Answer :
Sure, let's find the sum \( S_n \) for each geometric series case described.
The formula for the sum of the first \( n \) terms of a geometric series is:
[tex]\[ S_n = \begin{cases} a_1 \cdot \dfrac{1 - r^n}{1 - r} & \text{if } r \neq 1 \\ a_1 \cdot n & \text{if } r = 1 \end{cases} \][/tex]
### Case 1: \( a_1 = 2, a_n = 486, r = 3 \)
1. First, we need to find \( n \) (the number of terms):
[tex]\[ a_n = a_1 \cdot r^{n-1} \\ 486 = 2 \cdot 3^{n-1} \\ 243 = 3^{n-1} \][/tex]
Since \( 243 = 3^5 \), then \( n-1 = 5 \) and \( n = 6 \).
2. Now, calculate the sum \( S_6 \):
[tex]\[ S_6 = 2 \cdot \left(\frac{1 - 3^6}{1 - 3}\right) \\ = 2 \cdot \left(\frac{1 - 729}{1 - 3}\right) \\ = 2 \cdot \left(\frac{-728}{-2}\right) \\ = 2 \cdot 364 \\ = 728 \][/tex]
### Case 2: \( a_1 = 1200, a_n = 75, r = \frac{1}{2} \)
1. First, we need to find \( n \):
[tex]\[ a_n = a_1 \cdot r^{n-1} \\ 75 = 1200 \cdot \left(\frac{1}{2}\right)^{n-1} \\ \left(\frac{1}{2}\right)^{n-1} = \frac{75}{1200} \\ \left(\frac{1}{2}\right)^{n-1} = \frac{1}{16} \\ \left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^4 \][/tex]
Here, \( n-1 = 4 \) and \( n = 5 \).
2. Now, calculate the sum \( S_5 \):
[tex]\[ S_5 = 1200 \cdot \left(\frac{1 - \left(\frac{1}{2}\right)^5}{1 - \frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{1 - \frac{1}{32}}{\frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{\frac{31}{32}}{\frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{31}{16}\right) \\ = 1200 \cdot 1.9375 \\ = 2325 \][/tex]
### Case 4: \( a_1 = 3, r = \frac{1}{3}, n = 4 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_4 = 3 \cdot \left(\frac{1 - \left(\frac {1}{3}\right)^4}{1 - \frac{1}{3}}\right) \\ = 3 \cdot \left(\frac{1 - \frac{1}{81}}{\frac{2}{3}}\right) \\ = 3 \cdot \left(\frac{\frac{80}{81}}{\frac{2}{3}}\right) \\ = 3 \cdot \left(\frac{80}{81} \cdot \frac{3}{2}\right) \\ = 3 \cdot \frac {240}{162} \\ = 3 \cdot \frac{120}{81} \\ = 4.444 \approx 4.44 \][/tex]
### Case 5: \( a_1 = 2, r = 6, n = 4 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_4 = 2 \cdot \left(\frac{1 - 6^4}{1 - 6}\right) \\ = 2 \cdot \left(\frac{1 - 1296}{-5}\right) \\ = 2 \cdot \left(\frac{-1295}{-5}\right) \\ = 2 \cdot 259 \\ = 518 \][/tex]
### Case 7: \( a_1 = 100, r = - \frac{1}{2}, n = 5 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_5 = 100 \cdot \left(\frac{1 - \left(-\frac {1}{2}\right)^5}{1 - \left(-\frac {1}{2}\right)}\right) \\ = 100 \cdot \left(\frac{1 - \left(-\frac {1}{32}\right)}{1 + \frac{1}{2}}\right) \\ = 100 \cdot \left(\frac{1 - \left(-0.03125\right)}{1 + 0.5}\right) \\ = 100 \cdot \left(\frac{1 + 0.03125}{1.5}\right) \\ = 100 \cdot \left(\frac{1.03125}{1.5}\right) \\ \approx 68.75 \][/tex]
### Case 8: \( a_0 = 20 \) (initial term without ratio)
This case seems to refer a merely a single term geometric 0-th term series which has nothing further to sum just equal to 20
By keeping our interpretation \( a_{0}\ term \)
### Summarized Solutions:
1. \( S_1 = 728 \)
2. \( S_2 = 2325 \)
4. \( S_4 = 4.44 \)
5. \( S_5 = 518 \)
7. \( S_7 = 68.75 \)
8. \( S_{8}=20\
The formula for the sum of the first \( n \) terms of a geometric series is:
[tex]\[ S_n = \begin{cases} a_1 \cdot \dfrac{1 - r^n}{1 - r} & \text{if } r \neq 1 \\ a_1 \cdot n & \text{if } r = 1 \end{cases} \][/tex]
### Case 1: \( a_1 = 2, a_n = 486, r = 3 \)
1. First, we need to find \( n \) (the number of terms):
[tex]\[ a_n = a_1 \cdot r^{n-1} \\ 486 = 2 \cdot 3^{n-1} \\ 243 = 3^{n-1} \][/tex]
Since \( 243 = 3^5 \), then \( n-1 = 5 \) and \( n = 6 \).
2. Now, calculate the sum \( S_6 \):
[tex]\[ S_6 = 2 \cdot \left(\frac{1 - 3^6}{1 - 3}\right) \\ = 2 \cdot \left(\frac{1 - 729}{1 - 3}\right) \\ = 2 \cdot \left(\frac{-728}{-2}\right) \\ = 2 \cdot 364 \\ = 728 \][/tex]
### Case 2: \( a_1 = 1200, a_n = 75, r = \frac{1}{2} \)
1. First, we need to find \( n \):
[tex]\[ a_n = a_1 \cdot r^{n-1} \\ 75 = 1200 \cdot \left(\frac{1}{2}\right)^{n-1} \\ \left(\frac{1}{2}\right)^{n-1} = \frac{75}{1200} \\ \left(\frac{1}{2}\right)^{n-1} = \frac{1}{16} \\ \left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^4 \][/tex]
Here, \( n-1 = 4 \) and \( n = 5 \).
2. Now, calculate the sum \( S_5 \):
[tex]\[ S_5 = 1200 \cdot \left(\frac{1 - \left(\frac{1}{2}\right)^5}{1 - \frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{1 - \frac{1}{32}}{\frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{\frac{31}{32}}{\frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{31}{16}\right) \\ = 1200 \cdot 1.9375 \\ = 2325 \][/tex]
### Case 4: \( a_1 = 3, r = \frac{1}{3}, n = 4 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_4 = 3 \cdot \left(\frac{1 - \left(\frac {1}{3}\right)^4}{1 - \frac{1}{3}}\right) \\ = 3 \cdot \left(\frac{1 - \frac{1}{81}}{\frac{2}{3}}\right) \\ = 3 \cdot \left(\frac{\frac{80}{81}}{\frac{2}{3}}\right) \\ = 3 \cdot \left(\frac{80}{81} \cdot \frac{3}{2}\right) \\ = 3 \cdot \frac {240}{162} \\ = 3 \cdot \frac{120}{81} \\ = 4.444 \approx 4.44 \][/tex]
### Case 5: \( a_1 = 2, r = 6, n = 4 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_4 = 2 \cdot \left(\frac{1 - 6^4}{1 - 6}\right) \\ = 2 \cdot \left(\frac{1 - 1296}{-5}\right) \\ = 2 \cdot \left(\frac{-1295}{-5}\right) \\ = 2 \cdot 259 \\ = 518 \][/tex]
### Case 7: \( a_1 = 100, r = - \frac{1}{2}, n = 5 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_5 = 100 \cdot \left(\frac{1 - \left(-\frac {1}{2}\right)^5}{1 - \left(-\frac {1}{2}\right)}\right) \\ = 100 \cdot \left(\frac{1 - \left(-\frac {1}{32}\right)}{1 + \frac{1}{2}}\right) \\ = 100 \cdot \left(\frac{1 - \left(-0.03125\right)}{1 + 0.5}\right) \\ = 100 \cdot \left(\frac{1 + 0.03125}{1.5}\right) \\ = 100 \cdot \left(\frac{1.03125}{1.5}\right) \\ \approx 68.75 \][/tex]
### Case 8: \( a_0 = 20 \) (initial term without ratio)
This case seems to refer a merely a single term geometric 0-th term series which has nothing further to sum just equal to 20
By keeping our interpretation \( a_{0}\ term \)
### Summarized Solutions:
1. \( S_1 = 728 \)
2. \( S_2 = 2325 \)
4. \( S_4 = 4.44 \)
5. \( S_5 = 518 \)
7. \( S_7 = 68.75 \)
8. \( S_{8}=20\
