To solve this problem, we need to use the properties of arithmetic progressions (AP).
Let \( S_n \) be the sum of the first \( n \) terms of an AP, which can be expressed as:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where \( a \) is the first term and \( d \) is the common difference.
We are given:
- \( S_{16} = -504 \)
- \( S_{9} = -126 \)
Using the formula for the sum of the first \( n \) terms, we can write:
- For \( n = 16 \):
\[ S_{16} = \frac{16}{2} \times (2a + 15d) = 8(2a + 15d) = -504 \]
This simplifies to:
\[ 2a + 15d = \frac{-504}{8} = -63 \quad \text{(1)} \]
- For \( n = 9 \):
\[ S_{9} = \frac{9}{2} \times (2a + 8d) = \frac{9}{2}(2a + 8d) = -126 \]
Thus, we have:
\[ 2a + 8d = \frac{-126 \times 2}{9} = -28 \quad \text{(2)} \]
Now, we need to solve the two equations (1) and (2):
From equation (1):
\[ 2a + 15d = -63 \]
From equation (2):
\[ 2a + 8d = -28 \]
Subtract equation (2) from equation (1):
\[ (2a + 15d) - (2a + 8d) = -63 - (-28) \]
This simplifies to:
\[ 7d = -35 \]
Solving for \( d \):
\[ d = \frac{-35}{7} = -5 \]
Now substitute \( d = -5 \) back into equation (2) to find \( 2a \):
\[ 2a + 8(-5) = -28 \]
Simplifying:
\[ 2a - 40 = -28 \]
Adding 40 to both sides:
\[ 2a = 12 \]
Thus,
\[ a = 6 \]
Now we have \( a = 6 \) and \( d = -5 \).
We need to find the sum of the first 30 terms, \( S_{30} \):
\[ S_{30} = \frac{30}{2} \times (2a + 29d) \]
Simplifying \( S_{30} \):
\[ S_{30} = 15 \times (2 \cdot 6 + 29 \cdot -5) \]
Calculating inside the parentheses:
\[ 2 \cdot 6 = 12 \] \[ 29 \cdot -5 = -145 \]
Thus:
\[ S_{30} = 15 \times (12 - 145) = 15 \times (-133) = -1995 \]
So, the sum of the first 30 terms is:
\[ \boxed{-1995} \]