The following sequence forms a convergent geometric sequence. \( 7 x ; x^{2} ; \frac{x^{3}}{7} ; \ldots \) \( 6.1 \quad \) Determine the possible values of \( x \). \( 6.2 \quad \) If \( x=-4 \), calculate \( S_{\infty} \).

**Problem 6.1:** Determine the possible values of \( x \). **Solution:** The given sequence is: \[ 7x, \quad x^{2}, \quad \frac{x^{3}}{7}, \quad \ldots \] For the sequence to be geometric, the ratio between consecutive terms must be constant. Let's denote this common ratio by \( r \). 1. **First Ratio:** \[ r = \frac{\text{Term}_2}{\text{Term}_1} = \frac{x^{2}}{7x} = \frac{x}{7} \] 2. **Second Ratio:** \[ r = \frac{\text{Term}_3}{\text{Term}_2} = \frac{\frac{x^{3}}{7}}{x^{2}} = \frac{x}{7} \] Since both ratios are equal, the sequence is indeed geometric with a common ratio \( r = \frac{x}{7} \). For a geometric sequence to be **convergent**, the absolute value of the common ratio must be less than 1: \[ |r| = \left|\frac{x}{7}\right| < 1 \implies |x| < 7 \] **Answer to 6.1:** All real numbers with absolute value less than seven. That is, all real x with | x | < 7. --- **Problem 6.2:** If \( x = -4 \), calculate \( S_{\infty} \). **Solution:** Given \( x = -4 \), let's first determine the common ratio \( r \): \[ r = \frac{x}{7} = \frac{-4}{7} \] Since \( |r| = \left|\frac{-4}{7}\right| < 1 \), the infinite sum exists. The first term (\( a \)) of the geometric sequence is: \[ a = 7x = 7(-4) = -28 \] The formula for the sum of an infinite geometric series is: \[ S_{\infty} = \frac{a}{1 - r} \] Substituting the known values: \[ S_{\infty} = \frac{-28}{1 - \left(-\frac{4}{7}\right)} = \frac{-28}{1 + \frac{4}{7}} = \frac{-28}{\frac{11}{7}} = -28 \times \frac{7}{11} = \frac{-196}{11} \] **Answer to 6.2:** The infinite sum is –196⁄11.