The given series is:

$\frac{1}{8}, \frac{1}{4}, \frac{1}{2}, \ldots$

This is a geometric progression (GP) because each term is obtained by multiplying the previous term by a constant ratio. Let's identify the first term ($a$) and the common ratio ($r$):

• First term ($a$) = $\frac{1}{8}$
• Common ratio ($r$) = $\frac{\frac{1}{4}}{\frac{1}{8}} = 2$

The general formula for the $n$-th term of a geometric progression is:

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

We are looking for the 8th term ($T_8$):

$T_8 = \frac{1}{8} \cdot 2^{8-1} = \frac{1}{8} \cdot 2^7 = \frac{1}{8} \cdot 128 = \frac{128}{8} = 16$

Thus, the 8th term of the series is 16.

Would you like further details on this?

Here are 5 related questions for practice:

1. What is the 10th term of the same series?
2. What is the sum of the first 8 terms of this series?
3. Can you find the common ratio in a different way?
4. What would the first negative term be in this GP if extended?
5. How does the sum of infinite terms of a geometric series work?

Tip: In a geometric progression, knowing the first term and the common ratio allows you to calculate any term easily with the formula $T_n = a \cdot r^{n-1}$.