Find the geometric mean between

(i) $\dfrac{4}{9} \text{ and } \dfrac{9}{4}$.

(ii) 14 and $\dfrac{7}{32}$

(iii) 2a and 8a³

(i) Geometric mean between $\dfrac{4}{9} \text{ and } \dfrac{9}{4}$

$G = \sqrt{\dfrac{4}{9} \times \dfrac{9}{4}} \\[1em] = \sqrt{1} \\[1em] = 1.$

Hence, geometric mean = 1.

(ii) Geometric mean between 14 and $\dfrac{7}{32}$

$G = \sqrt{14 \times \dfrac{7}{32}} \\[1em] = \sqrt{\dfrac{49}{16}} \\[1em] = \dfrac{7}{4}.$

Hence, geometric mean = $\dfrac{7}{4} = 1\dfrac{3}{4}$.

(iii) Geometric mean between 2a and 8a³

$G = \sqrt{2a \times 8a^3} \\[1em] = \sqrt{16a^4} \\[1em] = 4a^2.$

Hence, geometric mean = 4a²