Answer
The first three terms of the geometric sequence are 3, 12, and 48 or 3, -15, and 75. The general term is \( T_n = 3 \times 4^{n-1} \) or \( T_n = 3 \times (-5)^{n-1} \). For the sequence 6, \( x \), \( y \), 27, the values are \( x = 12 \), \( y = 18 \) or \( x = 0.75 \), \( y = -4.5 \).
Solution
Let's solve the problem step by step. The problem consists of two distinct parts:
1. **Geometric Sequence Problem**
2. **Sequence with Arithmetic and Geometric Progressions**
---
## 1. Geometric Sequence Problem
**Given:**
- First term (\( a_1 \)) = 3
- Sum of the second and third terms (\( a_2 + a_3 \)) = 60
**Objective:**
- Determine the first three terms.
- Find the general term \( T_n \).
### **Solution:**
Let’s denote the common ratio of the geometric sequence by \( r \).
1. **Express the Terms:**
- \( a_1 = 3 \)
- \( a_2 = a_1 \times r = 3r \)
- \( a_3 = a_2 \times r = 3r^2 \)
2. **Set Up the Equation:**
\[
a_2 + a_3 = 60 \\
3r + 3r^2 = 60
\]
3. **Simplify the Equation:**
\[
r + r^2 = 20 \quad \text{(Divide both sides by 3)} \\
r^2 + r - 20 = 0
\]
4. **Solve the Quadratic Equation:**
\[
r = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm 9}{2}
\]
- **Case 1:** \( r = \frac{8}{2} = 4 \)
- **Case 2:** \( r = \frac{-10}{2} = -5 \)
5. **Find the First Three Terms for Each Case:**
- **Case 1 (r = 4):**
\[
a_1 = 3 \\
a_2 = 3 \times 4 = 12 \\
a_3 = 12 \times 4 = 48
\]
- **Case 2 (r = -5):**
\[
a_1 = 3 \\
a_2 = 3 \times (-5) = -15 \\
a_3 = (-15) \times (-5) = 75
\]
6. **Determine the General Term \( T_n \):**
\[
T_n = a_1 \times r^{n-1}
\]
- **Case 1 (r = 4):**
\[
T_n = 3 \times 4^{n-1}
\]
- **Case 2 (r = -5):**
\[
T_n = 3 \times (-5)^{n-1}
\]
### **Summary of Geometric Sequence:**
- **First Three Terms:**
- **Case 1:** 3, 12, 48
- **Case 2:** 3, -15, 75
- **General Term \( T_n \):**
- **Case 1:** \( T_n = 3 \times 4^{n-1} \)
- **Case 2:** \( T_n = 3 \times (-5)^{n-1} \)
---
## 2. Sequence with Arithmetic and Geometric Progressions
**Given:**
- A sequence: 6, \( x \), \( y \), 27
- **Condition 1:** The first three terms (6, \( x \), \( y \)) form an **Arithmetic Progression (AP)**.
- **Condition 2:** The last three terms (\( x \), \( y \), 27) form a **Geometric Progression (GP)**.
**Objective:**
- Determine the values of \( x \) and \( y \).
### **Solution:**
1. **Arithmetic Progression (AP) Condition:**
In an AP, the difference between consecutive terms is constant.
\[
x - 6 = y - x \\
\Rightarrow 2x = y + 6 \\
\Rightarrow y = 2x - 6 \quad \text{(Equation 1)}
\]
2. **Geometric Progression (GP) Condition:**
In a GP, the ratio between consecutive terms is constant.
\[
\frac{y}{x} = \frac{27}{y} \\
\Rightarrow y^2 = 27x \quad \text{(Equation 2)}
\]
3. **Substitute Equation 1 into Equation 2:**
\[
y = 2x - 6 \\
\Rightarrow (2x - 6)^2 = 27x \\
\]
\[
4x^2 - 24x + 36 = 27x \\
\]
\[
4x^2 - 51x + 36 = 0
\]
4. **Solve the Quadratic Equation:**
\[
x = \frac{51 \pm \sqrt{51^2 - 4 \times 4 \times 36}}{2 \times 4} \\
\]
\[
x = \frac{51 \pm \sqrt{2601 - 576}}{8} \\
\]
\[
x = \frac{51 \pm \sqrt{2025}}{8} \\
\]
\[
x = \frac{51 \pm 45}{8}
\]
- **Solution 1:** \( x = \frac{96}{8} = 12 \)
- **Solution 2:** \( x = \frac{6}{8} = 0.75 \)
5. **Find Corresponding \( y \) Values Using Equation 1:**
- **For \( x = 12 \):**
\[
y = 2(12) - 6 = 24 - 6 = 18
\]
- **For \( x = 0.75 \):**
\[
y = 2(0.75) - 6 = 1.5 - 6 = -4.5
\]
### **Verification:**
- **First Solution (x = 12, y = 18):**
- **AP Check:**
\[
6, 12, 18 \quad \text{with common difference} \ 6
\]
- **GP Check:**
\[
12, 18, 27 \quad \text{with common ratio} \ 1.5 \ (\frac{18}{12} = \frac{27}{18} = 1.5)
\]
- **Second Solution (x = 0.75, y = -4.5):**
- **AP Check:**
\[
6, 0.75, -4.5 \quad \text{with common difference} \ -5.25
\]
- **GP Check:**
\[
0.75, -4.5, 27 \quad \text{with common ratio} \ -6 \ (\frac{-4.5}{0.75} = \frac{27}{-4.5} = -6)
\]
Both solutions satisfy the given conditions.
### **Summary of Sequence Solutions:**
1. **First Solution:**
- \( x = 12 \)
- \( y = 18 \)
- **Sequence:** 6, 12, 18, 27
2. **Second Solution:**
- \( x = 0.75 \)
- \( y = -4.5 \)
- **Sequence:** 6, 0.75, -4.5, 27
---
Reviewed and approved by the UpStudy tutoring team
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