A) AP
B) HP
C) GP
D) None of these
Correct Answer: B
Solution :
Since, roots of the equation \[a(b-c){{x}^{2}}+b(c-a)x+c(a-b)=0\] are equal. Thus, \[{{B}^{2}}-4AC=0\] \[\Rightarrow \] \[{{b}^{2}}{{(c-a)}^{2}}-4a(b-c)c(a-b)=0\] \[\Rightarrow \] \[{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{b}^{2}}-2ac{{b}^{2}}\] \[-4ac(ab-{{b}^{2}}-ac+bc)=0\] \[\Rightarrow \] \[{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{b}^{2}}-2ac{{b}^{2}}-4{{a}^{2}}bc+4a{{b}^{2}}c\] \[+4{{a}^{2}}{{c}^{2}}-4ab{{c}^{2}}=0\] \[\Rightarrow \] \[{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+4{{a}^{2}}{{c}^{2}}+2a{{b}^{2}}c\] \[-4{{a}^{2}}bc-4ab{{c}^{2}}=0\] \[\Rightarrow \] \[{{(ab+bc-2ac)}^{2}}=0\] \[\Rightarrow \] \[ab+bc=2ac\] \[\Rightarrow \] \[b\frac{(a+c)}{ac}=2\] \[\Rightarrow \] \[\frac{1}{c}+\frac{1}{a}=\frac{2}{b}\] \[\Rightarrow \] a, b, c are in HP.
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