A) \[\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\]
B) a, b, c are in A.P
C) \[\sqrt{a},\sqrt{b},\sqrt{c}\] are in A.P D.
D) \[\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{a}}=+\frac{1}{\sqrt{c}}\]
Correct Answer: A
Solution :
\[a<b<c\]are radii of 3 circles, (say \[{{\text{c}}_{\text{1}}}\text{,}\]\[{{\text{c}}_{\text{2}}}\text{,}\]\[{{\text{c}}_{\text{3}}}\text{,}\]) touching each other externally |
Also \[x-\]axis is their common tangent then Sum of lengths of common tangents of \[{{\text{c}}_{\text{1}}}\,{{\text{c}}_{\text{2}}}\] these and \[{{\text{c}}_{2}}\,{{\text{c}}_{3}}\]= Lengths of the common tangent of \[{{\text{c}}_{2}}\,{{\text{c}}_{3}}\] |
\[\therefore \]\[\sqrt{{{(a+b)}^{2}}\,-\,{{(a-b)}^{2}}+\sqrt{{{(a+c)}^{2}}-{{(a-c)}^{2}}}}\]\[=\,\sqrt{{{(b+c)}^{2}}\,-\,(b-c)}\] |
\[\Rightarrow \] \[\sqrt{ab}+\sqrt{ac}\,=\,\sqrt{bc}\] |
\[\Rightarrow \] \[\frac{1}{\sqrt{c}}\,+\,\frac{1}{\sqrt{b}}\,=\,\frac{1}{\sqrt{a}}.\] |
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