Yes, yes: 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 that: SOOOOO!) want to know. Therefore, I will not torment you with long introductions and will get straight to the point.
First, a couple of examples. Let's look at 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 being one more than the previous one. In the second case, the difference between adjacent numbers is already five, but this difference is still constant. In the third case, there are roots altogether. 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 one differs from the previous one by exactly the same amount is called an arithmetic progression. The very amount by which the numbers differ is called the progression difference and is most often denoted by the letter $d$.
Notation: $\left(((a)_(n)) \right)$ is the progression itself, $d$ is its difference.
And just a couple of important notes. Firstly, progression is only considered ordered sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. Numbers cannot be rearranged or swapped.
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 infinite progression. The ellipsis after the four seems to hint that there are quite a few more numbers to come. Infinitely many, for example. :)
I would also like to note that progressions can be increasing or decreasing. We have already seen increasing ones - the same set (1; 2; 3; 4; ...). 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: the last example may seem overly complicated. But the rest, I think, you understand. Therefore, we introduce new definitions:
Definition. An arithmetic progression is called:
- increasing if each next element is greater 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; ...).
Only one question remains: how to distinguish an increasing progression from a decreasing one? Fortunately, everything here depends only on the sign of the number $d$, i.e. progression differences:
- If $d \gt 0$, then the progression increases;
- If $d \lt 0$, then the progression is obviously decreasing;
- Finally, there is the case $d=0$ - in this case the entire 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 we can see, in all three cases the difference actually 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 properties they have.
Progression terms and recurrence 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 a progression. They are indicated by a number: first member, second member, etc.
In addition, as we already know, neighboring terms 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 of a progression, you need to know the $n-1$th term and the difference $d$. This formula is called recurrent, because with its help you can find any number only by knowing the previous one (and in fact, all the previous ones). This is very inconvenient, so there is a more cunning formula that reduces any calculations to the first term and the difference:
\[((a)_(n))=((a)_(1))+\left(n-1 \right)d\]
You've probably already come across this formula. They like to give it in all sorts of reference books and solution books. And in any sensible mathematics textbook it is one of the first.
However, I suggest you practice a little.
Task No. 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 be substituted - the first term is already known to us. However, by substituting unity, we were convinced that even for the first term our formula works. In other cases, everything came down to banal arithmetic.
Task No. 2. Write down the first three terms of an arithmetic progression if its seventh term is equal to −40 and its seventeenth term is equal to −50.
Solution. Let's write the problem condition in familiar 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 system sign because these requirements must be met simultaneously. Now let’s 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 it is to find the progression difference! All that remains is 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 is solved.
Answer: (−34; −35; −36)
Notice the interesting property of 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 $n-m$ number:
\[((a)_(n))-((a)_(m))=d\cdot \left(n-m \right)\]
A simple but very useful property that you definitely need to know - with its help you can significantly speed up the solution of many progression problems. Here is a clear example of this:
Task No. 3. The fifth term of an 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 we need to find $((a)_(15))$, 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 $5d=6$, from which 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 didn’t need to create any systems of equations and calculate the first term and the difference - everything was solved in just a couple of lines.
Now let's look at another type of problem - searching for negative and positive terms of a progression. It is no secret that if a progression increases, and its first term is negative, then sooner or later positive terms will appear in it. And vice versa: the terms of a decreasing progression will sooner or later become negative.
At the same time, it is not always possible to find this moment “head-on” by sequentially going through the elements. Often, problems are written in such a way that without knowing the formulas, the calculations would take several sheets of paper—we would simply fall asleep while we found the answer. Therefore, let's try to solve these problems in a faster way.
Task No. 4. How many negative terms are there 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 indeed at some point we will stumble upon positive numbers. The only question is when this will happen.
Let's try to find out how long (i.e. up to what natural number $n$) the negativity of the terms remains:
\[\begin(align) & ((a)_(n)) \lt 0\Rightarrow ((a)_(1))+\left(n-1 \right)d \lt 0; \\ & -38.5+\left(n-1 \right)\cdot 2.7 \lt 0;\quad \left| \cdot 10 \right. \\ & -385+27\cdot \left(n-1 \right) \lt 0; \\ & -385+27n-27 \lt 0; \\ & 27n \lt 412; \\ & n \lt 15\frac(7)(27)\Rightarrow ((n)_(\max ))=15. \\ \end(align)\]
The last line requires some explanation. So we know that $n \lt 15\frac(7)(27)$. On the other hand, we are satisfied with only integer values of the number (moreover: $n\in \mathbb(N)$), so the largest permissible number is precisely $n=15$, and in no case 16.
Task No. 5. In arithmetic progression $(()_(5))=-150,(()_(6))=-147$. Find the number of the first positive term of this progression.
This would be exactly the same problem as the previous one, but we do not 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, let's try to express the fifth term through the first and the difference using 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. Let's find out at what point in our sequence positive numbers will appear:
\[\begin(align) & ((a)_(n))=-162+\left(n-1 \right)\cdot 3 \gt 0; \\ & -162+3n-3 \gt 0; \\ & 3n \gt 165; \\ & n \gt 55\Rightarrow ((n)_(\min ))=56. \\ \end(align)\]
The minimum integer solution to this inequality is the number 56.
Please note: in the last task everything came down to strict inequality, so the option $n=55$ will not suit us.
Now that we have learned how to solve simple problems, let's move on to more complex ones. But first, let's study another very useful property of arithmetic progressions, which will save us a lot of time and unequal cells in the future. :)
Arithmetic mean and equal indentations
Let's consider several consecutive terms of the increasing arithmetic progression $\left(((a)_(n)) \right)$. Let's try to mark them on the number line:
Terms of an arithmetic progression on the number lineI specifically marked arbitrary terms $((a)_(n-3)),...,((a)_(n+3))$, and not some $((a)_(1)) ,\ ((a)_(2)),\ ((a)_(3))$, etc. Because the rule that I’ll tell you about now works the same for any “segments”.
And the rule is very simple. Let's remember the recurrent formula and write it down for all marked terms:
\[\begin(align) & ((a)_(n-2))=((a)_(n-3))+d; \\ & ((a)_(n-1))=((a)_(n-2))+d; \\ & ((a)_(n))=((a)_(n-1))+d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n+1))+d; \\ \end(align)\]
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 $((a)_(n))$ at the same distance equal to $2d$. We can continue ad infinitum, but the meaning is well illustrated by the picture
The terms of the progression lie at the same distance from the center What does this mean for us? This means that $((a)_(n))$ can be found if the neighboring numbers are known:
\[((a)_(n))=\frac(((a)_(n-1))+((a)_(n+1)))(2)\]
We have derived an excellent statement: every term of an arithmetic progression is equal to the arithmetic mean of its neighboring terms! Moreover: we can step back from our $((a)_(n))$ to the left and to the right not by one step, but by $k$ steps - and the formula will still 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 tailored to use the arithmetic mean. Take a look:
Task No. 6. Find all values of $x$ for which the numbers $-6((x)^(2))$, $x+1$ and $14+4((x)^(2))$ are consecutive terms of an arithmetic progression (in in the order indicated).
Solution. Since these numbers are members of a progression, the arithmetic mean condition is satisfied for them: the central element $x+1$ can be expressed in terms of neighboring 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 $x=-3$ are the answers.
Answer: −3; 2.
Task No. 7. Find the values of $$ for which the numbers $-1;4-3;(()^(2))+1$ form an arithmetic progression (in that order).
Solution. Let us again express the middle term through the arithmetic mean of neighboring terms:
\[\begin(align) & 4x-3=\frac(x-1+((x)^(2))+1)(2); \\ & 4x-3=\frac(((x)^(2))+x)(2);\quad \left| \cdot 2 \right.; \\ & 8x-6=((x)^(2))+x; \\ & ((x)^(2))-7x+6=0. \\ \end(align)\]
Quadratic equation again. And again there are two roots: $x=6$ and $x=1$.
Answer: 1; 6.
If in the process of solving a problem you come up with some brutal numbers, or you are not entirely sure of the correctness of the answers found, then there is a wonderful technique that allows you to check: have we solved the problem correctly?
Let's say in problem No. 6 we received answers −3 and 2. How can we check that these answers are correct? Let's just plug them into the original 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. Let's substitute $x=-3$:
\[\begin(align) & x=-3\Rightarrow \\ & -6((x)^(2))=-54; \\ & x+1=-2; \\ & 14+4((x)^(2))=50. \end(align)\]
We got the numbers −54; −2; 50 that 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 was solved correctly. Those who wish 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 came across another interesting fact that also needs to be remembered:
If three numbers are such that the second is the arithmetic mean of the first and 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 conditions of the problem. But before we engage in such “construction”, we should pay attention to one more fact, which directly follows from what has already been discussed.
Grouping and summing elements
Let's return to the number axis again. Let us note there several members of the progression, between which, perhaps. is worth a lot of other members:
There are 6 elements marked on the number lineLet's try to express the “left tail” through $((a)_(n))$ and $d$, and the “right tail” through $((a)_(k))$ 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 amounts 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 number $S$, and then begin to step from these elements in opposite directions (toward 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 indentations give equal amounts Understanding this fact will allow us to solve problems of a fundamentally higher level of complexity than those we considered above. For example, these:
Task No. 8. Determine the difference of an 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 progression difference $d$. Actually, the entire 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 the total multiplier of 11 out of the second bracket. Thus, the desired 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, because 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 of the highest term is 11 - this is a positive number, so we are really dealing with a parabola with upward branches:
graph of a 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 using the standard scheme (there is the formula $((d)_(0))=(-b)/(2a)\;$), but it would be much more reasonable to note that the desired vertex lies on the axis symmetry of the parabola, therefore 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 particular hurry to open the brackets: in their 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 on the smallest value (by the way, we never calculated $((y)_(\min ))$ - this is not required of us). At the same time, this number is the difference of the original progression, i.e. we found the answer. :)
Answer: −36
Task No. 9. Between the numbers $-\frac(1)(2)$ and $-\frac(1)(6)$ insert three numbers so that together with these numbers they form an arithmetic progression.
Solution. Essentially, we need to make a sequence of five numbers, with the first and last number 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 the numbers $x$ and $z$, and from the numbers $-\frac(1)(2)$ and $-\frac(1)( 6)$. And if we currently cannot obtain $y$ from the numbers $x$ and $z$, then the situation is different with the ends of the progression. Let's remember the arithmetic mean:
Now, knowing $y$, we will find the remaining numbers. Note that $x$ lies between the numbers $-\frac(1)(2)$ and the $y=-\frac(1)(3)$ we just found. That's why
Using similar reasoning, we find the remaining number:
Ready! We found all three numbers. Let's write them in the answer in the order in which they should be inserted between the original numbers.
Answer: $-\frac(5)(12);\ -\frac(1)(3);\ -\frac(1)(4)$
Task No. 10. Between the numbers 2 and 42, insert several numbers that, 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 complex problem, 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 need to be inserted. Therefore, let us assume for definiteness 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 required arithmetic progression can be represented in the form:
\[\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. And 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:
\[\begin(align) & ((a)_(3))-((a)_(1))=12-2=10; \\ & ((a)_(3))-((a)_(1))=\left(3-1 \right)\cdot d=2d; \\ & 2d=10\Rightarrow d=5. \\ \end(align)\]
All that remains is to find the remaining terms:
\[\begin(align) & ((a)_(1))=2; \\ & ((a)_(2))=2+5=7; \\ & ((a)_(3))=12; \\ & ((a)_(4))=2+3\cdot 5=17; \\ & ((a)_(5))=2+4\cdot 5=22; \\ & ((a)_(6))=2+5\cdot 5=27; \\ & ((a)_(7))=2+6\cdot 5=32; \\ & ((a)_(8))=2+7\cdot 5=37; \\ & ((a)_(9))=2+8\cdot 5=42; \\ \end(align)\]
Thus, already at the 9th step we will arrive at 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 problems. Well, as simple as that: for most students who study mathematics at school and have not read what is written above, these problems may seem tough. Nevertheless, these are the types of problems that appear in the OGE and the Unified State Exam in mathematics, so I recommend that you familiarize yourself with them.
Task No. 11. The team produced 62 parts in January, and in each subsequent month they produced 14 more parts than in the previous month. How many parts did the team produce in November?
Solution. Obviously, the number of parts listed 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 $((a)_(11))$:
\[((a)_(11))=62+10\cdot 14=202\]
Therefore, 202 parts will be produced in November.
Task No. 12. The bookbinding workshop bound 216 books in January, and in each subsequent month it bound 4 more books than in the previous month. How many books did the workshop bind in December?
Solution. All the same:
$\begin(align) & ((a)_(1))=216;\quad d=4; \\ & ((a)_(n))=216+\left(n-1 \right)\cdot 4. \\ \end(align)$
December is the last, 12th month of the year, so we are looking for $((a)_(12))$:
\[((a)_(12))=216+11\cdot 4=260\]
This is the answer - 260 books will be bound in December.
Well, if you have read this far, I hasten to congratulate you: you have successfully completed the “young fighter’s course” in arithmetic progressions. You can safely move on to the next lesson, where we will study the formula for the sum of progression, as well as important and very useful consequences from it.
The concept of a number sequence implies that each natural number corresponds to some real value. Such a series of numbers can be either arbitrary or have certain properties - a progression. In the latter case, each subsequent element (member) of the sequence can be calculated using the previous one.
An arithmetic progression is a sequence of numerical values in which its neighboring members differ from each other by the same number (all elements of the series, starting from the 2nd, have a similar property). This number - the difference between the previous and subsequent terms - is constant and is called the progression difference.
Progression difference: definition
Consider a sequence consisting of j values A = a(1), a(2), a(3), a(4) ... a(j), j belongs to the set of natural numbers N. An arithmetic progression, according to its definition, is a sequence , in which a(3) – a(2) = a(4) – a(3) = a(5) – a(4) = … = a(j) – a(j-1) = d. The value d is the desired difference of this progression.
d = a(j) – a(j-1).
Highlight:
- An increasing progression, in which case d > 0. Example: 4, 8, 12, 16, 20, ...
- Decreasing progression, then d< 0. Пример: 18, 13, 8, 3, -2, …
Difference progression and its arbitrary elements
If 2 arbitrary terms of the progression are known (i-th, k-th), then the difference for a given sequence can be determined based on the relationship:
a(i) = a(k) + (i – k)*d, which means d = (a(i) – a(k))/(i-k).
Difference of progression and its first term
This expression will help determine an unknown value only in cases where the number of the sequence element is known.
Progression difference and its sum
The sum of a progression is the sum of its terms. To calculate the total value of its first j elements, use the appropriate formula:
S(j) =((a(1) + a(j))/2)*j, but since a(j) = a(1) + d(j – 1), then S(j) = ((a(1) + a(1) + d(j – 1))/2)*j=(( 2a(1) + d(– 1))/2)*j.
Online calculator.
Solving an arithmetic progression.
Given: a n , d, n
Find: a 1
This mathematical program finds \(a_1\) of an arithmetic progression based on user-specified numbers \(a_n, d\) and \(n\).
The numbers \(a_n\) and \(d\) can be specified not only as integers, but also as fractions. Moreover, the fractional number can be entered in the form of a decimal fraction (\(2.5\)) and in the form of an ordinary fraction (\(-5\frac(2)(7)\)).
The program not only gives the answer to the problem, but also displays the process of finding a solution.
This online calculator can be useful for high school students in secondary schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions.
In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.
If you are not familiar with the rules for entering numbers, we recommend that you familiarize yourself with them.
Rules for entering numbers
The numbers \(a_n\) and \(d\) can be specified not only as integers, but also as fractions.
The number \(n\) can only be a positive integer.
Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a period or a comma.
For example, you can enter decimal fractions like 2.5 or like 2.5
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
Input:
Result: \(-\frac(2)(3)\)
The whole part is separated from the fraction by the ampersand sign: &
Input:
Result: \(-1\frac(2)(3)\)
It was discovered that some scripts necessary to solve this problem were not loaded, and the program may not work.
You may have AdBlock enabled.
In this case, disable it and refresh the page.
For the solution to appear, you need to enable JavaScript.
Here are instructions on how to enable JavaScript in your browser.
Because There are a lot of people willing to solve the problem, your request has been queued.
In a few seconds the solution will appear below.
Please wait sec...
If you noticed an error in the solution, then you can write about this in the Feedback Form.
Do not forget indicate which task you decide what enter in the fields.
Our games, puzzles, emulators:
A little theory.
Number sequence
In everyday practice, numbering of various objects is often used to indicate the order in which they are arranged. For example, the houses on each street are numbered. In the library, reader's subscriptions are numbered and then arranged in the order of assigned numbers in special card files.
In a savings bank, using the depositor’s personal account number, you can easily find this account and see what deposit is on it. Let account No. 1 contain a deposit of a1 rubles, account No. 2 contain a deposit of a2 rubles, etc. It turns out number sequence
a 1 , a 2 , a 3 , ..., a N
where N is the number of all accounts. Here, each natural number n from 1 to N is associated with a number a n.
Also studied in mathematics infinite number sequences:
a 1 , a 2 , a 3 , ..., a n , ... .
The number a 1 is called first term of the sequence, number a 2 - second term of the sequence, number a 3 - third term of the sequence etc.
The number a n is called nth (nth) member of the sequence, and the natural number n is its number.
For example, in the sequence of squares of natural numbers 1, 4, 9, 16, 25, ..., n 2, (n + 1) 2, ... and 1 = 1 is the first term of the sequence; and n = n 2 is the nth term of the sequence; a n+1 = (n + 1) 2 is the (n + 1)th (n plus first) term of the sequence. Often a sequence can be specified by the formula of its nth term. For example, the formula \(a_n=\frac(1)(n), \; n \in \mathbb(N) \) defines the sequence \(1, \; \frac(1)(2) , \; \frac( 1)(3) , \; \frac(1)(4) , \dots,\frac(1)(n) , \dots \)
Arithmetic progression
The length of the year is approximately 365 days. A more accurate value is \(365\frac(1)(4)\) days, so every four years an error of one day accumulates.
To account for this error, a day is added to every fourth year, and the extended year is called a leap year.
For example, in the third millennium, leap years are the years 2004, 2008, 2012, 2016, ....
In this sequence, each member, starting from the second, is equal to the previous one, added to the same number 4. Such sequences are called arithmetic progressions.
Definition.
The number sequence a 1, a 2, a 3, ..., a n, ... is called arithmetic progression, if for all natural n the equality
\(a_(n+1) = a_n+d, \)
where d is some number.
From this formula it follows that a n+1 - a n = d. The number d is called the difference arithmetic progression.
By definition of an arithmetic progression we have:
\(a_(n+1)=a_n+d, \quad a_(n-1)=a_n-d, \)
where
\(a_n= \frac(a_(n-1) +a_(n+1))(2) \), where \(n>1 \)
Thus, each term of an arithmetic progression, starting from the second, is equal to the arithmetic mean of its two adjacent terms. This explains the name "arithmetic" progression.
Note that if a 1 and d are given, then the remaining terms of the arithmetic progression can be calculated using the recurrent formula a n+1 = a n + d. In this way it is not difficult to calculate the first few terms of the progression, however, for example, a 100 will already require a lot of calculations. Typically, the nth term formula is used for this. By definition of arithmetic progression
\(a_2=a_1+d, \)
\(a_3=a_2+d=a_1+2d, \)
\(a_4=a_3+d=a_1+3d \)
etc.
At all,
\(a_n=a_1+(n-1)d, \)
since the nth term of an arithmetic progression is obtained from the first term by adding (n-1) times the number d.
This formula is called formula for the nth term of an arithmetic progression.
Sum of the first n terms of an arithmetic progression
Find the sum of all natural numbers from 1 to 100.
Let's write this amount in two ways:
S = l + 2 + 3 + ... + 99 + 100,
S = 100 + 99 + 98 + ... + 2 + 1.
Let's add these equalities term by term:
2S = 101 + 101 + 101 + ... + 101 + 101.
This sum has 100 terms
Therefore, 2S = 101 * 100, hence S = 101 * 50 = 5050.
Let us now consider an arbitrary arithmetic progression
a 1 , a 2 , a 3 , ..., a n , ...
Let S n be the sum of the first n terms of this progression:
S n = a 1 , a 2 , a 3 , ..., a n
Then the sum of the first n terms of an arithmetic progression is equal to
\(S_n = n \cdot \frac(a_1+a_n)(2) \)
Since \(a_n=a_1+(n-1)d\), then replacing a n in this formula we get another formula for finding sum of the first n terms of an arithmetic progression:
\(S_n = n \cdot \frac(2a_1+(n-1)d)(2) \)
Important notes!
1. If you see gobbledygook instead of formulas, clear your cache. How to do this in your browser is written here:
2. Before you start reading the article, pay attention to our navigator for the most useful resources for
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 of them as you like (in our case, there are them). No matter how many numbers we write, we can always say which one is first, which one is second, and so on until 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 the same.
The number with number is called the th term of the sequence.
We usually call the entire sequence by 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 number 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 back in the 6th century and was understood in a broader sense as an infinite numerical sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which was studied by the ancient Greeks.
This is a number sequence, each member of which is equal to the previous one added to the same number. This number is called the difference of an arithmetic progression and is designated.
Try to determine which number sequences are an arithmetic progression and which are not:
a)
b)
c)
d)
Got it? Let's compare our answers:
Is 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 term. Exists two way to find it.
1. Method
We can add the progression number to the previous value until we reach the th term of the progression. It’s good that we don’t have much to summarize - only three values:
So, the th term 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 make mistakes when adding numbers.
Of course, mathematicians have come up with a way in which it is not necessary to add the difference of an arithmetic progression to the previous value. Take a closer look at the drawn picture... Surely you have already noticed a certain pattern, namely:
For example, let’s see what the value of the th term of this arithmetic progression consists of: 
In other words:
Try to find the value of a member of a given arithmetic progression yourself in this way.
Did you calculate? Compare your notes with the answer:
Please note that you got exactly the same number as in the previous method, when we sequentially added the terms of the arithmetic progression to the previous value.
Let’s try to “depersonalize” this formula - let’s put it in general form and get:
|
Arithmetic progression equation. |
Arithmetic progressions can be increasing or decreasing.
Increasing- progressions in which each subsequent value of the terms is greater than the previous one.
For example:
Descending- progressions in which each subsequent value of the terms is less than the previous one.
For example:
The derived formula is used in the calculation of 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 be if we use our formula to calculate it:
Since then:
Thus, we are convinced that the formula operates in both decreasing and increasing arithmetic progression.
Try to find the th and th terms of this arithmetic progression yourself.
Let's compare the results:
Arithmetic progression property
Let's complicate the problem - we will derive the property of 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, ah, 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 what if we are given numbers in the condition? Agree, there is a possibility of making a mistake in the calculations.
Now think about whether it is possible to solve this problem in one step using any formula? Of course yes, and that’s what we’ll try to bring out now.
Let us denote the required term of the arithmetic progression as, the formula for finding it is known to us - this is the same formula we derived at the beginning:
, Then:
- the previous term of the progression is:
- the next term of the progression is:
Let's sum up the previous and subsequent terms of the progression:
It turns out that the sum of the previous and subsequent terms of the progression is the double value of the progression term located between them. In other words, to find the value of a progression term with known previous and successive values, you need to add them and divide by.
That's right, we got the same number. Let's secure the material. Calculate the value for the progression yourself, it’s not at all difficult.
Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, was easily deduced by one of the greatest mathematicians of all time, the “king of mathematicians” - Karl Gauss...
When Carl Gauss was 9 years old, a teacher, busy checking the work of students in other classes, assigned the following task in class: “Calculate the sum of all natural numbers from to (according to other sources to) inclusive.” Imagine the teacher’s surprise when one of his students (this was Karl Gauss) a minute later gave the correct answer to the task, while most of the daredevil’s classmates, after long calculations, received the wrong result...
Young Carl Gauss noticed a certain pattern that you can easily notice too.
Let's say we have an arithmetic progression consisting of -th terms: We need to find the sum of these terms of the arithmetic progression. Of course, we can manually sum all the values, but what if the task requires finding the sum of its terms, as Gauss was looking for?
Let us depict the progression given to us. Take a closer look at the highlighted numbers and try to perform various mathematical operations with them. 
Have you tried it? What did you notice? Right! Their sums are equal
Now tell me, how many such pairs are there in total in the progression given to us? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two terms of an arithmetic progression is equal, and similar pairs are equal, we obtain that the total sum is equal to:
.
Thus, the formula for the sum of the first terms of any arithmetic progression will be:
In some problems we do not know the th term, but we know the difference of the progression. Try to substitute the formula of the th term into the sum formula.
What did you get?
Well done! Now let's return to the problem that was asked to Carl Gauss: calculate for yourself what the sum of numbers starting from the th is equal to and the sum of the numbers starting from the th.
How much did you get?
Gauss found that the sum of the terms is equal, and the sum of the terms. Is that what you decided?
In fact, the formula for the sum of the terms of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people made full use of the properties of the arithmetic progression.
For example, imagine Ancient Egypt and the largest construction project of that time - the construction of a pyramid... The picture shows one side of it. 
Where is the progression here, you say? Look carefully and find a pattern in the number of sand blocks in each row of the pyramid wall. 
Why not an arithmetic progression? Calculate how many blocks are needed to build one wall if block bricks are placed at the base. I hope you won’t count while moving your finger across the monitor, you remember the last formula and everything we said about arithmetic progression?
In this case, the progression looks like this: .
Arithmetic progression difference.
The number of terms of an arithmetic progression.
Let's substitute our data into the last formulas (calculate 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. Got it? Well done, you have mastered the sum of the nth terms of an 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:
Training
Tasks:
- Masha is getting in shape for summer. Every day she increases the number of squats by. How many times will Masha do squats in a week if she did squats at the first training session?
- What is the sum of all odd numbers contained in.
- When storing logs, loggers stack them in such a way that each top layer contains one log less than the previous one. How many logs are in one masonry, if the foundation of the masonry is logs?
Answers:
- Let us define the parameters of the arithmetic progression. In this case
(weeks = days).Answer: In two weeks, Masha should do squats once a day.
- First odd number, last number.
Arithmetic progression difference.
The number of odd numbers in is half, however, let’s check this fact using the formula for finding the th term of an arithmetic progression:Numbers do contain odd numbers.
Let's substitute the available data into the formula:Answer: The sum of all odd numbers contained in is equal.
- Let's remember the problem about pyramids. For our case, a , since each top layer is reduced by one log, then in total there are a bunch of layers, that is.
Let's substitute the data into the formula:Answer: There are logs in the masonry.
Let's sum it up
- - a number sequence in which the difference between adjacent numbers is the same and equal. It can be increasing or decreasing.
- Finding formula The th term of an arithmetic progression is written by the formula - , where is the number of numbers in the progression.
- Property of members of an arithmetic progression- - where is the number of numbers in progression.
- The sum of the terms 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 of them as you like. But we can always say which one is first, which one is 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 a unique one. And we will not assign this number to any other number from this set.
The number with number is called the th member of the sequence.
We usually call the entire sequence by some letter (for example,), and each member of this sequence is the same letter with an index equal to the number of this member: .
It is very convenient if the th term of the sequence can be specified by some formula. For example, the formula
sets 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 is). Or (, difference).
nth term formula
We call a formula recurrent in which, in order to find out the th term, you need to know the previous or several previous ones:
To find, for example, the th term of the progression using this formula, we will have to calculate the previous nine. For example, let it. Then:
Well, is it clear now what the formula is?
In each line we add to, multiplied by some number. Which one? 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? Here's what:
(This is why it is called difference because it is equal to the difference of successive terms of the progression).
So, the formula:
Then the hundredth term is equal to:
What is the sum of all natural numbers from to?
According to legend, the great mathematician Carl Gauss, as 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 penultimate is the same, the sum of the third and 3rd from the end is the same, and so on. How many such pairs are there in total? 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 will be:
Example:
Find the sum of all two-digit multiples.
Solution:
The first such number is this. Each subsequent number is obtained by adding to the previous number. Thus, the numbers we are interested in form an arithmetic progression with the first term and the difference.
Formula of the th term for this progression:
How many terms are there in the progression if they all have to be two-digit?
Very easy: .
The last term of the progression will be equal. Then the sum:
Answer: .
Now decide for yourself:
- Every day the athlete runs more meters than the previous day. How many total kilometers will he run in a week if he ran km m on the first day?
- A cyclist travels more kilometers every day than the previous day. On the first day he traveled km. How many days does he need to travel to cover a kilometer? How many kilometers will he travel during the last day of his journey?
- The price of a refrigerator in a store decreases by the same amount every year. Determine how much the price of a refrigerator decreased each year if, put up for sale for rubles, six years later it was sold for rubles.
Answers:
- The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
.
Answer: - Here it is given: , must be found.
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 path traveled over the last day using the formula of the th term:
(km).
Answer: - Given: . Find: .
It couldn't be simpler:
(rub).
Answer:
ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS
This is a number sequence in which the difference between adjacent numbers is the same and equal.
Arithmetic progression can be increasing () and decreasing ().
For example:
Formula for finding the nth term of an arithmetic progression
is written by the formula, where is the number of numbers in progression.
Property of members of an arithmetic progression
It allows you to easily find a term of a progression if its neighboring terms are known - where is the number of numbers in the progression.
Sum of terms of an arithmetic progression
There are two ways to find the amount:
Where is the number of values.
Where is the number of values.
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
Now the most important thing.
You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough...
For what?
For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.
I won’t convince you of anything, I’ll just say one thing...
People who have received a good education earn much more than those who have not received it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?
GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
You won't be asked for theory during the exam.
You will need solve problems against time.
And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.
It's like in sports - you need to repeat it many times to win for sure.
Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!
You can use our tasks (optional) and we, of course, recommend them.
In order to get better at using our tasks, you need to help extend the life of the YouClever textbook you are currently reading.
How? There are two options:
- Unlock all hidden tasks in this article -
- Unlock access to all hidden tasks in all 99 articles of the textbook - Buy a textbook - 499 RUR
Yes, we have 99 such articles in our textbook and access to all tasks and all hidden texts in them can be opened immediately.
Access to all hidden tasks is provided for the ENTIRE life of the site.
In conclusion...
If you don't like our tasks, find others. Just don't stop at theory.
“Understood” and “I can solve” are completely different skills. You need both.
Find problems and solve them!
Before we start deciding arithmetic progression problems, let's consider what a number sequence is, since an arithmetic progression is a special case of a number sequence.
A number sequence is a number set, each element of which has its own serial number. The elements of this set are called members of the sequence. The serial number of a sequence element is indicated by an index:
The first element of the sequence;
The fifth element of the sequence;
- the “nth” element of the sequence, i.e. element "standing in queue" at number n.
There is a relationship between the value of a sequence element and its sequence number. Therefore, we can consider a sequence as a function whose argument is the ordinal number of the element of the sequence. In other words, we can say that the sequence is a function of the natural argument:
The sequence can be set in three ways:
1 . The sequence can be specified using a table. In this case, we simply set the value of each member of the sequence.
For example, Someone decided to take up personal time management, and to begin with, count how much time he spends on VKontakte during the week. By recording the time in the table, he will receive a sequence consisting of seven elements:
The first line of the table indicates the number of the day of the week, the second - the time in minutes. We see that, that is, on Monday Someone spent 125 minutes on VKontakte, that is, on Thursday - 248 minutes, and, that is, on Friday only 15.
2 . The sequence can be specified using the nth term formula.
In this case, the dependence of the value of a sequence element on its number is expressed directly in the form of a formula.
For example, if , then
To find the value of a sequence element with a given number, we substitute the element number into the formula of the nth term.
We do the same thing if we need to find the value of a function if the value of the argument is known. We substitute the value of the argument into the function equation:
If, for example, 
, That
Let me note once again that in a sequence, unlike an arbitrary numerical function, the argument can only be a natural number.
3 . The sequence can be specified using a formula that expresses the dependence of the value of the sequence member number n on the values of the previous members. In this case, it is not enough for us to know only the number of the sequence member to find its value. We need to specify the first member or first few members of the sequence.
For example, consider the sequence 
, 
We can find the values of sequence members in sequence, starting from the third:
That is, every time, to find the value of the nth term of the sequence, we return to the previous two. This method of specifying a sequence is called recurrent, from the Latin word recurro- come back.
Now we can define an arithmetic progression. An arithmetic progression is a simple special case of a number sequence.
Arithmetic progression is a numerical sequence, each member of which, starting from the second, is equal to the previous one added to the same number.
The number is called difference of arithmetic progression. The difference of an arithmetic progression can be positive, negative, or equal to zero.
If title="d>0">, то каждый член арифметической прогрессии больше предыдущего, и прогрессия является !} increasing.
For example, 2; 5; 8; eleven;...
If , then each term of an arithmetic progression is less than the previous one, and the progression is decreasing.
For example, 2; -1; -4; -7;...
If , then all terms of the progression are equal to the same number, and the progression is stationary.
For example, 2;2;2;2;...
The main property of an arithmetic progression:
Let's look at the drawing.
We see that
, and at the same time
Adding these two equalities, we get:
.
Let's divide both sides of the equality by 2:
So, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the two neighboring ones:
Moreover, since
, and at the same time
, That
, and therefore
Each term of an arithmetic progression, starting with title="k>l">, равен среднему арифметическому двух равноотстоящих.
!}
Formula of the th term.
We see that the terms of the arithmetic progression satisfy the following relations:
and finally
We got formula of the nth term.
IMPORTANT! Any member of an arithmetic progression can be expressed through and. Knowing the first term and the difference of an arithmetic progression, you can find any of its terms.
The sum of n terms of an arithmetic progression.
In an arbitrary arithmetic progression, the sums of terms equidistant from the extreme ones are equal to each other:
Consider an arithmetic progression with n terms. Let the sum of n terms of this progression be equal to .
Let's arrange the terms of the progression first in ascending order of numbers, and then in descending order:
Let's add in pairs:
The sum in each bracket is , the number of pairs is n.
We get:
So, the sum of n terms of an arithmetic progression can be found using the formulas:
Let's consider solving arithmetic progression problems.
1 . The sequence is given by the formula of the nth term: . Prove that this sequence is an arithmetic progression.
Let us prove that the difference between two adjacent terms of the sequence is equal to the same number.
We found that the difference between two adjacent members of the sequence does not depend on their number and is a constant. Therefore, by definition, this sequence is an arithmetic progression.
2 . Given an arithmetic progression -31; -27;...
a) Find 31 terms of the progression.
b) Determine whether the number 41 is included in this progression.
A) We see that ;
Let's write down the formula for the nth term for our progression.
In general 
In our case 
, That's why 