Or arithmetic - this is a type of ordered number 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.
One thing worth considering 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 total number n terms.
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 represent the given segment from m to n of the progression as a new number series. In such m-th representation the 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, as well as knowing what numbers in the series they occupy, you can use the formula for the sum obtained in the previous paragraph. It will turn out:
S 5 12 = (12 - 5 + 1) * (8 + 29) / 2 = 148.
It is worth noting that this value could be obtained differently: first find the sum of the first 12 elements using the standard formula, then calculate the sum of the first 4 elements using the same formula, then subtract the second from the first sum.
When studying algebra in secondary school(9th grade) one of important topics is the study of number sequences, which include progressions - geometric and arithmetic. In this article we will look at an arithmetic progression and examples with solutions.
What is an arithmetic progression?
To understand this, it is necessary to define the progression in question, as well as provide the basic formulas that will be used later in solving problems.
An arithmetic or algebraic progression is a set of ordered rational numbers, each term of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of an ordered series of numbers and the difference, you can restore the entire arithmetic progression.
Let's give an example. The following sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the type of progression under consideration, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).
Important Formulas
Let us now present the basic formulas that will be needed to solve problems using arithmetic progression. Let us denote by the symbol a n nth term sequences where n is an integer. We denote the difference by the Latin letter d. Then the following expressions are valid:
- To determine the value of the nth term, the following formula is suitable: a n = (n-1)*d+a 1 .
- To determine the sum of the first n terms: S n = (a n +a 1)*n/2.
To understand any examples of arithmetic progression with solutions in 9th grade, it is enough to remember these two formulas, since any problems of the type under consideration are based on their use. You should also remember that the progression difference is determined by the formula: d = a n - a n-1.
Example #1: finding an unknown member
Let's give a simple example of an arithmetic progression and the formulas that need to be used to solve it.
Let the sequence 10, 8, 6, 4, ... be given, you need to find five terms in it.
From the conditions of the problem it already follows that the first 4 terms are known. The fifth can be defined in two ways:
- Let's first calculate the difference. We have: d = 8 - 10 = -2. Similarly, you could take any two other members standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d = a n - a n-1, then d = a 5 - a 4, from which we get: a 5 = a 4 + d. We substitute the known values: a 5 = 4 + (-2) = 2.
- The second method also requires knowledge of the difference of the progression in question, so you first need to determine it as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for the n number of the sequence. We have: a n = (n - 1) * d + a 1 = (n - 1) * (-2) + 10 = 12 - 2*n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.
As you can see, both solutions led to the same result. Note that in this example the progression difference d is a negative value. Such sequences are called decreasing, since each next term is less than the previous one.
Example #2: progression difference
Now let’s complicate the task a little, let’s give an example of how
It is known that in some the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.
Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . Let's substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 = 6 + 6 * d. From this expression you can easily calculate the difference: d = (18 - 6) /6 = 2. Thus, we have answered the first part of the problem.
To restore the sequence to the 7th term, you should use the definition of an algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d, and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2=8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16, a 7 = 18.
Example No. 3: drawing up a progression
Let's complicate the problem even more. Now we need to answer the question of how to find an arithmetic progression. The following example can be given: two numbers are given, for example - 4 and 5. It is necessary to create an algebraic progression so that three more terms are placed between these.
Before you start solving this problem, you need to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 = -4 and a 5 = 5. Having established this, we move on to the problem, which is similar to the previous one. Again, for the nth term we use the formula, we get: a 5 = a 1 + 4 * d. From: d = (a 5 - a 1)/4 = (5 - (-4)) / 4 = 2.25. What we got here is not an integer value of the difference, but it is a rational number, so the formulas for the algebraic progression remain the same.
Now let's add the found difference to a 1 and restore the missing terms of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 = 2.75 + 2.25 = 5, which coincided with the conditions of the problem.
Example No. 4: first term of progression
Let's continue to give examples of arithmetic progression with solutions. In all previous problems, the first number of the algebraic progression was known. Now let's consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find which number this sequence begins with.
The formulas used so far assume knowledge of a 1 and d. In the problem statement, nothing is known about these numbers. Nevertheless, we will write down expressions for each term about which information is available: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We received two equations in which there are 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.
The easiest way to solve this system is to express a 1 in each equation and then compare the resulting expressions. First equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 = a 43 - 42 * d = 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d = 37 - 42 * d, whence the difference d = (37 - 50) / (42 - 14) = - 0.464 (only 3 decimal places are given).
Knowing d, you can use any of the 2 expressions above for a 1. For example, first: a 1 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.
If you have doubts about the result obtained, you can check it, for example, determine the 43rd term of the progression, which is specified in the condition. We get: a 43 = a 1 + 42 * d = 56.496 + 42 * (- 0.464) = 37.008. The small error is due to the fact that rounding to thousandths was used in the calculations.
Example No. 5: amount
Now let's look at several examples with solutions for the sum of an arithmetic progression.
Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?
Thanks to the development of computer technology, it is possible to solve this problem, that is, add all the numbers sequentially, which Calculating machine will do as soon as the person presses the Enter key. However, the problem can be solved mentally if you pay attention that the presented series of numbers is an algebraic progression, and its difference is equal to 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.
It is interesting to note that this problem is called “Gaussian” because at the beginning of the 18th century the famous German, still only 10 years old, was able to solve it in his head in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add the numbers at the ends of the sequence in pairs, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer it is enough to multiply 50 by 101.
Example No. 6: sum of terms from n to m
Another typical example of the sum of an arithmetic progression is the following: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its terms from 8 to 14 will be equal to.
The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them sequentially. Since there are few terms, this method is not quite labor-intensive. Nevertheless, it is proposed to solve this problem using a second method, which is more universal.
The idea is to obtain a formula for the sum of the algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:
- S m = m * (a m + a 1) / 2.
- S n = n * (a n + a 1) / 2.
Since n > m, it is obvious that the 2nd sum includes the first. The last conclusion means that if we take the difference between these sums and add the term a m to it (in the case of taking the difference, it is subtracted from the sum S n), we will obtain the necessary answer to the problem. We have: S mn = S n - S m + a m =n * (a 1 + a n) / 2 - m *(a 1 + a m)/2 + a m = a 1 * (n - m) / 2 + a n * n/2 + a m * (1- m/2). It is necessary to substitute formulas for a n and a m into this expression. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d *(3 * m - m 2 - 2) / 2.
The resulting formula is somewhat cumbersome, however, the sum S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.
As can be seen from the above solutions, all problems are based on knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before starting to solve any of these problems, it is recommended that you carefully read the condition, clearly understand what you need to find, and only then proceed with the solution.
Another tip is to strive for simplicity, that is, if you can answer a question without using complex mathematical calculations, then you need to do just that, since in this case the likelihood of making a mistake is less. For example, in the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and divide the overall problem into separate subtasks (in this case, first find the terms a n and a m).
If you have doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. We found out how to find an arithmetic progression. If you figure it out, it's not that difficult.
Some people treat the word “progression” with caution, as a very complex term from the sections higher mathematics. And yet the simplest arithmetic progression- work of the taxi meter (where they still remain). And understanding the essence (and in mathematics there is nothing more important than “understanding the essence”) of an arithmetic sequence is not so difficult, having analyzed a few elementary concepts.
Mathematical number sequence
A numerical sequence is usually called a series of numbers, each of which has its own number.
a 1 is the first member of the sequence;
and 2 is the second term of the sequence;
and 7 is the seventh member of the sequence;
and n is the nth member of the sequence;
However, not any arbitrary set of numbers and numbers interests us. We will focus our attention on a numerical sequence in which the value of the nth term is related to its ordinal number by a relationship that can be clearly formulated mathematically. In other words: the numerical value of the nth number is some function of n.
a is the value of a member of a numerical sequence;
n is its serial number;
f(n) is a function, where the ordinal number in the numerical sequence n is the argument.
Definition
An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth term of an arithmetic sequence is as follows:
a n - the value of the current member of the arithmetic progression;
a n+1 - formula of the next number;
d - difference (certain number).
It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one and such an arithmetic progression will be increasing.
In the graph below it is easy to see why the number sequence is called “increasing”.
In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.
Specified member value
Sometimes it is necessary to determine the value of any arbitrary term a n of an arithmetic progression. This can be done by sequentially calculating the values of all members of the arithmetic progression, starting from the first to the desired one. However, this path is not always acceptable if, for example, it is necessary to find the value of the five-thousandth or eight-millionth term. Traditional calculations will take a lot of time. However, a specific arithmetic progression can be studied using certain formulas. There is also a formula for the nth term: the value of any term of an arithmetic progression can be determined as the sum of the first term of the progression with the difference of the progression, multiplied by the number of the desired term, reduced by one.
The formula is universal for increasing and decreasing progression.
An example of calculating the value of a given term
Let us solve the following problem of finding the value of the nth term of an arithmetic progression.
Condition: there is an arithmetic progression with parameters:
The first term of the sequence is 3;
The difference in the number series is 1.2.
Task: you need to find the value of 214 terms
Solution: to determine the value of a given term, we use the formula:
a(n) = a1 + d(n-1)
Substituting the data from the problem statement into the expression, we have:
a(214) = a1 + d(n-1)
a(214) = 3 + 1.2 (214-1) = 258.6
Answer: The 214th term of the sequence is equal to 258.6.
The advantages of this method of calculation are obvious - the entire solution takes no more than 2 lines.
Sum of a given number of terms
Very often, in a given arithmetic series, it is necessary to determine the sum of the values of some of its segments. To do this, there is also no need to calculate the values of each term and then add them up. This method is applicable if the number of terms whose sum needs to be found is small. In other cases, it is more convenient to use the following formula.
The sum of the terms of an arithmetic progression from 1 to n is equal to the sum of the first and nth terms, multiplied by the number of the term n and divided by two. If in the formula the value of the nth term is replaced by the expression from the previous paragraph of the article, we get:
Calculation example
For example, let’s solve a problem with the following conditions:
The first term of the sequence is zero;
The difference is 0.5.
The problem requires determining the sum of the terms of the series from 56 to 101.
Solution. Let's use the formula for determining the amount of progression:
s(n) = (2∙a1 + d∙(n-1))∙n/2
First, we determine the sum of the values of 101 terms of the progression by substituting the given conditions of our problem into the formula:
s 101 = (2∙0 + 0.5∙(101-1))∙101/2 = 2,525
Obviously, in order to find out the sum of the terms of the progression from the 56th to the 101st, it is necessary to subtract S 55 from S 101.
s 55 = (2∙0 + 0.5∙(55-1))∙55/2 = 742.5
Thus, the sum of the arithmetic progression for this example is:
s 101 - s 55 = 2,525 - 742.5 = 1,782.5
Example of practical application of arithmetic progression
At the end of the article, let's return to the example of an arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider this example.
Boarding a taxi (which includes 3 km of travel) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles/km. Travel distance is 30 km. Calculate the cost of the trip.
1. Let’s discard the first 3 km, the price of which is included in the cost of landing.
30 - 3 = 27 km.
2. Further calculation is nothing more than parsing an arithmetic number series.
Member number - the number of kilometers traveled (minus the first three).
The value of the member is the sum.
The first term in this problem will be equal to a 1 = 50 rubles.
Progression difference d = 22 r.
the number we are interested in is the value of the (27+1)th term of the arithmetic progression - the meter reading at the end of the 27th kilometer is 27.999... = 28 km.
a 28 = 50 + 22 ∙ (28 - 1) = 644
Calendar data calculations for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of the celestial body to the star. In addition, various number series are successfully used in statistics and other applied areas of mathematics.
Another type of number sequence is geometric
Geometric progression is characterized by greater rates of change compared to arithmetic progression. It is no coincidence that in politics, sociology, and medicine, in order to show the high speed of spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops in geometric progression.
The Nth term of the geometric number series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first term is 1, the denominator is correspondingly equal to 2, then:
n=1: 1 ∙ 2 = 2
n=2: 2 ∙ 2 = 4
n=3: 4 ∙ 2 = 8
n=4: 8 ∙ 2 = 16
n=5: 16 ∙ 2 = 32,
b n - the value of the current term of the geometric progression;
b n+1 - formula of the next term of the geometric progression;
q is the denominator of the geometric progression (a constant number).
If the graph of an arithmetic progression is a straight line, then a geometric progression paints a slightly different picture:
As in the case of arithmetic, geometric progression has a formula for the value of an arbitrary term. Any nth term of a geometric progression is equal to the product of the first term and the denominator of the progression to the power of n reduced by one:
Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Let's find the 5th term of the progression
b 5 = b 1 ∙ q (5-1) = 3 ∙ 1.5 4 = 15.1875
The sum of a given number of terms is also calculated using a special formula. The sum of the first n terms of a geometric progression is equal to the difference between the product of the nth term of the progression and its denominator and the first term of the progression, divided by the denominator reduced by one:
If b n is replaced using the formula discussed above, the value of the sum of the first n terms of the number series under consideration will take the form:
Example. The geometric progression starts with the first term equal to 1. The denominator is set to 3. Let's find the sum of the first eight terms.
s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280
For example, the sequence \(2\); \(5\); \(8\); \(eleven\); \(14\)... is an arithmetic progression, because each subsequent element differs from the previous one by three (can be obtained from the previous one by adding three):
In this progression, the difference \(d\) is positive (equal to \(3\)), and therefore each next term is greater than the previous one. Such progressions are called increasing.
However, \(d\) can also be a negative number. For example, in arithmetic progression \(16\); \(10\); \(4\); \(-2\); \(-8\)... the progression difference \(d\) is equal to minus six.
And in this case, each next element will be smaller than the previous one. These progressions are called decreasing.
Arithmetic progression notation
Progression is indicated by a small Latin letter.
Numbers that form a progression are called members(or elements).
They are denoted by the same letter as an arithmetic progression, but with a numerical index equal to the number of the element in order.
For example, the arithmetic progression \(a_n = \left\( 2; 5; 8; 11; 14…\right\)\) consists of the elements \(a_1=2\); \(a_2=5\); \(a_3=8\) and so on.
In other words, for the progression \(a_n = \left\(2; 5; 8; 11; 14…\right\)\)
Solving arithmetic progression problems
In principle, the information presented above is already enough to solve almost any arithmetic progression problem (including those offered at the OGE).
Example (OGE).
The arithmetic progression is specified by the conditions \(b_1=7; d=4\). Find \(b_5\).
Solution:
Answer: \(b_5=23\)
Example (OGE).
The first three terms of an arithmetic progression are given: \(62; 49; 36…\) Find the value of the first negative term of this progression..
Solution:
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We are given the first elements of the sequence and know that it is an arithmetic progression. That is, each element differs from its neighbor by the same number. Let's find out which one by subtracting the previous one from the next element: \(d=49-62=-13\). |
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Now we can restore our progression to the (first negative) element we need. |
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Ready. You can write an answer. |
Answer: \(-3\)
Example (OGE).
Given several consecutive elements of an arithmetic progression: \(…5; x; 10; 12.5...\) Find the value of the element designated by the letter \(x\).
Solution:
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To find \(x\), we need to know how much the next element differs from the previous one, in other words, the progression difference. Let's find it from two known neighboring elements: \(d=12.5-10=2.5\). |
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And now we can easily find what we are looking for: \(x=5+2.5=7.5\). |
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Ready. You can write an answer. |
Answer: \(7,5\).
Example (OGE).
The arithmetic progression is defined by the following conditions: \(a_1=-11\); \(a_(n+1)=a_n+5\) Find the sum of the first six terms of this progression.
Solution:
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We need to find the sum of the first six terms of the progression. But we do not know their meanings; we are given only the first element. Therefore, we first calculate the values one by one, using what is given to us: \(n=1\); \(a_(1+1)=a_1+5=-11+5=-6\) |
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\(S_6=a_1+a_2+a_3+a_4+a_5+a_6=\) |
The required amount has been found. |
Answer: \(S_6=9\).
Example (OGE).
In arithmetic progression \(a_(12)=23\); \(a_(16)=51\). Find the difference of this progression.
Solution:
Answer: \(d=7\).
Important formulas for arithmetic progression
As you can see, many problems on arithmetic progression can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each subsequent element in this chain is obtained by adding the same number to the previous one (the difference of the progression).
However, sometimes there are situations when deciding “head-on” is very inconvenient. For example, imagine that in the very first example we need to find not the fifth element \(b_5\), but the three hundred and eighty-sixth \(b_(386)\). Should we add four \(385\) times? Or imagine that in the penultimate example you need to find the sum of the first seventy-three elements. You'll be tired of counting...
Therefore, in such cases they do not solve things “head-on”, but use special formulas derived for arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum of \(n\) first terms.
Formula of the \(n\)th term: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first term of the progression;
\(n\) – number of the required element;
\(a_n\) – term of the progression with number \(n\).
This formula allows us to quickly find even the three-hundredth or the millionth element, knowing only the first and the difference of the progression.
Example.
The arithmetic progression is specified by the conditions: \(b_1=-159\); \(d=8.2\). Find \(b_(246)\).
Solution:
Answer: \(b_(246)=1850\).
Formula for the sum of the first n terms: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where
\(a_n\) – the last summed term;
Example (OGE).
The arithmetic progression is specified by the conditions \(a_n=3.4n-0.6\). Find the sum of the first \(25\) terms of this progression.
Solution:
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\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\) |
To calculate the sum of the first twenty-five terms, we need to know the value of the first and twenty-fifth terms. |
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\(n=1;\) \(a_1=3.4·1-0.6=2.8\) |
Now let's find the twenty-fifth term by substituting twenty-five instead of \(n\). |
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\(n=25;\) \(a_(25)=3.4·25-0.6=84.4\) |
Well, now we can easily calculate the required amount. |
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\(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25=\) |
The answer is ready. |
Answer: \(S_(25)=1090\).
For the sum \(n\) of the first terms, you can get another formula: you just need to \(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25\ ) instead of \(a_n\) substitute the formula for it \(a_n=a_1+(n-1)d\). We get:
Formula for the sum of the first n terms: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where
\(S_n\) – the required sum of \(n\) first elements;
\(a_1\) – the first summed term;
\(d\) – progression difference;
\(n\) – number of elements in total.
Example.
Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15.5\); \(14\)…
Solution:
Answer: \(S_(33)=-231\).
More complex arithmetic progression problems
Now you have all the information you need to solve almost any arithmetic progression problem. Let’s finish the topic by considering problems in which you not only need to apply formulas, but also think a little (in mathematics this can be useful ☺)
Example (OGE).
Find the sum of all negative terms of the progression: \(-19.3\); \(-19\); \(-18.7\)…
Solution:
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\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\) |
The task is very similar to the previous one. We begin to solve the same thing: first we find \(d\). |
|
|
\(d=a_2-a_1=-19-(-19.3)=0.3\) |
Now I would like to substitute \(d\) into the formula for the sum... and here a small nuance emerges - we do not know \(n\). In other words, we don’t know how many terms will need to be added. How to find out? Let's think. We will stop adding elements when we reach the first positive element. That is, you need to find out the number of this element. How? Let's write down the formula for calculating any element of an arithmetic progression: \(a_n=a_1+(n-1)d\) for our case. |
|
|
\(a_n=a_1+(n-1)d\) |
||
|
\(a_n=-19.3+(n-1)·0.3\) |
We need \(a_n\) to become greater than zero. Let's find out at what \(n\) this will happen. |
|
|
\(-19.3+(n-1)·0.3>0\) |
||
|
\((n-1)·0.3>19.3\) \(|:0.3\) |
We divide both sides of the inequality by \(0.3\). |
|
|
\(n-1>\)\(\frac(19.3)(0.3)\) |
We transfer minus one, not forgetting to change the signs |
|
|
\(n>\)\(\frac(19.3)(0.3)\) \(+1\) |
Let's calculate... |
|
|
\(n>65,333…\) |
...and it turns out that the first positive element will have the number \(66\). Accordingly, the last negative one has \(n=65\). Just in case, let's check this. |
|
|
\(n=65;\) \(a_(65)=-19.3+(65-1)·0.3=-0.1\) |
So we need to add the first \(65\) elements. |
|
|
\(S_(65)=\) \(\frac(2 \cdot (-19.3)+(65-1)0.3)(2)\)\(\cdot 65\) |
The answer is ready. |
Answer: \(S_(65)=-630.5\).
Example (OGE).
The arithmetic progression is specified by the conditions: \(a_1=-33\); \(a_(n+1)=a_n+4\). Find the sum from the \(26\)th to the \(42\) element inclusive.
Solution:
|
\(a_1=-33;\) \(a_(n+1)=a_n+4\) |
In this problem you also need to find the sum of elements, but starting not from the first, but from the \(26\)th. For such a case we do not have a formula. How to decide? |
|
|
For our progression \(a_1=-33\), and the difference \(d=4\) (after all, we add the four to the previous element to find the next one). Knowing this, we find the sum of the first \(42\)-y elements. |
|
\(S_(42)=\) \(\frac(2 \cdot (-33)+(42-1)4)(2)\)\(\cdot 42=\) |
Now the sum of the first \(25\) elements. |
|
\(S_(25)=\) \(\frac(2 \cdot (-33)+(25-1)4)(2)\)\(\cdot 25=\) |
And finally, we calculate the answer. |
|
\(S=S_(42)-S_(25)=2058-375=1683\) |
Answer: \(S=1683\).
For arithmetic progression, there are several more formulas that we did not consider in this article due to their low practical usefulness. However, you can easily find them.
If for every natural number n match real number a n , then they say that it is given number sequence :
a 1 , a 2 , a 3 , . . . , a n , . . . .
So, the number sequence is a function of the natural argument.
Number a 1 called first term of the sequence , number a 2 — second term of the sequence , number a 3 — third and so on. Number a n called nth term sequences , and a natural number n — his number .
From two adjacent members a n And a n +1 sequence member a n +1 called subsequent (towards a n ), A a n — previous (towards a n +1 ).
To define a sequence, you need to specify a method that allows you to find a member of the sequence with any number.
Often the sequence is specified using nth term formulas , that is, a formula that allows you to determine a member of a sequence by its number.
For example,
sequence of positive odd numbers can be given by the formula
a n= 2n- 1,
and the sequence of alternating 1 And -1 - formula
b n = (-1)n +1 . ◄
The sequence can be determined recurrent formula, that is, a formula that expresses any member of the sequence, starting with some, through the previous (one or more) members.
For example,
If a 1 = 1 , A a n +1 = a n + 5
a 1 = 1,
a 2 = a 1 + 5 = 1 + 5 = 6,
a 3 = a 2 + 5 = 6 + 5 = 11,
a 4 = a 3 + 5 = 11 + 5 = 16,
a 5 = a 4 + 5 = 16 + 5 = 21.
If a 1= 1, a 2 = 1, a n +2 = a n + a n +1 , then the first seven terms of the numerical sequence are established as follows:
a 1 = 1,
a 2 = 1,
a 3 = a 1 + a 2 = 1 + 1 = 2,
a 4 = a 2 + a 3 = 1 + 2 = 3,
a 5 = a 3 + a 4 = 2 + 3 = 5,
a 6 = a 4 + a 5 = 3 + 5 = 8,
a 7 = a 5 + a 6 = 5 + 8 = 13. ◄
Sequences can be final And endless .
The sequence is called ultimate , if it has a finite number of members. The sequence is called endless , if it has infinitely many members.
For example,
sequence of two-digit natural numbers:
10, 11, 12, 13, . . . , 98, 99
final.
Sequence of prime numbers:
2, 3, 5, 7, 11, 13, . . .
endless. ◄
The sequence is called increasing , if each of its members, starting from the second, is greater than the previous one.
The sequence is called decreasing , if each of its members, starting from the second, is less than the previous one.
For example,
2, 4, 6, 8, . . . , 2n, . . . — increasing sequence;
1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . — decreasing sequence. ◄
A sequence whose elements do not decrease as the number increases, or, conversely, do not increase, is called monotonous sequence .
Monotonic sequences, in particular, are increasing sequences and decreasing sequences.
Arithmetic progression
Arithmetic progression is a sequence in which each member, starting from the second, is equal to the previous one, to which the same number is added.
a 1 , a 2 , a 3 , . . . , a n, . . .
is an arithmetic progression if for any natural number n the condition is met:
a n +1 = a n + d,
Where d - a certain number.
Thus, the difference between the subsequent and previous terms of a given arithmetic progression is always constant:
a 2 - a 1 = a 3 - a 2 = . . . = a n +1 - a n = d.
Number d called difference of arithmetic progression.
To define an arithmetic progression, it is enough to indicate its first term and difference.
For example,
If a 1 = 3, d = 4 , then we find the first five terms of the sequence as follows:
a 1 =3,
a 2 = a 1 + d = 3 + 4 = 7,
a 3 = a 2 + d= 7 + 4 = 11,
a 4 = a 3 + d= 11 + 4 = 15,
a 5 = a 4 + d= 15 + 4 = 19. ◄
For an arithmetic progression with the first term a 1 and the difference d her n
a n = a 1 + (n- 1)d.
For example,
find the thirtieth term of the arithmetic progression
1, 4, 7, 10, . . .
a 1 =1, d = 3,
a 30 = a 1 + (30 - 1)d = 1 + 29· 3 = 88. ◄
a n-1 = a 1 + (n- 2)d,
a n= a 1 + (n- 1)d,
a n +1 = a 1 + nd,
then obviously
| a n=
| a n-1 + a n+1
|
| 2
|
Each member of an arithmetic progression, starting from the second, is equal to the arithmetic mean of the preceding and subsequent members.
the numbers a, b and c are successive terms of some arithmetic progression if and only if one of them is equal to the arithmetic mean of the other two.
For example,
a n = 2n- 7 , is an arithmetic progression.
Let's use the above statement. We have:
a n = 2n- 7,
a n-1 = 2(n- 1) - 7 = 2n- 9,
a n+1 = 2(n+ 1) - 7 = 2n- 5.
Hence,
| a n+1 + a n-1
| =
| 2n- 5 + 2n- 9
| = 2n- 7 = a n,
|
| 2
| 2
|
◄
Note that n The th term of an arithmetic progression can be found not only through a 1 , but also any previous a k
a n = a k + (n- k)d.
For example,
For a 5 can be written down
a 5 = a 1 + 4d,
a 5 = a 2 + 3d,
a 5 = a 3 + 2d,
a 5 = a 4 + d. ◄
a n = a n-k + kd,
a n = a n+k - kd,
then obviously
| a n=
| a n-k
+a n+k
|
| 2
|
any member of an arithmetic progression, starting from the second, is equal to half the sum of the equally spaced members of this arithmetic progression.
In addition, for any arithmetic progression the following equality holds:
a m + a n = a k + a l,
m + n = k + l.
For example,
in arithmetic progression
1) a 10 = 28 = (25 + 31)/2 = (a 9 + a 11 )/2;
2) 28 = a 10 = a 3 + 7d= 7 + 7 3 = 7 + 21 = 28;
3) a 10= 28 = (19 + 37)/2 = (a 7 + a 13)/2;
4) a 2 + a 12 = a 5 + a 9, because
a 2 + a 12= 4 + 34 = 38,
a 5 + a 9 = 13 + 25 = 38. ◄
S n= a 1 + a 2 + a 3 + . . .+ a n,
first n terms of an arithmetic progression is equal to the product of half the sum of the extreme terms and the number of terms:
From here, in particular, it follows that if you need to sum the terms
a k, a k +1 , . . . , a n,
then the previous formula retains its structure:
For example,
in arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, . . .
S 10 = 1 + 4 + . . . + 28 = (1 + 28) · 10/2 = 145;
10 + 13 + 16 + 19 + 22 + 25 + 28 = S 10 - S 3 = (10 + 28 ) · (10 - 4 + 1)/2 = 133. ◄
If an arithmetic progression is given, then the quantities a 1 , a n, d, n AndS n connected by two formulas:
Therefore, if the values of three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas, combined into a system of two equations with two unknowns.
An arithmetic progression is a monotonic sequence. Wherein:
- If d > 0 , then it is increasing;
- If d < 0 , then it is decreasing;
- If d = 0 , then the sequence will be stationary.
Geometric progression
Geometric progression is a sequence in which each member, starting from the second, is equal to the previous one multiplied by the same number.
b 1 , b 2 , b 3 , . . . , b n, . . .
is a geometric progression if for any natural number n the condition is met:
b n +1 = b n · q,
Where q ≠ 0 - a certain number.
Thus, the ratio of the subsequent term of a given geometric progression to the previous one is a constant number:
b 2 / b 1 = b 3 / b 2 = . . . = b n +1 / b n = q.
Number q called denominator of geometric progression.
To define a geometric progression, it is enough to indicate its first term and denominator.
For example,
If b 1 = 1, q = -3 , then we find the first five terms of the sequence as follows:
b 1 = 1,
b 2 = b 1 · q = 1 · (-3) = -3,
b 3 = b 2 · q= -3 · (-3) = 9,
b 4 = b 3 · q= 9 · (-3) = -27,
b 5 = b 4 · q= -27 · (-3) = 81. ◄
b 1 and denominator q her n The th term can be found using the formula:
b n = b 1 · qn -1 .
For example,
find the seventh term of the geometric progression 1, 2, 4, . . .
b 1 = 1, q = 2,
b 7 = b 1 · q 6 = 1 2 6 = 64. ◄
b n-1 = b 1 · qn -2 ,
b n = b 1 · qn -1 ,
b n +1 = b 1 · qn,
then obviously
b n 2 = b n -1 · b n +1 ,
each member of the geometric progression, starting from the second, is equal to the geometric mean (proportional) of the preceding and subsequent members.
Since the converse is also true, the following statement holds:
the numbers a, b and c are successive terms of some geometric progression if and only if the square of one of them is equal to the product of the other two, that is, one of the numbers is the geometric mean of the other two.
For example,
Let us prove that the sequence given by the formula b n= -3 2 n , is a geometric progression. Let's use the above statement. We have:
b n= -3 2 n,
b n -1 = -3 2 n -1 ,
b n +1 = -3 2 n +1 .
Hence,
b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) · (-3 · 2 n +1 ) = b n -1 · b n +1 ,
which proves the desired statement. ◄
Note that n The th term of a geometric progression can be found not only through b 1 , but also any previous member b k , for which it is enough to use the formula
b n = b k · qn - k.
For example,
For b 5 can be written down
b 5 = b 1 · q 4 ,
b 5 = b 2 · q 3,
b 5 = b 3 · q 2,
b 5 = b 4 · q. ◄
b n = b k · qn - k,
b n = b n - k · q k,
then obviously
b n 2 = b n - k· b n + k
the square of any term of a geometric progression, starting from the second, is equal to the product of the terms of this progression equidistant from it.
In addition, for any geometric progression the equality is true:
b m· b n= b k· b l,
m+ n= k+ l.
For example,
in geometric progression
1) b 6 2 = 32 2 = 1024 = 16 · 64 = b 5 · b 7 ;
2) 1024 = b 11 = b 6 · q 5 = 32 · 2 5 = 1024;
3) b 6 2 = 32 2 = 1024 = 8 · 128 = b 4 · b 8 ;
4) b 2 · b 7 = b 4 · b 5 , because
b 2 · b 7 = 2 · 64 = 128,
b 4 · b 5 = 8 · 16 = 128. ◄
S n= b 1 + b 2 + b 3 + . . . + b n
first n members of a geometric progression with denominator q ≠ 0 calculated by the formula:
And when q = 1 - according to the formula
S n= nb 1
Note that if you need to sum the terms
b k, b k +1 , . . . , b n,
then the formula is used:
| S n- S k -1 = b k + b k +1 + . . . + b n = b k · | 1 - qn -
k +1
| . |
| 1 - q
|
For example,
in geometric progression 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, . . .
S 10 = 1 + 2 + . . . + 512 = 1 · (1 - 2 10) / (1 - 2) = 1023;
64 + 128 + 256 + 512 = S 10 - S 6 = 64 · (1 - 2 10-7+1) / (1 - 2) = 960. ◄
If a geometric progression is given, then the quantities b 1 , b n, q, n And S n connected by two formulas:
Therefore, if the values of any three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas, combined into a system of two equations with two unknowns.
For a geometric progression with the first term b 1 and denominator q the following take place properties of monotonicity :
- progression is increasing if one of the following conditions is met:
b 1 > 0 And q> 1;
b 1 < 0 And 0 < q< 1;
- The progression is decreasing if one of the following conditions is met:
b 1 > 0 And 0 < q< 1;
b 1 < 0 And q> 1.
If q< 0 , then the geometric progression is alternating: its terms with odd numbers have the same sign as its first term, and terms with even numbers have the opposite sign. It is clear that an alternating geometric progression is not monotonic.
Product of the first n terms of a geometric progression can be calculated using the formula:
Pn= b 1 · b 2 · b 3 · . . . · b n = (b 1 · b n) n / 2 .
For example,
1 · 2 · 4 · 8 · 16 · 32 · 64 · 128 = (1 · 128) 8/2 = 128 4 = 268 435 456;
3 · 6 · 12 · 24 · 48 = (3 · 48) 5/2 = (144 1/2) 5 = 12 5 = 248 832.◄
Infinitely decreasing geometric progression
Infinitely decreasing geometric progression called an infinite geometric progression whose denominator modulus is less 1 , that is
|q| < 1 .
Note that an infinitely decreasing geometric progression may not be a decreasing sequence. It fits the occasion
1 < q< 0 .
With such a denominator, the sequence is alternating. For example,
1, - 1 / 2 , 1 / 4 , - 1 / 8 , . . . .
The sum of an infinitely decreasing geometric progression name the number to which the sum of the first ones approaches without limit n members of a progression with an unlimited increase in the number n . This number is always finite and is expressed by the formula
| S= b 1 + b 2 + b 3 + . . . = | b 1
| . |
| 1 - q
|
For example,
10 + 1 + 0,1 + 0,01 + . . . = 10 / (1 - 0,1) = 11 1 / 9 ,
10 - 1 + 0,1 - 0,01 + . . . = 10 / (1 + 0,1) = 9 1 / 11 . ◄
Relationship between arithmetic and geometric progressions
Arithmetic and geometric progression are closely related to each other. Let's look at just two examples.
a 1 , a 2 , a 3 , . . . d , That
b a 1 , b a 2 , b a 3 , . . . b d .
For example,
1, 3, 5, . . . - arithmetic progression with difference 2 And
7 1 , 7 3 , 7 5 , . . . - geometric progression with denominator 7 2 . ◄
b 1 , b 2 , b 3 , . . . - geometric progression with denominator q , That
log a b 1, log a b 2, log a b 3, . . . - arithmetic progression with difference log aq .
For example,
2, 12, 72, . . . - geometric progression with denominator 6 And
lg 2, lg 12, lg 72, . . . - arithmetic progression with difference lg 6 . ◄