A) AP
B) GP
C) HP
D) none of these
Correct Answer: A
Solution :
Let \[\alpha ,\beta \]be the roots of the equation \[a{{x}^{2}}+bx+c=0,\] then \[\alpha +\beta =-\frac{b}{a}\]\[\alpha \beta =\frac{c}{a}\] now, \[\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\frac{{{\beta }^{2}}+{{\alpha }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}\] \[=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{\alpha }^{2}}{{\beta }^{2}}}=\frac{\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{2c}{d}}{\frac{{{c}^{2}}}{{{a}^{2}}}}\] \[=\frac{{{b}^{2}}-2ac}{{{c}^{2}}}\] Since, it is given \[\alpha +\beta =\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}\] \[\Rightarrow \] \[-\frac{b}{a}=\frac{{{b}^{2}}-2ac}{{{c}^{2}}}\] \[\Rightarrow \] \[-b{{c}^{2}}=a{{b}^{2}}-2{{a}^{2}}c\] \[\Rightarrow \] \[2{{a}^{2}}c=a{{b}^{2}}+b{{c}^{2}}\] \[\Rightarrow \] \[b{{c}^{2}},c{{a}^{2}},a{{b}^{2}}\]are in AP
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