How to find the difference of an arithmetic progression. How to find the difference of an arithmetic progression: formulas and examples of solutions

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 will immediately get down to business.

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 an 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. Arithmetic progression 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 out 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, 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 its 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. until what natural number$ n $) the negativity of the members 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 needs some explanation. So, we know that $ n \ lt 15 \ frac (7) (27) $. On the other hand, we will only be satisfied with integer values of the number (moreover: $ n \ in \ mathbb (N) $), so the largest allowed number is exactly $ n = 15 $, and by no means 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 in the future will save us a lot of time and unequal cells. :)

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 line

I specifically noted arbitrary members $ ((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 recursion 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) \]

It turned out to be 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 the order shown).

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 into the initial condition 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 mean arithmetic 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 marked

Let'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 start walking 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 of a fundamentally higher level of complexity 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 arithmetic mean of the numbers −66 and −6:

\ [((d) _ (0)) = \ frac (-66-6) (2) = - 36 \]

What does the discovered number give us? With it, the required product takes 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. therefore

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, only 7 numbers had to be inserted: 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 passed 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 the progression, as well as important and very useful consequences from it.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very ..."
And for those who "very much ...")

An arithmetic progression is a series of numbers in which each number is greater (or less) than the previous one by the same amount.

This topic is often difficult and incomprehensible. Indexes for letters, nth term progressions, the difference in the progression - all this is somehow confusing, yes ... Let's figure out the meaning of the arithmetic progression and everything will work out right away.)

Arithmetic progression concept.

Arithmetic progression is a very simple and clear concept. Doubt? In vain.) See for yourself.

I'll write an unfinished series of numbers:

1, 2, 3, 4, 5, ...

Can you extend this row? What numbers will go next, after the five? Everyone ... uh-uh ..., in short, everyone will realize that the numbers 6, 7, 8, 9, etc. will go further.

Let's complicate the task. I give an unfinished series of numbers:

2, 5, 8, 11, 14, ...

You will be able to catch the pattern, extend the series, and name seventh row number?

If you figured out that this number is 20 - I congratulate you! Not only did you feel key points of the arithmetic progression, but also successfully used them in business! If you haven't figured it out, read on.

Now let's translate the key points from sensation to mathematics.)

First key point.

Arithmetic progression deals with series of numbers. This is confusing at first. We are used to solving equations, plotting graphs and all that ... And then extend the series, find the number of the series ...

It's OK. Just progressions are the first acquaintance with a new branch of mathematics. The section is called "Rows" and works with series of numbers and expressions. Get used to it.)

Second key point.

In an arithmetic progression, any number is different from the previous one by the same amount.

In the first example, this difference is one. Whatever number you take, it is one more than the previous one. In the second - three. Any number greater than the previous one by three. Actually, it is this moment that gives us the opportunity to catch the pattern and calculate the subsequent numbers.

The third key point.

This moment is not striking, yes ... But it is very, very important. There he is: each number in the progression stands in its place. There is the first number, there is the seventh, there is the forty-fifth, etc. If they are confused at random, the pattern will disappear. The arithmetic progression will also disappear. There will be just a row of numbers.

That's the whole point.

Of course, in new topic new terms and designations appear. You need to know them. Otherwise, you will not understand the task. For example, you have to decide something like:

Write out the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.

Does it inspire?) Letters, some indexes ... And the task, by the way - couldn't be easier. You just need to understand the meaning of terms and designations. Now we will master this business and return to the task.

Terms and designations.

Arithmetic progression is a series of numbers in which each number is different from the previous one by the same amount.

This quantity is called ... Let's deal with this concept in more detail.

Difference of arithmetic progression.

Difference of arithmetic progression is the amount by which any number of the progression more the previous one.

One important point. Please pay attention to the word "more". Mathematically, this means that each number in the progression is obtained adding the difference of the arithmetic progression to the previous number.

For calculation, let's say second number of the series, it is necessary to the first the number add this very difference of the arithmetic progression. For calculation fifth- the difference is necessary add To fourth, well, etc.

Difference of arithmetic progression may be positive, then each number of the row will turn out really more than the previous one. This progression is called increasing. For example:

8; 13; 18; 23; 28; .....

Here every number is obtained adding positive number, +5 to the previous one.

The difference can be negative, then each number in the row will turn out less than the previous one. Such a progression is called (you won't believe it!) decreasing.

For example:

8; 3; -2; -7; -12; .....

Here every number is obtained too adding to the previous, but already negative number, -5.

By the way, when working with a progression, it is very useful to immediately determine its nature - whether it is increasing or decreasing. It helps a lot to navigate the solution, to detect your mistakes and fix them before it's too late.

Difference of arithmetic progression denoted, as a rule, by the letter d.

How to find d? Very simple. It is necessary to subtract from any number of the series previous number. Subtract. By the way, the result of the subtraction is called the "difference".)

Let us define, for example, d for increasing arithmetic progression:

2, 5, 8, 11, 14, ...

We take any number of the row that we want, for example, 11. Subtract from it previous number, those. eight:

This is the correct answer. For this arithmetic progression, the difference is three.

You can take exactly any number of progression, since for a specific progression d -always the same. At least somewhere in the beginning of the row, at least in the middle, at least anywhere. You cannot take only the very first number. Just because at the very first number there is no previous one.)

By the way, knowing that d = 3, it is very easy to find the seventh number of this progression. Add 3 to the fifth number - we get the sixth, it will be 17. Add three to the sixth number, we get the seventh number - twenty.

We define d for a decreasing arithmetic progression:

8; 3; -2; -7; -12; .....

I remind you that, regardless of the signs, to determine d it is necessary from any number take away the previous one. We choose any number of the progression, for example -7. The previous one is -2. Then:

d = -7 - (-2) = -7 + 2 = -5

The difference of the arithmetic progression can be any number: whole, fractional, irrational, whatever.

Other terms and designations.

Each number in the series is called a member of an arithmetic progression.

Each member of the progression has its own number. The numbers are strictly in order, without any tricks. First, second, third, fourth, etc. For example, in the progression 2, 5, 8, 11, 14, ... two is the first term, five is the second, eleven is the fourth, well, you understand ...) Please understand clearly - the numbers themselves can be absolutely any, whole, fractional, negative, whatever, but numbering of numbers- strictly in order!

How to record a general progression? No problem! Each number in the row is written as a letter. As a rule, the letter is used to denote an arithmetic progression a... The member number is indicated by an index at the bottom right. We write members separated by commas (or semicolons), like this:

a 1, a 2, a 3, a 4, a 5, .....

a 1 is the first number, a 3- third, etc. Nothing tricky. You can briefly write this series like this: (a n).

Progressions are finite and endless.

The ultimate the progression has a limited number of members. Five, thirty-eight, whatever. But - a finite number.

Endless progression - has an infinite number of members, as you might guess.)

You can write the final progression through a series like this, all the members and a dot at the end:

a 1, a 2, a 3, a 4, a 5.

Or so, if there are many members:

a 1, a 2, ... a 14, a 15.

In a short entry, you will have to additionally indicate the number of members. For example (for twenty members), like this:

(a n), n = 20

An endless progression can be recognized by the ellipsis at the end of the row, as in the examples in this lesson.

Now you can solve tasks. The tasks are simple, purely for understanding the meaning of the arithmetic progression.

Examples of tasks on arithmetic progression.

Let's take a closer look at the task, which is given above:

1. Write down the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.

We translate the task into an understandable language. An infinite arithmetic progression is given. The second number of this progression is known: a 2 = 5. The difference in progression is known: d = -2.5. It is necessary to find the first, third, fourth, fifth and sixth members of this progression.

For clarity, I will write down a series according to the condition of the problem. The first six terms, where the second term is a five:

a 1, 5, a 3, a 4, a 5, a 6, ....

a 3 = a 2 + d

Substitute into expression a 2 = 5 and d = -2.5... Do not forget about the minus!

a 3=5+(-2,5)=5 - 2,5 = 2,5

The third term is smaller than the second. Everything is logical. If the number is greater than the previous one by negative value, then the number itself will turn out to be less than the previous one. The progression is decreasing. Okay, let's take it into account.) We consider the fourth member of our series:

a 4 = a 3 + d

a 4=2,5+(-2,5)=2,5 - 2,5 = 0

a 5 = a 4 + d

a 5=0+(-2,5)= - 2,5

a 6 = a 5 + d

a 6=-2,5+(-2,5)=-2,5 - 2,5 = -5

So, the terms from the third to the sixth are calculated. The result is such a series:

a 1, 5, 2.5, 0, -2.5, -5, ....

It remains to find the first term a 1 on famous second... This is a step in the other direction, to the left.) Hence, the difference of the arithmetic progression d need not add to a 2, but take away:

a 1 = a 2 - d

a 1=5-(-2,5)=5 + 2,5=7,5

That's all there is to it. Task answer:

7,5, 5, 2,5, 0, -2,5, -5, ...

Along the way, I will note that we solved this task recurrent way. This scary word only means searching for a member of the progression. by the previous (adjacent) number. We will consider other ways of working with progression later.

One important conclusion can be drawn from this simple task.

Remember:

If we know at least one term and the difference of an arithmetic progression, we can find any member of this progression.

Remember? This simple conclusion allows you to solve most of the tasks of the school course on this topic. All tasks revolve around three main parameters: member of arithmetic progression, difference of the progression, number of the member of the progression. Everything.

Of course, all the previous algebra is not canceled.) Inequalities, equations, and other things are attached to the progression. But by the very progression- everything revolves around three parameters.

Let's take a look at some of the popular assignments on this topic as an example.

2. Write down the final arithmetic progression as a series, if n = 5, d = 0.4, and a 1 = 3.6.

Everything is simple here. Everything has already been given. You need to remember how the members of an arithmetic progression are counted, count, and write them down. It is advisable not to miss the words in the condition of the task: "final" and " n = 5". Not to count until completely blue in the face.) There are only 5 (five) members in this progression:

a 2 = a 1 + d = 3.6 + 0.4 = 4

a 3 = a 2 + d = 4 + 0.4 = 4.4

a 4 = a 3 + d = 4.4 + 0.4 = 4.8

a 5 = a 4 + d = 4.8 + 0.4 = 5.2

It remains to write down the answer:

3,6; 4; 4,4; 4,8; 5,2.

Another task:

3. Determine whether the number 7 is a member of the arithmetic progression (a n), if a 1 = 4.1; d = 1.2.

Hmm ... Who knows? How to determine something?

How, how ... Yes, write down the progression in the form of a series and see if there will be a seven there or not! We consider:

a 2 = a 1 + d = 4.1 + 1.2 = 5.3

a 3 = a 2 + d = 5.3 + 1.2 = 6.5

a 4 = a 3 + d = 6.5 + 1.2 = 7.7

4,1; 5,3; 6,5; 7,7; ...

Now it is clearly visible that we are just a seven slipped through between 6.5 and 7.7! The seven did not get into our series of numbers, and, therefore, the seven will not be a member of the given progression.

The answer is no.

And here is a task based on real option GIA:

4. Several consecutive members of the arithmetic progression are written out:

...; fifteen; NS; nine; 6; ...

A row is written here without end and beginning. No member numbers, no difference d... It's OK. To solve the problem, it is enough to understand the meaning of the arithmetic progression. We look and think about what is possible to know from this series? What are the three main parameters?

Member numbers? There is not a single number here.

But there are three numbers and - attention! - word "consecutive" in the condition. This means that the numbers are strictly in order, without gaps. Are there two in this row neighboring known numbers? Yes, there is! These are 9 and 6. So we can calculate the difference of the arithmetic progression! We subtract from the six previous number, i.e. nine:

There are mere trifles left. What is the previous number for the X? Fifteen. This means that x can be easily found by simple addition. Add the difference of the arithmetic progression to 15:

That's all. Answer: x = 12

We solve the following problems ourselves. Note: these problems are not about formulas. Purely for understanding the meaning of an arithmetic progression.) We just write down a series with numbers-letters, look and think.

5. Find the first positive term of the arithmetic progression if a 5 = -3; d = 1.1.

6. It is known that the number 5.5 is a member of the arithmetic progression (a n), where a 1 = 1.6; d = 1.3. Determine the number n of this member.

7. It is known that in the arithmetic progression a 2 = 4; a 5 = 15.1. Find a 3.

8. Written out several consecutive members of the arithmetic progression:

...; 15.6; NS; 3.4; ...

Find the term in the progression indicated by the letter x.

9. The train started moving from the station, steadily increasing its speed by 30 meters per minute. What will the train speed be in five minutes? Give your answer in km / h.

10. It is known that in the arithmetic progression a 2 = 5; a 6 = -5. Find a 1.

Answers (in disarray): 7.7; 7.5; 9.5; nine; 0.3; 4.

Everything worked out? Amazing! You can master the arithmetic progression for more high level, in the following lessons.

Not everything worked out? No problem. In Special Section 555, all these problems are sorted out to pieces.) And, of course, a simple practical technique is described that immediately highlights the solution of such tasks clearly, clearly, as if in the palm of your hand!

By the way, in the puzzle about the train there are two problems that people often stumble on. One is purely in progression, and the second is common for any problems in mathematics, and physics too. This is a translation of dimensions from one to another. In it is shown how these problems should be solved.

In this lesson, we examined the elementary meaning of the arithmetic progression and its main parameters. This is enough to solve almost all problems on this topic. Add d to the numbers, write a series, everything will be decided.

The finger solution works well for very short pieces of a row, as in the examples in this lesson. If the row is longer, the calculations become more complicated. For example, if in problem 9 in the question, replace "five minutes" on the "thirty five minutes" the problem will become significantly angrier.)

And there are also tasks that are simple in essence, but incredible in terms of calculations, for example:

You are given an arithmetic progression (a n). Find a 121 if a 1 = 3 and d = 1/6.

And what, we will add many, many times by 1/6 ?! You can kill it !?

You can.) If you do not know the simple formula by which you can solve such tasks in a minute. This formula will be in the next lesson. And this problem is solved there. In a minute.)

If you like this site ...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Instant validation testing. Learning - with interest!)

you can get acquainted with functions and derivatives.

Instructions

An arithmetic progression is a sequence of the form a1, a1 + d, a1 + 2d ..., a1 + (n-1) d. D in steps progression It is obvious that the total of an arbitrary n-th term of the arithmetic progression has the form: An = A1 + (n-1) d. Then knowing one of the members progression, member progression and step progression, you can, that is, the number of the member of the progress. Obviously, it will be determined by the formula n = (An-A1 + d) / d.

Now let the mth term be known progression and another member progression- n-th, but n, as in the previous case, but it is known that n and m do not coincide. progression can be calculated by the formula: d = (An-Am) / (n-m). Then n = (An-Am + md) / d.

If the sum of several elements of the arithmetic is known progression, as well as its first and last, then the number of these elements can also be determined. progression will be equal to: S = ((A1 + An) / 2) n. Then n = 2S / (A1 + An) - chdenov progression... Using the fact that An = A1 + (n-1) d, this formula can be rewritten as: n = 2S / (2A1 + (n-1) d). From this one can express n by solving a quadratic equation.

An arithmetic sequence is an ordered set of numbers, each member of which, except for the first, differs from the previous one by the same amount. This constant value is called the difference of the progression or its step and can be calculated from the known members of the arithmetic progression.

Instructions

If the values of the first and second or any other pair of neighboring terms are known from the conditions of the problem, to calculate the difference (d), simply subtract the previous one from the next term. The resulting value can be either positive or negative, depending on whether the progression is increasing. In general form, write down the solution for an arbitrary pair (aᵢ and aᵢ₊₁) of adjacent members of the progression as follows: d = aᵢ₊₁ - aᵢ.

For a pair of members of such a progression, one of which is the first (a₁), and the other is any other arbitrarily chosen, it is also possible to compose a formula for finding the difference (d). However, in this case, the sequence number (i) of an arbitrary selected member of the sequence must be known. To calculate the difference, add both numbers, and divide the result by the ordinal number of an arbitrary term, reduced by one. In general, write this formula as follows: d = (a₁ + aᵢ) / (i-1).

If, in addition to an arbitrary member of the arithmetic progression with ordinal i, another member with ordinal u is known, change the formula from the previous step accordingly. In this case, the difference (d) of the progression will be the sum of these two terms divided by the difference of their ordinal numbers: d = (aᵢ + aᵥ) / (i-v).

The formula for calculating the difference (d) will become somewhat more complicated if the value of its first term (a₁) and the sum (Sᵢ) of a given number (i) of the first members of the arithmetic sequence are given in the problem conditions. To get the desired value, divide the amount by the number of members that make it up, subtract the value of the first number in the sequence, and double the result. Divide the resulting value by the number of members that make up the sum, reduced by one. In general, write down the formula for calculating the discriminant as follows: d = 2 * (Sᵢ / i-a₁) / (i-1).

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 \ (16 \); $10$; $4$; \ (- 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:

	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 \).
	Now we can restore our progression to the (first negative) element we need.
	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:

	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 \).
	And now we find the desired one without any problems: \ (x = 5 + 2.5 = 7.5 \).
	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:

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 \)
\ (n = 2 \); \ (a_ (2 + 1) = a_2 + 5 = -6 + 5 = -1 \)
\ (n = 3 \); \ (a_ (3 + 1) = a_3 + 5 = -1 + 5 = 4 \)
And having calculated the six elements we need, we find their sum.

\ (S_6 = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = \)
$=(-11)+(-6)+(-1)+4+9+14=9$

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:

\ (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. Our progression is given by the formula of the nth term depending on its number (see details). Let's calculate the first element by substituting one for \ (n \).
\ (n = 1; \) \ (a_1 = 3.4 1-0.6 = 2.8 \)		Now we find the twenty-fifth term, substituting twenty-five instead of \ (n \).
\ (n = 25; \) \ (a_ (25) = 3.4 25-0.6 = 84.4 \)		Well, now we can calculate the required amount without any problems.
\ (S_ (25) = \) \ (\ frac (a_1 + a_ (25)) (2) \) \ (\ cdot 25 = \) \ (= \) \ (\ frac (2.8 + 84.4) (2) \) \ (\ cdot 25 = \) \ (1090 \)		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 \).

What is an arithmetic progression?

To understand this, it is necessary to give a definition of the considered progression, as well as to give the basic formulas that will be further used in solving problems.

Arithmetic or is a set of ordered rational numbers, each term of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of the ordered series of numbers and the difference, you can restore the entire arithmetic progression.

Let's give an example. The following sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the considered type of progression, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).

Important formulas

Let us now give the basic formulas that will be needed to solve problems using an arithmetic progression. Let us denote by a n the nth term of the sequence, where n is an integer. The difference is denoted by the Latin letter d. Then the following expressions are valid:

To determine the value of the nth term, the formula is suitable: a n = (n-1) * d + a 1.
To determine the sum of the first n terms: S n = (a n + a 1) * n / 2.

To understand any examples of arithmetic progression with a solution in grade 9, it is enough to remember these two formulas, since any problems of the type under consideration are built on their use. You should also remember that the difference in progression is determined by the formula: d = a n - a n-1.

Example # 1: finding an unknown member

Let's give a simple example of an arithmetic progression and formulas that must be used to solve.

Let the sequence 10, 8, 6, 4, ... be given, it is necessary to find five terms in it.

It already follows from the problem statement that the first 4 terms are known. The fifth can be defined in two ways:

Let's calculate the difference first. We have: d = 8 - 10 = -2. Likewise, one could take any two other members standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d = a n - a n-1, then d = a 5 - a 4, whence we get: a 5 = a 4 + d. Substitute the known values: a 5 = 4 + (-2) = 2.
The second method also requires knowing the difference of the considered progression, so first you need to determine it, as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for n number of the sequence. We have: a n = (n - 1) * d + a 1 = (n - 1) * (-2) + 10 = 12 - 2 * n. Substituting n = 5 in the last expression, we get: a 5 = 12-2 * 5 = 2.

As you can see, both methods of solution led to the same result. Note that in this example, the difference d of the progression is negative. Such sequences are called decreasing because each next term is less than the previous one.

Example # 2: Progression Difference

Now let's complicate the task a little, we will give an example of how to find the difference of an arithmetic progression.

It is known that in some algebraic progression the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.

Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1. We substitute in it the known data from the condition, that is, the numbers a 1 and a 7, we have: 18 = 6 + 6 * d. From this expression, you can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, we have answered the first part of the problem.

To restore the sequence to 7 terms, you should use the definition algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2 = 8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16, a 7 = 18.

Example # 3: making a progression

Let us complicate the condition of the problem even more. Now it is necessary to answer the question of how to find the arithmetic progression. You can give the following example: given two numbers, for example - 4 and 5. It is necessary to make an algebraic progression so that three more terms fit between these.

Before starting to solve this problem, it is necessary to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 = -4 and a 5 = 5. Having established this, we proceed to the problem, which is similar to the previous one. Again, for the n-th term, we use the formula, we get: a 5 = a 1 + 4 * d. From where: d = (a 5 - a 1) / 4 = (5 - (-4)) / 4 = 2.25. Here we received not an integer value of the difference, but it is a rational number, so the formulas for the algebraic progression remain the same.

Now add the found difference to a 1 and restore the missing members of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 = 2.75 + 2.25 = 5, which coincided with the condition of the problem.

Example # 4: the first term of the progression

Let's continue to give examples of arithmetic progression with a solution. In all previous problems, the first number of the algebraic progression was known. Now consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find the number from which this sequence begins.

The formulas used so far assume knowledge of a 1 and d. Nothing is known about these numbers in the problem statement. Nevertheless, we write out expressions for each member about which there is information: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. Received two equations in which 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.

The easiest way to solve this system is to express a 1 in each equation, and then compare the resulting expressions. The first equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 = a 43 - 42 * d = 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d = 37 - 42 * d, whence the difference d = (37 - 50) / (42 - 14) = - 0.464 (only 3 decimal places are given).

Knowing d, you can use any of the 2 above expressions for a 1. For example, the first: a 1 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.

If you have doubts about the result, you can check it, for example, determine the 43 term of the progression, which is specified in the condition. We get: a 43 = a 1 + 42 * d = 56.496 + 42 * (- 0.464) = 37.008. A small error is due to the fact that the calculations used rounding to thousandths.

Example # 5: amount

Now let's look at some examples with solutions for the sum of an arithmetic progression.

Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How do you calculate the sum of these 100 numbers?

Thanks to the development of computer technology, it is possible to solve this problem, that is, to add up all the numbers sequentially, which the computer will do as soon as a person presses the Enter key. However, the problem can be solved in the mind, if we pay attention that the presented series of numbers is an algebraic progression, and its difference is 1. Applying the formula for the sum, we get: S n = n * (a 1 + an) / 2 = 100 * (1 + 100) / 2 = 5050.

It is curious to note that this problem is called "Gaussian", because at the beginning of the 18th century the famous German, while still only 10 years old, was able to solve it in his head in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add in pairs the numbers on the edges of the sequence, you always get one result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since of these amounts will be exactly 50 (100/2), then to get the correct answer, it is enough to multiply 50 by 101.

Example # 6: sum of members from n to m

Another typical example the sum of the arithmetic progression is as follows: given a number of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its members from 8 to 14 will equal.

The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then their sequential summation. Since there are few terms, this method is not laborious enough. Nevertheless, it is proposed to solve this problem by the second method, which is more universal.

The idea is to get a formula for the sum of the algebraic progression between the terms m and n, where n> m are integers. Let us write out two expressions for the sum for both cases:

S m = m * (a m + a 1) / 2.
S n = n * (a n + a 1) / 2.

Since n> m, it is obvious that the 2 sum includes the first. The last conclusion means that if we take the difference between these sums, and add to it the term a m (in the case of taking the difference, it is subtracted from the sum S n), then we get the necessary answer to the problem. We have: S mn = S n - S m + am = n * (a 1 + an) / 2 - m * (a 1 + am) / 2 + am = a 1 * (n - m) / 2 + an * n / 2 + am * (1- m / 2). In this expression it is necessary to substitute the formulas for a n and a m. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d * (3 * m - m 2 - 2) / 2.

The resulting formula is somewhat cumbersome; nevertheless, the sum of S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.

As can be seen from the solutions given, all problems are based on knowledge of the expression for the nth term and the formula for the sum of the set of the first terms. Before proceeding with the solution of any of these problems, it is recommended to carefully read the condition, clearly understand what is required to be found, and only then proceed to the solution.

Another tip is to strive for simplicity, that is, if you can answer a question without using complex mathematical calculations, then you need to do just that, since in this case the probability of making a mistake is less. For example, in the example of an arithmetic progression with solution # 6, one could stop at the formula S mn = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and split common task into separate subtasks (in this case, first find the terms a n and a m).

If there are doubts about the result obtained, it is recommended to check it, as it was done in some of the examples given. We figured out how to find the arithmetic progression. If you figure it out, it's not that difficult.

\ (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 \).
\ (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.
\ (a_n = a_1 + (n-1) d \)
\ (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.
\ (- 19.3+ (n-1) 0.3> 0 \)
\ ((n-1) 0.3> 19.3 \) \ (\|: 0.3 \)		We divide both sides of the inequality by \ (0,3 \).
\ (n-1> \) \ (\ frac (19,3) (0,3) \)		Move minus one, remembering to change signs
\ (n> \) \ (\ frac (19,3) (0,3) \) \ (+ 1 \)		We calculate ...
\ (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.
\ (n = 65; \) \ (a_ (65) = - 19.3+ (65-1) 0.3 = -0.1 \) \ (n = 66; \) \ (a_ (66) = - 19.3+ (66-1) 0.3 = 0.2 \)		Thus, we need to add the first \ (65 \) elements.
\ (S_ (65) = \) \ (\ frac (2 \ cdot (-19.3) + (65-1) 0.3) (2) \)\ (\ cdot 65 \) \ (S_ (65) = \) \ ((- 38.6 + 19.2) (2) \) \ (\ cdot 65 = -630.5 \)		The answer is ready.

\ (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? Easy - to get the sum from \ (26 \) th to \ (42 \) - oh, you must first find the sum from \ (1 \) - th to \ (42 \) - oh, and then subtract the sum from it first to \ (25 \) - th (see picture).
	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.
\ (S_ (42) = \) \ (\ frac (2 \ cdot (-33) + (42-1) 4) (2) \)\ (\ cdot 42 = \) \ (= \) \ (\ frac (-66 + 164) (2) \) \ (\ cdot 42 = 2058 \)	Now the sum of the first \ (25 \) - ty elements.
\ (S_ (25) = \) \ (\ frac (2 \ cdot (-33) + (25-1) 4) (2) \)\ (\ cdot 25 = \) \ (= \) \ (\ frac (-66 + 96) (2) \) \ (\ cdot 25 = 375 \)	Finally, we calculate the answer.
\ (S = S_ (42) -S_ (25) = 2058-375 = 1683 \)