To determine the sum of the first 12 terms of the series starting with the values 4, 10, 25, 62.5, we first analyze the pattern in the given terms.
The terms are:
- \( a_1 = 4 \)
- \( a_2 = 10 \)
- \( a_3 = 25 \)
- \( a_4 = 62.5 \)
We can express the series in terms of a relationship between each term and the previous term. Let's compute the ratio of consecutive terms:
\[ \frac{a_2}{a_1} = \frac{10}{4} = 2.5 \] \[ \frac{a_3}{a_2} = \frac{25}{10} = 2.5 \] \[ \frac{a_4}{a_3} = \frac{62.5}{25} = 2.5 \]
This suggests that the series has a common ratio of \( r = 2.5 \).
Now, we can express the general term \( a_n \) as:
\[ a_n = a_1 \cdot r^{n-1} = 4 \cdot (2.5)^{n-1} \]
To find the sum \( S_n \) of the first \( n \) terms of a geometric series, we use the formula:
\[ S_n = a_1 \frac{1 - r^n}{1 - r} \]
For our series, we have \( a_1 = 4 \), \( r = 2.5 \), and we want the sum of the first 12 terms \( S_{12} \):
\[ S_{12} = 4 \frac{1 - (2.5)^{12}}{1 - 2.5} \]
Calculating \( 1 - 2.5 \):
\[ 1 - 2.5 = -1.5 \]
Next, we need to calculate \( (2.5)^{12} \):
Using a calculator, we find that:
\[ (2.5)^{12} \approx 24414.0625 \]
Now substituting back into the sum formula:
\[ S_{12} = 4 \frac{1 - 24414.0625}{-1.5} = 4 \frac{-24413.0625}{-1.5} = 4 \cdot \frac{24413.0625}{1.5} \]
Calculating \( \frac{24413.0625}{1.5} \):
\[ \frac{24413.0625}{1.5} \approx 16275.375 \]
Now multiplying by 4:
\[ S_{12} \approx 4 \cdot 16275.375 \approx 65101.5 \]
Rounding this to the nearest whole number gives \( 65102 \).
Thus, the sum of the first 12 terms of the series, rounded to the nearest whole number, is:
\[ \boxed{65102} \]
(Note: While the provided answer options do not include 65102, based on the calculations, this is the derived result from the series.)