Solution:
According to the question, a, b and c are in GP
Making use of the property of geometric mean, we can write:
b2 = ac
Taking log on both sides with base m, we get:
$ lo{{g}_{m}}~{{b}^{2}}~=\text{ }lo{{g}_{m}}~aclo{{g}_{m}}~ $
$ {{b}^{2}}~=\text{ }lo{{g}_{m}}~a\text{ }+\text{ }lo{{g}_{m}}~c\text{ }\left\{ using\text{ }property\text{ }of\text{ }log \right\} $
$ 2lo{{g}_{m}}~b\text{ }=\text{ }lo{{g}_{m}}~a\text{ }+\text{ }lo{{g}_{m}}~c $
$ 2/lo{{g}_{b}}~m\text{ }=\text{ }1/lo{{g}_{a}}~m\text{ }+\text{ }1/lo{{g}_{c}}~m $
$ \therefore \text{ }1/lo{{g}_{a}}~m\text{ },\text{ }1/lo{{g}_{b}}~m,\text{ }1/lo{{g}_{c}}~m\text{ }are\text{ }in\text{ }A.P. $