This is an arithmetic series, where each term increases by a constant difference. Let's break it down to solve the sum.

Given:

- First term \( a = 4 \)
- Common difference \( d = 4 \)
- Last term \( l = 108 \)

We use the formula for the sum of an arithmetic series:

\[
S_n = \frac{n}{2} \times (a + l)
\]

First, we need to find \( n \), the number of terms. The formula for the \(n\)-th term of an arithmetic sequence is:

\[
l = a + (n-1) \times d
\]

Substitute the known values:

\[
108 = 4 + (n-1) \times 4
\]

Simplify:

\[
108 - 4 = (n-1) \times 4
\]
\[
104 = (n-1) \times 4
\]
\[
n - 1 = \frac{104}{4} = 26
\]
\[
n = 27
\]

Now that we have \( n = 27 \), we can find the sum:

\[
S_{27} = \frac{27}{2} \times (4 + 108) = \frac{27}{2} \times 112 = 27 \times 56 = 1512
\]

So, the sum is:

\[
S = 1512
\]

Would you like more details, or do you have any questions?

Here are 5 related questions to explore:

1. What is the general formula for the sum of an arithmetic series?
2. How do you determine the number of terms in an arithmetic progression?
3. What happens if the common difference is negative?
4. Can you derive the formula for the sum of an arithmetic series from first principles?
5. How can you use the concept of arithmetic series in real-life applications?

**Tip:** Always double-check the values of the first term, last term, and common difference to avoid calculation errors in series problems.

Learn how to find the sum of the arithmetic series 4 + 8 + 12 + ... + 108 using the arithmetic series sum formula. This problem explores key concepts such as sequences and summation, perfect for students in Grades 9-10.

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