1.     In a GP the **T3** is **18** and the **T6** is **-486**. Find the First term, the common ration and the 15th Term

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.

In a GP the T3 is 18 and the T6 is -486. Find the First term, the common ratio, and the 15th Term.

Solve a Geometric Progression problem where T3 = 18 and T6 = -486 to find the first term, common ratio, and 15th term. This detailed step-by-step solution explores the concept of geometric sequences, common ratios, and nth-term formulas. Suitable for Grades 9-12.

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