The quadratic equation ( (b-c)x^2 + (c-a)x + (a-b) = 0 ) has equal roots.
Discriminant ( D = 0 ): [ (c-a)^2 - 4(b-c)(a-b) = 0 ]
[ (c-a)^2 = 4(b-c)(a-b) ]
Expand and simplify: [ c^2 - 2ac + a^2 = 4\bigl( ba - b^2 - ca + cb \bigr) ]
After algebraic rearrangement one obtains [ 2b = a + c ]
which is the condition that ( a, b, c ) are in A.P.
Proved.