Let $S$ be the set of all $2\times 2$ real matrices \[M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\] whose entries $a,b,c,d$ (in that order) form an arithmetic progression. Find all matrices $M$ in $S$ for which there is some integer $k>1$ such that $M^k$ is also in $S.$

1. **Define the set \( S \) and the matrix \( M \):**
   Let \( S \) be the set of all \( 2 \times 2 \) real matrices \( M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) whose entries \( a, b, c, d \) form an arithmetic progression. This means there exists a common difference \( r \) such that:
   \[
   b = a + r, \quad c = a + 2r, \quad d = a + 3r
   \]
   Therefore, the matrix \( M \) can be written as:
   \[
   M = \begin{pmatrix} a & a + r \\ a + 2r & a + 3r \end{pmatrix}
   \]

2. **Condition for \( M^k \) to be in \( S \):**
   We need to find all matrices \( M \) in \( S \) for which there exists an integer \( k > 1 \) such that \( M^k \) is also in \( S \). 

3. **Check the determinant and invertibility:**
   Calculate the determinant of \( M \):
   \[
   \det(M) = a(a + 3r) - (a + r)(a + 2r) = a^2 + 3ar - (a^2 + 3ar + 2r^2) = -2r^2
   \]
   Since \( \det(M) = -2r^2 \), \( M \) is invertible if \( r \neq 0 \).

4. **Cayley-Hamilton theorem application:**
   By the Cayley-Hamilton theorem, \( M \) satisfies its characteristic polynomial. The characteristic polynomial of \( M \) is:
   \[
   \lambda^2 - (a + 3r + a) \lambda + \det(M) = \lambda^2 - (2a + 3r) \lambda - 2r^2 = 0
   \]
   Therefore, \( M^2 \) can be expressed as:
   \[
   M^2 = (2a + 3r)M + 2r^2 I
   \]

5. **General form of \( M^k \):**
   Repeated multiplication gives us that \( M^k = \alpha M + \beta I \) for some scalars \( \alpha \) and \( \beta \) which depend on \( k \).

6. **Condition for \( M^k \) to belong to \( S \):**
   If \( M^k \) belongs to \( S \), then it must at least belong to \( T \), where \( T \) is the set of all matrices \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) such that \( a + d = b + c \). Define \( f(M) = a + d - b - c \). For \( M \in S \), \( f(M) = 0 \).

7. **Linear function \( f \):**
   Since \( f \) is a linear function of \( M \), we have:
   \[
   f(\alpha M + \beta I) = \alpha f(M) + \beta f(I) = \alpha \cdot 0 + \beta \cdot 2 = 2\beta
   \]
   For \( M^k \in S \), \( f(M^k) = 0 \), which implies \( \beta = 0 \). Thus, \( M^k = \alpha M \).

8. **Eigenvalues and trace condition:**
   Since \( M^k = \alpha M \) and \( M \) is invertible, we have \( M^{k-1} = \alpha I \). The eigenvalues of \( M \) must be roots of the same real number \( \alpha \). Given \( \det(M) < 0 \), the eigenvalues are real and of opposite sign. The trace of \( M \) must be zero, leading to:
   \[
   a + 3r + a = 0 \implies 2a + 3r = 0 \implies a = -\frac{3r}{2}
   \]

9. **Form of \( M \):**
   Substituting \( a = -\frac{3r}{2} \) into the matrix \( M \):
   \[
   M = \begin{pmatrix} -\frac{3r}{2} & -\frac{3r}{2} + r \\ -\frac{3r}{2} + 2r & -\frac{3r}{2} + 3r \end{pmatrix} = \begin{pmatrix} -\frac{3r}{2} & -\frac{r}{2} \\ \frac{r}{2} & \frac{3r}{2} \end{pmatrix}
   \]
   This matrix is a scalar multiple of \( \begin{pmatrix} -3 & -1 \\ 1 & 3 \end{pmatrix} \).

10. **Verification:**
    Verify that \( M = \begin{pmatrix} -3 & -1 \\ 1 & 3 \end{pmatrix} \) satisfies the conditions:
    \[
    M^2 = \begin{pmatrix} -3 & -1 \\ 1 & 3 \end{pmatrix}^2 = \begin{pmatrix} 8 & 0 \\ 0 & 8 \end{pmatrix} = 8I
    \]
    \[
    M^3 = M \cdot M^2 = \begin{pmatrix} -3 & -1 \\ 1 & 3 \end{pmatrix} \cdot 8I = 8M
    \]
    This works for any odd power \( k \).

The final answer is \( \boxed{ M } \) could be any scalar multiple of \( \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \) with \( k \) being any power (greater than 1), or \( M \) could be any scalar multiple of \( \begin{pmatrix} -3 & -1 \\ 1 & 3 \end{pmatrix} \) with \( k \) being any odd power (greater than 1).