When studying algebra in a comprehensive school (grade 9), one of the important topics is the study of numerical sequences, which include progressions - geometric and arithmetic. In this article we will consider arithmetic progression and examples with solutions.
What is arithmetic progression?
To understand this, it is necessary to give a definition of the progression under consideration, as well as give the basic formulas that will be further used in solving problems.
Arithmetic or algebraic progression is such a set of ordered rational numbers, each member 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.
We 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 considered type of progression, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).
Important formulas
We now give the basic formulas that will be needed to solve problems using arithmetic progression. Denote by a n the nth member of the sequence, where n is an integer. The difference is denoted by the Latin letter d. Then the following expressions are valid:
- To determine the value of the nth term, the 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 a solution in the 9th grade, it is enough to remember these two formulas, since any tasks of the type under consideration are built on their use. Also, do not forget that the difference in progression is determined by the formula: d = a n - a n-1 .
Further, the article provides various examples of the use of these expressions.
Example No. 1: finding an unknown member
We give a simple example of the arithmetic progression and formulas that must be used for solving.
Let the sequence 10, 8, 6, 4, ... be given, it is necessary to find five members 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:
- We calculate for a start 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 , whence 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 in the considered progression, so first you need to determine it, as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for n numbers 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 obtain: a 5 = 12-2 * 5 = 2.
As you can see, both methods of solution led to the same result. Note that in this example, the difference d of the progression is a negative value. Such sequences are called decreasing, since each next term is less than the previous one.
Example No. 2: the difference of progression
Now let's complicate the task a bit, give an example of how to find the difference of the arithmetic progression.
It is known that in some algebraic progression the 1st term is 6, and the 7th term is 18. It is necessary to find the difference and restore this sequence to 7 members.
We use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . We 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, one can easily calculate the difference: d = (18 - 6) / 6 = 2. Thus, they answered the first part of the problem.
To restore the sequence to 7 terms, one should use the definition of 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: making progression
We complicate the problem condition even more. Now it is necessary to answer the question of how to find the arithmetic progression. You can give the following example: two numbers are given, for example, 4 and 5. It is necessary to compose an algebraic progression so that three more terms are placed between these.
Before you begin to solve this problem, you need to understand what place will be given numbers in a future progression. Since there will be three more terms between them, then a 1 = -4 and a 5 = 5. Having established this, we proceed 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. Where: d = (a 5 - a 1 ) / 4 = (5 - (-4)) / 4 = 2.25. They did not get the integer value of the difference, but it is a rational number, so the formulas for algebraic progression remain the same.
Now we 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 condition of the problem.
Example No. 4: the first member of the progression
We continue to give examples of arithmetic progression with a solution. In all previous problems, the first number of algebraic progression was known. Now 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 that have been used to date, require knowledge of a 1 and d. In the condition of the problem of these numbers, nothing is known. Nevertheless, we write out the expressions for each member about which information is available: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We obtained two equations in which 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.
The indicated system is most easily solved if a 1 is expressed in each equation, and then the obtained expressions are compared. The first equation: a 1 = a 15 - 14 * d = 50 - 14 * d; the 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 after the decimal point).
Knowing d, you can use any of the 2 above expressions for a 1 . For example, the first: a 1 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.
If there are doubts about the result, you can check it, for example, determine the 43 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. A small error is due to the fact that the calculations used rounding to thousandths.
Example No. 5: amount
Now consider a few examples with solutions in the amount of 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, this problem can be solved, that is, sequentially add up all the numbers that the computer will do as soon as a person presses the Enter key. However, the problem can be solved in the mind if you pay attention that the presented series of numbers is an algebraic progression, and its difference is 1. Using 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 XVIII century the famous German mathematician Gauss, still at the age of only 10 years, was able to solve it in his mind in a few seconds. The boy did not know the formula for the sum of algebraic progression, but he noticed that if you add the numbers at the edges of the sequence in pairs, you always get one result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since of these sums will be exactly 50 (100/2), then to get the correct answer, just multiply 50 by 101.
Example No. 6: the sum of members from n to m
Another typical example of the sum of an arithmetic progression is the following: given the following numbers: 3, 7, 11, 15, ..., you need to find what the sum of its members from 8 to 14 will be equal to.
The problem is solved in two ways. The first of them involves finding the unknown members from 8 to 14, and then their sequential summation. Since there are few terms, this method is not time-consuming. Nevertheless, it is proposed to solve this problem by the second method, which is more universal.
The idea is to obtain a formula for the sum of an algebraic progression between the terms m and n, where n> m are integers. For both cases, we write out 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 2 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 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). In this expression, it is necessary to substitute the formulas for a n and a m . 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 obtain: S mn = 301.
Some tips for solving problems with arithmetic progression
As can be seen from the above solutions, all tasks are based on knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before you begin 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 the question without using complex mathematical calculations, then you need to do just that, since in this case the probability of making a mistake is less. For example, in an example of 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 break down the general problem into separate subtasks (in this case, first find the terms a n and a m ).
If there are doubts about the result, it is recommended to check it, as was done in some of the examples given. How to find the arithmetic progression, found out. If you look, it’s not so difficult.