Yes, yes: arithmetic progression- these are not toys 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 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: every 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: 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 it! 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 us 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:
\[((a)_(n))-((a)_(m))=d\cdot \left(n-m \right)\]
Simple but very useful property, which 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 it! 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: until when (i.e. until what natural number$n$) the negativity of the terms is preserved:
\[\begin(align) & ((a)_(n)) \lt 0\Rightarrow ((a)_(1))+\left(n-1 \right)d \lt 0; \\ & -38.5+\left(n-1 \right)\cdot 2.7 \lt 0;\quad \left| \cdot 10 \right. \\ & -385+27\cdot \left(n-1 \right) \lt 0; \\ & -385+27n-27 \lt 0; \\ & 27n \lt 412; \\ & n \lt 15\frac(7)(27)\Rightarrow ((n)_(\max ))=15. \\ \end(align)\]
The last line requires 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 recurrence formula and write it down for all marked members:
\[\begin(align) & ((a)_(n-2))=((a)_(n-3))+d; \\ & ((a)_(n-1))=((a)_(n-2))+d; \\ & ((a)_(n))=((a)_(n-1))+d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n+1))+d; \\ \end(align)\]
However, these equalities can be rewritten differently:
\[\begin(align) & ((a)_(n-1))=((a)_(n))-d; \\ & ((a)_(n-2))=((a)_(n))-2d; \\ & ((a)_(n-3))=((a)_(n))-3d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n))+2d; \\ & ((a)_(n+3))=((a)_(n))+3d; \\ \end(align)\]
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:
\[\begin(align) & x+1=\frac(-6((x)^(2))+14+4((x)^(2)))(2); \\ & x+1=\frac(14-2((x)^(2)))(2); \\ & x+1=7-((x)^(2)); \\ & ((x)^(2))+x-6=0. \\ \end(align)\]
It turned out 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, which 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 in a fundamentally more high level difficulties 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 overall multiplier of 11 out of the second bracket. Thus, the required 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:
schedule quadratic function- parabola
Please note: this parabola takes its minimum value at its vertex with the abscissa $((d)_(0))$. Of course, we can calculate this abscissa 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 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 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 from 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 from the numbers $x$ and $z$ we are in at the moment we can’t get $y$, 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. 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 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 tasks. 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 one. How many books did the workshop bind in December?
Solution. Everything is 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.
In mathematics, any collection of numbers that follow each other, organized in some way, is called a sequence. Of all the existing sequences of numbers, two interesting cases are distinguished: algebraic and geometric progressions.
What is an arithmetic progression?
It should be said right away that algebraic progression is often called arithmetic, since its properties are studied by the branch of mathematics - arithmetic.
This progression is a sequence of numbers in which each next member differs from the previous one by a certain constant number. It is called the difference of an algebraic progression. For definiteness, we denote it by the Latin letter d.
An example of such a sequence could be the following: 3, 5, 7, 9, 11 ..., here you can see that the number is 5 more number 3 is 2, 7 is more than 5 is also 2, and so on. Thus, in the example presented, d = 5-3 = 7-5 = 9-7 = 11-9 = 2.
What are the types of arithmetic progressions?
The nature of these ordered sequences of numbers is largely determined by the sign of the number d. The following types of algebraic progressions are distinguished:
- increasing when d is positive (d>0);
- constant when d = 0;
- decreasing when d is negative (d<0).
The example given in the previous paragraph shows an increasing progression. An example of a decreasing sequence is the following sequence of numbers: 10, 5, 0, -5, -10, -15 ... A constant progression, as follows from its definition, is a collection of identical numbers.
nth term of progression
Due to the fact that each subsequent number in the progression under consideration differs by a constant d from the previous one, its nth term can be easily determined. To do this, you need to know not only d, but also a 1 - the first term of the progression. Using a recursive approach, one can obtain an algebraic progression formula for finding the nth term. It looks like: a n = a 1 + (n-1)*d. This formula is quite simple and can be understood intuitively.
It is also not difficult to use. For example, in the progression given above (d=2, a 1 =3), we define its 35th term. According to the formula, it will be equal to: a 35 = 3 + (35-1)*2 = 71.
Formula for amount
When given an arithmetic progression, the sum of its first n terms is a frequently encountered problem, along with determining the value of the nth term. The formula for the sum of an algebraic progression is written in the following form: ∑ n 1 = n*(a 1 +a n)/2, here the icon ∑ n 1 indicates that they are summed from 1st to nth term.
The above expression can be obtained by resorting to the properties of the same recursion, but there is an easier way to prove its validity. Let's write down the first 2 and last 2 terms of this sum, expressing them in numbers a 1, a n and d, and we get: a 1, a 1 +d,...,a n -d, a n. Now note that if we add the first term to the last, it will be exactly equal to the sum of the second and penultimate terms, that is, a 1 +a n. In a similar way, it can be shown that the same sum can be obtained by adding the third and penultimate terms, and so on. In the case of a pair of numbers in the sequence, we obtain n/2 sums, each of which is equal to a 1 +a n. That is, we obtain the above formula for the algebraic progression for the sum: ∑ n 1 = n*(a 1 +a n)/2.
For an unpaired number of terms n, a similar formula is obtained if you follow the described reasoning. Just remember to add the remaining term, which is in the center of the progression.
Let's show how to use the above formula using the example of a simple progression that was introduced above (3, 5, 7, 9, 11 ...). For example, it is necessary to determine the sum of its first 15 terms. First, let's define a 15. Using the formula for the nth term (see the previous paragraph), we get: a 15 = a 1 + (n-1)*d = 3 + (15-1)*2 = 31. Now we can apply the formula for the sum of an algebraic progression: ∑ 15 1 = 15*(3+31)/2 = 255.
It is interesting to cite an interesting historical fact. The formula for the sum of an arithmetic progression was first obtained by Carl Gauss (the famous German mathematician of the 18th century). When he was only 10 years old, the teacher asked the problem to find the sum of numbers from 1 to 100. They say that little Gauss solved this problem in a few seconds, noticing that by summing the numbers from the beginning and end of the sequence in pairs, you can always get 101, and since there are 50 such sums, he quickly gave the answer: 50*101 = 5050.
Example of problem solution
To complete the topic of algebraic progression, we will give an example of solving another interesting problem, thereby strengthening the understanding of the topic under consideration. Let a certain progression be given for which the difference d = -3 is known, as well as its 35th term a 35 = -114. It is necessary to find the 7th term of the progression a 7 .
As can be seen from the conditions of the problem, the value of a 1 is unknown, therefore it will not be possible to use the formula for the nth term directly. The recursion method is also inconvenient, which is difficult to implement manually, and there is a high probability of making a mistake. Let's proceed as follows: write out the formulas for a 7 and a 35, we have: a 7 = a 1 + 6*d and a 35 = a 1 + 34*d. Subtract the second from the first expression, we get: a 7 - a 35 = a 1 + 6*d - a 1 - 34*d. It follows: a 7 = a 35 - 28*d. It remains to substitute the known data from the problem statement and write down the answer: a 7 = -114 - 28*(-3) = -30.
Geometric progression
To reveal the topic of the article more fully, we provide a brief description of another type of progression - geometric. In mathematics, this name is understood as a sequence of numbers in which each subsequent term differs from the previous one by a certain factor. Let's denote this factor by the letter r. It is called the denominator of the type of progression under consideration. An example of this number sequence would be: 1, 5, 25, 125, ...
As can be seen from the above definition, algebraic and geometric progressions are similar in idea. The difference between them is that the first one changes more slowly than the second one.
Geometric progression can also be increasing, constant or decreasing. Its type depends on the value of the denominator r: if r>1, then there is an increasing progression, if r<1 - убывающая, наконец, если r = 1 - постоянная, которая в этом случае может также называться постоянной арифметической прогрессией.
Geometric progression formulas
As in the case of algebraic, the formulas of a geometric progression are reduced to determining its nth term and the sum of n terms. Below are these expressions:
- a n = a 1 *r (n-1) - this formula follows from the definition of geometric progression.
- ∑ n 1 = a 1 *(r n -1)/(r-1). It is important to note that if r = 1, then the above formula gives uncertainty, so it cannot be used. In this case, the sum of n terms will be equal to the simple product a 1 *n.
For example, let’s find the sum of only 10 terms of the sequence 1, 5, 25, 125, ... Knowing that a 1 = 1 and r = 5, we get: ∑ 10 1 = 1*(5 10 -1)/4 = 2441406. The resulting value is a clear example of how quickly the geometric progression grows.
Perhaps the first mention of this progression in history is the legend with the chessboard, when a friend of one Sultan, having taught him to play chess, asked for grain for his service. Moreover, the amount of grain should have been as follows: one grain must be placed on the first square of the chessboard, twice as much on the second as on the first, on the third twice as much as on the second, and so on. The Sultan willingly agreed to fulfill this request, but he did not know that he would have to empty all the bins of his country in order to keep his word.
Or arithmetic is a type of ordered numerical sequence, the properties of which are studied in a school algebra course. This article discusses in detail the question of how to find the sum of an arithmetic progression.
What kind of progression is this?
Before moving on to the question (how to find the sum of an arithmetic progression), it is worth understanding what we are talking about.
Any sequence of real numbers that is obtained by adding (subtracting) some value from each previous number is called an algebraic (arithmetic) progression. This definition, when translated into mathematical language, takes the form:
Here i - serial number element of the series a i . Thus, knowing just one starting number, you can easily restore the entire series. The parameter d in the formula is called the progression difference.
It can be easily shown that for the series of numbers under consideration the following equality holds:
a n = a 1 + d * (n - 1).
That is, to find the value of the nth element in order, you should add the difference d to the first element a 1 n-1 times.
What is the sum of an arithmetic progression: formula
Before giving the formula for the indicated amount, it is worth considering a simple special case. Given a progression of natural numbers from 1 to 10, you need to find their sum. Since there are few terms in the progression (10), it is possible to solve the problem head-on, that is, sum all the elements in order.
S 10 = 1+2+3+4+5+6+7+8+9+10 = 55.
It is worth considering one interesting thing: since each term differs from the next one by the same value d = 1, then the pairwise summation of the first with the tenth, the second with the ninth, and so on will give the same result. Really:
11 = 1+10 = 2+9 = 3+8 = 4+7 = 5+6.
As you can see, there are only 5 of these sums, that is, exactly two times less than the number of elements of the series. Then multiplying the number of sums (5) by the result of each sum (11), you will arrive at the result obtained in the first example.
If we generalize these arguments, we can write the following expression:
S n = n * (a 1 + a n) / 2.
This expression shows that it is not at all necessary to sum all the elements in a row; it is enough to know the value of the first a 1 and the last a n, as well as the total number of terms n.
It is believed that Gauss first thought of this equality when he was looking for a solution to a problem given by his school teacher: sum the first 100 integers.
Sum of elements from m to n: formula
The formula given in the previous paragraph answers the question of how to find the sum of an arithmetic progression (the first elements), but often in problems it is necessary to sum a series of numbers in the middle of the progression. How to do this?
The easiest way to answer this question is by considering the following example: let it be necessary to find the sum of terms from the m-th to the n-th. To solve the problem, you should present the given segment from m to n of the progression in the form of a new number series. In this representation, the mth term a m will be the first, and a n will be numbered n-(m-1). In this case, applying the standard formula for the sum, the following expression will be obtained:
S m n = (n - m + 1) * (a m + a n) / 2.
Example of using formulas
Knowing how to find the sum of an arithmetic progression, it is worth considering a simple example of using the above formulas.
Below is given number sequence, you should find the sum of its terms, starting from the 5th and ending with the 12th:
The given numbers indicate that the difference d is equal to 3. Using the expression for the nth element, you can find the values of the 5th and 12th terms of the progression. It turns out:
a 5 = a 1 + d * 4 = -4 + 3 * 4 = 8;
a 12 = a 1 + d * 11 = -4 + 3 * 11 = 29.
Knowing the values of the numbers at the ends of the algebraic progression under consideration, as well as knowing what numbers in the series they occupy, you can use the formula for the sum obtained in the previous paragraph. It will turn out:
S 5 12 = (12 - 5 + 1) * (8 + 29) / 2 = 148.
It is worth noting that this value could be obtained differently: first find the sum of the first 12 elements using the standard formula, then calculate the sum of the first 4 elements using the same formula, then subtract the second from the first sum.
For example, the sequence \(2\); \(5\); \(8\); \(11\); \(14\)... is an arithmetic progression, because each subsequent element differs from the previous one by three (can be obtained from the previous one by adding three):
In this progression, the difference \(d\) is positive (equal to \(3\)), and therefore each next term is greater than the previous one. Such progressions are called increasing.
However, \(d\) can also be a negative number. For example, in arithmetic progression \(16\); \(10\); \(4\); \(-2\); \(-8\)... the progression difference \(d\) is equal to minus six.
And in this case, each next element will be smaller than the previous one. These progressions are called decreasing.
Arithmetic progression notation
Progression is indicated by a small Latin letter.
Numbers that form a progression are called members(or elements).
They are denoted by the same letter as an arithmetic progression, but with a numerical index equal to the number of the element in order.
For example, the arithmetic progression \(a_n = \left\( 2; 5; 8; 11; 14...\right\)\) consists of the elements \(a_1=2\); \(a_2=5\); \(a_3=8\) and so on.
In other words, for the progression \(a_n = \left\(2; 5; 8; 11; 14…\right\)\)
Solving arithmetic progression problems
In principle, the information presented above is already enough to solve almost any arithmetic progression problem (including those offered at the OGE).
Example (OGE).
The arithmetic progression is specified by the conditions \(b_1=7; d=4\). Find \(b_5\).
Solution:
Answer: \(b_5=23\)
Example (OGE).
The first three terms of an arithmetic progression are given: \(62; 49; 36…\) Find the value of the first negative term of this progression..
Solution:
|
We are given the first elements of the sequence and know that it is an arithmetic progression. That is, each element differs from its neighbor by the same number. Let's find out which one by subtracting the previous one from the next element: \(d=49-62=-13\). |
|
|
Now we can restore our progression to the (first negative) element we need. |
|
|
Ready. You can write an answer. |
Answer: \(-3\)
Example (OGE).
Given several consecutive elements of an arithmetic progression: \(…5; x; 10; 12.5...\) Find the value of the element designated by the letter \(x\).
Solution:
|
|
To find \(x\), we need to know how much the next element differs from the previous one, in other words, the progression difference. Let's find it from two known neighboring elements: \(d=12.5-10=2.5\). |
|
|
And now we can easily find what we are looking for: \(x=5+2.5=7.5\). |
|
|
Ready. You can write an answer. |
Answer: \(7,5\).
Example (OGE).
The arithmetic progression is defined by the following conditions: \(a_1=-11\); \(a_(n+1)=a_n+5\) Find the sum of the first six terms of this progression.
Solution:
|
We need to find the sum of the first six terms of the progression. But we don’t know their meanings; we are given only the first element. Therefore, we first calculate the values one by one, using what is given to us: \(n=1\); \(a_(1+1)=a_1+5=-11+5=-6\) |
|
|
\(S_6=a_1+a_2+a_3+a_4+a_5+a_6=\) |
The required amount has been found. |
Answer: \(S_6=9\).
Example (OGE).
In arithmetic progression \(a_(12)=23\); \(a_(16)=51\). Find the difference of this progression.
Solution:
Answer: \(d=7\).
Important formulas for arithmetic progression
As you can see, many problems on arithmetic progression can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each subsequent element in this chain is obtained by adding the same number to the previous one (the difference of the progression).
However, sometimes there are situations when deciding “head-on” is very inconvenient. For example, imagine that in the very first example we need to find not the fifth element \(b_5\), but the three hundred and eighty-sixth \(b_(386)\). Do we need to add four \(385\) times? Or imagine that in the penultimate example you need to find the sum of the first seventy-three elements. You'll be tired of counting...
Therefore, in such cases they do not solve things “head-on”, but use special formulas derived for arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum of \(n\) first terms.
Formula of the \(n\)th term: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first term of the progression;
\(n\) – number of the required element;
\(a_n\) – term of the progression with number \(n\).
This formula allows us to quickly find even the three-hundredth or the millionth element, knowing only the first and the difference of the progression.
Example.
The arithmetic progression is specified by the conditions: \(b_1=-159\); \(d=8.2\). Find \(b_(246)\).
Solution:
Answer: \(b_(246)=1850\).
Formula for the sum of the first n terms: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where
\(a_n\) – the last summed term;
Example (OGE).
The arithmetic progression is specified by the conditions \(a_n=3.4n-0.6\). Find the sum of the first \(25\) terms of this progression.
Solution:
|
\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\) |
To calculate the sum of the first twenty-five terms, we need to know the value of the first and twenty-fifth terms. |
|
|
\(n=1;\) \(a_1=3.4·1-0.6=2.8\) |
Now let's find the twenty-fifth term by substituting twenty-five instead of \(n\). |
|
|
\(n=25;\) \(a_(25)=3.4·25-0.6=84.4\) |
Well, now we can easily calculate the required amount. |
|
|
\(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25=\) |
The answer is ready. |
Answer: \(S_(25)=1090\).
For the sum \(n\) of the first terms, you can get another formula: you just need to \(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25\ ) instead of \(a_n\) substitute the formula for it \(a_n=a_1+(n-1)d\). We get:
Formula for the sum of the first n terms: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where
\(S_n\) – the required sum of \(n\) first elements;
\(a_1\) – the first summed term;
\(d\) – progression difference;
\(n\) – number of elements in total.
Example.
Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15.5\); \(14\)…
Solution:
Answer: \(S_(33)=-231\).
More complex arithmetic progression problems
Now you have all the information you need to solve almost any arithmetic progression problem. Let’s finish the topic by considering problems in which you not only need to apply formulas, but also think a little (in mathematics this can be useful ☺)
Example (OGE).
Find the sum of all negative terms of the progression: \(-19.3\); \(-19\); \(-18.7\)…
Solution:
|
\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\) |
The task is very similar to the previous one. We begin to solve the same thing: first we find \(d\). |
|
|
\(d=a_2-a_1=-19-(-19.3)=0.3\) |
Now I would like to substitute \(d\) into the formula for the sum... and here a small nuance emerges - we do not know \(n\). In other words, we don’t know how many terms will need to be added. How to find out? Let's think. We will stop adding elements when we reach the first positive element. That is, you need to find out the number of this element. How? Let's write down the formula for calculating any element of an arithmetic progression: \(a_n=a_1+(n-1)d\) for our case. |
|
|
\(a_n=a_1+(n-1)d\) |
||
|
\(a_n=-19.3+(n-1)·0.3\) |
We need \(a_n\) to become greater than zero. Let's find out at what \(n\) this will happen. |
|
|
\(-19.3+(n-1)·0.3>0\) |
||
|
\((n-1)·0.3>19.3\) \(|:0.3\) |
We divide both sides of the inequality by \(0.3\). |
|
|
\(n-1>\)\(\frac(19.3)(0.3)\) |
We transfer minus one, not forgetting to change the signs |
|
|
\(n>\)\(\frac(19.3)(0.3)\) \(+1\) |
Let's calculate... |
|
|
\(n>65,333…\) |
...and it turns out that the first positive element will have the number \(66\). Accordingly, the last negative one has \(n=65\). Just in case, let's check this. |
|
|
\(n=65;\) \(a_(65)=-19.3+(65-1)·0.3=-0.1\) |
So we need to add the first \(65\) elements. |
|
|
\(S_(65)=\) \(\frac(2 \cdot (-19.3)+(65-1)0.3)(2)\)\(\cdot 65\) |
The answer is ready. |
Answer: \(S_(65)=-630.5\).
Example (OGE).
The arithmetic progression is specified by the conditions: \(a_1=-33\); \(a_(n+1)=a_n+4\). Find the sum from the \(26\)th to the \(42\) element inclusive.
Solution:
|
\(a_1=-33;\) \(a_(n+1)=a_n+4\) |
In this problem you also need to find the sum of elements, but starting not from the first, but from the \(26\)th. For such a case we do not have a formula. How to decide? |
|
|
For our progression \(a_1=-33\), and the difference \(d=4\) (after all, we add the four to the previous element to find the next one). Knowing this, we find the sum of the first \(42\)-y elements. |
|
\(S_(42)=\) \(\frac(2 \cdot (-33)+(42-1)4)(2)\)\(\cdot 42=\) |
Now the sum of the first \(25\) elements. |
|
\(S_(25)=\) \(\frac(2 \cdot (-33)+(25-1)4)(2)\)\(\cdot 25=\) |
And finally, we calculate the answer. |
|
\(S=S_(42)-S_(25)=2058-375=1683\) |
Answer: \(S=1683\).
For arithmetic progression, there are several more formulas that we did not consider in this article due to their low practical usefulness. However, you can easily find them.
Some people treat the word “progression” with caution, 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 exist). And understanding the essence (and in mathematics there is nothing more important than “getting the essence”) of an arithmetic sequence is not so difficult, having analyzed a few elementary concepts.
Mathematical number sequence
A numerical sequence is usually called a series of numbers, each of which has its own number.
a 1 is the first member of the sequence;
and 2 is the second term of the sequence;
and 7 is the seventh member of the sequence;
and n is the nth member of the sequence;
However, not any arbitrary set of numbers and numbers interests us. We will focus our attention on a numerical sequence in which the value of the nth term is related to its ordinal number by a relationship that can be clearly formulated mathematically. In other words: the numerical value of the nth number is some function of n.
a is the value of a member of a numerical sequence;
n is its serial number;
f(n) is a function, where the ordinal number in the numerical sequence n is the 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 term of an arithmetic sequence is as follows:
a n - the value of the current member of the arithmetic progression;
a n+1 - formula of the next number;
d - difference (certain number).
It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one and such an arithmetic progression will be increasing.
In the graph below it is easy to see why the number sequence is called “increasing”.
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.
Specified member value
Sometimes it is necessary to determine the value of any arbitrary term a n of an arithmetic progression. This can be done by sequentially calculating 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 value of the five thousandth or eight millionth term. Traditional calculations will take a lot of time. However, a specific arithmetic progression can be studied using certain formulas. There is also a formula for the nth term: the value of any term of an arithmetic progression can be determined as the sum of the first term of the progression with the difference of the progression, multiplied by the number of the desired term, reduced by one.
The formula is universal for increasing and decreasing progression.
An example of calculating the value of a given term
Let us 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 of the sequence is 3;
The difference in the number series is 1.2.
Task: you need to find the value of 214 terms
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 of the sequence is equal to 258.6.
The advantages of this method of calculation are obvious - the entire solution takes no more than 2 lines.
Sum of a given number of terms
Very often, in a given arithmetic series, it is necessary to determine the sum of the values of some of its segments. To do this, there is also no need to calculate the values of each term and then add them up. This method is applicable if the number of terms whose sum needs to be found is small. In other cases, it is more convenient to use the following formula.
The sum of the terms of an arithmetic progression from 1 to n is equal to the sum of the first and nth terms, multiplied by the number of the term 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 of the sequence is zero;
The difference is 0.5.
The problem requires determining the sum of the terms of the series from 56 to 101.
Solution. Let's use the formula for determining the amount of progression:
s(n) = (2∙a1 + d∙(n-1))∙n/2
First, we determine the sum of the values of 101 terms of the progression by substituting the given 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 terms of the progression from the 56th to the 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 is:
s 101 - s 55 = 2,525 - 742.5 = 1,782.5
Example of practical application of arithmetic progression
At the end of the article, let's return to the example of an arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider this example.
Boarding a taxi (which includes 3 km of travel) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles/km. Travel distance is 30 km. Calculate the cost of the trip.
1. Let’s discard the first 3 km, the price of which is included in the cost of landing.
30 - 3 = 27 km.
2. Further calculation is nothing more than parsing an arithmetic number series.
Member number - the number of kilometers traveled (minus the first three).
The value of the member is the sum.
The first term in this problem will be equal to a 1 = 50 rubles.
Progression difference d = 22 r.
the number we are interested in is the value of the (27+1)th term of the arithmetic progression - the meter reading at the end of the 27th kilometer is 27.999... = 28 km.
a 28 = 50 + 22 ∙ (28 - 1) = 644
Calendar data calculations 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 the celestial body to the star. In addition, various number series are successfully used in statistics and other applied areas of mathematics.
Another type of number sequence is geometric
Geometric progression is characterized by greater rates of change compared to arithmetic progression. It is no coincidence that in politics, sociology, and medicine, in order to show the high speed of spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops in geometric progression.
The Nth term of the geometric number 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 correspondingly equal to 2, 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 term of the geometric progression;
b n+1 - formula of the next term of the geometric progression;
q is the denominator of the geometric progression (a constant number).
If the graph of an arithmetic progression is a straight line, then a geometric progression paints a slightly different picture:
As in the case of arithmetic, geometric progression has a formula for the value of an arbitrary term. Any nth term of a geometric progression is equal to the product of the first term and 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. Let's 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 terms is also calculated 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 discussed above, the value of the sum of the first n terms of the number series under consideration will take the form:
Example. The geometric progression starts with the first term equal to 1. The denominator is set to 3. Let's find the sum of the first eight terms.
s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280
