of the sequence - the number 42. In total, it was necessary to insert only 7 numbers: 7; 12; 17; 22; 27; 32; 37.
Answer: 7; 12; 17; 22; 27; 32; 37
Word problems with progressions
In conclusion, I would like to consider a couple of relatively simple tasks... Well, how simple: for most students who study mathematics at school and have not read what is written above, these tasks may seem like a tin. Nevertheless, it is precisely such problems that come across in the OGE and USE in mathematics, so I recommend that you familiarize yourself with them.
Problem number 11. The brigade produced 62 parts in January, and in each next month it produced 14 more parts than in the previous one. How many parts did the team make in November?
Solution. Obviously, the number of parts, scheduled by month, will represent an increasing arithmetic progression. Moreover:
\ [\ begin (align) & ((a) _ (1)) = 62; \ quad d = 14; \\ & ((a) _ (n)) = 62+ \ left (n-1 \ right) \ cdot 14. \\ \ end (align) \]
November is the 11th month of the year, so we need to find $ ((a) _ (11)) $:
\ [((a) _ (11)) = 62 + 10 \ cdot 14 = 202 \]
Consequently, 202 parts will be manufactured in November.
Problem number 12. The bookbinding workshop bound 216 books in January, and each next month it bound 4 more books than the previous one. How many books did the workshop bind in December?
Solution. All the same:
$ \ begin (align) & ((a) _ (1)) = 216; \ quad d = 4; \\ & ((a) _ (n)) = 216+ \ left (n-1 \ right) \ cdot 4. \\ \ end (align) $
December is the last, 12th month of the year, so we are looking for $ ((a) _ (12)) $:
\ [((a) _ (12)) = 216 + 11 \ cdot 4 = 260 \]
This is the answer - 260 books will be bound in December.
Well, if you have read this far, I hasten to congratulate you: you have successfully completed the "Young Fighter Course" in arithmetic progressions. You can safely proceed to the next lesson, where we will study the formula for the sum of a progression, as well as important and very useful consequences from it.
Mathematics has its own beauty, just like painting and poetry.
Russian scientist, mechanic N.E. Zhukovsky
Very common tasks on entrance tests in mathematics are problems related to the concept of arithmetic progression. To successfully solve such problems, it is necessary to know well the properties of the arithmetic progression and have certain skills in their application.
We first recall the main properties of the arithmetic progression and present the most important formulas, related to this concept.
Definition. Numerical sequence, in which each subsequent term differs from the previous one by the same number, called arithmetic progression. Moreover, the numbercalled the difference in progression.
For an arithmetic progression, the following formulas are valid
, (1)
where . Formula (1) is called the formula for the general term of an arithmetic progression, and formula (2) is 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 considered progression is called "arithmetic".
The above formulas (1) and (2) are generalized as follows:
(3)
To calculate the amount the first members of the arithmetic progressionusually the formula is applied
(5) where and.
Taking into account the formula (1), then formula (5) implies
If we denote, then
where . Since, then formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).
In particular , from formula (5) it follows, what
The property of the arithmetic progression, formulated by means of the following theorem, is among the little-known to most students.
Theorem. If, then
Proof. If, then
The theorem is proved.
For example , using the theorem, it can be shown that
Let's move on to considering typical examples of solving problems on the topic "Arithmetic progression".
Example 1. Let and. Find .
Solution. Applying formula (6), we obtain. Since and, then or.
Example 2. Let it be three times more, and when dividing by in the quotient, we get 2 and remainder 8. Determine and.
Solution. The condition of the example implies the system of equations
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.By formula (5), we have
However, using the theorem, one can write
From this and 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 the system of equations
Solving which, we get and.
The natural root of the equation is an .
Example 7. Find if and.
Solution. Since by formula (3) we have that, then the problem statement implies the system of equations
If you 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 down and.
Hence follows 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 connection, from (12) it follows 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
Hence we get and. Taking into account formula (8), we write.
Example 10. Solve the equation.
Solution. From the given equation it follows that. Suppose that,, and. In this case .
According to formula (1), you can write or.
Since, then equation (13) has a single suitable root.
Example 11. Find the maximum value provided that and.
Solution. Since, the considered arithmetic progression is decreasing. In this regard, the expression takes on the maximum value when it is the number of the minimum positive term of the progression.
We use formula (1) and the fact, as. Then we get that or.
Since, then either ... However, in this inequalitygreatest natural number, therefore .
If the values, and are substituted in the formula (6), then we get.
Answer: .
Example 12. Determine the sum of all two-digit natural numbers, which, when divided by 6, give a remainder of 5.
Solution. Let us denote by the set of all two-digit natural numbers, i.e. ... Next, we construct a subset consisting of those elements (numbers) of the set that, when divided by 6, give the remainder 5.
It is not difficult to establish, what . Obviously , that the elements of the setform an arithmetic progression, in which and.
To establish the cardinality (number of elements) of a set, we assume that. Since and, then from formula (1) it follows or. Taking into account formula (5), we get.
The above examples of solving problems in no way can claim to be exhaustive. This article was written based on the analysis modern methods solving typical tasks on a given topic. For a deeper study of methods for solving problems associated with arithmetic progression, it is advisable to refer to the list of recommended literature.
1. Collection of problems in mathematics for applicants to technical colleges / Ed. M.I. Skanavi. - M .: Peace and Education, 2013 .-- 608 p.
2. Suprun V.P. Mathematics for high school students: additional sections school curriculum... - M .: Lenand / URSS, 2014 .-- 216 p.
3. Medynsky M.M. Complete course of elementary mathematics in problems and exercises. Book 2: Number sequences and progression. - M .: Edithus, 2015 .-- 208 p.
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