A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
As given, if \[\alpha ,\beta \] be the roots of the quadratic equation, then \[\alpha +\beta =\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{\alpha }^{2}}{{\beta }^{2}}}\] Þ \[-\frac{b}{a}=\frac{({{b}^{2}}/{{a}^{2}})-(2c/a)}{({{c}^{2}}/{{a}^{2}})}=\frac{{{b}^{2}}-2ac}{{{c}^{2}}}\] Þ \[\frac{2a}{c}=\frac{{{b}^{2}}}{{{c}^{2}}}+\frac{b}{a}=\frac{(a{{b}^{2}}+b{{c}^{2}})}{a{{c}^{2}}}\] Þ \[2{{a}^{2}}c=a{{b}^{2}}+b{{c}^{2}}\,\,\Rightarrow \frac{2a}{b}=\frac{b}{c}+\frac{c}{a}\] Þ \[\frac{c}{a},\frac{a}{b},\frac{b}{c}\]are in A.P. Þ \[\frac{a}{c},\frac{b}{a},\frac{c}{b}\]are in H.P.
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