To find the 10th term of a geometric sequence, you can use the formula:

$a_n = a_1 \times r^{(n-1)}$

where:

• $a_n$ is the $n$-th term of the sequence,
• $a_1$ is the first term,
• $r$ is the common ratio,
• $n$ is the term number.

Given:

• $a_1 = 8$,
• The common ratio $r = \frac{4}{8} = \frac{1}{2}$,
• $n = 10$.

Now, plug in the values:

$a_{10} = 8 \times \left(\frac{1}{2}\right)^{(10-1)} = 8 \times \left(\frac{1}{2}\right)^9 = 8 \times \frac{1}{512} = \frac{8}{512} = \frac{1}{64}$

So, the 10th term of the sequence is $\frac{1}{64}$.

Would you like further details or have any questions?

Here are some related questions:

1. How do you find the common ratio in a geometric sequence?
2. What is the sum of the first 10 terms of this geometric sequence?
3. How do you determine if a sequence is arithmetic or geometric?
4. What is the general formula for the sum of an infinite geometric series?
5. How does changing the common ratio affect the sequence?

Tip: In a geometric sequence, if the common ratio is a fraction between 0 and 1, the terms will get smaller as the sequence progresses.