If 1abc+loga,1abc+logb and 1abc+logc, are in H.P, where a,b,c∈R+, then a,b and c are in
A
A.P.
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B
G.P.
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C
H.P.
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D
None of these
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Solution
The correct option is B G.P. 1abc+loga,1abc+logb,1abc+logc are in H.P. ⇒abc+loga,abc+logb,abc+logc are in A.P. ⇒2(abc+logb)=(abc+loga)+(abc+logc) ⇒2logb=loga+logc ⇒logb2=logac⇒b2=ac