The given series is:

$48 + 45 + 42 + \dots + (-81) + (-84)$

This is an arithmetic series because the difference between consecutive terms is constant. Let's break it down:

1. First term (a): The first term is $48$.
2. Common difference (d): The common difference is $45 - 48 = -3$.
3. Last term (l): The last term is $-84$.

Step 1: Find the number of terms (n)

We use the formula for the $n$-th term of an arithmetic sequence:

$a_n = a + (n-1)d$

Substitute the known values:

$-84 = 48 + (n-1)(-3)$

Simplify:

$-84 = 48 - 3(n-1)$

$-84 = 48 - 3n + 3$

$-84 = 51 - 3n$

$-84 - 51 = -3n$

$-135 = -3n$

$n = 45$

So, there are 45 terms in the series.

Step 2: Sum of the arithmetic series

The sum of an arithmetic series is given by the formula:

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

Substitute the known values:

$S_{45} = \frac{45}{2} \times (48 + (-84))$

$S_{45} = \frac{45}{2} \times (-36)$

$S_{45} = 45 \times (-18)$

$S_{45} = -810$

Final Answer:

The sum of the series is $\boxed{-810}$.

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. What if the common difference in the arithmetic series was positive?
2. How do you find the sum of a geometric series?
3. Can you find the sum of the first 50 terms of a similar arithmetic sequence?
4. How would the sum change if the last term was different?
5. What is the difference between an arithmetic and a geometric progression?

Tip: In arithmetic series, knowing the number of terms is crucial to calculating the sum effectively!