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