In a geometric progression, T2 = 24, T3 = 12(x+1), and S3 = 76. Find x.

For the second term:

$$T_2 = ar = 24 \quad (1)$$

For the third term:

$$T_3 = ar^2 = 12(x+1) \quad (2)$$

Divide equation (2) by equation (1):

$$\frac{ar^2}{ar} = \frac{12(x+1)}{24}$$ $$r = \frac{x+1}{2}$$

From equation (1), $a = \frac{24}{r}$. Substitute the expression for $r$:

$$a = \frac{24}{\frac{x+1}{2}}$$ $$a = \frac{48}{x+1}$$

The first three terms are $T_1 = a$, $T_2 = ar$, and $T_3 = ar^2$.
We are given $T_1 + T_2 + T_3 = 76$.
Substitute the expressions for $a$, $ar$, and $ar^2$:

$$\frac{48}{x+1} + 24 + 12(x+1) = 76$$

Subtract 24 from both sides:

$$\frac{48}{x+1} + 12(x+1) = 76 - 24$$ $$\frac{48}{x+1} + 12(x+1) = 52$$

To simplify, let $y = x+1$. The equation becomes:

$$\frac{48}{y} + 12y = 52$$

Multiply the entire equation by $y$ (assuming $y \neq 0$, which means $x \neq -1$):

$$48 + 12y^2 = 52y$$

Rearrange into a standard quadratic form:

$$12y^2 - 52y + 48 = 0$$

Divide by 4 to simplify the coefficients:

$$3y^2 - 13y + 12 = 0$$

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$y = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(3)(12)}}{2(3)}$$ $$y = \frac{13 \pm \sqrt{169 - 144}}{6}$$ $$y = \frac{13 \pm \sqrt{25}}{6}$$ $$y = \frac{13 \pm 5}{6}$$

This gives two possible values for $y$:

$$y_1 = \frac{13 + 5}{6} = \frac{18}{6} = 3$$ $$y_2 = \frac{13 - 5}{6} = \frac{8}{6} = \frac{4}{3}$$

Since $y = x+1$, then $x = y-1$.
For $y_1 = 3$:

$$x_1 = 3 - 1 = 2$$

For $y_2 = \frac{4}{3}$:

$$x_2 = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}$$

Both values are valid solutions.

The value of $x$ is $\boxed{2 \text{ or } \frac{1}{3}}$.