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 arithmetic progression, its difference, as well as with typical tasks that schoolchildren may face.
The concept of algebraic progression
A numerical progression is a sequence of numbers in which each subsequent element can be obtained from the previous one, if you apply some mathematical law. Two simple types of progression are known: geometric and arithmetic, which is also called algebraic. Let us dwell on it in more detail.
Imagine a rational number, denote it by the symbol a 1 , where the index indicates its serial number in the series under consideration. Add to a 1 some other number, denote it by d. Then the second element of the series can be reflected as follows: a 2 = a 1 + d. Now add d again, we get: a 3 = a 2 + d. Continuing this mathematical operation, you can get a number of numbers, which will be called arithmetic progression.
As can be understood from the above, in order to find the nth element of this sequence, it is necessary to use the formula: a n = a 1 + (n-1) * d. Indeed, substituting n = 1 into the expression, we get a 1 = a 1 , if n = 2, then the formula implies: a 2 = a 1 + 1 * d, and so on.
For example, if the difference in arithmetic progression is 5, and a 1 = 1, then this means that the number series of the type in question has the form: 1, 6, 11, 16, 21, ... As you can see, each of its members is 5 times more than the previous one .
Arithmetic progression difference formulas
From the above definition of the considered series of numbers it follows that for its determination it is necessary to know two numbers: a 1 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, if d is negative, the numbers in the series will increase only in absolute value, but their absolute value will decrease with increasing number n.
What is the difference in arithmetic progression equal to? Consider the two main formulas that are used to calculate this quantity:
- d = a n + 1 -a n , this formula follows directly from the definition of the series of numbers in question.
- d = (-a 1 + a n ) / (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 ambiguous (0/0) if n = 1. This is due to the fact that knowledge of at least 2 elements of the series is necessary in order to determine its difference.
These two basic formulas are used to solve any problems of finding the difference of a progression. However, there is another formula that you also need to know about.
The sum of the first elements
The formula with which you can 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 XVIII century Karl Gauss. A German scientist, while still a boy in the elementary grades of a village school, noted that in order to add natural numbers in a row from 1 to 100, you must first add up the first element and the last (the obtained 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 sums, that is, by 50.
A formula that reflects the stated result in a particular example can be generalized to an arbitrary case. It will have the form: S n = n / 2 * (a n + a 1 ). Note that to find the indicated quantity, knowledge of the difference d is not required if two terms of the progression are known (a n and a 1 ).
Example No. 1. Determine the difference, knowing the two members of the series a1 and an
We show how to apply the formulas specified in the article. We give a simple example: the difference in arithmetic progression is unknown, it is necessary to determine what it will be equal to if a 13 = -5.6 and a 1 = -12.1.
Since we know the values of two elements of a 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 term with this serial number is known.
The resulting difference indicates that the progression is increasing, despite the fact that the data in the condition of the problem elements have a negative value. It can be seen that a 13 > a 1 , although | a 13 | <| a 1 |.
Example No. 2. Positive members of the progression in example No. 1
We will use the result obtained in the previous example to solve a new problem. It is formulated as follows: from which serial number do the progression elements in Example No. 1 begin to take positive values?
As has been shown, a progression in which a 1 = -12.1 and d = 0.54167 is increasing, therefore only positive values will begin to take numbers from some number. To determine this number n, it is necessary to solve a simple inequality, which is mathematically written as: a n > 0 or, using the appropriate formula, we rewrite the inequality: a 1 + (n-1) * d> 0. It is necessary to find the unknown n, we express it: n> -1 * a 1 / 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 inequality that any members of the series that have a number greater than 23 will be positive.
We verify the answer using the above formula to calculate the 23th and 24th elements of this arithmetic progression. We have: a 23 = -12.1 + 22 * 0.54167 = -0.18326 (negative number); a 24 = -12.1 + 23 * 0.54167 = 0.3584 (positive value). Thus, the result obtained is true: starting from n = 24, all members of the number series will be greater than zero.
Example No. 3. How many logs will fit?
Here is one curious problem: during logging, it was decided to lay down sawn logs on top of each other as shown in the figure below. How many logs can be stacked this way, knowing that a total of 10 rows will fit?
In this method of folding logs, one interesting thing can be noticed: each subsequent row will contain one log less than the previous one, that is, there is an algebraic progression, the difference of which is d = 1. Assuming that the number of logs of each row is a member of this progression, and also considering that a 1 = 1 (only one log will fit at the very top), we find the number a 10 . We have: a 10 = 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 by using the Gauss formula. We get: S 10 = 10/2 * (10 + 1) = 55 logs.