For which values of $$ x $$ will the geometric series $$ \sum_{i=1}^{\infty} 4(3-x)^{i} $$ converg

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To determine the values of $$ x $$ for which the geometric series

$$
\sum_{i=1}^{\infty} 4(3 - x)^{i}
$$

converges, let's analyze the series step by step.

### Understanding the Geometric Series

A geometric series has the general form:

$$
\sum_{i=0}^{\infty} ar^{i}
$$

where:
- $$ a $$ is the first term,
- $$ r $$ is the common ratio.

Such a series **converges** if and only if the absolute value of the common ratio is less than 1, i.e., $$ |r| < 1 $$. If this condition is met, the series converges to the sum:

$$
S = \frac{a}{1 - r}
$$

Otherwise, the series **diverges**.

### Applying to the Given Series

The given series is:

$$
\sum_{i=1}^{\infty} 4(3 - x)^{i}
$$

Notice that this series starts at $$ i = 1 $$ instead of $$ i = 0 $$. However, the convergence criteria remain the same, focusing on the common ratio $$ r $$.

**Identify the Common Ratio:**
- Here, the common ratio $$ r $$ is $$ (3 - x) $$.

**Convergence Condition:**
- For the series to converge, the absolute value of the common ratio must be less than 1:

$$
|3 - x| < 1
$$

### Solving the Inequality

Let's solve the inequality $$ |3 - x| < 1 $$:

1. **Break down the absolute value inequality:**

$$
-1 < 3 - x < 1
$$

2. **Solve the left part:**

$$
-1 < 3 - x \
-1 - 3 < -x \
-4 < -x \
x < 4 \quad (\text{after multiplying both sides by } -1 \text{ and reversing the inequality})
$$

3. **Solve the right part:**

$$
3 - x < 1 \
- x < 1 - 3 \
- x < -2 \
x > 2 \quad (\text{after multiplying both sides by } -1 \text{ and reversing the inequality})
$$

4. **Combine the two conditions:**

$$
2 < x < 4
$$

### Conclusion

The geometric series $$ \sum_{i=1}^{\infty} 4(3 - x)^{i} $$ **converges** for all real numbers $$ x $$ that satisfy:

$$
2 < x < 4
$$

**Summary:**
- **Converges:** $$ x $$ is between 2 and 4.
- **Diverges:** $$ x \leq 2 $$ or $$ x \geq 4 $$.

### Additional Note (Optional)

If you're interested in finding the sum of the series when it converges, you can use the formula for the sum of a geometric series starting at $$ i = 1 $$:

$$
S = \frac{ar}{1 - r}
$$

In this case:
- $$ a = 4 $$,
- $$ r = (3 - x) $$.

So,

$$
S = \frac{4(3 - x)}{1 - (3 - x)} = \frac{4(3 - x)}{x - 2}
$$

This formula is valid only for $$ 2 < x < 4 $$, where the series converges.
The geometric series $$ \sum_{i=1}^{\infty} 4(3 - x)^{i} $$ converges when $$ 2 < x < 4 $$.

For which values of \( x \) will the geometric series \( \sum_{i=1}^{\infty} 4(3-x)^{i} \) converg

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