To solve the problem, let's denote the first term of the arithmetic progression (AP) as \( a \) and the common difference as \( d \).
The formula for the \( n \)-th term of an AP is given by:
\[ a_n = a + (n-1)d \]
The sum of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} (2a + (n-1)d) \]
We are provided the following conditions:
- The sum of the first 8 terms is 160:
\[ S_8 = \frac{8}{2} (2a + 7d) = 4(2a + 7d) = 160 \]
This simplifies to:
\[ 2a + 7d = 40 \quad \text{(1)} \]
- The sum of the first 20 terms is 880:
\[ S_{20} = \frac{20}{2} (2a + 19d) = 10(2a + 19d) = 880 \]
This simplifies to:
\[ 2a + 19d = 88 \quad \text{(2)} \]
Now we have a system of two equations:
- \( 2a + 7d = 40 \)
- \( 2a + 19d = 88 \)
We can eliminate \( 2a \) by subtracting equation (1) from equation (2):
\[ (2a + 19d) - (2a + 7d) = 88 - 40 \]
This simplifies to:
\[ 12d = 48 \]
Solving for \( d \):
\[ d = 4 \]
Now, substituting \( d = 4 \) back into equation (1):
\[ 2a + 7(4) = 40 \]
\[ 2a + 28 = 40 \]
\[ 2a = 12 \]
\[ a = 6 \]
Now we have our values for \( a \) and \( d \):
- \( a = 6 \)
- \( d = 4 \)
Next, we can find the 43rd term, \( a_{43} \):
\[ a_{43} = a + (43-1)d = 6 + 42 \times 4 \]
Calculating it:
\[ a_{43} = 6 + 168 = 174 \]
Finally, we can find the sum of the first 12 terms, \( S_{12} \):
\[ S_{12} = \frac{12}{2} (2a + 11d) = 6(2a + 11d) \]
Substituting in our values for \( a \) and \( d \):
\[ S_{12} = 6(2(6) + 11(4)) = 6(12 + 44) = 6 \times 56 = 336 \]
So, the answers are:
- The 43rd term \( a_{43} = 174 \)
- The sum of the first 12 terms \( S_{12} = 336 \)