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Chapter 5: Problem 16

Write the first six terms of the arithmetic sequence with the first term, \(a_{1}\), and common difference, \(d\). \(a_{1}=\frac{3}{2}, d=-\frac{1}{4}\)

Short Answer

Expert verified
The first six terms of the arithmetic sequence are: \(\frac{3}{2}, \frac{5}{4}, 1, \frac{3}{4}, \frac{1}{2}, \frac{1}{4}\)

Step by step solution

01

Identify the first term and the common difference

The given sequence has a first term, \(a_{1}\) of \(\frac{3}{2}\), and a common difference, \(d\) of \(-\frac{1}{4}\).
02

Apply the formula for the second term

Using the arithmetic sequence formula, \(a_{n}= a_{1} + (n-1)*d\) , find the second term: \(a_{2}= \frac{3}{2} + (2-1)*(-\frac{1}{4}) = \frac{3}{2} - \frac{1}{4}= \frac{5}{4}\) .
03

Continue using the formula for subsequent terms

Continue applying the formula, substituting \(n\) for 3, then 4 and so on up to 6: \(a_{3}= a_{1} + (3-1)*d = \frac{3}{2} -2*-\frac{1}{4}= \frac{4}{4}=1, a_{4}= a_{1} + (4-1)*d = \frac{3}{2} - 3*-\frac{1}{4}= \frac{3}{4}, a_{5}= a_{1} + (5-1)*d = \frac{3}{2} - 4*-\frac{1}{4}= \frac{1}{2}, a_{6}= a_{1} + (6-1)*d = \frac{3}{2} - 5*-\frac{1}{4}= \frac{1}{4}\)

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Most popular questions from this chapter

Determine whether each sequence is arithmetic or geometric. Then find the next two terms. \(6,3, \frac{3}{2}, \frac{3}{4}, \ldots\)

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