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Chapter 8: Problem 8

Write the first six terms of each arithmetic sequence. $$a_{1}=\frac{3}{4}, d=-\frac{1}{4}$$

Short Answer

Expert verified
The first six terms of the arithmetic sequence are \(\frac{3}{4}, \frac{1}{2}, \frac{1}{4}, 0, -\frac{1}{4}, -\frac{1}{2}\).

Step by step solution

01

Calculate the first term

The first term \(a_{1}\) is given as \(\frac{3}{4}\). So, no calculation is needed for the first term.
02

Calculate the second term

For the second term, plug \(n = 2\), \(a_{1} = \frac{3}{4}\), and \(d = -\frac{1}{4}\) into the formula \(a_n = a_1 + (n - 1) * d\). This gives \(a_2 = \frac{3}{4} + (2 - 1) * -\frac{1}{4} = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\). So, the second term is \(\frac{1}{2}\).
03

Calculate the third term

For the third term, plug \(n = 3\), \(a_{1} = \frac{3}{4}\), and \(d = -\frac{1}{4}\) into the formula. This gives \(a_3 = \frac{3}{4} + (3 - 1) * -\frac{1}{4}= \frac{1}{2} - \frac{1}{4}= \frac{1}{4}\). So, the third term is \(\frac{1}{4}\).
04

Calculate the fourth term

For the fourth term, plug \(n = 4\), \(a_{1} = \frac{3}{4}\), and \(d = -\frac{1}{4}\) into the formula. This gives \(a_4 = \frac{3}{4} + (4 - 1) * -\frac{1}{4} = \frac{1}{4} - \frac{1}{4} = 0\). So, the fourth term is \(0\).
05

Calculate the fifth term

For the fifth term, plug \(n = 5\), \(a_{1} = \frac{3}{4}\), and \(d = -\frac{1}{4}\) into the formula. This gives \(a_5 = \frac{3}{4} + (5 - 1) * -\frac{1}{4} = 0 - \frac{1}{4} = -\frac{1}{4}\). So, the fifth term is \(-\frac{1}{4}\).
06

Calculate the sixth term

For the sixth term, plug \(n = 6\), \(a_{1} = \frac{3}{4}\), and \(d = -\frac{1}{4}\) into the formula. This gives \(a_6 = \frac{3}{4} + (6 - 1) * -\frac{1}{4} = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2}\). So, the sixth term is \(-\frac{1}{2}\).

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