Since, the given roots are equal
Now, $D=0$
$\Rightarrow (c-a)^{2}-4(a-b)(b-c)=0$
$\Rightarrow c^{2}+a^{2}-2\, c a-4\, a b+4 a c+4 b^{2}=0$
$\Rightarrow c^{2}+a^{2}+2\, a c+4 b^{2}-4 b(c+a)=0$
$\Rightarrow (c+a)^{2}+(2 b)^{2}-2 \cdot 2 b(c+a)=0$
$\Rightarrow [(c+a)-(2 b)]^{2}=0$
$\Rightarrow c+a-2\, b=0$
$\Rightarrow 2\, b=a+c$
Hence, $a, b$ and $c$ are in $A P$.