A geometric series has a sum to infinity of 3, and \( \frac{2}{3} \) as its second term. Determine the two possible values of the common ratio.
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To solve for the common ratio of the geometric series, we start with the formula for the sum to infinity of a geometric series given by \( S_{\infty} = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. Given that the sum to infinity is 3, we have: \[ \frac{a}{1 - r} = 3 \quad \text{(1)} \] We also know that the second term of the series can be expressed as \( ar \). Given that the second term is \( \frac{2}{3} \), we can write: \[ ar = \frac{2}{3} \quad \text{(2)} \] From equation (2), we can express \( a \) in terms of \( r \): \[ a = \frac{2}{3r} \] Now we can substitute this expression for \( a \) back into equation (1): \[ \frac{\frac{2}{3r}}{1 - r} = 3 \] Cross-multiplying gives us: \[ \frac{2}{3r} = 3(1 - r) \] \[ 2 = 9r(1 - r) \] \[ 2 = 9r - 9r^2 \] Rearranging gives us the quadratic equation: \[ 9r^2 - 9r + 2 = 0 \] We can solve this using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ r = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 9 \cdot 2}}{2 \cdot 9} \] \[ r = \frac{9 \pm \sqrt{81 - 72}}{18} \] \[ r = \frac{9 \pm \sqrt{9}}{18} \] \[ r = \frac{9 \pm 3}{18} \] This results in two possible solutions for \( r \): \[ r_1 = \frac{12}{18} = \frac{2}{3} \quad \text{and} \quad r_2 = \frac{6}{18} = \frac{1}{3} \] Thus, the two possible values of the common ratio are: \[ r = \frac{2}{3} \quad \text{and} \quad r = \frac{1}{3}. \]
