Certainly! Let's solve the problem step-by-step.
Given:
- The 4th term ([tex]\(a_n\)[/tex]) of a geometric progression is 8.
- The 5th term ([tex]\(a_m\)[/tex]) of the same geometric progression is [tex]\(\frac{64}{27}\)[/tex].
### Finding the Common Ratio [tex]\((r)\)[/tex]
In a geometric progression, the [tex]\(n\)[/tex]-th term is given by:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
where [tex]\(a\)[/tex] is the first term and [tex]\(r\)[/tex] is the common ratio.
Let's set up equations for the given terms:
1) For the 4th term ([tex]\(a_n = 8\)[/tex]):
[tex]\[ 8 = a \cdot r^{4-1} \][/tex]
[tex]\[ 8 = a \cdot r^3 \][/tex]
2) For the 5th term ([tex]\(a_m = \frac{64}{27}\)[/tex]):
[tex]\[ \frac{64}{27} = a \cdot r^{5-1} \][/tex]
[tex]\[ \frac{64}{27} = a \cdot r^4 \][/tex]
To find the common ratio [tex]\(r\)[/tex], we can divide the second equation by the first equation:
[tex]\[ \frac{\frac{64}{27}}{8} = \frac{a \cdot r^4}{a \cdot r^3} \][/tex]
[tex]\[ \frac{\frac{64}{27}}{8} = r \][/tex]
Simplifying the left side:
[tex]\[ \frac{64}{27} \cdot \frac{1}{8} = r \][/tex]
[tex]\[ \frac{64}{27} \cdot \frac{1}{8} = \frac{64}{216} = \frac{64}{6^3} = \frac{64}{666} = \frac{1}{6 \cdot 6} \approx 0.296 \][/tex]
Thus, the common ratio is:
[tex]\[ r \approx 0.296 \][/tex]
### Finding the First Term [tex]\((a)\)[/tex]
Now that we have the common ratio [tex]\(r\)[/tex], we can substitute it back into one of the equations to find the first term [tex]\(a\)[/tex].
Using the equation for the 4th term:
[tex]\[ 8 = a \cdot (0.296)^3 \][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[ a = \frac{8}{(0.296)^3} \][/tex]
[tex]\[ a \approx 307.546875 \][/tex]
So the first term [tex]\(a\)[/tex] is approximately:
[tex]\[ a \approx 307.55 \][/tex]
### Summary:
1) The common ratio [tex]\(r\)[/tex] is approximately [tex]\(0.296\)[/tex].
2) The first term [tex]\(a\)[/tex] is approximately [tex]\(307.55\)[/tex].
