9/4/2016 Arithmetic Sequences Arithmetic Sequences This lesson will work with arithmetic sequences, their recursive and explicit formulas and finding terms in a sequence. In this lesson, it is ass know what an arithmetic sequence is and can find a common difference. If you need to review these topics, click here. Let’s look at the arithmetic sequence 20, 24, 28, 32, 36, . . . This arithmetic sequence has a common difference of 4, meaning that we add 4 to a term in order to get the next term in the sequence. The recursive formula for an arithmetic sequence is written in the form For our particular sequence, since the common difference (d) is 4, we would write So once you know the common difference in an arithmetic sequence you can write the recursive form for that sequence. However, the recursive formula can become difficult to work with if we want to find the 50th term. Using the recursive formula, we would hav first 49 terms in order to find the 50th. This sounds like a lot of work. There must be an easier way. And there is! Rather than write a recursive formula, we can write an explicit formula. The explicit formula is also sometimes called the closed form. To w or closed form of an arithmetic sequence, we use an is the nth term of the sequence. When writing the general expression for an arithmetic sequence, you will not actually find a value fo will be part of your formula much in the same way x’s and y’s are part of algebraic equations. a1 is the first term in the sequence. To find the explicit formula, you will need to be given (or use computations to find out) the first ter that value in the formula. n is treated like the variable in a sequence. For example, when writing the general explicit formula, n is the variable and does not take value. But if you want to find the 12th term, then n does take on a value and it would be 12. d is the common difference for the arithmetic sequence. You will either be given this value or be given enough information to compute i must substitute a value for d into the formula. You must also simplify your formula as much as possible. Let's Practice: i. Write the explicit formula for the sequence that we were working with earlier. 20, 24, 28, 32, 36, . . . The first term in the sequence is 20 and the common difference is 4. This is enough information to write the explicit formula. Now we have to simplify this expression to obtain our final answer. So the explicit (or closed) formula for the arithmetic sequence is . Notice that an the and n terms did not take on numeric values. They are a part of the formula, again like x’s and y’s in algebraic expres If we wanted to find the 50th term of the sequence, we would use n = 50. Look at the example below to see what happens. ii. Given the sequence 20, 24, 28, 32, 36, . . . find the 50th term. To find the 50th term of any sequence, we would need to have an explicit formula for the sequence. Since we already found that in Exam we can use it here. If we do not already have an explicit form, we must find it first before finding any term in a sequence. Use the explicit formula http://www.algebralab.org/lessons/lesson.aspx?file=algebra_arithseq.xml and let n = 50. This will give us 1/4 9/4/2016 Arithmetic Sequences Notice how much easier it is to work with the explicit formula than with the recursive formula to find a particular term in a sequence. What happens if we know a particular term and the common difference, but not the entire sequence? Let’s see in the next example. iii. Find the explicit formula for a sequence where d = 3 and a12 = 58. The formula says that we need to know the first term and the common difference. We have d, but do not know a1. However, we have e information to find it. We know that when n = 12, the 12th term in the sequence is 58. If we simplify that equation, we can find a1. Now that we know the first term along with the d value given in the problem, we can find the explicit formula. Notice this example required making use of the general formula twice to get what we need. The first time we used the formula, we were work from an answer and the second time we were working forward to come up with the explicit formula. iv. Find the explicit formula for an arithmetic sequence where a1 = 4 and a2 = 10. In this situation, we have the first term, but do not know the common difference. However, we do know two consecutive terms which m can find the common difference by subtracting. 10 4 = 6 means that d = 6 Now we use the formula to get Notice that writing an explicit formula always requires knowing the first term and the common difference. If neither of those are given in the must take the given information and find them. v. Is 623 a term in the sequence 4, 10, 16, 22, . . . ? The way to solve this problem is to find the explicit formula and then see if 623 is a solution to that formula. We already found the explicit formula in the previous example to be in for an. . To find out if 623 is a term in the sequence, substitute What does this mean? Well, if 623 is a term in the sequence, when we solve the equation, we will get a whole number value for n. Since not get a whole number value, then 623 is not a term in the sequence. Look at it this way. There can be a 103rd term or a 104th in between. Examples Find the recursive formula for 15, 12, 9, 6, . . . What is your answer? Find the explicit formula for 15, 12, 9, 6, . . . What is your answer? Find the recursive formula for 5, 9, 13, 17, 21, . . . What is your answer? Find the explicit formula for 5, 9, 13, 17, 21, . . . What is your answer? http://www.algebralab.org/lessons/lesson.aspx?file=algebra_arithseq.xml 2/4 9/4/2016 Arithmetic Sequences Find a10, a35 and a82 for problem #4. What is your answer? Find the explicit formula when and d = 2. What is your answer? Find a10, a35 and a82 for problem #6. What is your answer? Find the explicit formula when and . What is your answer? Find a10, a35 and a82 for problem #8. What is your answer? Is 327 a term in the sequence 8, 13, 18, . . .? What is your answer? Is 852 a term in the sequence 8, 12, 16, 20, . . . ? What is your answer? Check Your Answers S Taylor Show Related AlgebraLab Documents http://www.algebralab.org/lessons/lesson.aspx?file=algebra_arithseq.xml 3/4 9/4/2016 Arithmetic Sequences ACT Math Practice Test Algebra Calculator Applications of Sequences and Series Start Download - View PD Ad chegg.com Ad homeworksimpli壩ed.com algebralab.org Ad fromdoctopdf.com Translating Word Problems into Equations Solving Two-Step Equations Introductory Calculus: Test #1 Preparation Applications of Exponent algebralab.org algebralab.org algebralab.org algebralab.org http://www.algebralab.org/lessons/lesson.aspx?file=algebra_arithseq.xml 4/4
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