A) A.P
B) GP.
C) H.P.
D) None of these
Correct Answer: C
Solution :
[c] We have \[2p\left| \frac{0+0-1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}} \right|\Rightarrow \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{4{{p}^{2}}}\] \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{2}{8{{p}^{2}}}\Rightarrow \frac{1}{{{a}^{2}}},\frac{1}{8{{p}^{2}}},\frac{1}{{{b}^{2}}}\] are in A.P. \[\Rightarrow {{a}^{2}},8{{p}^{2}},{{b}^{2}}\] are in H.P.
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