In mathematics, any combination of numbers that follow one after another, organized in some way, is called a sequence. Of all the existing sequences of numbers, two interesting cases are distinguished: algebraic and geometric progressions.
What is arithmetic progression?
It should immediately be said that algebraic progression is often called arithmetic, since its properties are studied by the branch of mathematics - arithmetic.
This progression is a sequence of numbers in which each of its next member differs from the previous one by some constant number. It is called the difference of algebraic progression. For definiteness, we denote it by the Latin letter d.
An example of such a sequence can be the following: 3, 5, 7, 9, 11 ..., it can be seen that the number 5 is greater than the number 3 by 2, 7 is greater than 5 also by 2, and so on. Thus, in the presented example, d = 5-3 = 7-5 = 9-7 = 11-9 = 2.
What are arithmetic progressions?
The nature of these ordered sequences of numbers is largely determined by the sign of d. The following types of algebraic progressions are distinguished:
- increasing when d is positive (d> 0);
- constant when d = 0;
- decreasing when d is negative (d <0).
The example in the previous paragraph shows an increasing progression. An example of a decreasing one is the following sequence of numbers: 10, 5, 0, -5, -10, -15 ... A constant progression, as follows from its definition, is a collection of identical numbers.
nth progression member
Due to the fact that each subsequent number in the considered progression differs by a constant d from the previous one, we can easily determine its nth term. To do this, you need to know not only d, but also a 1 - the first member of the progression. Using a recursive approach, we can obtain an algebraic progression formula for finding the nth term. It has the form: a n = a 1 + (n-1) * d. This formula is quite simple, and it can be understood on an intuitive level.
Also, its use is not difficult. For example, in the progression given above (d = 2, a 1 = 3), we define its 35th term. According to the formula, it will be equal to: a 35 = 3 + (35-1) * 2 = 71.
Formula for amount
When some arithmetic progression is given, the sum of its first n members is a frequently arising task, along with determining the value of the nth term. The formula for the sum of an algebraic progression is written in the following form: ∑ n 1 = n * (a 1 + a n ) / 2, here the icon n 1 indicates that they are summed from the 1st to the nth term.
The above expression can be obtained by resorting to the properties of the same recursion, however, there is an easier way to prove its validity. We write the first 2 and last 2 members of this sum, expressing them in the numbers a 1 , a n and d, and we get: a 1 , a 1 + d, ..., a n -d, a n . Now note that if you add the first term to the last, then it will be exactly equal to the sum of the second and penultimate terms, that is, a 1 + a n . In a similar way, it can be shown that the same amount can be obtained by adding the third and penultimate terms, and so on. In the case of a pair of numbers in the sequence, we get n / 2 sums, each of which is equal to a 1 + a n . That is, we obtain the above algebraic progression formula for the sum: ∑ n 1 = n * (a 1 + a n ) / 2.
For an unpaired number of members of n, a similar formula is obtained if we follow the described arguments. Just do not forget to add the remaining term, which is in the center of progression.
We show how to use the above formula as an example of a simple progression, which was introduced above (3, 5, 7, 9, 11 ...). For example, you need to determine the sum of the first 15 members. To begin with, we define a 15 . Using the formula for the nth term (see the previous paragraph), we get: a 15 = a 1 + (n-1) * d = 3 + (15-1) * 2 = 31. Now we can apply the algebraic progression sum formula: ∑ 15 1 = 15 * (3 + 31) / 2 = 255.
It is interesting to give an interesting historical fact. The formula for the sum of arithmetic progression was first obtained by Karl Gauss (the famous German mathematician of the XVIII century). When he was only 10 years old, the teacher asked the problem to find the sum of numbers from 1 to 100. They say that little Gauss solved this problem in a few seconds, noting that summing up the numbers in pairs from the beginning and end of the sequence, you can always get 101, and since there are 50 such amounts, he quickly issued an answer: 50 * 101 = 5050.
Problem solving example
To complete the topic of algebraic progression, we give an example of the solution of another interesting problem, thereby reinforcing the understanding of the topic under consideration. Let some progression be given for which the difference d = -3 is known, as well as its 35th term a 35 = -114. You must find the 7th member of the progression a 7 .
As can be seen from the conditions of the problem, the value of a 1 is unknown, so the formula for the nth term cannot be used directly. It is also an inconvenient way of recursion, which is difficult to implement manually, and it is likely to make a mistake. We proceed as follows: we write out the formulas for a 7 and a 35 , we have: a 7 = a 1 + 6 * d and a 35 = a 1 + 34 * d. Subtract the second from the first expression, we get: a 7 - a 35 = a 1 + 6 * d - a 1 - 34 * d. Whence follows: a 7 = a 35 - 28 * d. It remains to substitute the known data from the conditions of the problem and record the answer: a 7 = -114 - 28 * (- 3) = -30.
Geometric progression
To expand the topic of the article more fully, we give a brief description of another type of progression - geometric. In mathematics, this name is understood as a sequence of numbers in which each subsequent term differs from the previous one by a certain factor. Denote this factor by the letter r. It is called the denominator of the type of progression in question. An example of this sequence of numbers is the following: 1, 5, 25, 125, ...
As can be seen from the above definition, algebraic and geometric progressions are similar in their idea. The difference between the two is that the first changes more slowly than the second.
Geometric progression can also be increasing, constant and decreasing. Its type depends on the value of the denominator r: if r> 1, then there is an increasing progression, if r <1 is decreasing, finally, if r = 1 is a constant, which in this case can also be called constant arithmetic progression.
Geometric progression formulas
As in the case of algebraic, geometric progression formulas are reduced to the definition of its nth term and the sum of n terms. The following are these expressions:
- a n = a 1 * r (n-1) - this formula follows from the definition of geometric progression.
- ∑ n 1 = a 1 * (r n -1) / (r-1). It is important to note that if r = 1, then the above formula gives uncertainty, therefore it cannot be used. In this case, the sum of n members will be equal to the simple product a 1 * n.
For example, we find the sum of only 10 members of the sequence 1, 5, 25, 125, ... Knowing that a 1 = 1 and r = 5, we get: ∑ 10 1 = 1 * (5 10 -1) / 4 = 2441406. The resulting value is a good example of how fast geometric progression is growing.
Perhaps the first mention of this progression in history is a legend with a chessboard, when a friend of one sultan, having taught him how to play chess, asked for grain for his service. Moreover, the amount of grain should have been as follows: on the first cell of the chessboard it is necessary to put one grain, on the second two times more than on the first, on the third 2 times more than on the second and so on. The Sultan willingly agreed to fulfill this request, but he did not know that he would have to empty all the bins of his country in order to keep this word.