A) arithmetic progression
B) geometric progression
C) harmonic progression
D) arithmetico-geometric progression
Correct Answer: C
Solution :
If a, p be the roots of the equation \[\alpha {{x}^{2}}+bx+c+0\], then \[\alpha +\beta =-\frac{b}{\alpha }\] and \[\alpha \beta =\frac{c}{\alpha }\]. Given equation is \[a{{x}^{2}}+bx+c=0\] Let \[\alpha ,\beta \] be the roots of this equation. Then, \[\alpha +\beta =-\frac{b}{a}\] and \[\alpha \beta =\frac{c}{a}\] Also, \[\alpha +\beta =\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}\] \[=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}\] \[\Rightarrow \] \[\alpha +\beta =\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{({{\alpha }^{2}}\beta )}^{2}}}\] \[\Rightarrow \] \[\alpha +\beta ={{\left( \frac{\alpha +\beta }{\alpha \beta } \right)}^{2}}-\frac{2}{\alpha \beta }\] \[\Rightarrow \] \[\left( -\frac{b}{a} \right)={{\left( \frac{-b/a}{c/a} \right)}^{2}}-\frac{2}{c/a}\] \[\Rightarrow \] \[-\frac{b}{a}={{\left( \frac{b}{c} \right)}^{2}}-\frac{2a}{c}\] \[\Rightarrow \] \[\frac{2a}{c}={{\left( \frac{b}{c} \right)}^{2}}+\frac{b}{a}\] \[\Rightarrow \] \[\frac{2a}{c}=\frac{b}{c}\left[ \frac{b}{c}+\frac{c}{a} \right]\] \[\Rightarrow \] \[\frac{2a}{b}=\left( \frac{b}{c}+\frac{c}{a} \right)\] \[\Rightarrow \frac{c}{a},\frac{a}{b},\frac{b}{c}\] are in AP. \[\Rightarrow \frac{a}{c},\frac{b}{a},\frac{c}{b}\] are in HP.
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