Let's solve the given questions step by step while focusing only on what's required for the calculations.
### Question 5
The general term for the series is given by [tex]\( a_n = 3 - 5n \)[/tex]. We are asked to find the sum [tex]\( S_8 \)[/tex], which involves summing up the terms from [tex]\( a_1 \)[/tex] to [tex]\( a_8 \)[/tex]:
[tex]\[ S_8 = \sum_{k=1}^{8} a_k \][/tex]
Given that the final sum [tex]\( S_8 \)[/tex] results in:
[tex]\[ S_8 = -156 \][/tex]
Thus, the correct answer to Question 5 is:
[tex]\[ \boxed{-156} \][/tex]
### Question 6
We need to find the sum of the infinite series:
[tex]\[ \sum_{k=1}^{\infty} 5^{3-k} \][/tex]
Let's first work out the general form of the series. Notice that each term [tex]\( 5^{3-k} \)[/tex] can be written as:
[tex]\[ 5^{3-k} = 5^3 \cdot 5^{-k} = 125 \cdot (1/5)^k \][/tex]
This is a geometric series with the first term [tex]\( a = 125 \cdot (1/5) = 25 \)[/tex] and the common ratio [tex]\( r = 1/5 \)[/tex].
The sum of an infinite geometric series where [tex]\( |r| < 1 \)[/tex] is given by:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
For this series:
- [tex]\( a = 25 \)[/tex]
- [tex]\( r = 1/5 \)[/tex]
Plugging in these values, we have:
[tex]\[ S = \frac{25}{1 - 1/5} = \frac{25}{4/5} = \frac{25 \cdot 5}{4} = \frac{125}{4} \][/tex]
Thus, the correct answer to Question 6 is:
[tex]\[ \boxed{\frac{125}{4}} \][/tex]
