The third term of a geometric sequence is 6 and the sixth term is $$ \frac{3}{32} $$. Determine the first three terms and the general term $$ \mathrm{T}_{m} $$ The first term of a geometric sequence is 3 and the sum of the second and thir terms is 60 . Determine the first three terms and the general term $$ \mathrm{T}_{n} $$. The first three terms of the sequence $$ 6 ; x ; y ; 27 $$ form an arithmetic progression and the last three terms form a geometric progression. .1 Determine the values of $$ x $$ and $$ y $$.

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---

### **Problem 1**

**Given:**
- The third term of a geometric sequence is $$ T_3 = 6 $$.
- The sixth term is $$ T_6 = \frac{3}{32} $$.

**Find:**
- The first three terms.
- The general term $$ T_m $$.

**Solution:**

In a geometric sequence, each term is obtained by multiplying the previous term by a common ratio $$ r $$. The $$ m $$-th term is given by:
$$ T_m = a \cdot r^{m-1} $$
where $$ a $$ is the first term.

Given:
$$
\begin{cases}
T_3 = a \cdot r^{2} = 6 \
T_6 = a \cdot r^{5} = \frac{3}{32}
\end{cases}
$$

**Step 1: Find the common ratio $$ r $$.**

Divide the sixth term by the third term to eliminate $$ a $$:
$$
\frac{T_6}{T_3} = \frac{a \cdot r^{5}}{a \cdot r^{2}} = r^{3} = \frac{\frac{3}{32}}{6} = \frac{3}{32 \times 6} = \frac{1}{64}
$$
$$
r^{3} = \frac{1}{64} \implies r = \left( \frac{1}{64} \right)^{1/3} = \frac{1}{4}
$$

**Step 2: Find the first term $$ a $$.**

Using $$ T_3 = a \cdot r^{2} = 6 $$:
$$
a \cdot \left( \frac{1}{4} \right)^{2} = 6 \implies a \cdot \frac{1}{16} = 6 \implies a = 6 \times 16 = 96
$$

**Step 3: Determine the first three terms.**
$$
\begin{align*}
T_1 &= a = 96 \
T_2 &= a \cdot r = 96 \times \frac{1}{4} = 24 \
T_3 &= 6 \quad (\text{given})
\end{align*}
$$

**Step 4: Write the general term $$ T_m $$.**
$$
T_m = 96 \left( \frac{1}{4} \right)^{m-1} = 96 \cdot 4^{-(m-1)} = \frac{96}{4^{m-1}}
$$

**Answer:**
- **First Three Terms:** $$ 96, \ 24, \ 6 $$
- **General Term:** $$ T_m = 96 \left( \frac{1}{4} \right)^{m-1} $$

---

### **Problem 2**

**Given:**
- The first term of a geometric sequence is $$ a = 3 $$.
- The sum of the second and third terms is $$ T_2 + T_3 = 60 $$.

**Find:**
- The first three terms.
- The general term $$ T_n $$.

**Solution:**

In a geometric sequence:
$$
\begin{align*}
T_1 &= a = 3 \
T_2 &= a \cdot r \
T_3 &= a \cdot r^{2}
\end{align*}
$$
Given:
$$
T_2 + T_3 = 3r + 3r^{2} = 60
$$
Divide both sides by 3:
$$
r + r^{2} = 20 \implies r^{2} + r - 20 = 0
$$

**Step 1: Solve for $$ r $$.**

Using the quadratic formula:
$$
r = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm 9}{2}
$$
$$
r = \frac{8}{2} = 4 \quad \text{or} \quad r = \frac{-10}{2} = -5
$$

**Case 1:** $$ r = 4 $$
$$
\begin{align*}
T_1 &= 3 \
T_2 &= 3 \times 4 = 12 \
T_3 &= 3 \times 4^{2} = 48
\end{align*}
$$
**General Term:**
$$
T_n = 3 \cdot 4^{n-1}
$$

**Case 2:** $$ r = -5 $$
$$
\begin{align*}
T_1 &= 3 \
T_2 &= 3 \times (-5) = -15 \
T_3 &= 3 \times (-5)^{2} = 75
\end{align*}
$$
**General Term:**
$$
T_n = 3 \cdot (-5)^{n-1}
$$

**Answer:**
- **First Three Terms:**
  - If $$ r = 4 $$: $$ 3, \ 12, \ 48 $$
  - If $$ r = -5 $$: $$ 3, \ -15, \ 75 $$
- **General Term:**
  - If $$ r = 4 $$: $$ T_n = 3 \times 4^{n-1} $$
  - If $$ r = -5 $$: $$ T_n = 3 \times (-5)^{n-1} $$

---

### **Problem 3**

**Given:**
- A sequence of four terms: $$ 6; \ x; \ y; \ 27 $$
- The first three terms ($$ 6, x, y $$) form an **arithmetic progression (AP)**.
- The last three terms ($$ x, y, 27 $$) form a **geometric progression (GP)**.

**Find:**
- The values of $$ x $$ and $$ y $$.

**Solution:**

**Condition 1:** The first three terms form an AP.
$$
\text{In an AP}: x - 6 = y - x \implies 2x = y + 6 \implies y = 2x - 6 \quad \text{(Equation 1)}
$$

**Condition 2:** The last three terms form a GP.
$$
\text{In a GP}: \frac{y}{x} = \frac{27}{y} \implies y^{2} = 27x \quad \text{(Equation 2)}
$$

**Substitute Equation 1 into Equation 2:**
$$
(2x - 6)^{2} = 27x \
4x^{2} - 24x + 36 = 27x \
4x^{2} - 51x + 36 = 0
$$

**Solve the quadratic equation:**
$$
x = \frac{51 \pm \sqrt{51^{2} - 4 \cdot 4 \cdot 36}}{2 \cdot 4} = \frac{51 \pm \sqrt{2601 - 576}}{8} = \frac{51 \pm \sqrt{2025}}{8} = \frac{51 \pm 45}{8}
$$

**Possible Solutions:**
1. $$ x = \frac{51 + 45}{8} = \frac{96}{8} = 12 $$
2. $$ x = \frac{51 - 45}{8} = \frac{6}{8} = 0.75 $$

**Find corresponding $$ y $$ values:**

- **If $$ x = 12 $$:**
  $$
  y = 2(12) - 6 = 24 - 6 = 18
  $$
  **Verification in GP:**
  $$
  \frac{18}{12} = 1.5 \quad \text{and} \quad \frac{27}{18} = 1.5
  $$
  **Common Ratio:** $$ r = 1.5 $$

- **If $$ x = 0.75 $$:**
  $$
  y = 2(0.75) - 6 = 1.5 - 6 = -4.5
  $$
  **Verification in GP:**
  $$
  \frac{-4.5}{0.75} = -6 \quad \text{and} \quad \frac{27}{-4.5} = -6
  $$
  **Common Ratio:** $$ r = -6 $$

**Answer:**
- **Possible Values:**
  - $$ x = 12 $$ and $$ y = 18 $$
  - **OR**
  - $$ x = 0.75 $$ and $$ y = -4.5 $$

---
**Problem 1:** - **First Three Terms:** 96, 24, 6 - **General Term:** $$ T_m = 96 \left( \frac{1}{4} \right)^{m-1} $$ **Problem 2:** - **First Three Terms:** - If $$ r = 4 $$: 3, 12, 48 - If $$ r = -5 $$: 3, -15, 75 - **General Term:** - If $$ r = 4 $$: $$ T_n = 3 \times 4^{n-1} $$ - If $$ r = -5 $$: $$ T_n = 3 \times (-5)^{n-1} $$ **Problem 3:** - **Values of $$ x $$ and $$ y $$:** - $$ x = 12 $$, $$ y = 18 $$ - $$ x = 0.75 $$, $$ y = -4.5 $$

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