The topic "arithmetic progression" is studied in the general course of algebra in schools in the 9th grade. This topic is important for further in-depth study of the mathematics of number series. In this article, we will get acquainted with the arithmetic progression, its difference, as well as with typical tasks that schoolchildren may face.
The concept of algebraic progression
Numeric progression is a sequence of numbers in which each subsequent element can be obtained from the previous one, if some mathematical law is applied. There are two simple types of progression: geometric and arithmetic, which is also called algebraic. Let's dwell on it in more detail.
Let's imagine some rational number, denote it by the symbol a1, where the index indicates its ordinal number in the series under consideration. Let's add some other number to a1 , let's denote it d. Then the secondan element of a series can be reflected as follows: a2=a1+d. Now add d again, we get: a3=a2+d. Continuing this mathematical operation, you can get a whole series of numbers, which will be called an arithmetic progression.
As can be understood from the above, to find the n-th element of this sequence, you must use the formula: a =a1 + (n -1)d. Indeed, substituting n=1 into the expression, we get a1=a1, if n=2, then the formula implies: a 2=a1 + 1d, and so on.
For example, if the difference of an arithmetic progression is 5, and a1=1, then this means that the number series of the type in question looks like: 1, 6, 11, 16, 21, … As you can see, each of its terms is greater than the previous one by 5.
Formulas for the difference of arithmetic progression
From the above definition of the considered series of numbers, it follows that to determine it, you need to know two numbers: a1 and d. The latter is called the difference of this progression. It uniquely determines the behavior of the entire series. Indeed, if d is positive, then the number series will constantly increase, on the contrary, in the case of negative d, the numbers in the series will increase only modulo, while their absolute value will decrease with increasing number n.
What is the difference of the arithmetic progression? Consider the two main formulas that are used to calculate this value:
- d=an+1-a , this formula follows directly from the definition of the number series in question.
- d=(-a1+a)/(n-1), this expression is obtained by expressing d from the formula given in the previous paragraph of the article. Note that this expression becomes indeterminate (0/0) if n=1. This is due to the fact that it is necessary to know at least 2 elements of the series in order to determine its difference.
These two basic formulas are used to solve any problem of finding the progression difference. However, there is another formula that you also need to know about.
Sum of first elements
The formula that can be used to determine the sum of any number of members of an algebraic progression, according to historical evidence, was first obtained by the "prince" of mathematics of the 18th century, Carl Gauss. A German scientist, while still a boy in the elementary grades of a village school, noticed that in order to add natural numbers in the series from 1 to 100, you must first sum the first element and the last (the resulting value will be equal to the sum of the penultimate and second, penultimate and third elements, and so on), and then this number should be multiplied by the number of these amounts, that is, by 50.
The formula that reflects the stated result on a particular example can be generalized to an arbitrary case. It will look like this: S =n/2(a +a1). Note that in order to find the specified value, knowledge of the difference d is not required,if two terms of the progression are known (a and a1).
Example 1. Determine the difference, knowing the two terms of the series a1 and an
Let's show how to apply the formulas mentioned above in the article. Let's give a simple example: the difference of the arithmetic progression is unknown, it is necessary to determine what it will be equal to if a13=-5, 6 and a1=-12, 1.
Since we know the values of two elements of the numerical sequence, and one of them is the first number, we can use formula No. 2 to determine the difference d. We have: d=(-1(-12, 1)+(-5, 6))/12=0. 54167. In the expression, we used the value n=13, since the member with this serial number is known.
The resulting difference indicates that the progression is increasing, despite the fact that the elements given in the condition of the problem have a negative value. It can be seen that a13>a1, although |a13|<|a1 |.
Example 2. Positive members of the progression in example 1
Let's use the result obtained in the previous example to solve a new problem. It is formulated as follows: from what sequence number do the elements of the progression in example 1 begin to take positive values?
As shown, the progression in which a1=-12, 1 and d=0. 54167 is increasing, so from some number the numbers will start to take on only positive values. To determine this number n, one has to solve a simple inequality, which ismathematically written as follows: a >0 or, using the appropriate formula, we rewrite the inequality: a1 + (n-1)d>0. It is necessary to find the unknown n, let's express it: n>-1a1/d + 1. Now it remains to substitute the known values of the difference and the first member of the sequence. We get: n>-1(-12, 1) /0, 54167 + 1=23, 338 or n>23, 338. Since n can only take integer values, it follows from the resulting inequality that any members of the series that will have a number greater than 23 will be positive.
Check your answer by using the above formula to calculate the 23rd and 24th elements of this arithmetic progression. We have: a23=-12, 1 + 220, 54167=-0, 18326 (negative number); a24=-12, 1 + 230. 54167=0. 3584 (positive value). Thus, the result obtained is correct: starting from n=24, all members of the number series will be greater than zero.
Example 3. How many logs will fit?
Let's give one curious problem: during logging, it was decided to stack sawn logs on top of each other as shown in the figure below. How many logs can be stacked this way, knowing that 10 rows will fit in total?
In this way of stacking logs, you can notice one interesting thing: each subsequent row will contain one less log than the previous one, that is, there is an algebraic progression, the difference of which is d=1. Assuming that the number of logs in each row is a member of this progression,and also given that a1=1 (only one log will fit at the very top), we find the number a10. We have: a10=1 + 1(10-1)=10. That is, in the 10th row, which lies on the ground, there will be 10 logs.
The total amount of this "pyramidal" construction can be obtained using the Gauss formula. We get: S10=10/2(10+1)=55 logs.
