In a Geometric Progression (GP), the $n$-th term is given by:

$T_n = ar^{n-1}$

where:

• $a$ is the first term,
• $r$ is the common ratio,
• $T_n$ is the $n$-th term.

We are given the following:

• $T_3 = 18$,
• $T_6 = -486$.

We can set up two equations based on this information.

Step 1: Equation for $T_3$

From the formula for the $n$-th term:

$T_3 = ar^{3-1} = ar^2$

Substitute $T_3 = 18$:

$ar^2 = 18 \quad \text{(Equation 1)}$

Step 2: Equation for $T_6$

Similarly, for $T_6$:

$T_6 = ar^{6-1} = ar^5$

Substitute $T_6 = -486$:

$ar^5 = -486 \quad \text{(Equation 2)}$

Step 3: Solving for $r$

Now, divide Equation 2 by Equation 1 to eliminate $a$:

$\frac{ar^5}{ar^2} = \frac{-486}{18}$

Simplifying:

$r^3 = \frac{-486}{18} = -27$

Taking the cube root of both sides:

$r = \sqrt[3]{-27} = -3$

Step 4: Solving for $a$

Now substitute $r = -3$ into Equation 1:

$a(-3)^2 = 18$

$9a = 18$

$a = 2$

Step 5: Finding the 15th term

Now that we have $a = 2$ and $r = -3$, we can find the 15th term $T_{15}$:

$T_{15} = ar^{15-1} = ar^{14}$

$T_{15} = 2(-3)^{14}$

Since $(-3)^{14} = 4782969$, we get:

$T_{15} = 2 \times 4782969 = 9565938$

Final Answers:

• First term $a = 2$,
• Common ratio $r = -3$,
• 15th term $T_{15} = 9565938$.

Would you like further details or explanations?

Here are 5 related questions for practice:

1. Find the 10th term of the same geometric progression.
2. What is the sum of the first 6 terms of this GP?
3. If the common ratio was positive, how would the terms of the progression change?
4. Derive the formula for the sum of the first $n$ terms of a geometric series.
5. How would you find the first negative term in this GP?

Tip: When solving problems involving geometric progressions, it's useful to first express the terms algebraically, and then simplify by eliminating variables through division.