It's given system of equations 
$a_{11}x_1+a_{12}x_2+a_{1n}x_n=b_1$
$a_{21}x_1+a_{22}x_2+a_{2n}x_n=b_2$
                       ..........
$a_{n1}x_1+a_{n2}x_2+a_{nn}x_n=b_n$
such that $a_{11},a_{12},...,a_{1n},b_1,a_{21},a_{22},...,a_{2n},b_2,...,a_{n1},a_{n2},...,a_{nn},b_n,$ form an arithmetic sequence.If system has one solution find it
Given the system of equations:
\[
\begin{cases}
a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1 \\
a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2 \\
\vdots \\
a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n = b_n
\end{cases}
\]
where \(a_{11}, a_{12}, \ldots, a_{1n}, b_1, a_{21}, a_{22}, \ldots, a_{2n}, b_2, \ldots, a_{n1}, a_{n2}, \ldots, a_{nn}, b_n\) form an arithmetic sequence.

Let \(a + k\Delta\) be the arithmetic sequence with \(a = a_{11}\) and \(\Delta = a_{12} - a_{11}\).

1. **Case 1: \(a = \Delta = 0\)**
   - If \(a = 0\) and \(\Delta = 0\), then all coefficients and constants are zero. The system becomes:
     \[
     0 \cdot x_1 + 0 \cdot x_2 + \cdots + 0 \cdot x_n = 0
     \]
     This equation is always true for any \(x_i\). Therefore, any \(x_i\) are solutions.

2. **Case 2: \(a \neq 0\) and \(\Delta = 0\)**
   - If \(\Delta = 0\), all coefficients are equal to \(a\). The system becomes:
     \[
     a(x_1 + x_2 + \cdots + x_n) = b_1
     \]
     Since all rows are identical, the system reduces to:
     \[
     x_1 + x_2 + \cdots + x_n = \frac{b_1}{a}
     \]
     This system has infinitely many solutions unless additional constraints are provided.

3. **Case 3: \(\Delta \neq 0\) and \(n = 1\)**
   - If \(n = 1\), the system is:
     \[
     ax_1 = a + \Delta
     \]
     Solving for \(x_1\):
     \[
     x_1 = 1 + \frac{\Delta}{a}
     \]
     This is a unique solution if \(a \neq 0\).

4. **Case 4: \(\Delta \neq 0\) and \(n > 1\)**
   - Subtracting the first equation from the others, we get:
     \[
     (a_{21} - a_{11})x_1 + (a_{22} - a_{12})x_2 + \cdots + (a_{2n} - a_{1n})x_n = b_2 - b_1
     \]
     Since the coefficients form an arithmetic sequence, the differences \(a_{ij} - a_{1j}\) are constant. The system reduces to:
     \[
     \sum_{i=1}^n x_i = 1
     \]
     and
     \[
     \sum_{k=2}^n (k-1)x_k = n
     \]
     Solving these two equations, we get:
     \[
     x_1 + x_2 + \cdots + x_n = 1
     \]
     and
     \[
     x_2 + 2x_3 + \cdots + (n-1)x_n = n
     \]

     For \(n = 2\):
     \[
     x_1 + x_2 = 1
     \]
     \[
     x_2 = 2
     \]
     Solving these, we get:
     \[
     x_1 = -1, \quad x_2 = 2
     \]

Conclusion:
The system has a unique solution in the case of \(n = 1\) and \(a \neq 0\) and \(\Delta \neq 0\), or \(n = 2\) and \(\Delta \neq 0\).

The final answer is \( \boxed{ x_1 = 1 + \frac{\Delta}{a} } \) for \(n = 1\) and \(x_1 = -1, x_2 = 2\) for \(n = 2\).