A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
As given, \[{{b}^{2}}=ac\]Þ equation \[a{{x}^{2}}+2bx+c=0\]can be written as \[a{{x}^{2}}+2\sqrt{ac}x+c=0\] Þ \[{{(\sqrt{a}x+\sqrt{c})}^{2}}=0\] Þ \[x=-\sqrt{\frac{c}{a}}\] (repeated root) This must be the common root by hypothesis. So it must satisfy the equation \[d{{x}^{2}}+2ex+f=0\] Þ \[d\frac{c}{a}-2e\sqrt{\frac{c}{a}}+f=0\] Þ \[\frac{d}{a}+\frac{f}{c}=\frac{2e}{c}\sqrt{\frac{c}{a}}=\frac{2e}{b}\] Þ \[\frac{d}{a},\frac{e}{b},\frac{f}{c}\]are in A.P.
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