Question

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a geometric sequence has 1 as its third term and 8 as its sixth term.
what is the function that represents the sequence?
f(n) = 2^{n - 1}
f(n) = \frac{1}{4} \cdot 2^{n - 1}
f(n) = \frac{1}{4} \cdot 8^{n - 1}
f(n) = 8^{n - 1}

Explanation:

Step1: Recall geometric sequence formula

The nth term of a geometric sequence is $f(n) = a_1 r^{n-1}$, where $a_1$ is the first term, $r$ is the common ratio.

Step2: Set up equations for given terms

For the 3rd term: $a_1 r^{2} = 1$
For the 6th term: $a_1 r^{5} = 8$

Step3: Solve for common ratio $r$

Divide the 6th term equation by the 3rd term equation:
$\frac{a_1 r^{5}}{a_1 r^{2}} = \frac{8}{1}$
$r^3 = 8$
$r = \sqrt[3]{8} = 2$

Step4: Solve for first term $a_1$

Substitute $r=2$ into $a_1 r^{2} = 1$:
$a_1 (2)^2 = 1$
$4a_1 = 1$
$a_1 = \frac{1}{4}$

Step5: Write the sequence function

Substitute $a_1=\frac{1}{4}$ and $r=2$ into the nth term formula:
$f(n) = \frac{1}{4} \cdot 2^{n-1}$

Answer:

$f(n) = \frac{1}{4} \cdot 2^{n-1}$