The problem seems to involve a geometric sequence, where the third term is 8 and the sixth term is 64. We need to find the first term and the common ratio, and then use this information to find the sum of the first six terms.

Step 1: Identify the terms of the geometric sequence

Given:

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

For a geometric sequence, the $n$-th term is given by: $a_n = a_1 \cdot r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio.

Using the third term:

$a_3 = a_1 \cdot r^{3-1} = a_1 \cdot r^2 = 8 \quad \text{(Equation 1)}$

Using the sixth term:

$a_6 = a_1 \cdot r^{6-1} = a_1 \cdot r^5 = 64 \quad \text{(Equation 2)}$

Step 2: Solve for $r$ and $a_1$

We can divide Equation 2 by Equation 1 to eliminate $a_1$: $\frac{a_6}{a_3} = \frac{a_1 \cdot r^5}{a_1 \cdot r^2} = r^{5-2} = r^3$ $\frac{64}{8} = r^3 \quad \Rightarrow \quad 8 = r^3 \quad \Rightarrow \quad r = 2$

Now substitute $r = 2$ back into Equation 1 to find $a_1$: $a_1 \cdot 2^2 = 8 \quad \Rightarrow \quad a_1 \cdot 4 = 8 \quad \Rightarrow \quad a_1 = 2$

So, the first term $a_1 = 2$ and the common ratio $r = 2$.

Step 3: Find the sum of the first 6 terms

The sum $S_n$ of the first $n$ terms of a geometric sequence is given by: $S_n = a_1 \cdot \frac{r^n - 1}{r - 1}$

Substituting $n = 6$, $a_1 = 2$, and $r = 2$: $S_6 = 2 \cdot \frac{2^6 - 1}{2 - 1} = 2 \cdot \frac{64 - 1}{1} = 2 \cdot 63 = 126$

Final Answers:

1. The first term $a_1$ is 2 and the common ratio $r$ is 2.
2. The sum of the first 6 terms is 126.

Would you like further details or have any questions? Here are 5 related questions to expand your understanding:

1. How does the common ratio affect the growth of a geometric sequence?
2. What is the sum of the first 10 terms in this geometric sequence?
3. If the 4th term was known, how would that change the problem?
4. Can you explain the difference between arithmetic and geometric sequences?
5. How would you find the sum of an infinite geometric series?

Tip: In a geometric sequence, the ratio between consecutive terms is always constant, which makes it easier to predict the pattern and calculate sums over a range of terms.