The first three terms of the sequence $$ 6 ; x ; y ; 27 $$ form an arithmetic progression and the last three terms form a geometric progression. Determine the values of $$ x $$ and $$ y $$. State the arithmetic and geometric sequences for each of your solutions.

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Let's analyze the sequence $$ 6, x, y, 27 $$ with the given conditions:

1. **Arithmetic Progression (AP) for the first three terms:**
   - In an AP, the difference between consecutive terms is constant.
   - Thus, $$ x - 6 = y - x $$.
   - Simplifying, we get: $$ 2x = y + 6 $$ or $$ y = 2x - 6 $$.

2. **Geometric Progression (GP) for the last three terms:**
   - In a GP, the ratio between consecutive terms is constant.
   - Thus, $$ \frac{y}{x} = \frac{27}{y} $$.
   - Cross-multiplying, we get: $$ y^2 = 27x $$.
   - Substituting $$ y = 2x - 6 $$ from the AP into this equation:
     $$
     (2x - 6)^2 = 27x \
     4x^2 - 24x + 36 = 27x \
     4x^2 - 51x + 36 = 0
     $$
   - Solving the quadratic equation $$ 4x^2 - 51x + 36 = 0 $$:
     $$
     x = \frac{51 \pm \sqrt{51^2 - 4 \cdot 4 \cdot 36}}{8} = \frac{51 \pm \sqrt{2025}}{8} = \frac{51 \pm 45}{8}
     $$
     - **First solution:** $$ x = \frac{96}{8} = 12 $$.
     - **Second solution:** $$ x = \frac{6}{8} = \frac{3}{4} $$.

3. **Determining $$ y $$ for each $$ x $$:**
   - **For $$ x = 12 $$:**
     $$
     y = 2(12) - 6 = 24 - 6 = 18
     $$
   - **For $$ x = \frac{3}{4} $$:**
     $$
     y = 2\left(\frac{3}{4}\right) - 6 = \frac{3}{2} - 6 = -\frac{9}{2}
     $$

4. **Forming the Sequences:**
   - **First Solution ($$ x = 12 $$, $$ y = 18 $$):**
     - **Arithmetic Progression (AP):** $$ 6, 12, 18 $$ with a common difference of $$ 6 $$.
     - **Geometric Progression (GP):** $$ 12, 18, 27 $$ with a common ratio of $$ \frac{3}{2} $$.
   
   - **Second Solution ($$ x = \frac{3}{4} $$, $$ y = -\frac{9}{2} $$):**
     - **Arithmetic Progression (AP):** $$ 6, \frac{3}{4}, -\frac{9}{2} $$ with a common difference of $$ -\frac{21}{4} $$.
     - **Geometric Progression (GP):** $$ \frac{3}{4}, -\frac{9}{2}, 27 $$ with a common ratio of $$ -6 $$.

**Final Answer:**
There are two possible solutions:

1. $$ x = 12 $$ and $$ y = 18 $$
   - Arithmetic sequence: 6, 12, 18
   - Geometric sequence: 12, 18, 27

2. $$ x = \tfrac{3}{4} $$ and $$ y = -\tfrac{9}{2} $$
   - Arithmetic sequence: 6, ¾, –9⁄2
   - Geometric sequence: ¾, –9⁄2, 27
There are two solutions: 1. $$ x = 12 $$ and $$ y = 18 $$ - Arithmetic sequence: 6, 12, 18 - Geometric sequence: 12, 18, 27 2. $$ x = \frac{3}{4} $$ and $$ y = -\frac{9}{2} $$ - Arithmetic sequence: 6, 0.75, -4.5 - Geometric sequence: 0.75, -4.5, 27

The first three terms of the sequence form an arithmetic progression and the last three terms form a geometric progression.
Determine the values of and .
State the arithmetic and geometric sequences for each of your solutions.

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The first three terms of the sequence form an arithmetic progression and the last three terms form a geometric progression. Determine the values of and . State the arithmetic and geometric sequences for each of your solutions.