We are given that the third term of a Geometric Progression (GP) is 3, and we need to find the product of the first five terms of the GP.

Let’s denote the terms of the GP:

• Let the first term be $a$.
• Let the common ratio be $r$.

The general formula for the $n$-th term of a GP is: $T_n = a \cdot r^{n-1}$

Given: $T_3 = a \cdot r^2 = 3$

Now, the product of the first five terms of the GP is: $P = T_1 \cdot T_2 \cdot T_3 \cdot T_4 \cdot T_5$

Substituting the terms: $P = (a \cdot a \cdot r) \cdot (a \cdot r^2) \cdot (a \cdot r^3) \cdot (a \cdot r^4)$ This simplifies to: $P = a^5 \cdot r^{10}$

Now, we already know: $a \cdot r^2 = 3$ Therefore: $a^5 \cdot r^{10} = (a \cdot r^2)^5 = 3^5 = 243$

Thus, the product of the first five terms is $243$.

The correct answer is:

$\boxed{243}$

Would you like further details on this, or do you have any questions?

Here are 5 related questions:

1. How do you find the common ratio of a GP if the first term and second term are known?
2. What is the general formula for the sum of the first $n$ terms of a GP?
3. How does the common ratio affect the growth or decay of a GP?
4. Can you solve a GP problem where only the sum of certain terms is given?
5. How does an infinite geometric series differ from a finite geometric series?

Tip: In a GP, the ratio between consecutive terms remains constant, making the geometric sequence predictable when terms are known.