Sure, let's solve the problem step-by-step.
We are given the following information:
1. [tex]\(128 \times \text{first term} = 2 \times \text{seventh term}\)[/tex]
2. The [tex]\(3^4\)[/tex] term is 8.
Let's denote the first term by [tex]\( a \)[/tex] and the common ratio by [tex]\( r \)[/tex].
The [tex]\(\text{n}^{\text{th}}\)[/tex] term of a geometric sequence is given by the formula:
[tex]\[ T_n = a \cdot r^{n-1} \][/tex]
From the first piece of information, [tex]\(128 \times \text{first term} = 2 \times \text{seventh term}\)[/tex]:
[tex]\[ 128a = 2 \cdot a \cdot r^6 \][/tex]
Now, divide both sides by 2:
[tex]\[ 64a = a \cdot r^6 \][/tex]
Assuming [tex]\( a \neq 0 \)[/tex], we can divide both sides by [tex]\( a \)[/tex]:
[tex]\[ 64 = r^6 \][/tex]
Solving for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt[6]{64} \][/tex]
[tex]\[ r = 2 \][/tex]
Now we know the common ratio [tex]\( r = 2 \)[/tex].
From the second piece of information, the [tex]\(3^4\)[/tex] term is 8. Given that this should actually reference the 4th term (since [tex]\(3^4\)[/tex] would be considered 81st which is irrelevant here), let’s start by interpreting it correctly as:
[tex]\[ T_4 = a \cdot r^3 = 8 \][/tex]
With [tex]\( r = 2 \)[/tex]:
[tex]\[ a \cdot 2^3 = 8 \][/tex]
[tex]\[ a \cdot 8 = 8 \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ a = 1 \][/tex]
Now we need to find the 10th term ([tex]\( T_{10} \)[/tex]):
[tex]\[ T_{10} = a \cdot r^9 \][/tex]
Substitute [tex]\( a = 1 \)[/tex] and [tex]\( r = 2 \)[/tex]:
[tex]\[ T_{10} = 1 \cdot 2^9 \][/tex]
[tex]\[ T_{10} = 2^9 \][/tex]
[tex]\[ T_{10} = 512 \][/tex]
Therefore, the 10th term is [tex]\( 512 \)[/tex].
