The problem states that the second term of a geometric progression (GP) is 12, and the fourth term is 27. We are tasked with finding the common ratio and the first term.

Step-by-Step Solution:

1. General Formula for a GP: The $n$-th term of a GP is given by: $T_n = a \cdot r^{n-1}$ where:

   • $a$ is the first term,
   • $r$ is the common ratio,
   • $n$ is the term number.

2. Using the given information:

   • The second term ($T_2$) is 12: $T_2 = a \cdot r = 12$ (Equation 1)

   • The fourth term ($T_4$) is 27: $T_4 = a \cdot r^3 = 27$ (Equation 2)

3. Divide Equation 2 by Equation 1: $\frac{T_4}{T_2} = \frac{a \cdot r^3}{a \cdot r} = \frac{27}{12}$ Simplify: $r^2 = \frac{27}{12} = \frac{9}{4}$ Therefore: $r = \sqrt{\frac{9}{4}} = \frac{3}{2}$ (Since all terms are positive, we take the positive root.)

4. Substitute $r = \frac{3}{2}$ into Equation 1: $a \cdot \frac{3}{2} = 12$ Solve for $a$: $a = 12 \cdot \frac{2}{3} = 8$

Final Answer:

• The first term ($a$) is: $\boxed{8}$
• The common ratio ($r$) is: $\boxed{\frac{3}{2}}$

Would you like a more detailed explanation or assistance with related concepts?

Related Questions:

1. What is the general formula for the $n$-th term of a geometric progression?
2. How do we derive the sum of the first $n$ terms of a geometric progression?
3. What happens to the GP if the common ratio is negative?
4. How can we verify that $a = 8$ and $r = \frac{3}{2}$ work for all terms of the sequence?
5. Can we solve for the $n$-th term of this GP where $n > 4$?

Tip:

Always verify your solutions by substituting them back into the original problem to ensure they satisfy all given conditions!