Knowing that in a BP the 2nd term is 9 and the 11th term is 45, write this PA: We can write the terms of this PA according to the first as follows: a) In the archethetic progression (2.7,…) we have that the first term is 2. The ratio is equal to 7 – 2 = 5. However, the fourth term of these APs is different because a4 = 13 and b4 = 14. This is because the first term of these progressions is different. Thus, the first term affects the value of the term we want to find, which is represented by one. The general concept of arithmetic progression is defined by year = aư + (n – 1).r, which means: resolution Let`s write these terms and r: Note that this formula requires three pieces of information to use: the position of the term you want to discover, represented by the letter n; the PA`s first mandate and its reason. Note the following example, which resolves in two different ways. Keep in mind that the first term of an AP is a1 and the next one a2, a3, . This is the archethretic progression formed from all even numbers of 2. So the first term is 2, the ratio is 2 and the number of terms is 100 because we want to know the hundredth term.
See: This formula can be obtained from an analysis of BP terms. To do this, it is necessary to be familiar with some elements and properties of arithmetic progressions, which are briefly discussed below. Arrhythmetic progression (AP) is a numerical sequence that has the following definition: The difference between two consecutive terms is always equal to a constant, usually referred to as the BP ratio. It is possible to find the value of any term from the first term and the reason for an AP. This calculation depends on its position in the numerical sequence and can be done using the formula of the general term of an AP, which will be discussed later in this article. Before that, however, it`s important to be familiar with the concept that defines a PA. The number that multiplies the ratio is always a unit smaller than the position of the term we calculate. Therefore, we can write the following expressions: The formula of the general term of each AP: a) 5n – 3; (b) 6n – 7.
So, if we rewrite the last equality and reorganize the terms of the last member, we have: this is the formula used to find any term, that is, the general term of PA given as an example. So the general term of this archethetic progression is: Since we know that each term of a BP is equal to its previous term added to a constant, we can write bp terms according to the first term. In progression A = (1, 3, 5, 7, 9, 11, 13, . for example, we will have: From the previous conclusion, we can begin to think about a way to get any term of an AP. The ratio of an arithmetic progression is equal to ₋, where n is > 1, that is, it is the difference between a term and its predecessor. Since we know that an represents any term of an AP, we can try to find the general concept of an arithmetic progression, the terms of which are unknown. To do this, consider a PA with n terms. Know that a1 is the first, the last and the ratio r. The difference between two consecutive terms (ratio) is 1. The ellipses indicate that the list of numbers continues, that is, the next term is always equal to the previous one added with a ratio of 1. The reason for this is therefore the number to which each term must be added in order to obtain the next.
Determine the general term of BP (-19, -15, -11, …): therefore, the 16th term of BP (3, 9, 15, …) 93. An anitmetic progression is a sequence of numbers in which each term is the result of the sum of its predecessor with a constant called reason. The terms of an AP are specified by indexes, so that each index determines the position of each progression element. Here is an example: The general term arithmetic progression (AP) is a formula used to find any term of an AP that is indicated by one if its first term (a1), the ratio (r) and the number of terms (n) that PA a are known. In b we have a1=−1a_1=-1a1=−1 and r=5−(−1)=5+1=6r=5-(-1)=5+1=6=5−(−1)=5+1=6 By replacing in the general formula remains: We can therefore imagine that each term (to) is obtained by the sum of the first term (a1) with the product between n – 1 and r. Therefore, the formula of the general concept of an AP is as follows: the first term of this BP is 2 and the ratio is 3. In the formula of the general term we have: Note that we can write all the terms of a PA as a function of and r: (UFRGS) In an archetic progression, in which the first term is 23 and the ratio – is 6, the position that the element – 13 occupies is: With the formula of the general term of an AP, we have: Therefore, the general term of BP is given by the formula: This example is not a PA because the difference between the first and second terms is equal to 1, but the difference between the fifth and fourth terms is equal to 2. Where (a_1=2) and (a_2=7), then for the general term: If we know this, we will write some terms of the first AP according to the first.
Note: To find the tenth term of this AP, simply add the reason to the last term until you find it. The AP obtained will be: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. More information about the archaeological course: brainly.com.br/tarefa/3523769. . (Enem) The graph, which comes from data from the Ministry of environment, shows the growth in the number of threatened species of Brazilian wildlife. If the growth trend shown in the graph is maintained for the next few years, the number of threatened species in 2011 will be the same: follow the answer to your problem and enjoy it. Also follow the Matematicanalitica, which has several free courses and is available on this platform. . Pay attention only to the first and last part of equality.
First, note that the following two arithmetic progressions have the same reason: If a – on – 1 = k for all n, then the above sequence is an arithmetic progression. We conclude that there are between 101 and 999,179 multiples of 5. A sequence of numbers is a set in which the numbers are in a certain order. In the case of PA, what determines this order is the reason. The next sequence of numbers is a PA. Note:. . In a ratio of a1=2a_1=2a1=2 and r=7−2=5r=7-2=5r=7−2=5 found, it is enough to interpolate the arithmetic means: (-8, -5, -2, 1, 4, 7, 10, 13). an=a1+(n−1)∗ra_n=a_1+(n-1)*ran=a1+(n−1)∗r I hope the answer helped, if they liked it and the answer with the calculation helped you, follow me in direct passage and if possible my materials in direct passage. .
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