If the roots of $$a(b-c)x^2+b(c-a)x+c(a-b)=0$$​ are equal, then $$a, b, c$$​ are:

Correct option is A

Given:
The quadratic equation $$a(b - c)x^2 + b(c - a)x + c(a - b) = 0$$​  has equal roots.
Concept Used:
For a quadratic equation $$Ax^2 + Bx + C = 0$$​, the condition for equal roots is:
Discriminant (D) = $$B^2 - 4AC = 0$$​​
Solution:
A = a(b - c), 
B = b(c - a), 
C = c(a - b).
​$$D = B^2 - 4AC = \left[b(c - a)\right]^2 - 4 \cdot a(b - c) \cdot c(a - b)$$​
​$$D=b^2c^2 - 2ab^2c + a^2b^2 - 4a^2bc + 4ab^2c + 4a^2c^2 - 4abc^2$$​
​$$D=a^2b^2 + b^2c^2 + 4a^2c^2 + 2ab^2c - 4a^2bc - 4abc^2$$

D = (ab + bc -2ac)² 

As we know, 

ab + bc - 2ac = 0

ab + bc = 2ac 

Dividing by abc 

​$$\frac 1c + \frac 1a = \frac 2b$$  (Since, a,b,c $$\not =0)$$

This shows a, b , c are in harmonic progression

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