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. $