A) \[AP\]
B) \[GP\]
C) \[HP\]
D) None of these
Correct Answer: C
Solution :
Given, equation is \[a{{x}^{2}}+bx+c=0\] If \[\alpha \] and \[\beta \] be the roots of this equation. Then, according to question \[\alpha +\beta =\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}\] \[=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{\alpha }^{2}}{{\beta }^{2}}}\] \[\Rightarrow \] \[\frac{-b}{a}=\frac{\frac{{{b}^{2}}}{{{a}^{2}}}-2\frac{c}{a}}{\frac{{{c}^{2}}}{{{a}^{2}}}}\] \[\Rightarrow \] \[\frac{-b}{a}=\frac{{{b}^{2}}-2ac}{{{c}^{2}}}\] \[\Rightarrow \] \[\frac{2a}{c}=\frac{{{b}^{2}}}{{{c}^{2}}}+\frac{b}{a}=\frac{a{{b}^{2}}+b{{c}^{2}}}{a{{c}^{2}}}\] \[\Rightarrow \] \[2{{a}^{2}}c=a{{b}^{2}}+b{{c}^{2}}\] \[\Rightarrow \] \[\frac{2a}{b}=\frac{b}{c}+\frac{c}{a}\] \[\Rightarrow \] \[\frac{c}{a},\,\,\,\frac{a}{b}\]and\[\frac{b}{c}\]are in\[AP\]. \[\Rightarrow \] \[\frac{a}{c},\,\,\,\frac{b}{a}\]and\[\frac{c}{b}\]are in\[HP\].
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