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Write a function for the geometric sequence with a starting term of 128 and a common ratio of [tex]\( \frac{1}{2} \)[/tex]. Find the fifth term.

A. [tex]\( f(n) = 128 \cdot \left( \frac{1}{2} \right)^{n-1} \)[/tex], the fifth term is 8
B. [tex]\( f(n) = -\frac{1}{2} \cdot 8^{n-1} \)[/tex], the fifth term is -2,048
C. [tex]\( f(n) = 128 \cdot \left( \frac{1}{2} \right)^{n-1} \)[/tex], the fifth term is 4
D. [tex]\( f(n) = 128 \cdot \left( -\frac{1}{2} \right)^{n-1} \)[/tex], the fifth term is -8



Answer :

GinnyAnswer
To address this question, we first need to verify the form of the function and then calculate the fifth term using the correct function.

1. The correct form of the geometric sequence provided is:
[tex]\[ f(n) = 128 \cdot \left(\frac{1}{2}\right)^{n-1} \][/tex]
Here, [tex]\(128\)[/tex] is the starting term (first term) of the sequence, and [tex]\(\frac{1}{2}\)[/tex] is the common ratio.

2. To find the fifth term, we substitute [tex]\(n = 5\)[/tex] into the function:
[tex]\[ f(5) = 128 \cdot \left(\frac{1}{2}\right)^{5-1} \][/tex]
[tex]\[ f(5) = 128 \cdot \left(\frac{1}{2}\right)^4 \][/tex]

3. Evaluating [tex]\(\left(\frac{1}{2}\right)^4\)[/tex]:

[tex]\[ \left(\frac{1}{2}\right)^4 = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16} \][/tex]

4. Now, multiply the starting term [tex]\(128\)[/tex] by [tex]\(\frac{1}{16}\)[/tex]:
[tex]\[ 128 \cdot \frac{1}{16} = 8 \][/tex]

Thus, the fifth term of the geometric sequence is [tex]\(8.0\)[/tex].

Therefore, the correct option from the provided list is:
[tex]\[ f(n) = 128 \cdot \left(\frac{1}{2}\right)^{n-1}, \text{ the fifth term is } 8 \][/tex]

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