3.2 Given the convergent geometric series: $$ 3^{\prime}+9^{\prime}+27^{\prime}+81^{\prime}+\ldots $$ 3.2 .1 Write down the common ratio in terms of $$ x $$. 3.2 .2 Calculate the value of $$ x $$, if $$ S_{-}=\frac{1}{2} $$.

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### **3.2 Given the convergent geometric series:**
$$ 3^{\prime} + 9^{\prime} + 27^{\prime} + 81^{\prime} + \ldots $$

*Assumption:* The primes (′) indicate exponents involving $$ x $$. Therefore, we can interpret the series as:
$$ 3^x + 9^x + 27^x + 81^x + \ldots $$
This is a geometric series where each term is $$ 3^x $$ times the previous term.

---

### **3.2.1 Write down the common ratio in terms of $$ x $$.**

**Solution:**

A geometric series has the form:
$$ a + ar + ar^2 + ar^3 + \ldots $$
where:
- $$ a $$ is the first term,
- $$ r $$ is the common ratio.

In our series:
- The first term $$ a = 3^x $$.
- The second term is $$ 9^x = (3^2)^x = 3^{2x} $$.

To find the common ratio $$ r $$:
$$
r = \frac{\text{Second term}}{\text{First term}} = \frac{9^x}{3^x} = \left( \frac{9}{3} \right)^x = 3^x
$$

**Answer:**
$$
\boxed{r = 3^{\,x}}
$$

---

### **3.2.2 Calculate the value of $$ x $$, if $$ S_{-} = \frac{1}{2} $$.**

**Given:**
$$ S = \frac{a}{1 - r} = \frac{1}{2} $$
where $$ S $$ is the sum of the infinite series.

**Solution:**

Using the sum formula for a convergent geometric series:
$$
\frac{a}{1 - r} = \frac{1}{2}
$$

Substitute $$ a = 3^x $$ and $$ r = 3^x $$:
$$
\frac{3^x}{1 - 3^x} = \frac{1}{2}
$$

Cross-multiply to solve for $$ 3^x $$:
$$
2 \cdot 3^x = 1 - 3^x
$$
$$
2 \cdot 3^x + 3^x = 1
$$
$$
3 \cdot 3^x = 1
$$
$$
3^{x + 1} = 1
$$

Since $$ 3^0 = 1 $$:
$$
x + 1 = 0 \quad \Rightarrow \quad x = -1
$$

**Answer:**
$$
\boxed{x = -1}
$$

---

### **Verification:**

To ensure our solution is correct, let's verify by substituting $$ x = -1 $$ back into the series.

**Series with $$ x = -1 $$:**
$$
3^{-1} + 9^{-1} + 27^{-1} + 81^{-1} + \ldots = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \ldots
$$

This is a geometric series with:
- $$ a = \frac{1}{3} $$
- $$ r = \frac{1}{3} $$

Sum:
$$
S = \frac{a}{1 - r} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}
$$

Which matches the given $$ S = \frac{1}{2} $$.

---

**Final Answers:**

1. **Common Ratio:**
   $$
   \boxed{r = 3^{\,x}}
   $$

2. **Value of $$ x $$:**
   $$
   \boxed{x = -1}
   $$
**Common Ratio:** $$ r = 3^{x} $$ **Value of $$ x $$:** $$ x = -1 $$

3.2 Given the convergent geometric series: \( 3^{\prime}+9^{\prime}+27^{\prime}+81^{\prime}+\ldots \) 3.2 .1 Write down the common ratio in terms of \( x \). 3.2 .2 Calculate the value of \( x \), if \( S_{-}=\frac{1}{2} \).

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