The given series is:

$48, 24, 12, \dots$

This is a geometric series where the first term $a_1 = 48$ and the common ratio $r = \frac{24}{48} = \frac{1}{2}$.

The sum of the first $n$ terms of a geometric series is given by the formula:

$S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$

For this series, we need to find the sum of the first 12 terms ($S_{12}$):

$S_{12} = 48 \cdot \frac{1 - \left(\frac{1}{2}\right)^{12}}{1 - \frac{1}{2}}$

Let's compute this step by step:

1. $r^{12} = \left(\frac{1}{2}\right)^{12} = \frac{1}{4096}$
2. $1 - \left(\frac{1}{2}\right)^{12} = 1 - \frac{1}{4096} = \frac{4096 - 1}{4096} = \frac{4095}{4096}$
3. $S_{12} = 48 \cdot \frac{\frac{4095}{4096}}{\frac{1}{2}} = 48 \cdot 2 \cdot \frac{4095}{4096} = 96 \cdot \frac{4095}{4096} = \frac{393120}{4096} \approx 95.76$

So, the sum of the first 12 terms is approximately:

$\boxed{95.76}$

Would you like more details on geometric series or have any further questions?

Here are 5 related questions for practice:

1. What is the sum of the first 10 terms of the series $100, 50, 25, \dots$?
2. How can you find the sum of an infinite geometric series?
3. What is the sum of the first 5 terms of the series $3, 6, 12, \dots$?
4. How do you find the common ratio in a geometric series?
5. What is the sum of the first 15 terms of the series $80, 40, 20, \dots$?

Tip: When calculating the sum of geometric series, be mindful of whether the common ratio is greater than or less than 1. It affects convergence for infinite series!