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 is the serial number of the element of the row 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 it?
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 a numerical 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, and also 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.
Problems on arithmetic progression existed already in ancient times. They appeared and demanded a solution because they had a practical need.
Thus, one of the papyri of Ancient Egypt that has mathematical content, the Rhind papyrus (19th century BC), contains the following task: divide ten measures of bread among ten people, provided that the difference between each of them is one eighth of the measure.”
And in the mathematical works of the ancient Greeks there are elegant theorems related to arithmetic progression. Thus, Hypsicles of Alexandria (2nd century, who compiled many interesting problems and added the fourteenth book to Euclid’s Elements), formulated the idea: “In an arithmetic progression that has an even number of terms, the sum of the terms of the 2nd half is greater than the sum of the terms of the 1st on the square 1/ 2 numbers of members."
The sequence is denoted by an. The numbers of a sequence are called its members and are usually designated by letters with indices that indicate the serial number of this member (a1, a2, a3 ... read: “a 1st”, “a 2nd”, “a 3rd” and so on ).
The sequence can be infinite or finite.
What is an arithmetic progression? By it we mean the one obtained by adding the previous term (n) with the same number d, which is the difference of the progression.
If d<0, то мы имеем убывающую прогрессию. Если d>0, then this progression is considered increasing.
An arithmetic progression is called finite if only its first few terms are taken into account. With a very large number of members, this is already an endless progression.
Any arithmetic progression is defined by the following formula:
an =kn+b, while b and k are some numbers.
The opposite statement is absolutely true: if a sequence is given by a similar formula, then it is exactly an arithmetic progression that has the properties:
- Each term of the progression is the arithmetic mean of the previous term and the subsequent one.
- Converse: if, starting from the 2nd, each term is the arithmetic mean of the previous term and the subsequent one, i.e. if the condition is met, then this sequence is an arithmetic progression. This equality is also a sign of progression, which is why it is usually called a characteristic property of progression.
In the same way, the theorem that reflects this property is true: a sequence is an arithmetic progression only if this equality is true for any of the terms of the sequence, starting with the 2nd.
The characteristic property for any four numbers of an arithmetic progression can be expressed by the formula an + am = ak + al, if n + m = k + l (m, n, k are progression numbers).
In an arithmetic progression, any necessary (Nth) term can be found using the following formula:
For example: the first term (a1) in an arithmetic progression is given and equal to three, and the difference (d) is equal to four. You need to find the forty-fifth term of this progression. a45 = 1+4(45-1)=177
The formula an = ak + d(n - k) allows you to determine the nth term of an arithmetic progression through any of its kth terms, provided that it is known.
The sum of the terms of an arithmetic progression (meaning the first n terms of a finite progression) is calculated as follows:
Sn = (a1+an) n/2.
If the 1st term is also known, then another formula is convenient for calculation:
Sn = ((2a1+d(n-1))/2)*n.
The sum of an arithmetic progression that contains n terms is calculated as follows:
The choice of formulas for calculations depends on the conditions of the problems and the initial data.
The natural series of any numbers, such as 1,2,3,...,n,..., is the simplest example of an arithmetic progression.
In addition to the arithmetic progression, there is also a geometric progression, which has its own properties and characteristics.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
An arithmetic progression is a series of numbers in which each number is greater (or less) than the previous one by the same amount.
This topic often seems complex and incomprehensible. The indices of the letters, the nth term of the progression, the difference of the progression - all this is somehow confusing, yes... Let's figure out the meaning of the arithmetic progression and everything will get better right away.)
The concept of arithmetic progression.
Arithmetic progression is a very simple and clear concept. Do you have any doubts? In vain.) See for yourself.
I'll write an unfinished series of numbers:
1, 2, 3, 4, 5, ...
Can you extend this series? What numbers will come next, after the five? Everyone... uh..., in short, everyone will realize that the numbers 6, 7, 8, 9, etc. will come next.
Let's complicate the task. I give you an unfinished series of numbers:
2, 5, 8, 11, 14, ...
You will be able to catch the pattern, extend the series, and name seventh row number?
If you realized that this number is 20, congratulations! Not only did you feel key points of arithmetic progression, but also successfully used them in business! If you haven’t figured it out, read on.
Now let’s translate the key points from sensations into mathematics.)
First key point.
Arithmetic progression deals with series of numbers. This is confusing at first. We are used to solving equations, drawing graphs and all that... But here we extend the series, find the number of the series...
It's OK. It's just that progressions are the first acquaintance with a new branch of mathematics. The section is called "Series" and works specifically with series of numbers and expressions. Get used to it.)
Second key point.
In an arithmetic progression, any number is different from the previous one by the same amount.
In the first example, this difference is one. Whatever number you take, it is one more than the previous one. In the second - three. Any number is three more than the previous one. Actually, it is this moment that gives us the opportunity to grasp the pattern and calculate subsequent numbers.
Third key point.
This moment is not striking, yes... But it is very, very important. Here he is: Each progression number is in its place. There is the first number, there is the seventh, there is the forty-fifth, etc. If you mix them up at random, the pattern will disappear. Arithmetic progression will also disappear. What's left is just a series of numbers.
That's the whole point.
Of course, new terms and designations appear in a new topic. You need to know them. Otherwise you won’t understand the task. For example, you will have to decide something like:
Write down the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.
Inspiring?) Letters, some indexes... And the task, by the way, couldn’t be simpler. You just need to understand the meaning of the terms and designations. Now we will master this matter and return to the task.
Terms and designations.
Arithmetic progression is a series of numbers in which each number is different from the previous one by the same amount.
This quantity is called . Let's look at this concept in more detail.
Arithmetic progression difference.
Arithmetic progression difference is the amount by which any progression number more previous one.
One important point. Please pay attention to the word "more". Mathematically, this means that each progression number is by adding difference of arithmetic progression to the previous number.
To calculate, let's say second numbers of the series, you need to first number add this very difference of an arithmetic progression. For calculation fifth- the difference is necessary add To fourth, well, etc.
Arithmetic progression difference May be positive, then each number in the series will turn out to be real more than the previous one. This progression is called increasing. For example:
8; 13; 18; 23; 28; .....
Here each number is obtained by adding positive number, +5 to the previous one.
The difference may be negative, then each number in the series will be less than the previous one. This progression is called (you won’t believe it!) decreasing.
For example:
8; 3; -2; -7; -12; .....
Here each number is also obtained by adding to the previous one, but already a negative number, -5.
By the way, when working with progression, it is very useful to immediately determine its nature - whether it is increasing or decreasing. This helps a lot to navigate the decision, spot your mistakes and correct them before it’s too late.
Arithmetic progression difference usually denoted by the letter d.
How to find d? Very simple. It is necessary to subtract from any number in the series previous number. Subtract. By the way, the result of subtraction is called "difference".)
Let us define, for example, d for increasing arithmetic progression:
2, 5, 8, 11, 14, ...
We take any number in the series that we want, for example, 11. We subtract from it previous number those. 8:
This is the correct answer. For this arithmetic progression, the difference is three.
You can take it any progression number, because for a specific progression d-always the same. At least somewhere at the beginning of the row, at least in the middle, at least anywhere. You cannot take only the very first number. Simply because the very first number no previous one.)
By the way, knowing that d=3, finding the seventh number of this progression is very simple. Let's add 3 to the fifth number - we get the sixth, it will be 17. Let's add three to the sixth number, we get the seventh number - twenty.
Let's define d for descending arithmetic progression:
8; 3; -2; -7; -12; .....
I remind you that, regardless of the signs, to determine d need from any number take away the previous one. Choose any progression number, for example -7. His previous number is -2. Then:
d = -7 - (-2) = -7 + 2 = -5
The difference of an arithmetic progression can be any number: integer, fractional, irrational, any number.
Other terms and designations.
Each number in the series is called member of an arithmetic progression.
Each member of the progression has its own number. The numbers are strictly in order, without any tricks. First, second, third, fourth, etc. For example, in the progression 2, 5, 8, 11, 14, ... two is the first term, five is the second, eleven is the fourth, well, you understand...) Please clearly understand - the numbers themselves can be absolutely anything, whole, fractional, negative, whatever, but numbering of numbers- strictly in order!
How to write a progression in general form? No problem! Each number in a series is written as a letter. To denote an arithmetic progression, the letter is usually used a. The member number is indicated by an index at the bottom right. We write terms separated by commas (or semicolons), like this:
a 1, a 2, a 3, a 4, a 5, .....
a 1- this is the first number, a 3- third, etc. Nothing fancy. This series can be written briefly like this: (a n).
Progressions happen finite and infinite.
Ultimate the progression has a limited number of members. Five, thirty-eight, whatever. But it's a finite number.
Infinite progression - has an infinite number of members, as you might guess.)
You can write the final progression through a series like this, all terms and a dot at the end:
a 1, a 2, a 3, a 4, a 5.
Or like this, if there are many members:
a 1, a 2, ... a 14, a 15.
In the short entry you will have to additionally indicate the number of members. For example (for twenty members), like this:
(a n), n = 20
An infinite progression can be recognized by the ellipsis at the end of the row, as in the examples in this lesson.
Now you can solve the tasks. The tasks are simple, purely for understanding the meaning of an arithmetic progression.
Examples of tasks on arithmetic progression.
Let's look at the task given above in detail:
1. Write out the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.
We translate the task into understandable language. An infinite arithmetic progression is given. The second number of this progression is known: a 2 = 5. The progression difference is known: d = -2.5. We need to find the first, third, fourth, fifth and sixth terms of this progression.
For clarity, I will write down a series according to the conditions of the problem. The first six terms, where the second term is five:
a 1, 5, a 3, a 4, a 5, a 6,....
a 3 = a 2 + d
Substitute into expression a 2 = 5 And d = -2.5. Don't forget about the minus!
a 3=5+(-2,5)=5 - 2,5 = 2,5
The third term turned out to be smaller than the second. Everything is logical. If the number is greater than the previous one negative value, which means the number itself will be less than the previous one. Progression is decreasing. Okay, let's take it into account.) We count the fourth term of our series:
a 4 = a 3 + d
a 4=2,5+(-2,5)=2,5 - 2,5 = 0
a 5 = a 4 + d
a 5=0+(-2,5)= - 2,5
a 6 = a 5 + d
a 6=-2,5+(-2,5)=-2,5 - 2,5 = -5
So, terms from the third to the sixth were calculated. The result is the following series:
a 1, 5, 2.5, 0, -2.5, -5, ....
It remains to find the first term a 1 according to the well-known second. This is a step in the other direction, to the left.) So, the difference of the arithmetic progression d should not be added to a 2, A take away:
a 1 = a 2 - d
a 1=5-(-2,5)=5 + 2,5=7,5
That's it. Assignment answer:
7,5, 5, 2,5, 0, -2,5, -5, ...
In passing, I would like to note that we solved this task recurrent way. This terrible word means only the search for a member of the progression according to the previous (adjacent) number. We'll look at other ways to work with progression below.
One important conclusion can be drawn from this simple task.
Remember:
If we know at least one term and the difference of an arithmetic progression, we can find any term of this progression.
Do you remember? This simple conclusion allows you to solve most of the problems of the school course on this topic. All tasks revolve around three main parameters: member of an arithmetic progression, difference of a progression, number of a member of the progression. All.
Of course, all previous algebra is not canceled.) Inequalities, equations, and other things are attached to progression. But according to the progression itself- everything revolves around three parameters.
As an example, let's look at some popular tasks on this topic.
2. Write the finite arithmetic progression as a series if n=5, d = 0.4, and a 1 = 3.6.
Everything is simple here. Everything has already been given. You need to remember how the members of an arithmetic progression are counted, count them, and write them down. It is advisable not to miss the words in the task conditions: “final” and “ n=5". So as not to count until you are completely blue in the face.) There are only 5 (five) members in this progression:
a 2 = a 1 + d = 3.6 + 0.4 = 4
a 3 = a 2 + d = 4 + 0.4 = 4.4
a 4 = a 3 + d = 4.4 + 0.4 = 4.8
a 5 = a 4 + d = 4.8 + 0.4 = 5.2
It remains to write down the answer:
3,6; 4; 4,4; 4,8; 5,2.
Another task:
3. Determine whether the number 7 will be a member of the arithmetic progression (a n), if a 1 = 4.1; d = 1.2.
Hmm... Who knows? How to determine something?
How-how... Write down the progression in the form of a series and see whether there will be a seven there or not! We count:
a 2 = a 1 + d = 4.1 + 1.2 = 5.3
a 3 = a 2 + d = 5.3 + 1.2 = 6.5
a 4 = a 3 + d = 6.5 + 1.2 = 7.7
4,1; 5,3; 6,5; 7,7; ...
Now it is clearly visible that we are just seven slipped through between 6.5 and 7.7! Seven did not fall into our series of numbers, and, therefore, seven will not be a member of the given progression.
Answer: no.
And here is a problem based on a real version of the GIA:
4. Several consecutive terms of the arithmetic progression are written out:
...; 15; X; 9; 6; ...
Here is a series written without end and beginning. No member numbers, no difference d. It's OK. To solve the problem, it is enough to understand the meaning of an arithmetic progression. Let's look and see what's possible to know from this series? What are the three main parameters?
Member numbers? There is not a single number here.
But there are three numbers and - attention! - word "consistent" in condition. This means that the numbers are strictly in order, without gaps. Are there two in this row? neighboring known numbers? Yes, I have! These are 9 and 6. Therefore, we can calculate the difference of the arithmetic progression! Subtract from six previous number, i.e. nine:
What remains are mere trifles. What number will be the previous one for X? Fifteen. This means that X can be easily found by simple addition. Add the difference of the arithmetic progression to 15:
That's all. Answer: x=12
We solve the following problems ourselves. Note: these problems are not based on formulas. Purely to understand the meaning of an arithmetic progression.) We just write down a series of numbers and letters, look and figure it out.
5. Find the first positive term of the arithmetic progression if a 5 = -3; d = 1.1.
6. It is known that the number 5.5 is a member of the arithmetic progression (a n), where a 1 = 1.6; d = 1.3. Determine the number n of this term.
7. It is known that in arithmetic progression a 2 = 4; a 5 = 15.1. Find a 3 .
8. Several consecutive terms of the arithmetic progression are written out:
...; 15.6; X; 3.4; ...
Find the term of the progression indicated by the letter x.
9. The train began moving from the station, uniformly increasing speed by 30 meters per minute. What will be the speed of the train in five minutes? Give your answer in km/hour.
10. It is known that in arithmetic progression a 2 = 5; a 6 = -5. Find a 1.
Answers (in disarray): 7.7; 7.5; 9.5; 9; 0.3; 4.
Everything worked out? Amazing! You can master arithmetic progression at a higher level in the following lessons.
Didn't everything work out? No problem. In Special Section 555, all these problems are sorted out piece by piece.) And, of course, a simple practical technique is described that immediately highlights the solution to such tasks clearly, clearly, at a glance!
By the way, in the train puzzle there are two problems that people often stumble over. One is purely in terms of progression, and the second is general for any problems in mathematics, and physics too. This is a translation of dimensions from one to another. It shows how these problems should be solved.
In this lesson we looked at the elementary meaning of an arithmetic progression and its main parameters. This is enough to solve almost all problems on this topic. Add d to the numbers, write a series, everything will be solved.
The finger solution works well for very short pieces of a row, as in the examples in this lesson. If the series is longer, the calculations become more complicated. For example, if in problem 9 in the question we replace "five minutes" on "thirty-five minutes" the problem will become significantly worse.)
And there are also tasks that are simple in essence, but absurd in terms of calculations, for example:
An arithmetic progression (a n) is given. Find a 121 if a 1 =3 and d=1/6.
So what, are we going to add 1/6 many, many times?! You can kill yourself!?
You can.) If you don’t know a simple formula by which you can solve such tasks in a minute. This formula will be in the next lesson. And this problem is solved there. In a minute.)
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
What is the main essence of the formula?
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 also not to forget at the right moment, yes...) 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 give 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. Such tasks are often encountered in the State Academy of Sciences.
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 next to 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 tell from the 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. All that remains is 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, how to be, how to be... Turn on your creative abilities!)
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:
An 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 of the State Examination or Unified State Examination, you have forgotten the useful formula for the 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’s 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.
Arithmetic and geometric progressions
Theoretical information
Theoretical information
Arithmetic progression |
Geometric progression |
|
Definition |
Arithmetic progression a n is a sequence in which each member, starting from the second, is equal to the previous member added to the same number d (d- progression difference) |
Geometric progression b n is a sequence of non-zero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression) |
Recurrence formula |
For any natural n |
For any natural n |
Formula nth term |
a n = a 1 + d (n – 1) |
b n = b 1 ∙ q n - 1 , b n ≠ 0 |
| Characteristic property |  |
|
| Sum of the first n terms |  |
 |
Examples of tasks with comments
Exercise 1
In arithmetic progression ( a n) a 1 = -6, a 2
According to the formula of the nth term:
a 22 = a 1+ d (22 - 1) = a 1+ 21 d
By condition:
a 1= -6, then a 22= -6 + 21 d .
It is necessary to find the difference of progressions:
d = a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = - 48.
Answer : a 22 = -48.
Task 2
Find the fifth term of the geometric progression: -3; 6;....
1st method (using the n-term formula)
According to the formula for the nth term of a geometric progression:
b 5 = b 1 ∙ q 5 - 1 = b 1 ∙ q 4.
Because b 1 = -3,
2nd method (using recurrent formula)
Since the denominator of the progression is -2 (q = -2), then:
b 3 = 6 ∙ (-2) = -12;
b 4 = -12 ∙ (-2) = 24;
b 5 = 24 ∙ (-2) = -48.
Answer : b 5 = -48.
Task 3
In arithmetic progression ( a n ) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.
For an arithmetic progression, the characteristic property has the form 
.
Therefore:
.
Let's substitute the data into the formula:
Answer: 95.
Task 4
In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.
To find the sum of the first n terms of an arithmetic progression, two formulas are used:
.
Which of them is more convenient to use in this case?
By condition, the formula for the nth term of the original progression is known ( a n) a n= 3n - 4. You can immediately find a 1, And a 16 without finding d. Therefore, we will use the first formula.
Answer: 368.
Task 5
In arithmetic progression( a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.
According to the formula of the nth term:
a 22 = a 1 + d (22 – 1) = a 1+ 21d.
By condition, if a 1= -6, then a 22= -6 + 21d . It is necessary to find the difference of progressions:
d = a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = -48.
Answer : a 22 = -48.
Task 6
Several consecutive terms of the geometric progression are written:
Find the term of the progression indicated by x.
When solving, we will use the formula for the nth term b n = b 1 ∙ q n - 1 for geometric progressions. The first term of the progression. To find the denominator of the progression q, you need to take any of the given terms of the progression and divide by the previous one. In our example, we can take and divide by. We obtain that q = 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.
Substituting the found values into the formula, we get:
.
Answer : .
Task 7
From the arithmetic progressions given by the formula of the nth term, select the one for which the condition is satisfied a 27 > 9:
Since the given condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:
.
Answer: 4.
Task 8
In arithmetic progression a 1= 3, d = -1.5. Specify the largest value of n for which the inequality holds a n > -6.
