A) A. P.
B) G.P.
C) H. P.
D) None of these
Correct Answer: C
Solution :
We have \[2p=\left| \,\frac{0+0-1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}}\, \right|\Rightarrow \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{4{{p}^{2}}}\] Þ \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{2}{8{{p}^{2}}}\Rightarrow \frac{1}{{{a}^{2}}},\frac{1}{8{{p}^{2}}},\frac{1}{{{b}^{2}}}\]are in A. P. Þ \[{{a}^{2}},8{{p}^{2}},{{p}^{2}}\]are in H.P .You need to login to perform this action.
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