Understanding many topics in mathematics and physics is associated with knowledge of the properties of number series. Students in grade 9 when studying the subject "Algebra" consider one of the important sequences of numbers - arithmetic progression. We give the basic formulas of arithmetic progression (Grade 9), as well as examples of their use for solving problems.
Algebraic or arithmetic progression
The number series that will be considered in this article is called in two different ways, presented in the title of this paragraph. So, by arithmetic progression in mathematics we mean a number series in which any two numbers standing next to each other differ by the same amount, called the difference. The numbers in this series are usually denoted by letters with a lower integer index, for example, a 1 , a 2 , a 3 and so on, where the index indicates the number of the element of the series.
Given the above definition of arithmetic progression, we can write the following equality: a 2 -a 1 = ... = a n -a n-1 = d, here d is the difference between the algebraic progression and n is any integer. If d> 0, then we can expect that each subsequent member of the series will be larger than the previous one, in this case they speak of an increasing progression. If d <0, then the previous term will be greater than the next, that is, the series will decrease. A special case arises when d = 0, that is, a series is a sequence in which a 1 = a 2 = ... = a n .
Arithmetic progression formulas (school grade 9)
The considered series of numbers, since it is ordered and obeys some mathematical law, has two properties important for its use:
- First, knowing only two numbers a 1 and d, you can find any member of the sequence. This is done using the following formula: a n = a 1 + (n-1) * d.
- Secondly, to calculate the sum of the n members of the first, it is not necessary to add them in order, because you can use the following formula: S n = n * (a n + a 1 ) / 2.
The first formula is easy to understand, since it is a direct consequence of the fact that each member of the series in question differs from his neighbor by the same difference.
The second formula of arithmetic progression can be obtained if we pay attention to the fact that the sum a 1 + a n turns out to be equivalent to the sums a 2 + a n-1 , a 3 + a n-2 and so on. Indeed, since a 2 = d + a 1 , a n-2 = -2 * d + a n , a 3 = 2 * d + a 1 , and a n-1 = -d + a n , then substituting these expressions in the appropriate amount, we get that they will be the same. The factor n / 2 in the 2nd formula (for S n ) appears due to the fact that sums of the type a i + 1 + a ni turn out to be exactly n / 2, here i is an integer running through the values from 0 to n / 2 -1.
According to the preserved historical evidence, the formula for the sum S n was first obtained by Karl Gauss (the famous German mathematician), when he was tasked with the school teacher to add the first 100 numbers.
Further in the article, we consider examples with arithmetic progression formulas.
Example of task number 1: find the difference
Tasks in which the question is posed as follows: knowing the formulas of arithmetic progression, how to find d (d), are the simplest that can only be for this topic.
We give an example: a numerical sequence is given -5, -2, 1, 4, ..., it is necessary to determine its difference, that is, d.
Making it easier is simple: you need to take two elements and subtract the smaller one from the larger one. In this case, we have: d = -2 - (-5) = 3.
To be sure of the answer obtained, it is recommended to check the remaining differences, since the presented sequence may not satisfy the algebraic progression condition. We have: 1 - (- 2) = 3 and 4-1 = 3. These data indicate that we obtained the correct result (d = 3) and proved that the number of numbers in the condition of the problem really represents an algebraic progression.
Example of task number 2: find the difference, knowing the two members of the progression
Consider another interesting problem that poses the question of how to find the difference. The arithmetic progression formula in this case must be used for the nth term. So, the task: the first and fifth numbers of the series are given, which correspond to all the properties of algebraic progression, for example, these are numbers a 1 = 8 and a 5 = -10. How to find the difference d?
To begin the solution of this problem, one should write down the general form of the formula for the nth element: a n = a 1 + d * (- 1 + n). Now you can go in two ways: either substitute the numbers at once and work with them already, or express d, and then go to the specific a 1 and a 5 . We use the last method, we get: a 5 = a 1 + d * (- 1 + 5) or a 5 = 4 * d + a 1 , which implies that d = (a 5 -a 1 ) / 4. Now you can safely substitute the known data from the condition and get the final answer: d = (-10-8) / 4 = -4.5.
Note that in this case the progression difference turned out to be negative, that is, a decreasing sequence of numbers takes place. It is necessary to pay attention to this fact when solving problems, so as not to confuse the signs "+" and "-". All the formulas above are universal, so you should always follow them regardless of the sign of the numbers with which operations are carried out.
An example of solving problem 3: find a1, knowing the difference and the element
Change the condition of the problem a little. Let there be two numbers: the difference d = 6 and the 9th element of the progression a 9 = 10. How to find a1? The arithmetic progression formulas remain unchanged, we will use them. For the number a 9, we have the following expression: a 1 + d * (9-1) = a 9 . From where we easily get the first element of the series: a 1 = a 9 -8 * d = 10 - 8 * 6 = -38.
An example of solving problem 4: find a1, knowing two elements
This version of the problem is a complicated version of the previous one. The essence is the same, it is necessary to calculate a 1 , but now the difference d is not known, and instead another element of the progression is given.
An example of this type of task is the following: find the first number of a sequence for which it is known that it is an arithmetic progression, and that its 15th and 23rd elements are 7 and 12, respectively.
It is necessary to solve this problem by writing an expression for the nth term for each element known from the condition, we have: a 15 = d * (15-1) + a 1 and a 23 = d * (23-1) + a 1 . As you can see, we got two linear equations that need to be solved with respect to a 1 and d. We do this: we subtract the first from the second equation, then we get the expression: a 23 -a 15 = 22 * d - 14 * d = 8 * d. Upon receipt of the last equation, the values of a 1 were omitted, since they are reduced by subtraction. Substituting the known data, we find the difference: d = (a 23 -a 15 ) / 8 = (12-7) / 8 = 0.625.
The value of d must be substituted into any formula for a known element in order to obtain the first member of the sequence: a 15 = 14 * d + a 1 , whence: a 1 = a 15 -14 * d = 7-14 * 0.625 = -1.75.
Check the result, for this we find a 1 through the second expression: a 23 = d * 22 + a 1 or a 1 = a 23 -d * 22 = 12 - 0.625 * 22 = -1.75.
An example of solving problem 5: find the sum of n elements
As you can see, up to this point, only one formula of arithmetic progression (grade 9) was used for the solution. Now we present a problem whose solutions require knowledge of the second formula, that is, for the sum S n .
There is the following ordered series of numbers -1,1, -2,1, -3,1, ..., you need to calculate the sum of its 11 first elements.
From this series it is clear that it is decreasing, and a 1 = -1.1. Its difference is: d = -2.1 - (-1.1) = -1. Now we define the 11th term: a 11 = 10 * d + a 1 = -10 + (-1,1) = -11,1. Having completed the preparatory calculations, we can use the formula for the sum noted above, we have: S 11 = 11 * (- 1.1 + (- 11.1)) / 2 = -67.1. Since all terms were negative numbers, their sum also has the corresponding sign.
An example of solving problem 6: find the sum of elements from n to m
Perhaps this type of task is the most difficult for most students. We give a typical example: a number of numbers 2, 4, 6, 8 ... is given, it is necessary to find the sum from the 7th to the 13th members.
The arithmetic progression formulas (Grade 9) are used exactly the same as in all problems earlier. This task is recommended to be solved in stages:
- First find the sum of 13 members using the standard formula.
- Then calculate this amount for the first 6 elements.
- After that, subtract the 2nd from the 1st amount.
Let's get down to the solution. As in the previous case, we carry out preparatory calculations: a 6 = 5 * d + a 1 = 10 + 2 = 12, a 13 = 12 * d + a 1 = 24 + 2 = 26.
We calculate two sums: S 13 = 13 * (2 + 26) / 2 = 182, S 6 = 6 * (2 + 12) / 2 = 42. We take the difference and get the desired answer: S 7-13 = S 13 - S 6 = 182-42 = 140. Note that when obtaining this value, the sum of 6 elements of the progression was used as the deductible, since the 7th term is included in the sum of S 7-13 .