Consider the geometric series: $$ 5-10 x+20 x^{2}-40 x^{3}+\ldots $$ 1.5.1 Determine the value of $$ x $$ that will ensure that the sum conver 1.5.2 Determine the value $$ x $$ for which $$ S_{\infty}=100 $$.

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To solve the problem, we will analyze the given geometric series:

$$
S = 5 - 10x + 20x^2 - 40x^3 + \ldots
$$

This series can be recognized as a geometric series with the first term $$ a = 5 $$ and a common ratio $$ r = -2x $$.

### Step 1: Determine the value of $$ x $$ that will ensure that the sum converges.

A geometric series converges if the absolute value of the common ratio is less than 1:

$$
|r| < 1
$$

Substituting the common ratio:

$$
|-2x| < 1
$$

This simplifies to:

$$
2|x| < 1
$$

Dividing both sides by 2:

$$
|x| < \frac{1}{2}
$$

Thus, the values of $$ x $$ that ensure convergence are:

$$
-\frac{1}{2} < x < \frac{1}{2}
$$

### Step 2: Determine the value of $$ x $$ for which $$ S_{\infty} = 100 $$.

The sum of an infinite geometric series can be calculated using the formula:

$$
S_{\infty} = \frac{a}{1 - r}
$$

Substituting the values of $$ a $$ and $$ r $$:

$$
S_{\infty} = \frac{5}{1 - (-2x)} = \frac{5}{1 + 2x}
$$

We want to find $$ x $$ such that:

$$
\frac{5}{1 + 2x} = 100
$$

To solve for $$ x $$, we first cross-multiply:

$$
5 = 100(1 + 2x)
$$

Expanding the right side:

$$
5 = 100 + 200x
$$

Rearranging the equation gives:

$$
200x = 5 - 100
$$

$$
200x = -95
$$

Dividing both sides by 200:

$$
x = -\frac{95}{200} = -\frac{19}{40}
$$

### Summary of Results

1. The values of $$ x $$ that ensure convergence are $$ -\frac{1}{2} < x < \frac{1}{2} $$.
2. The value of $$ x $$ for which $$ S_{\infty} = 100 $$ is $$ x = -\frac{19}{40} $$.
1. The series converges when $$ -\frac{1}{2} < x < \frac{1}{2} $$. 2. When $$ S_{\infty} = 100 $$, $$ x = -\frac{19}{40} $$.

Consider the geometric series: \( 5-10 x+20 x^{2}-40 x^{3}+\ldots \) 1.5.1 Determine the value of \( x \) that will ensure that the sum conver 1.5.2 Determine the value \( x \) for which \( S_{\infty}=100 \).

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