A geometric sequence has a third term 8 and a fifth term 128. a) find the common ration and the first term  b) find the eight term 

In a geometric sequence, each term is given by the formula:

\[
T_n = ar^{n-1}
\]

where:
- \( T_n \) is the \(n\)-th term,
- \( a \) is the first term,
- \( r \) is the common ratio, and
- \( n \) is the term number.

### Given:
- Third term \( T_3 = 8 \),
- Fifth term \( T_5 = 128 \).

#### a) Find the common ratio \(r\) and the first term \(a\).

We can use the formula for the terms of the geometric sequence.

For the third term:
\[
T_3 = ar^2 = 8
\]

For the fifth term:
\[
T_5 = ar^4 = 128
\]

Now, divide the equation for \( T_5 \) by the equation for \( T_3 \):
\[
\frac{ar^4}{ar^2} = \frac{128}{8}
\]
\[
r^2 = 16
\]
\[
r = \pm 4
\]

Now, substitute \( r = 4 \) (we will first try the positive value) into the equation for \( T_3 \):
\[
ar^2 = 8
\]
\[
a(4^2) = 8
\]
\[
16a = 8
\]
\[
a = \frac{8}{16} = \frac{1}{2}
\]

So, the first term \( a = \frac{1}{2} \) and the common ratio \( r = 4 \).

#### b) Find the eighth term.

The eighth term \( T_8 \) is given by:
\[
T_8 = ar^7
\]
Substitute \( a = \frac{1}{2} \) and \( r = 4 \):
\[
T_8 = \frac{1}{2} \times 4^7
\]
\[
T_8 = \frac{1}{2} \times 16384
\]
\[
T_8 = 8192
\]

### Final Answers:
- The first term \( a = \frac{1}{2} \),
- The common ratio \( r = 4 \),
- The eighth term \( T_8 = 8192 \).

Would you like more details or further clarifications? 

Here are some related questions to consider:
1. How does the sign of \( r \) affect the sequence?
2. What happens if the common ratio is negative?
3. Can you derive the formula for the sum of the first \(n\) terms of a geometric sequence?
4. How would you handle this if given the sum of terms instead of individual terms?
5. What is the general behavior of a geometric sequence when \(r > 1\) compared to when \(0 < r < 1\)?

**Tip:** In geometric sequences, knowing two terms can often help you solve for both the common ratio and the first term by setting up a system of equations.

A geometric sequence has a third term 8 and a fifth term 128. a) Find the common ratio and the first term. b) Find the eighth term.

Solve a geometric sequence problem with given third and fifth terms (8 and 128). Learn how to find the common ratio, first term, and the eighth term using key algebraic formulas. This solution is suitable for students in Grades 9-11.

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