Calculating determinant of a symmetric matrix where the $k$th row is given by $[a_{k-1},a_k,...,a_0,a_1,...,a_{n-(k-1)}]$

Subtracting from each row the one above it, we shall obtain

$$ \begin{pmatrix} a_{0} & a_{1} & a_{2} & ... & a_{n}\\ d & -d & -d & ... & -d\\ d & d & -d & .... & -d\\ \ldots & \ldots & \ldots & \ldots & \ldots\\ d & d & d & ... & -d \end{pmatrix}$$

Now, subtracting from each column the one before it, we shall obtain

$$ \begin{pmatrix} a_{0} & d & d & ... & d\\ d & -2d & 0 & ... & 0\\ d & 0 & -2d & .... & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots\\ d & 0 & 0 & ... & -2d \end{pmatrix}$$

Next, multiplying columns 2,3,$\ldots$,n+1 by $\frac{1}{2}$ and adding to the first we obtain:

$$ \begin{pmatrix} a_{0}+\frac{nd}{2} & d & d & ... & d\\ 0 & -2d & 0 & ... & 0\\ 0 & 0 & -2d & .... & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots\\ 0 & 0 & 0 & ... & -2d \end{pmatrix}$$

Finally, these row and column operations don't change the determinat of your matrix. Therefore $\det(A)=(a_o+\frac{nd}{2})(-2d)^n$.