Question

Find the 9th term of the geometric sequence​

Answer 1

Answer:

The answer to your question is:

9th  term = $\frac{1}{781250}$

Step-by-step explanation:

The factor is 1/5, then:

$\frac{1}{2} , \frac{1}{10} , \frac{1}{50} , \frac{1}{250} , \frac{1}{1250}$

$\frac{1}{6250} , \frac{1}{31250} , \frac{1}{156250}, \frac{1}{781250}$

Answer 2

Answer:

$\frac{1}{781250}$

The $n^{\text{th}}$ term is given by $A_n=\frac{1}{2} \cdot (\frac{1}{5})^{n-1}$.

Step-by-step explanation:

Geometric sequences have this explicit form:

$a \cdot r^{n-1}$

where $a$ is the first term and $r$ is the common ratio:

We are given the first term is $a=\frac{1}{2}$.

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:

$\frac{\frac{1}{10}}{\frac{1}{2}}=\frac{\frac{1}{50}}{\frac{1}{10}}=\frac{\frac{1}{250}}{\frac{1}{50}}=\frac{\frac{1}{1250}}{\frac{1}{250}}$

Let's check it in our calculators (since I'm feeling lazy):

$0.2=0.2=0.2=0.2$

So all four of those fractions gave me 0.2 which tells me it is indeed geometric and that the common ratio is $r=0.2 \text{ or } \frac{2}{10}=\frac{1}{5}$.

So the explicit form for this sequence given is:

$\frac{1}{2} \cdot (\frac{1}{5})^{n-1}$

We want to evaluate this for $n=9$:

$\frac{1}{2} \cdot (\frac{1}{5})^{9-1}$

$\frac{1}{2} \cdot (\frac{1}{5})^{8}$

$\frac{1}{2} \cdot \frac{1}{390625}$

$\frac{1}{781250}$