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

The correct option is A A.P.

We have,

$(b-c)x^2+(c-a)x+(a-b)=0$

Comparing with the quadratic equation

$A{x}^2+Bx+C=0$

$A=(b-c),B=c-a,C=a-b$

Discriminate when roots are equal

$D=B^2-4AC=0$

$D=(c-a{)}^2-4(b-c\left)\right(a-b)=0$

$D=(c^2+a^2-2ac)-4(ba-ac-b^2+bc)=0$

$D=c^2+a^2-2ac-4ab+4ac+4b^2-4bc=0$

$c^2+a^2+2ac-4b(a+c)+4b^2=0$

$(a+c{)}^2-4b(a+c)+4b^2=0$

$\left[\right(a+c)-2b{]}^2=0$

$a+c=2b$

So,

$a,b,c$ are in $A.P$

Hence, this is the answer.