A) A.P.
B) H.R
C) G.P.
D) None of these
Correct Answer: A
Solution :
[a] \[\because a,b,c\]be in G.P. \[\therefore {{b}^{2}}=ca.\] \[\because a{{x}^{2}}+2bx+c=0\] \[a{{x}^{2}}+2\sqrt{ca}.x+c=0\] \[\Rightarrow \]\[{{(\sqrt{a}x)}^{2}}+2.\sqrt{c}.\sqrt{a}x+{{(\sqrt{c})}^{2}}=0\] \[{{\left( \sqrt{a}x+\sqrt{c} \right)}^{2}}=0\therefore x=-\sqrt{\frac{c}{a}}\] According to question, \[a{{x}^{2}}+2bx+c=0\]and \[d{{x}^{2}}+2ex+f=0\] have common root. \[\therefore x=-\sqrt{\frac{c}{a}}\]must satisfy the equation \[d{{x}^{2}}+2ex+f=0\therefore d\left( \frac{c}{a} \right)+2.e\left( -\sqrt{\frac{c}{a}} \right)+f=0\]Dividing by c on both sides, we have \[\Rightarrow \frac{d}{a}-\frac{2e}{\sqrt{ca}}+\frac{f}{c}=0\] \[\Rightarrow \frac{d}{a}-2\frac{e}{b}+\frac{f}{c}=0\Rightarrow 2\frac{e}{b}=\frac{f}{c}+\frac{d}{a}\] Hence \[\frac{d}{a},\frac{e}{b},\frac{f}{c}\]be in A.P. Hence, option [a] is correct
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