Find the value of x such that

(i) $-\dfrac{2}{7}, x, -\dfrac{7}{2}$ are three consecutive terms of a G.P.

(ii) x + 9, x - 6 and 4 are three consecutive terms of a G.P.

(iii) x, x + 3, x + 9 are first three terms of a G.P.

(i) Since, $-\dfrac{2}{7}, x, -\dfrac{7}{2}$ are three consecutive terms of a G.P.

So,

$\Rightarrow \dfrac{x}{-\dfrac{2}{7}} = r = \dfrac{-\dfrac{7}{2}}{x} \\[1em] \Rightarrow \dfrac{x}{-\dfrac{2}{7}} = \dfrac{-\dfrac{7}{2}}{x} \\[1em] \Rightarrow x^2 = -\dfrac{7}{2} \times -\dfrac{2}{7} \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x^2 - 1 = 0 \\[1em] \Rightarrow (x - 1)(x + 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x + 1 = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = -1.$

Hence, the value of x = 1 or -1.

(ii) Since, x + 9, x - 6 and 4 are three consecutive terms of a G.P.

So,

$\Rightarrow \dfrac{x - 6}{x + 9} = r = \dfrac{4}{x - 6} \\[1em] \Rightarrow \dfrac{x - 6}{x + 9} = \dfrac{4}{x - 6} \\[1em] \Rightarrow (x - 6)^2 = 4(x + 9) \\[1em] \Rightarrow x^2 + 36 - 12x = 4x + 36 \\[1em] \Rightarrow x^2 - 12x - 4x + 36 - 36 = 0 \\[1em] \Rightarrow x^2 - 16x = 0 \\[1em] \Rightarrow x(x - 16) = 0 \\[1em] \Rightarrow x = 0 \text{ or } x - 16 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = 16.$

Hence, the value of x = 0 or 16.

(iii) Since, x, x + 3 and x + 9 are first three terms of a G.P.

So,

$\Rightarrow \dfrac{x + 3}{x} = r = \dfrac{x + 9}{x + 3} \\[1em] \Rightarrow \dfrac{x + 3}{x} = \dfrac{x + 9}{x + 3} \\[1em] \Rightarrow (x + 3)^2 = x(x + 9) \\[1em] \Rightarrow x^2 + 9 + 6x = x^2 + 9x \\[1em] \Rightarrow x^2 - x^2 + 9 + 6x - 9x = 0 \\[1em] \Rightarrow 9 - 3x = 0 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = 3. \\[1em]$

Hence, the value of x = 3.