The zeros of $x^3-12x^2+39x-28$ are in arithmetic progression. The middle zero is:

Write the zeros as $m-d,m$ and $m+d$. Their sum is $3m=12$, so the middle zero is $m=4$ (and indeed $p(4)=64-192+156-28=0$). The product $4(16-d^2)=28$ then gives $d^2=9$, so the zeros are 1, 4 and 7.