A) \[\frac{1}{\sqrt{b}}\,\,=\,\,\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\,\]
B) a, b, c are in A. P.
C) \[\sqrt{a},\,\,\sqrt{b},\,\,\sqrt{c}\,\,are\,\,in\,\,A.P.\]
D) \[\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\]
Correct Answer: D
Solution :
\[AB=AC+CB\] \[\sqrt{{{(b+c)}^{2}}\,+\,{{(b-c)}^{2}}}\,\,\,=\,\,\,\sqrt{{{(b-a)}^{2}}\,-\,{{(b-a)}^{2}}\,+}\] \[\sqrt{{{(a+c)}^{2}}\,-\,{{(a-c)}^{2}}}\,\,\,\] Solve \[\frac{1}{\sqrt{a}}\,\,=\,\,\frac{1}{\sqrt{b}}\,\,=\,\,\frac{1}{\sqrt{c}}\,\]You need to login to perform this action.
You will be redirected in
3 sec