The first three terms of the sequence $$ 6 ; x ; y ; 16 $$ form an arithmetic progression and the last three terms form a geometric progression. Determine $$ x $$ and $$ y $$. State the arithmetic and geometric sequences for each of your solutions.

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To determine the values of $$ x $$ and $$ y $$ in the sequence $$ 6 ; x ; y ; 16 $$, we need to satisfy two conditions:

1. **Arithmetic Progression (AP):** The first three terms $$ 6, x, y $$ form an arithmetic progression.
2. **Geometric Progression (GP):** The last three terms $$ x, y, 16 $$ form a geometric progression.

### Step 1: Establish the Equations

**Arithmetic Progression:**
For $$ 6, x, y $$ to be in AP, the difference between consecutive terms must be equal.
$$
x - 6 = y - x \implies 2x = y + 6 \implies y = 2x - 6
$$

**Geometric Progression:**
For $$ x, y, 16 $$ to be in GP, the ratio of consecutive terms must be equal.
$$
\frac{y}{x} = \frac{16}{y} \implies y^2 = 16x
$$

### Step 2: Solve the System of Equations

Substitute $$ y = 2x - 6 $$ into $$ y^2 = 16x $$:
$$
(2x - 6)^2 = 16x \
4x^2 - 24x + 36 = 16x \
4x^2 - 40x + 36 = 0 \
x^2 - 10x + 9 = 0
$$
Solving the quadratic equation:
$$
x = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm 8}{2}
$$
Thus, we have two possible solutions:
$$
x = 9 \quad \text{or} \quad x = 1
$$

Now, find the corresponding $$ y $$ for each $$ x $$:

- **If $$ x = 9 $$:**
  $$
  y = 2(9) - 6 = 12
  $$

- **If $$ x = 1 $$:**
  $$
  y = 2(1) - 6 = -4
  $$

### Step 3: Define the Sequences for Each Solution

**1. First Solution: $$ x = 9 $$ and $$ y = 12 $$**

- **Arithmetic Progression (AP):**
  $$
  6, \ 9, \ 12 \quad \text{with common difference} \quad d = 3
  $$
  
- **Geometric Progression (GP):**
  $$
  9, \ 12, \ 16 \quad \text{with common ratio} \quad r = \frac{4}{3}
  $$
  Verification:
  $$
  \frac{12}{9} = \frac{16}{12} = \frac{4}{3}
  $$

**2. Second Solution: $$ x = 1 $$ and $$ y = -4 $$**

- **Arithmetic Progression (AP):**
  $$
  6, \ 1, \ -4 \quad \text{with common difference} \quad d = -5
  $$
  
- **Geometric Progression (GP):**
  $$
  1, \ -4, \ 16 \quad \text{with common ratio} \quad r = -4
  $$
  Verification:
  $$
  \frac{-4}{1} = \frac{16}{-4} = -4
  $$

### Conclusion

There are two possible solutions for $$ x $$ and $$ y $$:

1. $$ x = 9 $$ and $$ y = 12 $$
   - **Arithmetic Progression:** $$ 6, 9, 12 $$ with common difference $$ 3 $$
   - **Geometric Progression:** $$ 9, 12, 16 $$ with common ratio $$ \frac{4}{3} $$

2. $$ x = 1 $$ and $$ y = -4 $$
   - **Arithmetic Progression:** $$ 6, 1, -4 $$ with common difference $$ -5 $$
   - **Geometric Progression:** $$ 1, -4, 16 $$ with common ratio $$ -4 $$

**Final Answer:**

There are two possibilities:

• x = 9 and y = 12
 Arithmetic sequence: 6, 9, 12
 Geometric sequence: 9, 12, 16

• x = 1 and y = –4
 Arithmetic sequence: 6, 1, –4
 Geometric sequence: 1, –4, 16
There are two possible solutions: 1. $$ x = 9 $$ and $$ y = 12 $$ - **Arithmetic Sequence:** 6, 9, 12 - **Geometric Sequence:** 9, 12, 16 2. $$ x = 1 $$ and $$ y = -4 $$ - **Arithmetic Sequence:** 6, 1, -4 - **Geometric Sequence:** 1, -4, 16

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