Calculating determinant of a symmetric matrix where the $k$th row is given by $[a_{k-1},a_k,...,a_0,a_1,...,a_{n-(k-1)}]$ 
For $j = 0,...,n$ set $a_{j} = a_{0} + jd$, where $a_{0}, d$ are fixed real numbers. Calculate the determinant of the $(n+1)\times (n+1)$ matrix
  $$A = \begin{pmatrix}
    a_{0}   & a_{1} & a_{2} &  \dots & a_{n}\\
    a_{1} & a_{0}  & a_{1} & \dots & a_{n-1}\\
    a_{2} & a_{1}  & a_{0} & \dots & a_{n-2}\\
    \ldots & \ldots & \ldots & \ldots & \ldots\\
    a_{n} & a_{n-1}  & a_{n-2} & \dots & a_{0} 
\end{pmatrix}.$$

How to calculate that? I haven't found any property of determinant of symmetric matrix which could help.
I've tried to use Gaussian elimination (subtracting each row from the row above it), but it didn't work
Gaussian elimination(subtracting each row from the row above it)  brings to the matrix:
$$\begin{pmatrix}
    -d   & d & d &  ... & d\\
    -d & -d  & d & ... & d\\
    -d & -d  & -d & .... & d\\
    \ldots & \ldots & \ldots & \ldots & \ldots\\
    a_{n} & a_{n-1}  & a_{n-2} & ... & a_{0} 
\end{pmatrix} = d^{n-1} \cdot \begin{pmatrix}
    -1   & 1 & 1 &  ... & 1\\
    -1 & -1  & 1 & ... & 1\\
    -1 & -1  & -1 & .... & 1\\
    \ldots & \ldots & \ldots & \ldots & \ldots\\
    a_{n} & a_{n-1}  & a_{n-2} & ... & a_{0} 
\end{pmatrix}$$

Partial Answer: If we use the matrix determinant lemma, then it suffices to consider the case of $a_0 = 0$ and $d = 1$. 
With that said, let $A_n$ denote the matrix defined with your formula, with $a_0 = 0$ and $d = 1$.   Let $R$ denote an $n \times n$ identity matrix, where the first column has been replaced by $(1,0,\dots,0,n,1-n)$.  Note that 
$$
RA_nR^T = \pmatrix{-2n(n-1) & 0 & 2n\\
0 & A_{n-1} &-\\
2n & |}.
$$
Conclude that $\det(A_n) = -2n(n-1)\det(A_{n-1}) + 4n^2 \det(A_{n-2})$.
