Sure! To solve the problem of finding the sum of the first 20 terms of an arithmetic series where the sum of the first 6 terms is 75 and the sum of the first 12 terms is 390, we can follow these steps:
1. Identify the formula for the sum of the first \(\( n \)\) terms of an arithmetic series.
The sum \(\( S_n \)\) of the first \(\( n \)\) terms of an arithmetic series can be given by:
\(\[
S_n = \frac{n}{2} (2a + (n-1)d)
\]\)
where \(\( a \)\) is the first term and \(\( d \)\) is the common difference.
2. Set up the equations from the given information.
Given:
- Sum of the first 6 terms, \(\( S_6 = 75 \)\):
\(\[
75 = \frac{6}{2} (2a + 5d)
\]\)
Simplifying, we get:
\(\[
75 = 3 (2a + 5d) \implies 2a + 5d = 25
\]\)
- Sum of the first 12 terms, \(\( S_12 = 390 \)\):
\(\[
390 = \frac{12}{2} (2a + 11d)
\]\)
Simplifying, we get:
\(\[
390 = 6 (2a + 11d) \implies 2a + 11d = 65
\]\)
3. Solve the system of equations to find \(\( a \)\) and \(\( d \)\).
We have the system of equations:
\(\[
\begin{cases}
2a + 5d = 25 \\
2a + 11d = 65
\end{cases}
\]\)
Subtract the first equation from the second:
\(\[
(2a + 11d) - (2a + 5d) = 65 - 25 \implies 6d = 40 \implies d = \frac{40}{6} = \frac{20}{3}
\]\)
Substitute \(\( d = \frac{20}{3} \)\) back into the first equation:
\(\[
2a + 5 \left(\frac{20}{3}\right) = 25 \implies 2a + \frac{100}{3} = 25
\]\)
Multiply everything by 3 to clear the fraction:
\(\[
6a + 100 = 75 \implies 6a = 75 - 100 \implies 6a = -25 \implies a = \frac{-25}{6}
\]\)
4. Calculate the sum of the first 20 terms using \(\( a \)\) and \(\( d \)\).
Use the formula for the sum of the first 20 terms:
\(\[
S_{20} = \frac{20}{2} (2a + 19d) = 10 (2a + 19d)
\]\)
Substitute \(\( a = \frac{-25}{6} \)\) and \(\( d = \frac{20}{3} \)\):
\(\[
S_{20} = 10 \left(2 \left(\frac{-25}{6}\right) + 19 \left(\frac{20}{3}\right)\right)
\]\)
Simplify inside the parentheses:
\(\[
2 \left(\frac{-25}{6}\right) = \frac{-50}{6} = \frac{-25}{3}
\]\)
\(\[
19 \left(\frac{20}{3}\right) = \frac{380}{3}
\]\)
Combine the terms:
\(\[
2a + 19d = \frac{-25}{3} + \frac{380}{3} = \frac{355}{3}
\]\)
Therefore:
\(\[
S_{20} = 10 \left(\frac{355}{3}\right) = \frac{3550}{3} \approx 1183.33
\]\)
5. Conclusion:
The sum of the first 20 terms of the arithmetic series is \(\( \boxed{1183.33} \)\).
