Discussion
The sum of the first two terms of a geometric series is 12 and the third term is 16. Then the common ratio \(r \gt 0\) of the geometric progression is
Answer: C
The common ratio is 2.
With first term \(a\) and ratio \(r\):
\[ \begin{aligned} a + ar &= 12, \quad ar^{2} = 16 \\ \frac{r^{2}}{1 + r} &= \frac{16}{12} = \frac{4}{3} \\ 3r^{2} - 4r - 4 &= 0 \\ (3r + 2)(r - 2) &= 0 \end{aligned} \]
Since \(r \gt 0\), \(r = 2\), which gives \(a = 4\) and the terms 4, 8, 16. Check: \(4 + 8 = 12\) and the third term is 16. The other root, \(-\frac{2}{3}\), is excluded by the condition \(r \gt 0\), and none of the other positive options satisfies the quadratic.
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