x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P. Find the value of G.

Given,

x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P.

$\therefore \dfrac{2x + 2}{x} = \dfrac{3x + 3}{2x + 2} \\[1em] \Rightarrow (2x + 2)^2 = x(3x + 3) \\[1em] \Rightarrow (2x)^2 + 2^2 + 2 \times 2x \times 2 = 3x^2 + 3x \\[1em] \Rightarrow 4x^2 + 4 + 8x = 3x^2 + 3x \\[1em] \Rightarrow 4x^2 - 3x^2 + 8x - 3x + 4 = 0 \\[1em] \Rightarrow x^2 + 5x + 4 = 0 \\[1em] \Rightarrow x^2 + 4x + x + 4 = 0 \\[1em] \Rightarrow x(x + 4) + 1(x + 4) = 0 \\[1em] \Rightarrow (x + 1)(x + 4) = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x + 4 = 0 \\[1em] \Rightarrow x = -1 \text{ or } x = -4.$

Substituting value of x = -1, in terms we get :

Terms : -1, 2(-1) + 2, 3(-1) + 3, G

= -1, -2 + 2, -3 + 3, G

= -1, 0, 0, G

This is not possible as in this case common ratio is different.

Substituting value of x = -4, in terms we get :

Terms : -4, 2(-4) + 2, 3(-4) + 3, G

= -4, -8 + 2, -12 + 3, G

= -4, -6, -9, G

Here, common difference = $\dfrac{-6}{-4} = \dfrac{3}{2}$.

G = $-9 \times \dfrac{3}{2} = -\dfrac{27}{2}$.

Hence, G = $-\dfrac{27}{2}$.