3.2 Consider the infinite geometric series: \( 2(k-5)+2(k-5)^{2}+2(k-5)^{3}+\cdots \) \( 3.2 .1 \quad \) For which value(s) of \( k \), is the series convergent? \( 3.2 .2 \quad \) If \( k=4 \frac{1}{2} \), calculate \( S_{\infty} \) 3.3 Three numbers are in the ratio \( 1: 3: 10 \). If 20 is subtracted from the third number, the numbers form a geometric sequence. Determine the three numbers.

For the infinite geometric series to converge, the common ratio must lie between -1 and 1. In your series, the first term is \( 2(k-5) \) and the common ratio is \( (k-5) \). Therefore, to find the values of \( k \) that make the series convergent, we need the condition: \( -1 < k - 5 < 1 \). Solving this gives us the range \( 4 < k < 6 \). Plugging \( k = 4 \frac{1}{2} \) into the formula, we find \( k - 5 = -\frac{1}{2} \). The first term becomes \( 2(-\frac{1}{2}) = -1 \). The sum of the infinite series is given by the formula \( S_{\infty} = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. Thus, \( S_{\infty} = \frac{-1}{1 - (-\frac{1}{2})} = \frac{-1}{\frac{3}{2}} = -\frac{2}{3}. \) To find the three numbers that are in the ratio \( 1:3:10 \), we can assign the numbers as \( x \), \( 3x \), and \( 10x \). According to the problem, if 20 is subtracted from the third number, the new set of numbers forms a geometric sequence, thus we have \( x, 3x, 10x - 20 \). In a geometric sequence, the square of the middle term should equal the product of the other two terms: \( (3x)^2 = x(10x - 20) \). Solving \( 9x^2 = 10x^2 - 20x \) simplifies to \( x^2 - 20x = 0 \) leading to \( x(x - 20) = 0 \), giving us \( x = 0 \) or \( x = 20 \). Hence, the three numbers are \( 20, 60, \) and \( 200 \).