A) GP
B) HP
C) AP
D) None of these
Correct Answer: D
Solution :
\[\because a,b,c\]are in HP, then b \[=\frac{2ac}{(a+c)}\] (i) \[{{4}^{-b}}\ne \frac{1}{2}({{4}^{-a}}+{{4}^{-c}})\] \[{{4}^{-a}},{{4}^{-b}},{{4}^{-c}}\] are not in AP. (ii) \[{{4}^{-2b}}\ne {{4}^{-a}}\times {{4}^{-c}}\] \[\Rightarrow \] \[2b\ne (a+c)\] \[{{4}^{-a}},{{4}^{-b}},{{4}^{-c}}\] are not in GP. \[{{4}^{-b}}\ne \frac{2\cdot {{4}^{-a}}\cdot {{4}^{-c}}}{{{4}^{-a}}+{{4}^{-c}}}\] \[{{4}^{-a}},{{4}^{-b}},{{4}^{-c}}\]are not in HP. Hence, \[{{4}^{-a}},{{4}^{-b}},{{4}^{-c}}\] are neither in AP nor in GP nor in HP.
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