A) AP
B) GP
C) HP
D) None of these
Correct Answer: C
Solution :
Since, roots of the equation are equal. \[\therefore \] Discriminant = 0 i.e., \[{{b}^{2}}{{(c-a)}^{2}}-4ac(b-c)(a-b)=0\] \[\Rightarrow \]\[{{b}^{2}}{{(c+a)}^{2}}-4abc(a+c)+4{{a}^{2}}{{c}^{2}}=0\] \[\Rightarrow \]\[{{(b(a+c)-2ac)}^{2}}=0\,\,\Rightarrow \,\,\,b=\frac{2ac}{a+c}\] \[\Rightarrow \]a, b, c are in HP.You need to login to perform this action.
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