Problems of arithmetic progression existed already in ancient times. They appeared and demanded a solution because they had a practical need.

So, in one of the papyri of Ancient Egypt, which has a mathematical content - the Rhind papyrus (XIX century BC) - contains the following problem: divide ten measures of bread into ten people, provided that the difference between each of them is one-eighth of a measure. "

And in the mathematical works of the ancient Greeks, there are elegant theorems related to arithmetic progression. So, Hypsicles of Alexandria (II century, who made up many interesting problems and added the fourteenth book to Euclid's "Principles", formulated the idea: “In an arithmetic progression with an even number of members, the sum of the members of the second half is greater than the sum of the members of the first half per square 1 / 2 number of members ".

The sequence is denoted by an. The numbers of the sequence are called its members and are usually denoted by letters with indices that indicate the ordinal number of this member (a1, a2, a3 ... read: "a 1st", "a 2nd", "a 3rd" and so on ).

The sequence can be endless or finite.

What is it arithmetic progression? It is understood as the one obtained by adding the previous term (n) with the same number d, which is the difference of the progression.

If d<0, то мы имеем убывающую прогрессию. Если d>0, then this progression is considered ascending.

An arithmetic progression is called finite if only a few of its first members are taken into account. With very a large number members is already an endless progression.

Any arithmetic progression is specified by the following formula:

an = kn + b, while b and k are some numbers.

The opposite statement is absolutely true: if a sequence is given by a similar formula, then it is exactly an arithmetic progression that has the following properties:

1. Each member of the progression is the arithmetic mean of the previous member and the next.
2. The opposite: if, starting from the 2nd, each term is the arithmetic mean of the previous term and the next, i.e. if the condition is met, then this sequence is an arithmetic progression. This equality is also a sign of progression, therefore it is usually called the characteristic property of progression.
   In the same way, the theorem that reflects this property is true: a sequence is an arithmetic progression only if this equality is true for any of the members of the sequence, starting from the 2nd.

The characteristic property for any four numbers of an arithmetic progression can be expressed by the formula an + am = ak + al, if n + m = k + l (m, n, k are the numbers of the progression).

In an arithmetic progression, any necessary (Nth) term can be found using the following formula:

For example: the first term (a1) in the arithmetic progression is given and equal to three, and the difference (d) is equal to four. You need to find the forty-fifth term of this progression. a45 = 1 + 4 (45-1) = 177

The formula an = ak + d (n - k) allows us to determine nth term arithmetic progression through any of its k-th terms, provided that it is known.

The sum of the members of the arithmetic progression (meaning the 1st n members of the final progression) is calculated as follows:

Sn = (a1 + an) n / 2.

If the 1st term is also known, then another formula is convenient for the calculation:

Sn = ((2a1 + d (n-1)) / 2) * n.

The sum of an arithmetic progression that contains n members is calculated as follows:

The choice of formulas for calculations depends on the conditions of the problems and the initial data.

Natural series of any numbers such as 1,2,3, ..., n, ...- simplest example arithmetic progression.

In addition to the arithmetic progression, there is also a geometric one, which has its own properties and characteristics.

Arithmetic progression called a sequence of numbers (members of a progression)

In which each subsequent term differs from the previous one by a new term, which is also called step or progression difference.

Thus, setting the step of the progression and its first term, you can find any of its elements by the formula

Arithmetic progression properties

1) Each member of the arithmetic progression, starting from the second number, is the arithmetic mean of the previous and next member of the progression

The converse is also true. If the arithmetic mean of adjacent odd (even) members of the progression is equal to the term between them, then this sequence of numbers is an arithmetic progression. This statement makes it very easy to check any sequence.

Also, by the property of arithmetic progression, the above formula can be generalized to the following

This is easy to verify if we write out the terms to the right of the equal sign

It is often used in practice to simplify computations in problems.

2) The sum of the first n terms of the arithmetic progression is calculated by the formula

Remember well the formula for the sum of an arithmetic progression, it is indispensable for calculations and is quite common in simple life situations.

3) If you need to find not the entire sum, but part of the sequence starting from the k -th term, then the following sum formula will come in handy

4) It is of practical interest to find the sum of n terms of an arithmetic progression starting from the kth number. To do this, use the formula

This concludes the theoretical material and move on to solving common problems in practice.

Example 1. Find the fortieth term of the arithmetic progression 4; 7; ...

Solution:

According to the condition, we have

Determine the step of the progression

Using the well-known formula, we find the fortieth term of the progression

Example 2. The arithmetic progression is given by its third and seventh terms. Find the first term of the progression and the sum of ten.

Solution:

Let's write out the given elements of the progression using the formulas

We subtract the first from the second equation, as a result, we find the step of the progression

We substitute the found value into any of the equations to find the first term of the arithmetic progression

We calculate the sum of the first ten members of the progression

Without using complicated calculations, we found all the required quantities.

Example 3. An arithmetic progression is given by the denominator and one of its members. Find the first member of the progression, the sum of its 50 members starting with 50 and the sum of the first 100.

Solution:

Let's write the formula for the hundredth element of the progression

and find the first

Based on the first, we find the 50 term of the progression

Find the sum of the part of the progression

and the sum of the first 100

The total of the progression is 250.

Example 4.

Find the number of members of an arithmetic progression if:

a3-a1 = 8, a2 + a4 = 14, Sn = 111.

Solution:

We write the equations in terms of the first term and the step of the progression and define them

We substitute the obtained values ​​into the sum formula to determine the number of members in the sum

Performing simplifications

and we decide quadratic equation

Of the two values ​​found for the problem condition, only the number 8 is suitable. Thus, the sum of the first eight members of the progression is 111.

Example 5.

Solve the equation

1 + 3 + 5 + ... + x = 307.

Solution: This equation is the sum of an arithmetic progression. Let's write out its first term and find the difference in progression

Yes, yes: the arithmetic progression is not a toy for you :)

Well, friends, if you are reading this text, then the internal cap-evidence 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 row standing numbers 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 do not be afraid that this number is irrational).

So: all such sequences are just 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. Firstly, 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; ... $

Okay, okay: this last example might seem overly complicated. But the rest, I think, is clear to you. Therefore, we will introduce new definitions:

Definition. An arithmetic progression is called:

1. ascending if each next element is larger than the previous one;
2. 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:

1. If $ d \ gt 0 $, then the progression is increasing;
2. If $ d \ lt 0 $, then the progression is obviously decreasing;
3. 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 adjacent 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 down 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, 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 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 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 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 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 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 terms $ ((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 have deduced 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 arithmetic mean condition is satisfied for them: 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 $ form 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 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 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 in a 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, 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 for 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

Note: minimum value this parabola takes at its vertex with abscissa $ ((d) _ (0)) $. Of course, we can calculate this abscissa by standard scheme(there is a formula $ ((d) _ (0)) = (- b) / (2a) \; $), but it would be much more reasonable to notice that the desired vertex lies on the axis of symmetry of the parabola, so the point $ ((d) _ (0)) $ 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 the smallest value (by the way, we haven’t counted $ ((y) _ (\ min)) $ - this is not required from us). At the same time, this number is the difference between the original 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 at the moment we cannot get $ y $ from the numbers $ x $ and $ z $, 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 desired 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 problems. 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 binding workshop bound 216 books in January, and each 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 look 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 the progression, as well as important and very useful consequences from it.

First level

Arithmetic progression. Detailed theory with examples (2019)

Number sequence

So let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many as you like (in our case, them). No matter how many numbers we write, we can always say which one is the first, which is the second, and so on to the last, that is, we can number them. This is an example of a number sequence:

Number sequence
For example, for our sequence:

The assigned number is specific to only one number in the sequence. In other words, there are no three second numbers in the sequence. The second number (like the -th number) is always one.
The number with the number is called the th member of the sequence.

We usually call the entire sequence some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member:.

In our case:

Let's say we have a numerical sequence in which the difference between adjacent numbers is the same and equal.
For example:

etc.
This number sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius in the 6th century and was understood in more broad sense like an endless number sequence. The name "arithmetic" was carried over from the theory of continuous proportions, which the ancient Greeks were engaged in.

This is a numerical sequence, each term of which is equal to the previous one, added to the same number. This number is called the difference of the arithmetic progression and is denoted by.

Try to determine which number sequences are arithmetic progression and which are not:

a)
b)
c)
d)

Understood? Let's compare our answers:
Is an arithmetic progression - b, c.
Is not arithmetic progression - a, d.

Let's return to the given progression () and try to find the value of its th member. Exists two the way to find it.

1. Method

We can add to the previous value of the number of the progression until we get to the th term of the progression. It's good that we don't have much left to summarize - only three values:

So, the th member of the described arithmetic progression is equal to.

2. Method

What if we needed to find the value of the th term of the progression? The summation would take us more than one hour, and it is not a fact that we would not be mistaken when adding numbers.
Of course, mathematicians have come up with a way in which you do not need to add the difference of the arithmetic progression to the previous value. Look closely at the drawing you have drawn ... Surely you have already noticed a certain pattern, namely:

For example, let's see how the value of the th member of this arithmetic progression is added:


In other words:

Try to independently find the value of a member of a given arithmetic progression in this way.

Calculated? Compare your notes to the answer:

Pay attention that you got exactly the same number as in the previous method, when we successively added the members of the arithmetic progression to the previous value.
Let's try to "depersonalize" this formula - we will bring it into general form and get:

Arithmetic progression equation.

Arithmetic progressions are ascending and sometimes decreasing.

Ascending- progressions in which each subsequent value of the members is greater than the previous one.
For example:

Decreasing- progressions in which each subsequent value of the members is less than the previous one.
For example:

The derived formula is used in calculating the terms in both increasing and decreasing terms of an arithmetic progression.
Let's check this in practice.
We are given an arithmetic progression consisting of the following numbers: Let's check what the th number of this arithmetic progression will turn out if we use our formula to calculate it:

Since, then:

Thus, we made sure that the formula works in both decreasing and increasing arithmetic progression.
Try to find the th and th terms of this arithmetic progression on your own.

Let's compare the results obtained:

Arithmetic progression property

Let's complicate the task - we will derive the property of the arithmetic progression.
Let's say we are given the following condition:
- arithmetic progression, find the value.
Easy, you say and start counting according to the formula you already know:

Let, a, then:

Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but if we are given numbers in the condition? Admit it, there is a chance of making a mistake in the calculations.
Now think about whether it is possible to solve this problem in one action using any formula? Of course, yes, and it is her that we will try to withdraw now.

Let's denote the required term of the arithmetic progression as, we know the formula for finding it - this is the same formula we derived at the beginning:
, then:

• the previous member of the progression is:
• the next member of the progression is:

Let's summarize the previous and subsequent members of the progression:

It turns out that the sum of the previous and subsequent members of the progression is the doubled value of the member of the progression located between them. In other words, to find the value of a member of the progression with known previous and consecutive values, it is necessary to add them up and divide by.

That's right, we got the same number. Let's fix the material. Calculate the value for the progression yourself, because it's not difficult at all.

Well done! You know almost everything about progression! There is only one formula left to learn, which, according to legend, was easily deduced for himself by one of the greatest mathematicians of all time, the "king of mathematicians" - Karl Gauss ...

When Karl Gauss was 9 years old, the teacher, who was busy checking the work of students in other grades, asked the following problem in the lesson: “Count the sum of all natural numbers from to (according to other sources up to) inclusive ". Imagine the teacher's surprise when one of his students (it was Karl Gauss) gave the correct answer to the problem in a minute, while most of the daredevil's classmates, after long calculations, received the wrong result ...

Young Karl Gauss noticed a certain pattern that you can easily notice.
Let's say we have an arithmetic progression consisting of -th members: We need to find the sum of the given members of the arithmetic progression. Of course, we can manually sum all the values, but what if in the task it is necessary to find the sum of its members, as Gauss was looking for?

Let's draw a given progression. Look closely at the highlighted numbers and try to perform various mathematical operations with them.


Have you tried it? What have you noticed? Right! Their sums are equal

Now tell me, how many such pairs are there in the given progression? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two members of an arithmetic progression is equal, and similar equal pairs, we get that the total sum is:
.
Thus, the formula for the sum of the first terms of any arithmetic progression will be as follows:

In some problems, we do not know the th term, but we know the difference in the progression. Try to substitute in the formula for the sum, the formula for the th term.
What did you do?

Well done! Now let's return to the problem that was given to Karl Gauss: calculate yourself what is the sum of the numbers starting from the -th, and the sum of the numbers starting from the -th.

How much did you get it?
Gauss found that the sum of the members is equal, and the sum of the members. Is that how you decided?

In fact, the formula for the sum of the members of an arithmetic progression was proved by the ancient Greek scientist Diophantus in the 3rd century, and throughout this time, witty people were using the properties of an arithmetic progression to the utmost.
For example, imagine Ancient Egypt and the most ambitious construction site of that time - the construction of the pyramid ... The figure shows one side of it.

Where is the progression here you say? Look closely and find a pattern in the number of sand blocks in each row of the pyramid wall.


Isn't it an arithmetic progression? Calculate how many blocks are needed to build one wall if block bricks are placed in the base. I hope you won't count by running your finger across the monitor, do you remember the last formula and everything we said about the arithmetic progression?

In this case, the progression looks like this:.
Difference of arithmetic progression.
The number of members of the arithmetic progression.
Let's substitute our data into the last formulas (we will count the number of blocks in 2 ways).

Method 1.

Method 2.

And now you can calculate on the monitor: compare the obtained values ​​with the number of blocks that are in our pyramid. Did it come together? Well done, you have mastered the sum of the terms of the arithmetic progression.
Of course, you can't build a pyramid from blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:

Workout

Tasks:

1. Masha is getting in shape by summer. Every day she increases the number of squats by. How many times will Masha squat in weeks, if at the first workout she did squats.
2. What is the sum of all the odd numbers contained in.
3. When storing logs, lumberjacks stack them in such a way that each upper layer contains one log less than the previous one. How many logs are in one masonry, if logs serve as the basis of the masonry.

Answers:

1. Let's define the parameters of the arithmetic progression. In this case
   (weeks = days).

   Answer: After two weeks, Masha should squat once a day.

2. First odd number, last number.
   Difference of arithmetic progression.
   The number of odd numbers in is half, however, we will check this fact using the formula for finding the -th term of an arithmetic progression:

   The numbers do contain odd numbers.
   Substitute the available data into the formula:

   Answer: The sum of all odd numbers contained in is equal to.

3. Let's remember the pyramid problem. For our case, a, since each top layer is reduced by one log, then only in a bunch of layers, that is.
   Let's substitute the data into the formula:

   Answer: There are logs in the masonry.

Let's summarize

1. - a numerical sequence in which the difference between adjacent numbers is the same and equal. It can be increasing and decreasing.
2. Finding formula The th member of the arithmetic progression is written by the formula -, where is the number of numbers in the progression.
3. Property of members of an arithmetic progression- - where is the number of numbers in the progression.
4. The sum of the members of an arithmetic progression can be found in two ways:
   
   , where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

Let's sit down and start writing some numbers. For example:

You can write any numbers, and there can be as many as you like. But you can always say which one is the first, which is the second, and so on, that is, we can number them. This is an example of a number sequence.

Number sequence is a set of numbers, each of which can be assigned a unique number.

In other words, each number can be associated with a certain natural number, and the only one. And we will not assign this number to any other number from this set.

The number with the number is called the th member of the sequence.

We usually call the entire sequence some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member:.

It is very convenient if the th term of the sequence can be given by some formula. For example, the formula

specifies the sequence:

And the formula is the following sequence:

For example, an arithmetic progression is a sequence (the first term here is equal, and the difference). Or (, difference).

Nth term formula

We call recurrent a formula in which to find out the th member, you need to know the previous or several previous ones:

To find, for example, the th term of the progression using such a formula, we will have to calculate the previous nine. For example, let. Then:

Well, what is the formula now?

In each line we add to, multiplied by some number. For what? Very simple: this is the number of the current member minus:

Much more convenient now, right? We check:

Decide for yourself:

In an arithmetic progression, find the formula for the nth term and find the hundredth term.

Solution:

The first term is equal. What is the difference? And here's what:

(it is because it is called the difference, which is equal to the difference of the consecutive members of the progression).

So the formula is:

Then the hundredth term is:

What is the sum of all natural numbers from to?

According to legend, the great mathematician Karl Gauss, being a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last numbers is equal, the sum of the second and the last but one is the same, the sum of the third and third from the end is the same, and so on. How many such pairs will there be? That's right, exactly half the number of all numbers, that is. So,

The general formula for the sum of the first terms of any arithmetic progression would be:

Example:
Find the sum of all two-digit numbers, multiples.

Solution:

The first such number is. Each next is obtained by adding to the previous number. Thus, the numbers of interest to us form an arithmetic progression with the first term and the difference.

The th term formula for this progression is:

How many members are in the progression if they all have to be double digits?

Very easy: .

The last term in the progression will be equal. Then the sum:

Answer: .

Now decide for yourself:

1. Every day, the athlete runs more m than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
2. A cyclist drives more kilometers every day than the previous one. On the first day, he drove km. How many days does he need to travel to cover the km? How many kilometers will he travel in the last day of the journey?
3. The price of a refrigerator in a store decreases by the same amount every year. Determine how much the price of the refrigerator has decreased every year, if, put up for sale for rubles, six years later it was sold for rubles.

Answers:

1. The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first members of this progression:
   .
   Answer:
2. It is given here:, it is necessary to find.
   Obviously, you need to use the same sum formula as in the previous problem:
   .
   Substitute the values:

   The root obviously doesn't fit, so the answer is.
   Let's calculate the distance traveled for the last day using the th term formula:
   (km).
   Answer:

3. Given:. Find: .
   It couldn't be easier:
   (rub).
   Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN

This is a numerical sequence in which the difference between adjacent numbers is the same and equal.

Arithmetic progression can be ascending () and decreasing ().

For example:

The formula for finding the n-th term of an arithmetic progression

written by the formula, where is the number of numbers in the progression.

Property of members of an arithmetic progression

It allows you to easily find a member of the progression if its neighboring members are known - where is the number of numbers in the progression.

The sum of the members of an arithmetic progression

There are two ways to find the amount:

Where is the number of values.

Where is the number of values.

Someone is wary of the word "progression", as a very complex term from the branches of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi meter (where they still remain). And to understand the essence (and in mathematics there is nothing more important than "understanding the essence") of the arithmetic sequence is not so difficult, having analyzed several elementary concepts.

Math number sequence

It is customary to name a series of numbers by a numerical sequence, each of which has its own number.

a 1 - the first member of the sequence;

and 2 is the second member of the sequence;

and 7 is the seventh member of the sequence;

and n is the n-th member of the sequence;

However, we are not interested in any arbitrary set of numbers and numbers. We will focus our attention on the numerical sequence, in which the value of the nth term is associated with its ordinal number by a dependence that can be clearly formulated mathematically. In other words: the numerical value of the n-th number is some function of n.

a - value of a member of a numerical sequence;

n is its serial number;

f (n) is a function where the ordinal in the numerical sequence n is an argument.

Definition

An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth member of an arithmetic sequence is as follows:

a n - the value of the current member of the arithmetic progression;

a n + 1 - the formula for the next number;

d - difference (a certain number).

It is easy to determine that if the difference is positive (d> 0), then each subsequent term of the series under consideration will be larger than the previous one, and such an arithmetic progression will be increasing.

In the graph below, it’s easy to see why the number sequence was called “ascending”.

In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.

The value of the specified member

Sometimes it is necessary to determine the value of any arbitrary member a n of an arithmetic progression. You can do this by calculating sequentially the values ​​of all members of the arithmetic progression, starting from the first to the desired one. However, this path is not always acceptable if, for example, it is necessary to find the meaning of the five-thousandth or eight-millionth member. The traditional calculation will take a long time. However, a specific arithmetic progression can be investigated using specific formulas. There is also a formula for the nth term: the value of any member of an arithmetic progression can be defined as the sum of the first term of the progression with the difference of the progression, multiplied by the number of the desired term, decreased by one.

The formula is universal for both increasing and decreasing progression.

An example of calculating the value of a given member

Let's solve the following problem of finding the value of the nth term of an arithmetic progression.

Condition: there is an arithmetic progression with parameters:

The first term in the sequence is 3;

The difference in the number series is 1.2.

Assignment: you need to find the value of 214 members

Solution: to determine the value of a given term, we use the formula:

a (n) = a1 + d (n-1)

Substituting the data from the problem statement into the expression, we have:

a (214) = a1 + d (n-1)

a (214) = 3 + 1.2 (214-1) = 258.6

Answer: The 214th term in the sequence is 258.6.

The advantages of this method of calculation are obvious - the whole solution takes no more than 2 lines.

Sum of a given number of members

Very often, in a given arithmetic series, it is required to determine the sum of the values ​​of a certain segment of it. This also does not require calculating the values ​​of each term and then summing. This method is applicable if the number of terms, the sum of which must be found, is small. In other cases, it is more convenient to use the following formula.

The sum of the members of the arithmetic progression from 1 to n is equal to the sum of the first and n-th members, multiplied by the number of the member n and divided by two. If in the formula the value of the nth term is replaced by the expression from the previous paragraph of the article, we get:

Calculation example

For example, let's solve a problem with the following conditions:

The first term in the sequence is zero;

The difference is 0.5.

In the problem, you need to determine the sum of the members of the series from 56th to 101.

Solution. Let's use the formula for determining the sum of the progression:

s (n) = (2 ∙ a1 + d ∙ (n-1)) ∙ n / 2

First, we determine the sum of the values ​​of 101 members of the progression, substituting the data of their conditions of our problem into the formula:

s 101 = (2 ∙ 0 + 0.5 ∙ (101-1)) ∙ 101/2 = 2 525

Obviously, in order to find out the sum of the members of the progression from 56th to 101st, it is necessary to subtract S 55 from S 101.

s 55 = (2 ∙ 0 + 0.5 ∙ (55-1)) ∙ 55/2 = 742.5

Thus, the sum of the arithmetic progression for this example:

s 101 - s 55 = 2,525 - 742.5 = 1,782.5

An example of practical application of arithmetic progression

At the end of the article, let's go back to the example of the arithmetic sequence given in the first paragraph - the taximeter (the counter of a taxi car). Let's consider an example.

Boarding a taxi (which includes 3 km of run) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles / km. Travel distance 30 km. Calculate the cost of the trip.

1. Let's discard the first 3 km, the price of which is included in the landing price.

30 - 3 = 27 km.

2. Further calculation is nothing more than an analysis of an arithmetic number series.

Member number - the number of kilometers traveled (minus the first three).

The member value is the sum.

The first term in this problem will be equal to a 1 = 50 p.

Difference in progression d = 22 p.

the number we are interested in is the value of the (27 + 1) -th term of the arithmetic progression - the counter reading at the end of the 27th kilometer is 27.999 ... = 28 km.

a 28 = 50 + 22 ∙ (28 - 1) = 644

Calculations of calendar data for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of a celestial body to a luminary. In addition, various numerical series are successfully used in statistics and other applied branches of mathematics.

Another type of number sequence is geometric

Geometric progression is characterized by large, in comparison with arithmetic, rates of change. It is no coincidence that in politics, sociology, medicine, they often say that the process develops exponentially in order to show the high rate of spread of this or that phenomenon, for example, a disease during an epidemic.

The Nth term of the geometric numerical series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first term is 1, the denominator is 2, respectively, then:

n = 1: 1 ∙ 2 = 2

n = 2: 2 ∙ 2 = 4

n = 3: 4 ∙ 2 = 8

n = 4: 8 ∙ 2 = 16

n = 5: 16 ∙ 2 = 32,

b n - the value of the current member of the geometric progression;

b n + 1 - the formula of the next term of the geometric progression;

q is the denominator of a geometric progression (constant number).

If the graph of the arithmetic progression is a straight line, then the geometric one paints a slightly different picture:

As in the case of arithmetic, a geometric progression has a formula for the value of an arbitrary term. Any n-th term of the geometric progression is equal to the product of the first term by the denominator of the progression to the power of n, reduced by one:

Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Find the 5th term of the progression

b 5 = b 1 ∙ q (5-1) = 3 ∙ 1.5 4 = 15.1875

The sum of a given number of members is calculated in the same way using a special formula. The sum of the first n terms of a geometric progression is equal to the difference between the product of the nth term of the progression and its denominator and the first term of the progression, divided by the denominator reduced by one:

If b n is replaced using the formula considered above, the value of the sum of the first n terms of the considered numerical series will take the form:

Example. The geometric progression starts with the first term equal to 1. The denominator is set equal to 3. Find the sum of the first eight terms.

s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280