A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
\[\alpha +\beta =\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\]\[\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{(\alpha \,\beta )}^{2}}}\] \[=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{(\alpha \,\beta )}^{2}}}\] ?..(i) \[\alpha +\beta =-b/a\]and \[\alpha \beta =c/a\] Putting these value in (i) Þ \[\left( \frac{-b}{a} \right)\,\left( \frac{{{c}^{2}}}{{{a}^{2}}} \right)=\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{2c}{a}\] or \[-b{{c}^{2}}=a{{b}^{2}}-2c{{a}^{2}}\]or \[2c\,{{a}^{2}}=a{{b}^{2}}+b{{c}^{2}}\] Dividing by abc we get, \[\frac{2a}{b}=\frac{b}{c}+\frac{c}{a}\] Þ \[\frac{c}{a},\frac{a}{b},\frac{b}{c}\] are in A.P.
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