Explanation

(a)
Since, roots of the equation
$(a - b)x^{2} + (c - a)x + (b - c) = 0\ are\ equal.$
$\therefore\$ Discriminant, $B^{2} - 4AC = 0$
$\Rightarrow (c - a)^{2} - 4(a - b)(b - c) = 0$
$\Rightarrow a^{2} + 4b^{2} + c^{2} + 2ac - 4ab - 4bc = 0$
$\Rightarrow (a + c - 2b)^{2} = 0$
$\Rightarrow a + c = 2b$
Hence, $a,b,c$ are in AP.