Problems on arithmetic progression existed already in ancient times. They appeared and demanded a solution because they had a practical need.
So, in one of the papyri Ancient Egypt", which has a 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, which amounted to a lot interesting tasks and who added the fourteenth book to Euclid’s Elements, formulated the thought: “In an arithmetic progression, which has even number members, sum of members of the 2nd half more than the amount members of the 1st on the square of 1/2 the number of members.”
The sequence is denoted by an. The numbers of a sequence are called its members and are usually denoted by letters with indices that indicate serial number this member (a1, a2, a3 ... reads: “a 1st”, “a 2nd”, “a 3rd” and so on).
The sequence can be infinite or finite.
What is it 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. At very large quantities members it's already 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 the sequence is given similar formula, then this 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 given sequence- arithmetic progression. This equality is also a sign of progression, which is why it is usually called characteristic property 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 us to determine nth term 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 1st n terms finite progression) is calculated in the following way:
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.
Natural series of any numbers, such as 1,2,3,...,n,...- simplest example arithmetic progression.
In addition to the arithmetic progression, there is also a geometric progression, which has its own properties and characteristics.
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 terms from the fifth to the twentieth - direct application formulas 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 is correct definition elements of the formula.
The task writers encrypt these same 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's bring similar ones, we get new formula sums of terms of an arithmetic progression:
 |
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. Is it possible in right moment it’s easy to display it, 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 are 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 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. quite applicable to them standard formula amounts. 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?) Will it help? additional formula from task 2.
Answers (in disarray): 7, 3240, 6.
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.
Some people treat the word “progression” with caution, as a very complex term from sections higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi meter (where they still exist). And understand the essence (and in mathematics there is nothing more important than “getting the essence”) arithmetic sequence It's not that difficult once you understand a few basic concepts.
Mathematical number sequence
A numerical sequence is usually called a series of numbers, each of which has its own number.
a 1 is the first member of the sequence;
and 2 is the second term of the sequence;
and 7 is the seventh member of the sequence;
and n is the nth member of the sequence;
However, not any arbitrary set of numbers and numbers interests us. We will focus our attention on a numerical sequence in which the value of the nth term is related to its ordinal number by a relationship that can be clearly formulated mathematically. In other words: numerical value The nth number is some function of n.
a is the value of a member of a numerical sequence;
n is its serial number;
f(n) is a function, where the ordinal number in the numerical sequence n is the argument.
Definition
An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth term of an arithmetic sequence is as follows:
a n - the value of the current member of the arithmetic progression;
a n+1 - formula of the next number;
d - difference (certain number).
It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one and such an arithmetic progression will be increasing.
In the graph below it is easy to see why number sequence called "increasing".
In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.
Specified member value
Sometimes it is necessary to determine the value of any arbitrary term a n of an arithmetic progression. This can be done by sequentially calculating the values of all members of the arithmetic progression, starting from the first to the desired one. However, this path is not always acceptable if, for example, it is necessary to find the value of the five-thousandth or eight-millionth term. Traditional calculations will take a lot of time. However, a specific arithmetic progression can be studied using certain formulas. There is also a formula for the nth term: the value of any term of an arithmetic progression can be determined as the sum of the first term of the progression with the difference of the progression, multiplied by the number of the desired term, reduced by one.
The formula is universal for increasing and decreasing progression.
An example of calculating the value of a given term
Let us solve the following problem of finding the value of the nth term of an arithmetic progression.
Condition: there is an arithmetic progression with parameters:
The first term of the sequence is 3;
The difference in the number series is 1.2.
Task: you need to find the value of 214 terms
Solution: to determine the value of a given term, we use the formula:
a(n) = a1 + d(n-1)
Substituting the data from the problem statement into the expression, we have:
a(214) = a1 + d(n-1)
a(214) = 3 + 1.2 (214-1) = 258.6
Answer: The 214th term of the sequence is equal to 258.6.
The advantages of this method of calculation are obvious - the entire solution takes no more than 2 lines.
Sum of a given number of terms
Very often, in a given arithmetic series, it is necessary to determine the sum of the values of some of its segments. To do this, there is also no need to calculate the values of each term and then add them up. This method is applicable if the number of terms whose sum needs to be found is small. In other cases, it is more convenient to use the following formula.
The sum of the terms of an arithmetic progression from 1 to n is equal to the sum of the first and nth terms, multiplied by the number of the term n and divided by two. If in the formula the value of the nth term is replaced by the expression from the previous paragraph of the article, we get:
Calculation example
For example, let’s solve a problem with the following conditions:
The first term of the sequence is zero;
The difference is 0.5.
The problem requires determining the sum of the terms of the series from 56 to 101.
Solution. Let's use the formula for determining the amount of progression:
s(n) = (2∙a1 + d∙(n-1))∙n/2
First, we determine the sum of the values of 101 terms of the progression by substituting the given conditions of our problem into the formula:
s 101 = (2∙0 + 0.5∙(101-1))∙101/2 = 2,525
Obviously, in order to find out the sum of the terms of the progression from the 56th to the 101st, it is necessary to subtract S 55 from S 101.
s 55 = (2∙0 + 0.5∙(55-1))∙55/2 = 742.5
Thus, the sum of the arithmetic progression for this example is:
s 101 - s 55 = 2,525 - 742.5 = 1,782.5
Example of practical application of arithmetic progression
At the end of the article, let's return to the example of an arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider this example.
Boarding a taxi (which includes 3 km of travel) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles/km. Travel distance is 30 km. Calculate the cost of the trip.
1. Let’s discard the first 3 km, the price of which is included in the cost of landing.
30 - 3 = 27 km.
2. Further calculation is nothing more than parsing an arithmetic number series.
Member number - the number of kilometers traveled (minus the first three).
The value of the member is the sum.
The first term in this problem will be equal to a 1 = 50 rubles.
Progression difference d = 22 r.
the number we are interested in is the value of the (27+1)th term of the arithmetic progression - the meter reading at the end of the 27th kilometer is 27.999... = 28 km.
a 28 = 50 + 22 ∙ (28 - 1) = 644
Calendar data calculations for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of the celestial body to the star. In addition, various number series are successfully used in statistics and other applied areas of mathematics.
Another type of number sequence is geometric
Geometric progression is characterized by greater rates of change compared to arithmetic progression. It is no coincidence that in politics, sociology, and medicine, in order to show the high speed of spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops in geometric progression.
The Nth term of the geometric number series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first term is 1, the denominator is correspondingly equal to 2, then:
n=1: 1 ∙ 2 = 2
n=2: 2 ∙ 2 = 4
n=3: 4 ∙ 2 = 8
n=4: 8 ∙ 2 = 16
n=5: 16 ∙ 2 = 32,
b n - the value of the current term of the geometric progression;
b n+1 - formula of the next term of the geometric progression;
q is the denominator of the geometric progression (a constant number).
If the graph of an arithmetic progression is a straight line, then a geometric progression paints a slightly different picture:
As in the case of arithmetic, geometric progression has a formula for the value of an arbitrary term. Any nth term of a geometric progression is equal to the product of the first term and the denominator of the progression to the power of n reduced by one:
Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Let's find the 5th term of the progression
b 5 = b 1 ∙ q (5-1) = 3 ∙ 1.5 4 = 15.1875
The sum of a given number of terms is also calculated using a special formula. The sum of the first n terms of a geometric progression is equal to the difference between the product of the nth term of the progression and its denominator and the first term of the progression, divided by the denominator reduced by one:
If b n is replaced using the formula discussed above, the value of the sum of the first n terms of the number series under consideration will take the form:
Example. The geometric progression starts with the first term equal to 1. The denominator is set to 3. Let's find the sum of the first eight terms.
s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280
Arithmetic progression name a sequence of numbers (terms of a progression)
In which each subsequent term differs from the previous one by a new term, which is also called step or progression difference.
Thus, by specifying the progression step and its first term, you can find any of its elements using the formula
Properties of an arithmetic progression
1) Each member of an arithmetic progression, starting from the second number, is the arithmetic mean of the previous and next members of the progression
The converse is also true. If the arithmetic mean of adjacent odd (even) terms of a progression is equal to the term that stands between them, then this sequence of numbers is an arithmetic progression. Using this statement, it is very easy to check any sequence.
Also, by the property of arithmetic progression, the above formula can be generalized to the following
This is easy to verify if you write the terms to the right of the equal sign
It is often used in practice to simplify calculations in problems.
2) The sum of the first n terms of an arithmetic progression is calculated using the formula
Remember well the formula for the sum of an arithmetic progression; it is indispensable in calculations and is quite often found in simple life situations.
3) If you need to find not the whole sum, but part of the sequence starting from its kth term, then the following sum formula will be useful to you
4) Of practical interest is finding the sum of n terms of an arithmetic progression starting from the kth number. To do this, use the formula
This concludes the theoretical material and moves on to solving common problems in practice.
Example 1. Find the fortieth term of the arithmetic progression 4;7;...
Solution:
According to the condition we have
Let's determine the progression step
Using a well-known formula, we find the fortieth term of the progression
Example 2. An arithmetic progression is given by its third and seventh terms. Find the first term of the progression and the sum of ten.
Solution:
Let us write down the given elements of the progression using the formulas
We subtract the first from the second equation, as a result we find the progression step
We substitute the found value into any of the equations to find the first term of the arithmetic progression
We calculate the sum of the first ten terms of the progression
Without using complex calculations, we found all the required quantities.
Example 3. An arithmetic progression is given by the denominator and one of its terms. Find the first term of the progression, the sum of its 50 terms starting from 50 and the sum of the first 100.
Solution:
Let's write down the formula for the hundredth element of the progression
and find the first one
Based on the first, we find the 50th term of the progression
Finding the sum of the part of the progression
and the sum of the first 100
The progression amount is 250.
Example 4.
Find the number of terms of an arithmetic progression if:
a3-a1=8, a2+a4=14, Sn=111.
Solution:
Let's write the equations in terms of the first term and the progression step and determine them
We substitute the obtained values into the sum formula to determine the number of terms in the sum
We carry out simplifications
and solve the quadratic equation
Of the two values found, only the number 8 fits the problem conditions. Thus, the sum of the first eight terms of the progression is 111.
Example 5.
Solve the equation
1+3+5+...+x=307.
Solution: This equation is the sum of an arithmetic progression. Let's write out its first term and find the difference in progression
Yes, yes: arithmetic progression is not a toy for you :) Well, friends, if you are reading this text, then the internal cap-evidence tells me that you do not yet know what an arithmetic progression is, but you really (no, like that: SOOOOO!) want to know. Therefore, I will not torment you with long introductions and will get straight to the point.
First, a couple of examples. Let's look at several sets of numbers:
- 1; 2; 3; 4; ...
- 15; 20; 25; 30; ...
- $\sqrt(2);\ 2\sqrt(2);\ 3\sqrt(2);...$
What do all these sets have in common? At first glance, nothing. But actually there is something. Namely: each next element differs from the previous one by the same number.
Judge for yourself. The first set is simply consecutive numbers, each next being one more than the previous one. In the second case, the difference between adjacent numbers is already five, but this difference is still constant. In the third case, there are roots altogether. However, $2\sqrt(2)=\sqrt(2)+\sqrt(2)$, and $3\sqrt(2)=2\sqrt(2)+\sqrt(2)$, i.e. and in this case, each next element simply increases by $\sqrt(2)$ (and don’t be afraid that this number is irrational).
So: all such sequences are called arithmetic progressions. Let's give a strict definition:
Definition. A sequence of numbers in which each next one differs from the previous one by exactly the same amount is called an arithmetic progression. The very amount by which the numbers differ is called the progression difference and is most often denoted by the letter $d$.
Notation: $\left(((a)_(n)) \right)$ is the progression itself, $d$ is its difference.
And just a couple of important notes. Firstly, progression is only considered ordered sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. Numbers cannot be rearranged or swapped.
Secondly, the sequence itself can be either finite or infinite. For example, the set (1; 2; 3) is obviously a finite arithmetic progression. But if you write something in the spirit (1; 2; 3; 4; ...) - this is already an infinite progression. The ellipsis after the four seems to hint that there are quite a few more numbers to come. Infinitely many, for example. :)
I would also like to note that progressions can be increasing or decreasing. We have already seen increasing ones - the same set (1; 2; 3; 4; ...). Here are examples of decreasing progressions:
- 49; 41; 33; 25; 17; ...
- 17,5; 12; 6,5; 1; −4,5; −10; ...
- $\sqrt(5);\ \sqrt(5)-1;\ \sqrt(5)-2;\ \sqrt(5)-3;...$
Okay, okay: the last example may seem overly complicated. But the rest, I think, you understand. Therefore, we introduce new definitions:
Definition. An arithmetic progression is called:
- increasing if each next element is greater than the previous one;
- decreasing if, on the contrary, each subsequent element is less than the previous one.
In addition, there are so-called “stationary” sequences - they consist of the same repeating number. For example, (3; 3; 3; ...).
Only one question remains: how to distinguish an increasing progression from a decreasing one? Fortunately, everything here depends only on the sign of the number $d$, i.e. progression differences:
- If $d \gt 0$, then the progression increases;
- If $d \lt 0$, then the progression is obviously decreasing;
- Finally, there is the case $d=0$ - in this case the entire progression is reduced to a stationary sequence of identical numbers: (1; 1; 1; 1; ...), etc.
Let's try to calculate the difference $d$ for the three decreasing progressions given above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract the number on the left from the number on the right. It will look like this:
- 41−49=−8;
- 12−17,5=−5,5;
- $\sqrt(5)-1-\sqrt(5)=-1$.
As we can see, in all three cases the difference actually turned out to be negative. And now that we have more or less figured out the definitions, it’s time to figure out how progressions are described and what properties they have.
Progression terms and recurrence formula
Since the elements of our sequences cannot be swapped, they can be numbered:
\[\left(((a)_(n)) \right)=\left\( ((a)_(1)),\ ((a)_(2)),((a)_(3 )),... \right\)\]
The individual elements of this set are called members of a progression. They are indicated by a number: first member, second member, etc.
In addition, as we already know, neighboring terms of the progression are related by the formula:
\[((a)_(n))-((a)_(n-1))=d\Rightarrow ((a)_(n))=((a)_(n-1))+d \]
In short, to find the $n$th term of a progression, you need to know the $n-1$th term and the difference $d$. This formula is called recurrent, because with its help you can find any number only by knowing the previous one (and in fact, all the previous ones). This is very inconvenient, so there is a more cunning formula that reduces any calculations to the first term and the difference:
\[((a)_(n))=((a)_(1))+\left(n-1 \right)d\]
You've probably already come across this formula. They like to give it in all sorts of reference books and solution books. And in any sensible mathematics textbook it is one of the first.
However, I suggest you practice a little.
Task No. 1. Write down the first three terms of the arithmetic progression $\left(((a)_(n)) \right)$ if $((a)_(1))=8,d=-5$.
Solution. So, we know the first term $((a)_(1))=8$ and the difference of the progression $d=-5$. Let's use the formula just given and substitute $n=1$, $n=2$ and $n=3$:
\[\begin(align) & ((a)_(n))=((a)_(1))+\left(n-1 \right)d; \\ & ((a)_(1))=((a)_(1))+\left(1-1 \right)d=((a)_(1))=8; \\ & ((a)_(2))=((a)_(1))+\left(2-1 \right)d=((a)_(1))+d=8-5= 3; \\ & ((a)_(3))=((a)_(1))+\left(3-1 \right)d=((a)_(1))+2d=8-10= -2. \\ \end(align)\]
Answer: (8; 3; −2)
That's all! Please note: our progression is decreasing.
Of course, $n=1$ could not be substituted - the first term is already known to us. However, by substituting unity, we were convinced that even for the first term our formula works. In other cases, everything came down to banal arithmetic.
Task No. 2. Write down the first three terms of an arithmetic progression if its seventh term is equal to −40 and its seventeenth term is equal to −50.
Solution. Let's write the problem condition in familiar terms:
\[((a)_(7))=-40;\quad ((a)_(17))=-50.\]
\[\left\( \begin(align) & ((a)_(7))=((a)_(1))+6d \\ & ((a)_(17))=((a) _(1))+16d \\ \end(align) \right.\]
\[\left\( \begin(align) & ((a)_(1))+6d=-40 \\ & ((a)_(1))+16d=-50 \\ \end(align) \right.\]
I put the system sign because these requirements must be met simultaneously. Now let’s note that if we subtract the first from the second equation (we have the right to do this, since we have a system), we get this:
\[\begin(align) & ((a)_(1))+16d-\left(((a)_(1))+6d \right)=-50-\left(-40 \right); \\ & ((a)_(1))+16d-((a)_(1))-6d=-50+40; \\&10d=-10; \\&d=-1. \\ \end(align)\]
That's how easy it is to find the progression difference! All that remains is to substitute the found number into any of the equations of the system. For example, in the first:
\[\begin(matrix) ((a)_(1))+6d=-40;\quad d=-1 \\ \Downarrow \\ ((a)_(1))-6=-40; \\ ((a)_(1))=-40+6=-34. \\ \end(matrix)\]
Now, knowing the first term and the difference, it remains to find the second and third terms:
\[\begin(align) & ((a)_(2))=((a)_(1))+d=-34-1=-35; \\ & ((a)_(3))=((a)_(1))+2d=-34-2=-36. \\ \end(align)\]
Ready! The problem is solved.
Answer: (−34; −35; −36)
Notice the interesting property of progression that we discovered: if we take the $n$th and $m$th terms and subtract them from each other, we get the difference of the progression multiplied by the $n-m$ number:
\[((a)_(n))-((a)_(m))=d\cdot \left(n-m \right)\]
A simple but very useful property that you definitely need to know - with its help you can significantly speed up the solution of many progression problems. Here is a clear example of this:
Task No. 3. The fifth term of an arithmetic progression is 8.4, and its tenth term is 14.4. Find the fifteenth term of this progression.
Solution. Since $((a)_(5))=8.4$, $((a)_(10))=14.4$, and we need to find $((a)_(15))$, we note following:
\[\begin(align) & ((a)_(15))-((a)_(10))=5d; \\ & ((a)_(10))-((a)_(5))=5d. \\ \end(align)\]
But by condition $((a)_(10))-((a)_(5))=14.4-8.4=6$, therefore $5d=6$, from which we have:
\[\begin(align) & ((a)_(15))-14,4=6; \\ & ((a)_(15))=6+14.4=20.4. \\ \end(align)\]
Answer: 20.4
That's all! We didn’t need to create any systems of equations and calculate the first term and the difference - everything was solved in just a couple of lines.
Now let's look at another type of problem - searching for negative and positive terms of a progression. It is no secret that if a progression increases, and its first term is negative, then sooner or later positive terms will appear in it. And vice versa: the terms of a decreasing progression will sooner or later become negative.
At the same time, it is not always possible to find this moment “head-on” by sequentially going through the elements. Often, problems are written in such a way that without knowing the formulas, the calculations would take several sheets of paper—we would simply fall asleep while we found the answer. Therefore, let's try to solve these problems in a faster way.
Task No. 4. How many negative terms are there in the arithmetic progression −38.5; −35.8; ...?
Solution. So, $((a)_(1))=-38.5$, $((a)_(2))=-35.8$, from where we immediately find the difference:
Note that the difference is positive, so the progression increases. The first term is negative, so indeed at some point we will stumble upon positive numbers. The only question is when this will happen.
Let's try to find out how long (i.e. up to what natural number $n$) the negativity of the terms remains:
\[\begin(align) & ((a)_(n)) \lt 0\Rightarrow ((a)_(1))+\left(n-1 \right)d \lt 0; \\ & -38.5+\left(n-1 \right)\cdot 2.7 \lt 0;\quad \left| \cdot 10 \right. \\ & -385+27\cdot \left(n-1 \right) \lt 0; \\ & -385+27n-27 \lt 0; \\ & 27n \lt 412; \\ & n \lt 15\frac(7)(27)\Rightarrow ((n)_(\max ))=15. \\ \end(align)\]
The last line requires some explanation. So we know that $n \lt 15\frac(7)(27)$. On the other hand, we are satisfied with only integer values of the number (moreover: $n\in \mathbb(N)$), so the largest permissible number is precisely $n=15$, and in no case 16.
Task No. 5. In arithmetic progression $(()_(5))=-150,(()_(6))=-147$. Find the number of the first positive term of this progression.
This would be exactly the same problem as the previous one, but we do not know $((a)_(1))$. But the neighboring terms are known: $((a)_(5))$ and $((a)_(6))$, so we can easily find the difference of the progression:
In addition, let's try to express the fifth term through the first and the difference using the standard formula:
\[\begin(align) & ((a)_(n))=((a)_(1))+\left(n-1 \right)\cdot d; \\ & ((a)_(5))=((a)_(1))+4d; \\ & -150=((a)_(1))+4\cdot 3; \\ & ((a)_(1))=-150-12=-162. \\ \end(align)\]
Now we proceed by analogy with the previous task. Let's find out at what point in our sequence positive numbers will appear:
\[\begin(align) & ((a)_(n))=-162+\left(n-1 \right)\cdot 3 \gt 0; \\ & -162+3n-3 \gt 0; \\ & 3n \gt 165; \\ & n \gt 55\Rightarrow ((n)_(\min ))=56. \\ \end(align)\]
The minimum integer solution to this inequality is the number 56.
Please note: in the last task everything came down to strict inequality, so the option $n=55$ will not suit us.
Now that we have learned how to solve simple problems, let's move on to more complex ones. But first, let's study another very useful property of arithmetic progressions, which will save us a lot of time and unequal cells in the future. :)
Arithmetic mean and equal indentations
Let's consider several consecutive terms of the increasing arithmetic progression $\left(((a)_(n)) \right)$. Let's try to mark them on the number line:
Terms of an arithmetic progression on the number lineI specifically marked arbitrary terms $((a)_(n-3)),...,((a)_(n+3))$, and not some $((a)_(1)) ,\ ((a)_(2)),\ ((a)_(3))$, etc. Because the rule that I’ll tell you about now works the same for any “segments”.
And the rule is very simple. Let's remember the recurrent formula and write it down for all marked terms:
\[\begin(align) & ((a)_(n-2))=((a)_(n-3))+d; \\ & ((a)_(n-1))=((a)_(n-2))+d; \\ & ((a)_(n))=((a)_(n-1))+d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n+1))+d; \\ \end(align)\]
However, these equalities can be rewritten differently:
\[\begin(align) & ((a)_(n-1))=((a)_(n))-d; \\ & ((a)_(n-2))=((a)_(n))-2d; \\ & ((a)_(n-3))=((a)_(n))-3d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n))+2d; \\ & ((a)_(n+3))=((a)_(n))+3d; \\ \end(align)\]
Well, so what? And the fact that the terms $((a)_(n-1))$ and $((a)_(n+1))$ lie at the same distance from $((a)_(n)) $. And this distance is equal to $d$. The same can be said about the terms $((a)_(n-2))$ and $((a)_(n+2))$ - they are also removed from $((a)_(n))$ at the same distance equal to $2d$. We can continue ad infinitum, but the meaning is well illustrated by the picture
The terms of the progression lie at the same distance from the center What does this mean for us? This means that $((a)_(n))$ can be found if the neighboring numbers are known:
\[((a)_(n))=\frac(((a)_(n-1))+((a)_(n+1)))(2)\]
We have derived an excellent statement: every term of an arithmetic progression is equal to the arithmetic mean of its neighboring terms! Moreover: we can step back from our $((a)_(n))$ to the left and to the right not by one step, but by $k$ steps - and the formula will still be correct:
\[((a)_(n))=\frac(((a)_(n-k))+((a)_(n+k)))(2)\]
Those. we can easily find some $((a)_(150))$ if we know $((a)_(100))$ and $((a)_(200))$, because $(( a)_(150))=\frac(((a)_(100))+((a)_(200)))(2)$. At first glance, it may seem that this fact does not give us anything useful. However, in practice, many problems are specially tailored to use the arithmetic mean. Take a look:
Task No. 6. Find all values of $x$ for which the numbers $-6((x)^(2))$, $x+1$ and $14+4((x)^(2))$ are consecutive terms of an arithmetic progression (in in the order indicated).
Solution. Since these numbers are members of a progression, the arithmetic mean condition is satisfied for them: the central element $x+1$ can be expressed in terms of neighboring elements:
\[\begin(align) & x+1=\frac(-6((x)^(2))+14+4((x)^(2)))(2); \\ & x+1=\frac(14-2((x)^(2)))(2); \\ & x+1=7-((x)^(2)); \\ & ((x)^(2))+x-6=0. \\ \end(align)\]
The result is a classic quadratic equation. Its roots: $x=2$ and $x=-3$ are the answers.
Answer: −3; 2.
Task No. 7. Find the values of $$ for which the numbers $-1;4-3;(()^(2))+1$ form an arithmetic progression (in that order).
Solution. Let us again express the middle term through the arithmetic mean of neighboring terms:
\[\begin(align) & 4x-3=\frac(x-1+((x)^(2))+1)(2); \\ & 4x-3=\frac(((x)^(2))+x)(2);\quad \left| \cdot 2 \right.; \\ & 8x-6=((x)^(2))+x; \\ & ((x)^(2))-7x+6=0. \\ \end(align)\]
Quadratic equation again. And again there are two roots: $x=6$ and $x=1$.
Answer: 1; 6.
If in the process of solving a problem you come up with some brutal numbers, or you are not entirely sure of the correctness of the answers found, then there is a wonderful technique that allows you to check: have we solved the problem correctly?
Let's say in problem No. 6 we received answers −3 and 2. How can we check that these answers are correct? Let's just plug them into the original condition and see what happens. Let me remind you that we have three numbers ($-6(()^(2))$, $+1$ and $14+4(()^(2))$), which must form an arithmetic progression. Let's substitute $x=-3$:
\[\begin(align) & x=-3\Rightarrow \\ & -6((x)^(2))=-54; \\ & x+1=-2; \\ & 14+4((x)^(2))=50. \end(align)\]
We got the numbers −54; −2; 50 that differ by 52 is undoubtedly an arithmetic progression. The same thing happens for $x=2$:
\[\begin(align) & x=2\Rightarrow \\ & -6((x)^(2))=-24; \\ & x+1=3; \\ & 14+4((x)^(2))=30. \end(align)\]
Again a progression, but with a difference of 27. Thus, the problem was solved correctly. Those who wish can check the second problem on their own, but I’ll say right away: everything is correct there too.
In general, while solving the last problems, we came across another interesting fact that also needs to be remembered:
If three numbers are such that the second is the arithmetic mean of the first and last, then these numbers form an arithmetic progression.
In the future, understanding this statement will allow us to literally “construct” the necessary progressions based on the conditions of the problem. But before we engage in such “construction”, we should pay attention to one more fact, which directly follows from what has already been discussed.
Grouping and summing elements
Let's return to the number axis again. Let us note there several members of the progression, between which, perhaps. is worth a lot of other members:
There are 6 elements marked on the number lineLet's try to express the “left tail” through $((a)_(n))$ and $d$, and the “right tail” through $((a)_(k))$ and $d$. It's very simple:
\[\begin(align) & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n))+2d; \\ & ((a)_(k-1))=((a)_(k))-d; \\ & ((a)_(k-2))=((a)_(k))-2d. \\ \end(align)\]
Now note that the following amounts are equal:
\[\begin(align) & ((a)_(n))+((a)_(k))=S; \\ & ((a)_(n+1))+((a)_(k-1))=((a)_(n))+d+((a)_(k))-d= S; \\ & ((a)_(n+2))+((a)_(k-2))=((a)_(n))+2d+((a)_(k))-2d= S. \end(align)\]
Simply put, if we consider as a start two elements of the progression, which in total are equal to some number $S$, and then begin to step from these elements in opposite directions (toward each other or vice versa to move away), then the sums of the elements that we will stumble upon will also be equal$S$. This can be most clearly represented graphically:
Equal indentations give equal amounts Understanding this fact will allow us to solve problems of a fundamentally higher level of complexity than those we considered above. For example, these:
Task No. 8. Determine the difference of an arithmetic progression in which the first term is 66, and the product of the second and twelfth terms is the smallest possible.
Solution. Let's write down everything we know:
\[\begin(align) & ((a)_(1))=66; \\&d=? \\ & ((a)_(2))\cdot ((a)_(12))=\min . \end(align)\]
So, we do not know the progression difference $d$. Actually, the entire solution will be built around the difference, since the product $((a)_(2))\cdot ((a)_(12))$ can be rewritten as follows:
\[\begin(align) & ((a)_(2))=((a)_(1))+d=66+d; \\ & ((a)_(12))=((a)_(1))+11d=66+11d; \\ & ((a)_(2))\cdot ((a)_(12))=\left(66+d \right)\cdot \left(66+11d \right)= \\ & =11 \cdot \left(d+66 \right)\cdot \left(d+6 \right). \end(align)\]
For those in the tank: I took the total multiplier of 11 out of the second bracket. Thus, the desired product is a quadratic function with respect to the variable $d$. Therefore, consider the function $f\left(d \right)=11\left(d+66 \right)\left(d+6 \right)$ - its graph will be a parabola with branches up, because if we expand the brackets, we get:
\[\begin(align) & f\left(d \right)=11\left(((d)^(2))+66d+6d+66\cdot 6 \right)= \\ & =11(( d)^(2))+11\cdot 72d+11\cdot 66\cdot 6 \end(align)\]
As you can see, the coefficient of the highest term is 11 - this is a positive number, so we are really dealing with a parabola with upward branches:
graph of a quadratic function - parabola
Please note: this parabola takes its minimum value at its vertex with the abscissa $((d)_(0))$. Of course, we can calculate this abscissa using the standard scheme (there is the formula $((d)_(0))=(-b)/(2a)\;$), but it would be much more reasonable to note that the desired vertex lies on the axis symmetry of the parabola, therefore the point $((d)_(0))$ is equidistant from the roots of the equation $f\left(d \right)=0$:
\[\begin(align) & f\left(d \right)=0; \\ & 11\cdot \left(d+66 \right)\cdot \left(d+6 \right)=0; \\ & ((d)_(1))=-66;\quad ((d)_(2))=-6. \\ \end(align)\]
That is why I was in no particular hurry to open the brackets: in their original form, the roots were very, very easy to find. Therefore, the abscissa is equal to the arithmetic mean of the numbers −66 and −6:
\[((d)_(0))=\frac(-66-6)(2)=-36\]
What does the discovered number give us? With it, the required product takes on the smallest value (by the way, we never calculated $((y)_(\min ))$ - this is not required of us). At the same time, this number is the difference of the original progression, i.e. we found the answer. :)
Answer: −36
Task No. 9. Between the numbers $-\frac(1)(2)$ and $-\frac(1)(6)$ insert three numbers so that together with these numbers they form an arithmetic progression.
Solution. Essentially, we need to make a sequence of five numbers, with the first and last number already known. Let's denote the missing numbers by the variables $x$, $y$ and $z$:
\[\left(((a)_(n)) \right)=\left\( -\frac(1)(2);x;y;z;-\frac(1)(6) \right\ )\]
Note that the number $y$ is the “middle” of our sequence - it is equidistant from the numbers $x$ and $z$, and from the numbers $-\frac(1)(2)$ and $-\frac(1)( 6)$. And if we currently cannot obtain $y$ from the numbers $x$ and $z$, then the situation is different with the ends of the progression. Let's remember the arithmetic mean:
Now, knowing $y$, we will find the remaining numbers. Note that $x$ lies between the numbers $-\frac(1)(2)$ and the $y=-\frac(1)(3)$ we just found. That's why
Using similar reasoning, we find the remaining number:
Ready! We found all three numbers. Let's write them in the answer in the order in which they should be inserted between the original numbers.
Answer: $-\frac(5)(12);\ -\frac(1)(3);\ -\frac(1)(4)$
Task No. 10. Between the numbers 2 and 42, insert several numbers that, together with these numbers, form an arithmetic progression, if you know that the sum of the first, second and last of the inserted numbers is 56.
Solution. An even more complex problem, which, however, is solved according to the same scheme as the previous ones - through the arithmetic mean. The problem is that we don’t know exactly how many numbers need to be inserted. Therefore, let us assume for definiteness that after inserting everything there will be exactly $n$ numbers, and the first of them is 2, and the last is 42. In this case, the required arithmetic progression can be represented in the form:
\[\left(((a)_(n)) \right)=\left\( 2;((a)_(2));((a)_(3));...;(( a)_(n-1));42 \right\)\]
\[((a)_(2))+((a)_(3))+((a)_(n-1))=56\]
Note, however, that the numbers $((a)_(2))$ and $((a)_(n-1))$ are obtained from the numbers 2 and 42 at the edges by one step towards each other, i.e. . to the center of the sequence. And this means that
\[((a)_(2))+((a)_(n-1))=2+42=44\]
But then the expression written above can be rewritten as follows:
\[\begin(align) & ((a)_(2))+((a)_(3))+((a)_(n-1))=56; \\ & \left(((a)_(2))+((a)_(n-1)) \right)+((a)_(3))=56; \\ & 44+((a)_(3))=56; \\ & ((a)_(3))=56-44=12. \\ \end(align)\]
Knowing $((a)_(3))$ and $((a)_(1))$, we can easily find the difference of the progression:
\[\begin(align) & ((a)_(3))-((a)_(1))=12-2=10; \\ & ((a)_(3))-((a)_(1))=\left(3-1 \right)\cdot d=2d; \\ & 2d=10\Rightarrow d=5. \\ \end(align)\]
All that remains is to find the remaining terms:
\[\begin(align) & ((a)_(1))=2; \\ & ((a)_(2))=2+5=7; \\ & ((a)_(3))=12; \\ & ((a)_(4))=2+3\cdot 5=17; \\ & ((a)_(5))=2+4\cdot 5=22; \\ & ((a)_(6))=2+5\cdot 5=27; \\ & ((a)_(7))=2+6\cdot 5=32; \\ & ((a)_(8))=2+7\cdot 5=37; \\ & ((a)_(9))=2+8\cdot 5=42; \\ \end(align)\]
Thus, already at the 9th step we will arrive at the left end of the sequence - the number 42. In total, only 7 numbers had to be inserted: 7; 12; 17; 22; 27; 32; 37.
Answer: 7; 12; 17; 22; 27; 32; 37
Word problems with progressions
In conclusion, I would like to consider a couple of relatively simple problems. Well, as simple as that: for most students who study mathematics at school and have not read what is written above, these problems may seem tough. Nevertheless, these are the types of problems that appear in the OGE and the Unified State Exam in mathematics, so I recommend that you familiarize yourself with them.
Task No. 11. The team produced 62 parts in January, and in each subsequent month they produced 14 more parts than in the previous month. How many parts did the team produce in November?
Solution. Obviously, the number of parts listed by month will represent an increasing arithmetic progression. Moreover:
\[\begin(align) & ((a)_(1))=62;\quad d=14; \\ & ((a)_(n))=62+\left(n-1 \right)\cdot 14. \\ \end(align)\]
November is the 11th month of the year, so we need to find $((a)_(11))$:
\[((a)_(11))=62+10\cdot 14=202\]
Therefore, 202 parts will be produced in November.
Task No. 12. The bookbinding workshop bound 216 books in January, and in each subsequent month it bound 4 more books than in the previous month. How many books did the workshop bind in December?
Solution. All the same:
$\begin(align) & ((a)_(1))=216;\quad d=4; \\ & ((a)_(n))=216+\left(n-1 \right)\cdot 4. \\ \end(align)$
December is the last, 12th month of the year, so we are looking for $((a)_(12))$:
\[((a)_(12))=216+11\cdot 4=260\]
This is the answer - 260 books will be bound in December.
Well, if you have read this far, I hasten to congratulate you: you have successfully completed the “young fighter’s course” in arithmetic progressions. You can safely move on to the next lesson, where we will study the formula for the sum of progression, as well as important and very useful consequences from it.