The problem you've provided appears to involve a geometric sequence, where:

• The third term is $a_3 = 8$
• The sixth term is $a_6 = -64$

Let the first term of the sequence be $a_1 = a$ and the common ratio be $r$.

(i) Finding the first term $a$ and the common ratio $r$

For a geometric sequence, the $n$-th term is given by:

$a_n = a_1 \cdot r^{n-1}$

So, we have:

$a_3 = a \cdot r^{3-1} = a \cdot r^2 = 8$

$a_6 = a \cdot r^{6-1} = a \cdot r^5 = -64$

Now, we can solve these two equations:

1. $a \cdot r^2 = 8$
2. $a \cdot r^5 = -64$

To find $r$, divide the second equation by the first:

$\frac{a \cdot r^5}{a \cdot r^2} = \frac{-64}{8}$

Simplifying, we get:

$r^3 = -8$

Taking the cube root of both sides:

$r = -2$

Now substitute $r = -2$ back into the first equation:

$a \cdot (-2)^2 = 8$

$a \cdot 4 = 8$

$a = 2$

So, the first term $a$ is 2 and the common ratio $r$ is -2.

(ii) Finding the sum of the first 6 terms

The sum $S_n$ of the first $n$ terms of a geometric sequence is given by:

S_n = a \cdot \frac{1 - r^n}{1 - r} \quad \text{(for r \neq 1)}

For the first 6 terms:

$S_6 = 2 \cdot \frac{1 - (-2)^6}{1 - (-2)}$

Simplify the expression:

$S_6 = 2 \cdot \frac{1 - 64}{1 + 2}$

$S_6 = 2 \cdot \frac{-63}{3}$

$S_6 = 2 \cdot (-21)$

$S_6 = -42$

So, the sum of the first 6 terms is $-42$.

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Would you like more details or have any questions? Here are some related problems to consider:

1. What is the 7th term of the sequence?
2. How would the sequence change if the common ratio was positive?
3. What is the sum of the first 10 terms of this sequence?
4. Can you determine the sum to infinity if the sequence was convergent?
5. How would the solution change if the third term was 16 instead of 8?

Tip: Understanding the properties of geometric sequences, especially how the common ratio affects the terms, can help in solving a variety of related problems more efficiently.