If the roots of the equation $x^3-12x^2+39x-28=0$ are in A.P., then common difference is

$S_1=\alpha -\beta +\alpha +\alpha +\beta =12$

$\begin{array}{l}3\alpha =12\\ \therefore \alpha =4\\ \alpha ({\alpha }^2-{\beta }^2)=28\end{array}$

Take $4(16-{\beta }^2)=28\Rightarrow {\beta }^2=9\Rightarrow \beta =\pm 3$

Take  $\alpha =4,\beta =3\Rightarrow$the roots are $1,4,7$

$\Rightarrow$common difference is 3