If b is positive, what is the value of b in the geometric sequence 16, \(a, 12, b\)?
+9519 Since b is positive, the common ratio \(\dfrac b{12}\) is also positive. Then all the numbers in the sequence are positive.
Using geometric mean, \(a = \sqrt{16 \cdot 12} = 8\sqrt3\).
We can find the common ratio by calculating the second term divided by the first term, so the common ratio is \(\dfrac{8\sqrt 3}{16} = \dfrac{\sqrt 3}2\).
Therefore, \(\dfrac b{12} = \dfrac{\sqrt 3}2\). Multiply both sides by 12 to get \(b = 6\sqrt 3\).
