To find the 8th term of the sequence [tex]\(1, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}\)[/tex], we first need to identify the pattern and rule governing the sequence.
We see the first term is [tex]\(1\)[/tex], the second term is [tex]\(-\frac{1}{2}\)[/tex], the third term is [tex]\(\frac{1}{4}\)[/tex], and the fourth term is [tex]\(-\frac{1}{8}\)[/tex].
Notice the following pattern:
- The first term [tex]\(a_1\)[/tex] = [tex]\(1\)[/tex]
- The second term [tex]\(a_2\)[/tex] = [tex]\(-\frac{1}{2}\)[/tex]
- The third term [tex]\(a_3\)[/tex] = [tex]\(\frac{1}{4}\)[/tex]
- The fourth term [tex]\(a_4\)[/tex] = [tex]\(-\frac{1}{8}\)[/tex]
We recognize that this is a geometric sequence. In a geometric sequence, each term is obtained by multiplying the previous term by a constant called the common ratio [tex]\(r\)[/tex].
From the first term to the second term:
[tex]\[ -\frac{1}{2} = 1 \times \left(-\frac{1}{2}\right) \][/tex]
So, the common ratio [tex]\(r\)[/tex] is [tex]\(-\frac{1}{2}\)[/tex].
To find the 8th term of a geometric sequence, we use the formula for the [tex]\(n\)[/tex]-th term of a geometric sequence:
[tex]\[ a_n = a_1 \times r^{(n-1)} \][/tex]
Given:
- [tex]\(a_1 = 1\)[/tex]
- [tex]\(r = -\frac{1}{2}\)[/tex]
- [tex]\(n = 8\)[/tex]
Plug these values into the formula:
[tex]\[ a_8 = 1 \times \left(-\frac{1}{2}\right)^{(8-1)} \][/tex]
[tex]\[ a_8 = 1 \times \left(-\frac{1}{2}\right)^7 \][/tex]
[tex]\[ a_8 = \left(-\frac{1}{2}\right)^7 \][/tex]
[tex]\[ a_8 = - \frac{1}{128} \][/tex]
Thus, the 8th term of this sequence is
[tex]\[ -\frac{1}{128} \][/tex]
So, the correct answer is:
D. [tex]\(-\frac{1}{128}\)[/tex]
