To solve this problem, let's define the variables and use the formulas related to geometric series:

Given:

• The first term $a = 2$,
• The nth term $T_n = 486$,
• The sum of the n terms $S_n = 728$.

Step 1: General Formula for the nth Term

The nth term of a geometric series is given by: $T_n = a \cdot r^{n-1}$ where $r$ is the common ratio, and $n$ is the number of terms.

Using $T_n = 486$, we have: $486 = 2 \cdot r^{n-1}$ $r^{n-1} = \frac{486}{2} = 243$ $r^{n-1} = 243$

Step 2: Sum of the First n Terms Formula

The sum of the first $n$ terms of a geometric series is given by: $S_n = \frac{a(1 - r^n)}{1 - r} \quad \text{if} \, r \neq 1$ Substitute the values $S_n = 728$ and $a = 2$: $728 = \frac{2(1 - r^n)}{1 - r}$ Simplify: $364 = \frac{1 - r^n}{1 - r}$

Now, we have two equations:

1. $r^{n-1} = 243$
2. $364 = \frac{1 - r^n}{1 - r}$

Step 3: Solving the System of Equations

Let's solve these two equations to find $r$ and $n$.

I'll calculate this step now.The solution to the system of equations is $r = 3$ and $n = 6$.

Summary:

• The common ratio $r = 3$,
• The number of terms $n = 6$.

If you'd like, I can explain any step in more detail or show how these results fit into the original formulas.

Here are 5 related questions to explore:

1. How would the solution change if the first term were different?
2. What is the general impact of the common ratio $r$ on the sum $S_n$?
3. How can the sum formula be applied to an infinite geometric series?
4. What happens to the geometric series when $r = 1$?
5. Can a geometric series have a negative common ratio?

Tip: When working with geometric series, always check if the common ratio is less than 1, as this leads to special cases, especially for sums to infinity.