If (x − 2), (x + 2), (2x + 1) and (2x + 19) are in proportion, find the value of x.

Let,

∴ x - 2 : x + 2 :: 2x + 1 : 2x + 19

$\Rightarrow \dfrac{x - 2}{x + 2} = \dfrac{2x + 1}{2x + 19} \\[1em] \Rightarrow (x - 2)(2x + 19) = (2x + 1)(x + 2) \\[1em] \Rightarrow 2x^2 + 19x - 4x - 38 = 2x^2 + 4x + x + 2 \\[1em] \Rightarrow 2x^2 + 15x - 38 = 2x^2 + 5x + 2 \\[1em] \Rightarrow 2x^2 - 2x^2 + 15x - 5x = 2 + 38 \\[1em] \Rightarrow 10x = 40 \\[1em] \Rightarrow x = 4.$

Hence, the value of x = 4.