A) AP
B) GP
C) HP
D) None of these
Correct Answer: A
Solution :
Since, roots of the equation \[(a-b){{x}^{2}}+(c-a)x+(b-c)=0\]are equal \[\therefore \] \[D={{b}^{2}}-4ac=0\] \[\Rightarrow \] \[{{(c-a)}^{2}}-4(a-b)(b-c)=0\] \[\Rightarrow \] \[{{a}^{2}}+4{{b}^{2}}+{{c}^{2}}+2ac-4ab-4bc=0\] \[\Rightarrow \] \[{{(a+c-2b)}^{2}}=0\] \[\Rightarrow \] \[a+c=2b\] Hence a, b, c are in AP.You need to login to perform this action.
You will be redirected in
3 sec