An explicit rule of a sequence is a function rule where the input is a term's position, and the output is the value of the term. This is opposed to recursive rules, where a term's value is found using previous terms. Explicit rules are often written as an=f(n), where f(n) is the function rule. An example would be the sequence 2,3,4,…, which can be described by the explicit rule an=n+1,
where n is the position of a term in the sequence. It says that the first term, a1, has the value 2, the second term, a2, has the value 3, and so on.All arithmetic sequences have some common difference, d. Using this common difference, and the value of the first term, a1, it's possible to find an explicit rule that describes the sequence. By thinking of the terms in a sequence using a1 and d, a pattern emerges.
n | an | Using a1 and d |
---|---|---|
1 | a1 | a1+0d |
2 | a2 | a1+1d |
3 | a3 | a1+2d |
4 | a4 | a1+3d |
When n increases by 1, the coefficient of d increases by 1 as well. Due to this, and that the coefficient is 0 when n is 1, the coefficient is always 1 less than n. Expressing this in a general form gives the explicit rule.
an=a1+(n−1)d
Thus, knowing a1 and d is enough to write the explicit rule of an arithmetic sequence. Note that an arithmetic sequence is a linear function where the domain is the positive integers. The difference d is then the slope, and (1,a1) is a point on the graph. Substituting this into the point-slope form is an alternative way of finding the rule: an−a1=d(n−1).
This equality can be rearranged into the explicit rule previously stated.The first five terms of an arithmetic sequence are 4,7,10,13,and 16. Find an explicit rule describing the arithmetic sequence. Then, use the rule to find the twelfth term of the sequence.
To find the rule, we first need to find the common difference, d, of the sequence. We can do this by subtracting one term from the next: d=7−4=3. We also know that the first term of the sequence is 4. Substituting these pieces of information into the general form of the explicit rule gives the desired rule.
To find the twelfth term of the sequence, we can now substitute n=12 into the rule.
The twelfth term is 37.
Pelle is an avid collector of pellets. During his spare time, he likes to arrange his pellets in different patterns. Today, he's chosen to place them in the shape of a triangle. The top row consists of one pellet, the second of three pellets, the third of five pellets, and so on. Write a rule an=f(n), where an is the amount of pellets in row n. Then, use the rule to find which row has 53 pellets.
To begin, we can make sense of the given information. For every row, the amount of pellets increase by 2. Thus, we know that d=2. It is also given that the first row consists of 1 pellet, a1=1. Using this information, we can find the rule.
Thus, the explicit rule is an=-1+2n. Next, we can find the row that contains 53 pellets. In other words, we are looking for the n that gives an=53. This is done by substituting an=53 into the rule and solving the resulting equation for n.Row 27 has 53 pellets.
For an arithmetic sequence, a3=15anda6=0. Write an explicit rule of the sequence and give its first six terms.
To begin, we must determine the common difference, d. Since we do not know the values of two consecutive terms, we cannot directly find d. However, the terms a3 and a6 are 3 positions apart, so they must differ by 3d.
This gives the equation a6−a3=3d, which we can solve for d.
Now that we know the common difference, we have to find the first term of the sequence, a1, to write the explicit rule. Using a3 and d, we can find a1. Knowing one term, a subsequent term can be found by adding d. Similarly, a previous term is found by subtracting d. a2=a3−d⇒a2=15−(-5)=20 Repeating this, we find a1. a1=a2−d⇒a1=20−(-5)=25 We now know both a1 and d, so we can find the explicit rule.
Thus, the explicit rule is an=30−5n. We have already found the terms a1, a2, a3, and a6. Finding the last two can be done either by using the explicit rule, or by adding d to a3 and then to a4. For simplicity's sake, let's add d to a3 to find a4. a4=a3+d⇒a4=15+(-5)=10 Then, a5 is found using a4. a5=a4+d⇒a5=10+(-5)=5 Thus, the first six terms of the sequence are 25,20,15,10,5,and 0.