Let's denote the first term of the arithmetic progression (AP) as \( a_1 = 21 \) and the last term (or \( n \)-th term) as \( a_n = -47 \).
The sum of the first \( n \) terms of an AP can be calculated with the formula:
\[ S_n = \frac{n}{2} (a_1 + a_n) \]
According to the problem, the sum of the series is given as \( S_n = -234 \). Substituting the values we know:
\[ -234 = \frac{n}{2} (21 + (-47)) \]
Calculating \( 21 + (-47) \):
\[ 21 - 47 = -26 \]
Now substituting back into the equation:
\[ -234 = \frac{n}{2} (-26) \]
Multiplying both sides by \( -2 \):
\[ 468 = n \cdot 26 \]
Now, divide by 26:
\[ n = \frac{468}{26} = 18 \]
So, the number of terms \( n = 18 \).
Next, we need to find the common difference \( d \). The formula for the \( n \)-th term of an AP is given by:
\[ a_n = a_1 + (n-1) d \]
Substituting the values we have:
\[ -47 = 21 + (18 - 1) d \]
This simplifies to:
\[ -47 = 21 + 17d \]
Now, isolate \( d \):
\[ -47 - 21 = 17d \] \[ -68 = 17d \] \[ d = \frac{-68}{17} = -4 \]
Thus, the common difference \( d = -4 \).
Finally, we need to calculate the sum of the first 18 terms, which we have already calculated as \( S_{18} = -234 \).
To summarize:
- The number of terms \( n = 18 \)
- The common difference \( d = -4 \)
- The sum of the first 18 terms \( S_{18} = -234 \)