3.2 The sum of an infinite, geometric series is \( 13,5(r \neq 1) \). The sum of the same series, calculated from the third term is 1,5 . 3.2.1 Calculate \( r \) if \( r>0 \). 3.2.2 Hence, determine the first THREE terms of the series.

To find \( r \), we start with the formula for the sum of an infinite geometric series: \[ S = \frac{a}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio. From the problem, we have: \[ \frac{a}{1 - r} = 13.5 \quad \text{(1)} \] According to the problem, the sum from the third term onward is given as 1.5. The sum from the third term can be calculated as: \[ S' = \frac{ar^2}{1 - r} = 1.5 \quad \text{(2)} \] Now we have two equations (1) and (2): 1. \( \frac{a}{1 - r} = 13.5 \) 2. \( \frac{ar^2}{1 - r} = 1.5 \) From equation (1), we can express \( a \) as: \[ a = 13.5(1 - r) \quad \text{(3)} \] Substituting (3) into (2): \[ \frac{13.5(1 - r)r^2}{1 - r} = 1.5 \] This simplifies to: \[ 13.5r^2 = 1.5 \] Dividing both sides by 1.5 gives: \[ 9r^2 = 1 \] Thus, \[ r^2 = \frac{1}{9} \quad \Rightarrow \quad r = \frac{1}{3} \quad \text{(since \( r > 0 \))} \] Now that we have \( r \), we can determine the first three terms. Using equation (3): \[ a = 13.5(1 - \frac{1}{3}) = 13.5 \cdot \frac{2}{3} = 9 \] Now, the first three terms of the series are: - First term: \( a = 9 \) - Second term: \( ar = 9 \cdot \frac{1}{3} = 3 \) - Third term: \( ar^2 = 9 \cdot \left(\frac{1}{3}\right)^2 = 1 \) Thus, the first three terms of the series are \( 9, 3, 1 \).