A) G.P.
B) H.P.
C) A.P.
D) None of these
Correct Answer: C
Solution :
[c] Let \[\alpha \] and\[\beta \] be the roots of equation \[a{{x}^{2}}+bx+c=0\] According to question, \[\alpha +\beta =\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}\] \[\Rightarrow \alpha +\beta =\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{\alpha }^{2}}.{{\beta }^{2}}}=\frac{{{\left( \alpha +\beta \right)}^{2}}-2\alpha .\beta }{{{\left( \alpha .\beta \right)}^{2}}}\] \[\Rightarrow \frac{-b}{a}=\frac{{{\left( \frac{-b}{a} \right)}^{2}}-2\left( \frac{c}{a} \right)}{{{\left( \frac{c}{a} \right)}^{2}}}\] \[=\frac{\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{2c}{a}}{\frac{{{c}^{2}}}{{{a}^{2}}}}\Rightarrow \frac{-b}{a}=\frac{{{b}^{2}}-2ac}{{{c}^{2}}}\] \[\Rightarrow -b{{c}^{2}}=a{{b}^{2}}-2{{a}^{2}}c\] \[\Rightarrow a{{b}^{2}}+b{{e}^{2}}=2{{a}^{2}}c\] \[\Rightarrow b{{c}^{2}},c{{a}^{2}}and\text{ }a{{b}^{2}}are\text{ }in\text{ }A.P.\]
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