$(x + 3), (x + 6)$ and $(x + 10)$ are in continued proportion. The value of $x$ is:

Given,

$(x + 3), (x + 6)$ and $(x + 10)$ are in continued proportion.

We know that,

Three quantities $a, b$ and $c$ are said to be in continued proportion if the ratio of the first to the second is equal to the ratio of the second to the third.

i.e. $a : b = b : c$

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{c} \\[1em] \Rightarrow b^2 = ac.\end{aligned}$$

So, the square of the middle term is equal to the product of the first and the last terms.

Here,

$a = (x + 3), b = (x + 6)$ and $c = (x + 10)$

∴ $(x + 6)^2 = (x + 3)(x + 10)$

Expanding both sides,

$\Rightarrow x^2 + 12x + 36 = x^2 + 13x + 30$

Cancelling $x^2$ from both sides,

$\Rightarrow 12x + 36 = 13x + 30$

Transposing the like terms,

$$\begin{aligned}\Rightarrow 36 - 30 = 13x - 12x \\[1em] \Rightarrow 6 = x \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, option 2 is the correct option.