A) \[AP\]
B) \[GP\]
C) \[HP\]
D) None of these
Correct Answer: B
Solution :
\[a,\,\,\,b,\,\,\,c\] are in\[GP\], therefore\[{{b}^{2}}=ac\] Now, \[a{{x}^{2}}+2bx+c=0\] \[\Rightarrow \] \[a{{x}^{2}}+2\sqrt{ac}x+c=0\] \[\Rightarrow \] \[{{(\sqrt{ax}+\sqrt{c})}^{2}}=0\] \[\Rightarrow \] \[x=-\frac{\sqrt{c}}{\sqrt{a}}\] Putting\[x=-\sqrt{\frac{c}{a}}\]in\[d{{x}^{2}}+2ex+f=0\], we get \[d\frac{c}{a}-2e\sqrt{\frac{c}{a}}+f=0\] \[\Rightarrow \] \[\frac{d}{a}-2e\cdot \frac{1}{\sqrt{ac}}+\frac{f}{c}=0\] \[(\because \]dividing both sides by\[c)\]. \[\Rightarrow \] \[\frac{d}{a}-\frac{2e}{b}+\frac{f}{c}=0\] \[[\because \,\,{{b}^{2}}=ac]\] \[\Rightarrow \] \[\frac{d}{a}+\frac{f}{c}=\frac{2e}{b}\] \[\Rightarrow \] \[\frac{d}{a},\,\,\frac{e}{b},\,\,\frac{f}{c}\]are in \[GP\].
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