\( \begin{array}{ll}3.2 & \text { The following sequence forms a convergent geometric sequence: } \\ \begin{array}{ll}5 x ; & x^{2} ; \frac{x^{3}}{5} ; \ldots \\ 3.2 .1 & \text { Determine the possible value of } x \\ 3.2 .2 & \text { If } x=2, \text { calculate the } S_{\infty}\end{array}\end{array} \)

To find the possible value of \( x \) in the geometric sequence \( 5x, x^2, \frac{x^3}{5} \), we need to find a common ratio \( r \). Thus, we can set up the following relationships: 1. \( r = \frac{x^2}{5x} = \frac{x}{5} \) 2. \( r = \frac{\frac{x^3}{5}}{x^2} = \frac{x}{5} \) This shows that \( \frac{x}{5} \) is indeed the common ratio. For the sequence to converge, the common ratio's absolute value must be less than 1, leading us to the inequality \( \left|\frac{x}{5}\right| < 1 \). Therefore, we have the condition \( |x| < 5 \). Next, if \( x = 2 \), we can calculate the sum of the infinite series \( S_{\infty} \). The first term \( a = 5 \times 2 = 10 \) and the common ratio \( r = \frac{2}{5} \). The formula for the sum of a convergent geometric series is: \[ S_{\infty} = \frac{a}{1 - r} = \frac{10}{1 - \frac{2}{5}} = \frac{10}{\frac{3}{5}} = \frac{10 \times 5}{3} = \frac{50}{3} \approx 16.67 \] So, \( S_{\infty} \) when \( x = 2 \) is \( \frac{50}{3} \).