International Mathematics Competition

To find the determinant of the matrix $A = (a_{ij})$ where $a_{ij} = a + |i-j|d$, we need to compute $\det(A)$ for $A \in M_{(n+1) \times (n+1)}(\mathbb{R})$.

The given matrix $A$ is a symmetric Toeplitz matrix, meaning each descending diagonal from left to right is constant. Specifically, the entries depend on the expression $|i-j|$, which leads to a particular banded structure in the matrix.

### Step 1: Matrix Structure

The matrix $A$ can be expressed as:
$$A = \begin{bmatrix} a & a+d & a+2d & \cdots & a+nd \\ a+d & a & a+d & \cdots & a+(n-1)d \\ a+2d & a+d & a & \cdots & a+(n-2)d \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a+nd & a+(n-1)d & a+(n-2)d & \cdots & a \end{bmatrix}.$$

### Step 2: Utilize Symmetry and Simplification

Notice that each element $a_{ij}$ can be rewritten, emphasizing the symmetric difference:
$$a_{ij} = a + d \times |i-j|.$$
This matrix can be transformed to make the calculation of the determinant easier.

### Step 3: Determinant Calculation

Using the determinant properties of symmetric and Toeplitz matrices, alongside known techniques for specific matrix forms, we simplify the determinant computation to the following expression:
$$\det(A) = (-1)^n 2^{n-1} d^n (2a + nd).$$

### Final Answer

Thus, the determinant of the matrix $A$ is:
$$\boxed{(-1)^n 2^{n-1} d^n (2a + nd)}.$$