Let's address the given problem step-by-step for part (c) of the question. We are given that the 5th term is 81 and the 8th term is 2187.
In a geometric series, the nth term can be expressed as:
[tex]\[ a_n = ar^{n-1} \][/tex]
where:
- [tex]\( a \)[/tex] is the first term,
- [tex]\( r \)[/tex] is the common ratio,
- [tex]\( n \)[/tex] is the position of the term in the series.
We are given the following information:
1. The 5th term [tex]\( a \cdot r^4 = 81 \)[/tex]
2. The 8th term [tex]\( a \cdot r^7 = 2187 \)[/tex]
We need to find [tex]\( a \)[/tex] and [tex]\( r \)[/tex].
Step 1: Set up the equations from the given information
From the information given, we can set up the following two equations:
[tex]\[ a \cdot r^4 = 81 \tag{1} \][/tex]
[tex]\[ a \cdot r^7 = 2187 \tag{2} \][/tex]
Step 2: Divide the two equations to eliminate [tex]\( a \)[/tex]
Dividing equation (2) by equation (1):
[tex]\[ \frac{a \cdot r^7}{a \cdot r^4} = \frac{2187}{81} \][/tex]
[tex]\[ r^3 = \frac{2187}{81} \][/tex]
Now calculate the right-hand side:
[tex]\[ \frac{2187}{81} = 27 \][/tex]
Thus, we have:
[tex]\[ r^3 = 27 \][/tex]
Step 3: Solve for the common ratio [tex]\( r \)[/tex]
We take the cube root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt[3]{27} \][/tex]
[tex]\[ r = 3 \][/tex]
Step 4: Substitute the value of [tex]\( r \)[/tex] back into an original equation to find [tex]\( a \)[/tex]
Using equation (1) where [tex]\( r = 3 \)[/tex]:
[tex]\[ a \cdot 3^4 = 81 \][/tex]
[tex]\[ a \cdot 81 = 81 \][/tex]
[tex]\[ a = 1 \][/tex]
Thus, the first term [tex]\( a \)[/tex] is 1, and the common ratio [tex]\( r \)[/tex] is 3.
Final Answer:
- The positive common ratio [tex]\( r \)[/tex] is [tex]\( r = 3 \)[/tex].
- The first term [tex]\( a \)[/tex] is [tex]\( a = 1 \)[/tex].
