Answer:
[tex]\frac{1}{781250}[/tex]
The [tex]n^{\text{th}}[/tex] term is given by [tex]A_n=\frac{1}{2} \cdot (\frac{1}{5})^{n-1}[/tex].
Step-by-step explanation:
Geometric sequences have this explicit form:
[tex]a \cdot r^{n-1}[/tex]
where [tex]a[/tex] is the first term and [tex]r[/tex] is the common ratio:
We are given the first term is [tex]a=\frac{1}{2}[/tex].
The common ratio is the result of taking a term and dividing by it's previous term. So all of these should lead to the same result since we are given the sequence is geometric:
[tex]\frac{\frac{1}{10}}{\frac{1}{2}}=\frac{\frac{1}{50}}{\frac{1}{10}}=\frac{\frac{1}{250}}{\frac{1}{50}}=\frac{\frac{1}{1250}}{\frac{1}{250}}[/tex]
Let's check it in our calculators (since I'm feeling lazy):
[tex]0.2=0.2=0.2=0.2[/tex]
So all four of those fractions gave me 0.2 which tells me it is indeed geometric and that the common ratio is [tex]r=0.2 \text{ or } \frac{2}{10}=\frac{1}{5}[/tex].
So the explicit form for this sequence given is:
[tex]\frac{1}{2} \cdot (\frac{1}{5})^{n-1}[/tex]
We want to evaluate this for [tex]n=9[/tex]:
[tex]\frac{1}{2} \cdot (\frac{1}{5})^{9-1}[/tex]
[tex]\frac{1}{2} \cdot (\frac{1}{5})^{8}[/tex]
[tex]\frac{1}{2} \cdot \frac{1}{390625}[/tex]
[tex]\frac{1}{781250}[/tex]
