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 . 4.2.1 Calculate \( r \) if \( r>0 \). 4.2.2 Hence, determine the first THREE terms of the series.

1. We are given that the sum of the infinite geometric series is \[ S = \frac{a}{1-r} = 13.5, \] and the sum from the third term onward is \[ S' = \frac{a\,r^2}{1-r} = 1.5. \] 2. Divide the sum from the third term by the total sum: \[ \frac{S'}{S} = \frac{\frac{a\,r^2}{1-r}}{\frac{a}{1-r}} = r^2. \] Thus, we have \[ r^2 = \frac{1.5}{13.5} = \frac{1}{9}. \] 3. Since \( r > 0 \), we take the positive square root: \[ r = \frac{1}{3}. \] 4. Next, we determine the first term \( a \) from the sum formula: \[ 13.5 = \frac{a}{1 - \frac{1}{3}} = \frac{a}{\frac{2}{3}}. \] Solving for \( a \): \[ a = 13.5 \times \frac{2}{3} = 9. \] 5. Finally, the first three terms of the series are: - First term: \( a = 9 \), - Second term: \( ar = 9 \times \frac{1}{3} = 3 \), - Third term: \( ar^2 = 9 \times \left(\frac{1}{3}\right)^2 = 9 \times \frac{1}{9} = 1 \).