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; ...
- $\sqrt(5);\ \sqrt(5)-1;\ \sqrt(5)-2;\ \sqrt(5)-3;...$
OK OK: last example may seem overly complicated. But the rest, I think, you understand. Therefore, we introduce new definitions:
Definition. Arithmetic progression 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:
\[((a)_(n))-((a)_(m))=d\cdot \left(n-m \right)\]
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 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, 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 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:
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 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 from the numbers $x$ and $z$ we are in this 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 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.
What the main point formulas?
This formula allows you to find any BY HIS NUMBER " n" .
Of course, you also need to know the first term a 1 and progression difference d, well, without these parameters you can’t write down a specific progression.
Memorizing (or cribing) this formula is not enough. You need to understand its essence and apply the formula in various problems. And don’t forget in right moment, but how not forget- I don't know. And here how to remember If necessary, I will definitely advise you. For those who complete the lesson to the end.)
So, let's look at the formula for the nth term of an arithmetic progression.
What is a formula in general? By the way, take a look if you haven’t read it. Everything is simple there. It remains to figure out what it is nth term.
Progression in general can be written as a series of numbers:
a 1, a 2, a 3, a 4, a 5, .....
a 1- denotes the first term of an arithmetic progression, a 3- third member, a 4- the fourth, and so on. If we are interested in the fifth term, let's say we are working with a 5, if one hundred and twentieth - s a 120.
How can we define it in general terms? any term of an arithmetic progression, with any number? Very simple! Like this:
a n
That's what it is nth term of an arithmetic progression. The letter n hides all the member numbers at once: 1, 2, 3, 4, and so on.
And what does such a record give us? Just think, instead of a number they wrote down a letter...
This notation gives us a powerful tool for working with arithmetic progression. Using the notation a n, we can quickly find any member any arithmetic progression. And solve a bunch of other progression problems. You'll see for yourself further.
In the formula for the nth term of an arithmetic progression:
| a n = a 1 + (n-1)d |
a 1- the first term of an arithmetic progression;
n- member number.
The formula connects the key parameters of any progression: a n ; a 1 ; d And n. All progression problems revolve around these parameters.
The nth term formula can also be used to write a specific progression. For example, the problem may say that the progression is specified by the condition:
a n = 5 + (n-1) 2.
Such a problem can be a dead end... There is neither a series nor a difference... But, comparing the condition with the formula, it is easy to understand that in this progression a 1 =5, and d=2.
And it can be even worse!) If we take the same condition: a n = 5 + (n-1) 2, Yes, open the parentheses and bring similar ones? We get a new formula:
a n = 3 + 2n.
This Just not general, but for a specific progression. This is where the pitfall lurks. Some people think that the first term is a three. Although in reality the first term is five... A little lower we will work with such a modified formula.
In progression problems there is another notation - a n+1. This is, as you guessed, the “n plus first” term of the progression. Its meaning is simple and harmless.) This is a member of the progression whose number is greater than number n by one. For example, if in some problem we take a n fifth term then a n+1 will be the sixth member. Etc.
Most often the designation a n+1 found in recurrence formulas. Don't be afraid of this scary word!) This is just a way of expressing a member of an arithmetic progression through the previous one. Let's say we are given an arithmetic progression in this form, using a recurrent formula:
a n+1 = a n +3
a 2 = a 1 + 3 = 5+3 = 8
a 3 = a 2 + 3 = 8+3 = 11
The fourth - through the third, the fifth - through the fourth, and so on. How can we immediately count, say, the twentieth term? a 20? But there’s no way!) Until we find out the 19th term, we can’t count the 20th. This is the fundamental difference between the recurrent formula and the formula of the nth term. Recurrent works only through previous term, and the formula of the nth term is through first and allows straightaway find any member by its number. Without calculating the entire series of numbers in order.
In an arithmetic progression, it is easy to turn a recurrent formula into a regular one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write the formula in its usual form, and work with it. In the State Academy of Sciences, such tasks are often encountered.
Application of the formula for the nth term of an arithmetic progression.
First, let's look at the direct application of the formula. At the end of the previous lesson there was a problem:
An arithmetic progression (a n) is given. Find a 121 if a 1 =3 and d=1/6.
This problem can be solved without any formulas, simply based on the meaning of an arithmetic progression. Add and add... An hour or two.)
And according to the formula, the solution will take less than a minute. You can time it.) Let's decide.
The conditions provide all the data for using the formula: a 1 =3, d=1/6. It remains to figure out what is equal n. No problem! We need to find a 121. So we write:
Please pay attention! Instead of an index n a specific number appeared: 121. Which is quite logical.) We are interested in the member of the arithmetic progression number one hundred twenty one. This will be ours n. This is the meaning n= 121 we will substitute further into the formula, in brackets. We substitute all the numbers into the formula and calculate:
a 121 = 3 + (121-1) 1/6 = 3+20 = 23
That's it. Just as quickly one could find the five hundred and tenth term, and the thousand and third, any one. We put instead n the desired number in the index of the letter " a" and in brackets, and we count.
Let me remind you the point: this formula allows you to find any arithmetic progression term BY HIS NUMBER " n" .
Let's solve the problem in a more cunning way. Let us come across the following problem:
Find the first term of the arithmetic progression (a n), if a 17 =-2; d=-0.5.
If you have any difficulties, I will tell you the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Write down with your hands, right in your notebook:
| a n = a 1 + (n-1)d |
And now, looking at the letters of the formula, we understand what data we have and what is missing? Available d=-0.5, there is a seventeenth member... Is that it? If you think that’s it, then you won’t solve the problem, yes...
We still have a number n! In condition a 17 =-2 hidden two parameters. This is both the value of the seventeenth term (-2) and its number (17). Those. n=17. This “trifle” often slips past the head, and without it, (without the “trifle”, not the head!) the problem cannot be solved. Although... and without a head too.)
Now we can simply stupidly substitute our data into the formula:
a 17 = a 1 + (17-1)·(-0.5)
Oh yes, a 17 we know it's -2. Okay, let's substitute:
-2 = a 1 + (17-1)·(-0.5)
That's basically all. It remains to express the first term of the arithmetic progression from the formula and calculate it. The answer will be: a 1 = 6.
This technique - writing down a formula and simply substituting known data - is a great help in simple tasks. Well, of course, you must be able to express a variable from a formula, but what to do!? Without this skill, you may not study mathematics at all...
Another popular puzzle:
Find the difference of the arithmetic progression (a n), if a 1 =2; a 15 =12.
What are we doing? You will be surprised, we are writing the formula!)
| a n = a 1 + (n-1)d |
Let's consider what we know: a 1 =2; a 15 =12; and (I’ll especially highlight!) n=15. Feel free to substitute this into the formula:
12=2 + (15-1)d
We do the arithmetic.)
12=2 + 14d
d=10/14 = 5/7
This is the correct answer.
So, the tasks for a n, a 1 And d decided. All that remains is to learn how to find the number:
The number 99 is a member of the arithmetic progression (a n), where a 1 =12; d=3. Find this member's number.
We substitute the quantities known to us into the formula of the nth term:
a n = 12 + (n-1) 3
At first glance, there are two unknown quantities here: a n and n. But a n- this is some member of the progression with a number n...And we know this member of the progression! It's 99. We don't know its number. n, So this number is what you need to find. We substitute the term of the progression 99 into the formula:
99 = 12 + (n-1) 3
We express from the formula n, we think. We get the answer: n=30.
And now a problem on the same topic, but more creative):
Determine whether the number 117 is a member of the arithmetic progression (a n):
-3,6; -2,4; -1,2 ...
Let's write the formula again. What, there are no parameters? Hm... Why are we given eyes?) Do we see the first term of the progression? We see. This is -3.6. You can safely write: a 1 = -3.6. Difference d can you determine from a series? It’s easy if you know what the difference of an arithmetic progression is:
d = -2.4 - (-3.6) = 1.2
So, we did the simplest thing. It remains to deal with the unknown number n and the incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know... What to do!? Well, what to do, what to do... Turn on Creative skills!)
We suppose that 117 is, after all, a member of our progression. With an unknown number n. And, just like in the previous problem, let's try to find this number. Those. we write the formula (yes, yes!)) and substitute our numbers:
117 = -3.6 + (n-1) 1.2
Again we express from the formulan, we count and get:
Oops! The number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion can we draw? Yes! Number 117 is not member of our progression. It is somewhere between the one hundred and first and one hundred and second terms. If the number turned out natural, i.e. is a positive integer, then the number would be a member of the progression with the number found. And in our case, the answer to the problem will be: No.
A task based on a real version of the GIA:
The arithmetic progression is given by the condition:
a n = -4 + 6.8n
Find the first and tenth terms of the progression.
Here the progression is set in an unusual way. Some kind of formula... It happens.) However, this formula (as I wrote above) - also the formula for the nth term of an arithmetic progression! She also allows find any member of the progression by its number.
We are looking for the first member. The one who thinks. that the first term is minus four is fatally mistaken!) Because the formula in the problem is modified. The first term of the arithmetic progression in it hidden. It’s okay, we’ll find it now.)
Just as in previous problems, we substitute n=1 into this formula:
a 1 = -4 + 6.8 1 = 2.8
Here! The first term is 2.8, not -4!
We look for the tenth term in the same way:
a 10 = -4 + 6.8 10 = 64
That's it.
And now, for those who have read to these lines, the promised bonus.)
Suppose, in a difficult combat situation, State Examination or Unified State Examination, you forgot useful formula nth term of an arithmetic progression. I remember something, but somehow uncertainly... Or n there, or n+1, or n-1... How to be!?
Calm! This formula is easy to derive. It’s not very strict, but it’s definitely enough for confidence and the right decision!) To make a conclusion, it’s enough to remember the elementary meaning of an arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.
Draw a number line and mark the first one on it. second, third, etc. members. And we note the difference d between members. Like this:
We look at the picture and think: what does the second term equal? Second one d:
a 2 =a 1 + 1 d
What is the third term? Third term equals first term plus two d.
a 3 =a 1 + 2 d
Do you get it? It’s not for nothing that I highlight some words in bold. Okay, one more step).
What is the fourth term? Fourth term equals first term plus three d.
a 4 =a 1 + 3 d
It's time to realize that the number of gaps, i.e. d, Always one less than the number of the member you are looking for n. That is, to the number n, number of spaces will n-1. Therefore, the formula will be (without variations!):
| a n = a 1 + (n-1)d |
In general, visual pictures are very helpful in solving many problems in mathematics. Don't neglect the pictures. But if it’s difficult to draw a picture, then... only a formula!) In addition, the formula of the nth term allows you to connect the entire powerful arsenal of mathematics to the solution - equations, inequalities, systems, etc. You can't insert a picture into the equation...
Tasks for independent solution.
To warm up:
1. In arithmetic progression (a n) a 2 =3; a 5 =5.1. Find a 3 .
Hint: according to the picture, the problem can be solved in 20 seconds... According to the formula, it turns out more difficult. But for mastering the formula, it is more useful.) In Section 555, this problem is solved using both the picture and the formula. Feel the difference!)
And this is no longer a warm-up.)
2. In arithmetic progression (a n) a 85 =19.1; a 236 =49, 3. Find a 3 .
What, you don’t want to draw a picture?) Of course! Better according to the formula, yes...
3. The arithmetic progression is given by the condition:a 1 = -5.5; a n+1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.
In this task, the progression is specified in a recurrent manner. But counting to the one hundred and twenty-fifth term... Not everyone is capable of such a feat.) But the formula of the nth term is within the power of everyone!
4. Given an arithmetic progression (a n):
-148; -143,8; -139,6; -135,4, .....
Find the number of the smallest positive term of the progression.
5. According to the conditions of task 4, find the sum of the smallest positive and largest negative terms of the progression.
6. The product of the fifth and twelfth terms of an increasing arithmetic progression is equal to -2.5, and the sum of the third and eleventh terms is equal to zero. Find a 14 .
Not the easiest task, yes...) The “fingertip” method won’t work here. You will have to write formulas and solve equations.
Answers (in disarray):
3,7; 3,5; 2,2; 37; 2,7; 56,5
Happened? It's nice!)
Not everything works out? Happens. By the way, there is one subtle point in the last task. Care will be required when reading the problem. And logic.
The solution to all these problems is discussed in detail in Section 555. And the element of fantasy for the fourth, and the subtle point for the sixth, and general approaches for solving any problems involving the formula of the nth term - everything is described. I recommend.
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
When studying algebra in secondary school(9th grade) one of important topics is the study of number sequences, which include progressions - geometric and arithmetic. In this article we will look at an arithmetic progression and examples with solutions.
What is an arithmetic progression?
To understand this, it is necessary to define the progression in question, as well as provide the basic formulas that will be used later in solving problems.
An arithmetic or algebraic progression is a set of ordered rational numbers, each term of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of an ordered series of numbers and the difference, you can restore the entire arithmetic progression.
Let's give an example. The following sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the type of progression under consideration, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).
Important Formulas
Let us now present the basic formulas that will be needed to solve problems using arithmetic progression. Let us denote by the symbol a n the nth member of the sequence, where n is an integer. We denote the difference by the Latin letter d. Then the following expressions are valid:
- To determine the value of the nth term, the following formula is suitable: a n = (n-1)*d+a 1 .
- To determine the sum of the first n terms: S n = (a n +a 1)*n/2.
To understand any examples of arithmetic progression with solutions in 9th grade, it is enough to remember these two formulas, since any problems of the type under consideration are based on their use. You should also remember that the progression difference is determined by the formula: d = a n - a n-1.
Example #1: finding an unknown member
Let's give a simple example of an arithmetic progression and the formulas that need to be used to solve it.
Let the sequence 10, 8, 6, 4, ... be given, you need to find five terms in it.
From the conditions of the problem it already follows that the first 4 terms are known. The fifth can be defined in two ways:
- Let's first calculate the difference. We have: d = 8 - 10 = -2. Similarly, you could take any two other members standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d = a n - a n-1, then d = a 5 - a 4, from which we get: a 5 = a 4 + d. We substitute the known values: a 5 = 4 + (-2) = 2.
- The second method also requires knowledge of the difference of the progression in question, so you first need to determine it as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for the n number of the sequence. We have: a n = (n - 1) * d + a 1 = (n - 1) * (-2) + 10 = 12 - 2*n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.
As you can see, both solutions led to the same result. Note that in this example the progression difference d is a negative value. Such sequences are called decreasing, since each next term is less than the previous one.
Example #2: progression difference
Now let’s complicate the task a little, let’s give an example of how
It is known that in some the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.
Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . Let's substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 = 6 + 6 * d. From this expression you can easily calculate the difference: d = (18 - 6) /6 = 2. Thus, we have answered the first part of the problem.
To restore the sequence to the 7th term, you should use the definition of an algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d, and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2=8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16, a 7 = 18.
Example No. 3: drawing up a progression
Let's complicate the problem even more. Now we need to answer the question of how to find an arithmetic progression. The following example can be given: two numbers are given, for example - 4 and 5. It is necessary to create an algebraic progression so that three more terms are placed between these.
Before you start solving this problem, you need to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 = -4 and a 5 = 5. Having established this, we move on to the problem, which is similar to the previous one. Again, for the nth term we use the formula, we get: a 5 = a 1 + 4 * d. From: d = (a 5 - a 1)/4 = (5 - (-4)) / 4 = 2.25. What we got here is not an integer value of the difference, but it is rational number, so the formulas for the algebraic progression remain the same.
Now let's add the found difference to a 1 and restore the missing terms of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 = 2.75 + 2.25 = 5, which coincided with the conditions of the problem.
Example No. 4: first term of progression
Let's continue to give examples of arithmetic progression with solutions. In all previous problems, the first number of the algebraic progression was known. Now let's consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find which number this sequence begins with.
The formulas used so far assume knowledge of a 1 and d. In the problem statement, nothing is known about these numbers. Nevertheless, we will write down expressions for each term about which information is available: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We received two equations in which there are 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.
The easiest way to solve this system is to express a 1 in each equation and then compare the resulting expressions. First equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 = a 43 - 42 * d = 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d = 37 - 42 * d, whence the difference d = (37 - 50) / (42 - 14) = - 0.464 (only 3 decimal places are given).
Knowing d, you can use any of the 2 expressions above for a 1. For example, first: a 1 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.
If you have doubts about the result obtained, you can check it, for example, determine the 43rd term of the progression, which is specified in the condition. We get: a 43 = a 1 + 42 * d = 56.496 + 42 * (- 0.464) = 37.008. The small error is due to the fact that rounding to thousandths was used in the calculations.
Example No. 5: amount
Now let's look at several examples with solutions for the sum of an arithmetic progression.
Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?
Thanks to development computer technology you can solve this problem, that is, add all the numbers sequentially, which the computer will do immediately as soon as a person presses the Enter key. However, the problem can be solved mentally if you pay attention that the presented series of numbers is an algebraic progression, and its difference is equal to 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.
It is interesting to note that this problem is called "Gaussian" because in early XVIII century, the famous German, while still only 10 years old, was able to solve it in his head in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add the numbers at the ends of the sequence in pairs, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer it is enough to multiply 50 by 101.
Example No. 6: sum of terms from n to m
Another typical example of the sum of an arithmetic progression is the following: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its terms from 8 to 14 will be equal to.
The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them sequentially. Since there are few terms, this method is not quite labor-intensive. Nevertheless, it is proposed to solve this problem using a second method, which is more universal.
The idea is to obtain a formula for the sum of the algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:
- S m = m * (a m + a 1) / 2.
- S n = n * (a n + a 1) / 2.
Since n > m, it is obvious that the 2nd sum includes the first. The last conclusion means that if we take the difference between these sums and add the term a m to it (in the case of taking the difference, it is subtracted from the sum S n), we will obtain the necessary answer to the problem. We have: S mn = S n - S m + a m =n * (a 1 + a n) / 2 - m *(a 1 + a m)/2 + a m = a 1 * (n - m) / 2 + a n * n/2 + a m * (1- m/2). It is necessary to substitute formulas for a n and a m into this expression. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d *(3 * m - m 2 - 2) / 2.
The resulting formula is somewhat cumbersome, however, the sum S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.
As can be seen from the above solutions, all problems are based on knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before starting to solve any of these problems, it is recommended that you carefully read the condition, clearly understand what you need to find, and only then proceed with the solution.
Another tip is to strive for simplicity, that is, if you can answer a question without using complex mathematical calculations, then you need to do just that, since in this case the likelihood of making a mistake is less. For example, in the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and break common task into separate subtasks (in this case, first find the terms a n and a m).
If you have doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. We found out how to find an arithmetic progression. If you figure it out, it's not that difficult.
Mathematics has its own beauty, just like painting and poetry.
Russian scientist, mechanic N.E. Zhukovsky
Very common tasks in entrance examinations in mathematics are problems related to the concept of arithmetic progression. To successfully solve such problems, you must have a good knowledge of the properties of arithmetic progression and have certain skills in their application.
Let us first recall the basic properties of an arithmetic progression and present the most important formulas, associated with this concept.
Definition. Number sequence, in which each subsequent term differs from the previous one by the same number, called an arithmetic progression. In this case, the numbercalled the progression difference.
For an arithmetic progression, the following formulas are valid:
, (1)
Where . Formula (1) is called the formula of the general term of an arithmetic progression, and formula (2) represents the main property of an arithmetic progression: each term of the progression coincides with the arithmetic mean of its neighboring terms and .
Note that it is precisely because of this property that the progression under consideration is called “arithmetic”.
The above formulas (1) and (2) are generalized as follows:
(3)
To calculate the amount first terms of an arithmetic progressionthe formula is usually used
(5) where and .
If we take into account the formula (1), then from formula (5) it follows
If we denote , then
Where . Since , formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).
In particular , from formula (5) it follows, What
Little known to most students is the property of arithmetic progression, formulated through the following theorem.
Theorem. If , then
Proof. If , then
The theorem has been proven.
For example , using the theorem, it can be shown that
Let's move on to consider typical examples of solving problems on the topic “Arithmetic progression”.
Example 1. Let it be. Find .
Solution. Applying formula (6), we obtain . Since and , then or .
Example 2. Let it be three times greater, and when divided by the quotient, the result is 2 and the remainder is 8. Determine and .
Solution. From the conditions of the example, the system of equations follows
Since , , and , then from the system of equations (10) we obtain
The solution to this system of equations is and .
Example 3. Find if and .
Solution. According to formula (5) we have or . However, using property (9), we obtain .
Since and , then from the equality the equation follows or .
Example 4. Find if .
Solution.According to formula (5) we have
However, using the theorem, we can write
From here and from formula (11) we obtain .
Example 5. Given: . Find .
Solution. Since, then. However, therefore.
Example 6. Let , and . Find .
Solution. Using formula (9), we obtain . Therefore, if , then or .
Since and then here we have a system of equations
Solving which, we get and .
Natural root of the equation is .
Example 7. Find if and .
Solution. Since according to formula (3) we have that , then the system of equations follows from the problem conditions
If we substitute the expressioninto the second equation of the system, then we get or .
Roots quadratic equation are And .
Let's consider two cases.
1. Let , then . Since and , then .
In this case, according to formula (6), we have
2. If , then , and
Answer: and.
Example 8. It is known that and. Find .
Solution. Taking into account formula (5) and the condition of the example, we write and .
This implies the system of equations
If we multiply the first equation of the system by 2 and then add it to the second equation, we get
According to formula (9) we have. In this regard, it follows from (12) or .
Since and , then .
Answer: .
Example 9. Find if and .
Solution. Since , and by condition , then or .
From formula (5) it is known, What . Since, then.
Hence , here we have a system of linear equations
From here we get and . Taking into account formula (8), we write .
Example 10. Solve the equation.
Solution. From given equation follows that . Let us assume that , , and . In this case .
According to formula (1), we can write or .
Since , then equation (13) has the only suitable root .
Example 11. Find the maximum value provided that and .
Solution. Since , then the arithmetic progression under consideration is decreasing. In this regard, the expression takes on its maximum value when it is the number of the minimum positive term of the progression.
Let us use formula (1) and the fact, that and . Then we get that or .
Since , then or . However, in this inequalitylargest natural number, That's why .
If the values of , and are substituted into formula (6), we get .
Answer: .
Example 12. Determine the sum of all two-digit natural numbers, which when divided by 6 leaves a remainder of 5.
Solution. Let us denote by the set of all two-digit natural numbers, i.e. . Next, we will construct a subset consisting of those elements (numbers) of the set that, when divided by the number 6, give a remainder of 5.
Easy to install, What . Obviously , that the elements of the setform an arithmetic progression, in which and .
To establish the cardinality (number of elements) of the set, we assume that . Since and , it follows from formula (1) or . Taking into account formula (5), we obtain .
The above examples of problem solving can by no means claim to be exhaustive. This article is written based on the analysis modern methods solving typical problems on a given topic. For a more in-depth study of methods for solving problems related to arithmetic progression, it is advisable to refer to the list of recommended literature.
1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace and Education, 2013. – 608 p.
2. Suprun V.P. Mathematics for high school students: additional sections school curriculum. – M.: Lenand / URSS, 2014. – 216 p.
3. Medynsky M.M. Full course elementary mathematics in problems and exercises. Book 2: Number sequences and progression. – M.: Editus, 2015. – 208 p.
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Sum of an arithmetic progression.
The sum of an arithmetic progression is a simple thing. Both in meaning and in formula. But there are all sorts of tasks on this topic. From basic to quite solid.
First, let's understand the meaning and formula of the amount. And then we'll decide. For your own pleasure.) The meaning of the amount is as simple as a moo. To find the sum of an arithmetic progression, you just need to carefully add all its terms. If these terms are few, you can add without any formulas. But if there is a lot, or a lot... addition is annoying.) In this case, the formula comes to the rescue.
The formula for the amount is simple:
Let's figure out what kind of letters are included in the formula. This will clear things up a lot.
S n - the sum of an arithmetic progression. Addition result everyone members, with first By last. It is important. They add up exactly All members in a row, without skipping or skipping. And, precisely, starting from first. In problems like finding the sum of the third and eighth terms, or the sum of the fifth to twentieth terms, direct application of the formula will disappoint.)
a 1 - first member of the progression. Everything is clear here, it's simple first row number.
a n- last member of the progression. The last number of the series. Not a very familiar name, but when applied to the amount, it’s very suitable. Then you will see for yourself.
n - number of the last member. It is important to understand that in the formula this number coincides with the number of added terms.
Let's define the concept last member a n. Tricky question: which member will be the last one if given endless arithmetic progression?)
To answer confidently, you need to understand the elementary meaning of arithmetic progression and... read the task carefully!)
In the task of finding the sum of an arithmetic progression, the last term always appears (directly or indirectly), which should be limited. Otherwise, a final, specific amount simply doesn't exist. For the solution, it does not matter whether the progression is given: finite or infinite. It doesn’t matter how it is given: a series of numbers, or a formula for the nth term.
The most important thing is to understand that the formula works from the first term of the progression to the term with number n. Actually, the full name of the formula looks like this: the sum of the first n terms of an arithmetic progression. The number of these very first members, i.e. n, is determined solely by the task. In a task, all this valuable information is often encrypted, yes... But never mind, in the examples below we reveal these secrets.)
Examples of tasks on the sum of an arithmetic progression.
First of all, helpful information:
The main difficulty in tasks involving the sum of an arithmetic progression lies in the correct determination of the elements of the formula.
The task writers encrypt these very elements with boundless imagination.) The main thing here is not to be afraid. Understanding the essence of the elements, it is enough to simply decipher them. Let's look at a few examples in detail. Let's start with a task based on a real GIA.
1. The arithmetic progression is given by the condition: a n = 2n-3.5. Find the sum of its first 10 terms.
Good job. Easy.) To determine the amount using the formula, what do we need to know? First member a 1, last term a n, yes the number of the last member n.
Where can I get the last member's number? n? Yes, right there, on condition! It says: find the sum first 10 members. Well, what number will it be with? last, tenth member?) You won’t believe it, his number is tenth!) Therefore, instead of a n We will substitute into the formula a 10, and instead n- ten. I repeat, the number of the last member coincides with the number of members.
It remains to determine a 1 And a 10. This is easily calculated using the formula for the nth term, which is given in the problem statement. Don't know how to do this? Attend the previous lesson, without this there is no way.
a 1= 2 1 - 3.5 = -1.5
a 10=2·10 - 3.5 =16.5
S n = S 10.
We have found out the meaning of all elements of the formula for the sum of an arithmetic progression. All that remains is to substitute them and count:
That's it. Answer: 75.
Another task based on the GIA. A little more complicated:
2. Given an arithmetic progression (a n), the difference of which is 3.7; a 1 =2.3. Find the sum of its first 15 terms.
We immediately write the sum formula:
This formula allows us to find the value of any term by its number. We look for a simple substitution:
a 15 = 2.3 + (15-1) 3.7 = 54.1
It remains to substitute all the elements into the formula for the sum of an arithmetic progression and calculate the answer:
Answer: 423.
By the way, if in the sum formula instead of a n We simply substitute the formula for the nth term and get:
Let us present similar ones and obtain a new formula for the sum of terms of an arithmetic progression:
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As you can see, the nth term is not required here a n. In some problems this formula helps a lot, yes... You can remember this formula. Or you can simply display it at the right time, like here. After all, you always need to remember the formula for the sum and the formula for the nth term.)
Now the task in the form of a short encryption):
3. Find the sum of all positive double digit numbers, multiples of three.
Wow! Neither your first member, nor your last, nor progression at all... How to live!?
You will have to think with your head and pull out all the elements of the sum of the arithmetic progression from the condition. We know what two-digit numbers are. They consist of two numbers.) What two-digit number will be first? 10, presumably.) A last thing double digit number? 99, of course! The three-digit ones will follow him...
Multiples of three... Hm... These are numbers that are divisible by three, here! Ten is not divisible by three, 11 is not divisible... 12... is divisible! So, something is emerging. You can already write down a series according to the conditions of the problem:
12, 15, 18, 21, ... 96, 99.
Will this series be an arithmetic progression? Certainly! Each term differs from the previous one by strictly three. If you add 2 or 4 to a term, say, the result, i.e. the new number is no longer divisible by 3. You can immediately determine the difference of the arithmetic progression: d = 3. It will come in handy!)
So, we can safely write down some progression parameters:
What will the number be? n last member? Anyone who thinks that 99 is fatally mistaken... The numbers always go in a row, but our members jump over three. They don't match.
There are two solutions here. One way is for the super hardworking. You can write down the progression, the entire series of numbers, and count the number of members with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If we apply the formula to our problem, we find that 99 is the thirtieth term of the progression. Those. n = 30.
Let's look at the formula for the sum of an arithmetic progression:
We look and rejoice.) We pulled out from the problem statement everything necessary to calculate the amount:
a 1= 12.
a 30= 99.
S n = S 30.
All that remains is elementary arithmetic. We substitute the numbers into the formula and calculate:
Answer: 1665
Another type of popular puzzle:
4. Given an arithmetic progression:
-21,5; -20; -18,5; -17; ...
Find the sum of terms from twentieth to thirty-four.
We look at the formula for the amount and... we get upset.) The formula, let me remind you, calculates the amount from the first member. And in the problem you need to calculate the sum since the twentieth... The formula won't work.
You can, of course, write out the entire progression in a series, and add terms from 20 to 34. But... it’s somehow stupid and takes a long time, right?)
There is a more elegant solution. Let's divide our series into two parts. The first part will be from the first term to the nineteenth. Second part - from twenty to thirty-four. It is clear that if we calculate the sum of the terms of the first part S 1-19, let's add it with the sum of the terms of the second part S 20-34, we get the sum of the progression from the first term to the thirty-fourth S 1-34. Like this:
S 1-19 + S 20-34 = S 1-34
From this we can see that find the sum S 20-34 can be done by simple subtraction
S 20-34 = S 1-34 - S 1-19
Both amounts on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Let's get started?
We extract the progression parameters from the problem statement:
d = 1.5.
a 1= -21,5.
To calculate the sums of the first 19 and first 34 terms, we will need the 19th and 34th terms. We calculate them using the formula for the nth term, as in problem 2:
a 19= -21.5 +(19-1) 1.5 = 5.5
a 34= -21.5 +(34-1) 1.5 = 28
There's nothing left. From the sum of 34 terms subtract the sum of 19 terms:
S 20-34 = S 1-34 - S 1-19 = 110.5 - (-152) = 262.5
Answer: 262.5
One important note! There is a very useful trick in solving this problem. Instead of direct calculation what you need (S 20-34), we counted something that would seem not to be needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the complete result. This kind of “feint with your ears” often saves you in wicked problems.)
In this lesson, we looked at problems for which it is enough to understand the meaning of the sum of an arithmetic progression. Well, you need to know a couple of formulas.)
When solving any problem involving the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.
Formula for the nth term:
These formulas will immediately tell you what to look for and in what direction to think in order to solve the problem. Helps.
And now the tasks for independent solution.
5. Find the sum of all two-digit numbers that are not divisible by three.
Cool?) The hint is hidden in the note to problem 4. Well, problem 3 will help.
6. The arithmetic progression is given by the condition: a 1 = -5.5; a n+1 = a n +0.5. Find the sum of its first 24 terms.
Unusual?) This recurrence formula. You can read about it in the previous lesson. Don’t ignore the link, such problems are often found in the State Academy of Sciences.
7. Vasya saved up money for the holiday. As much as 4550 rubles! And I decided to give my favorite person (myself) a few days of happiness). Live beautifully without denying yourself anything. Spend 500 rubles on the first day, and on each subsequent day spend 50 rubles more than the previous one! Until the money runs out. How many days of happiness did Vasya have?
Is it difficult?) The additional formula from problem 2 will help.
Answers (in disarray): 7, 3240, 6.
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