If $${a_1},{a_2},{a_3},........,{a_n},.....$$ are in G.P., then the determinant
$$$\Delta \left| {\matrix{
{\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr
{\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr
{\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr
} } \right|$$$
is equal to
Explanation
As $$\,\,\,\,{a_1},{a_2},{a_3},.........$$ are in $$G.P.$$
$$\therefore$$ Using $${a_n} = a{r^{n - 1}},\,\,\,$$ we get the given determinant,
as $$\,\,\,\,\,\,\,\left| {\matrix{
{\log a{r^{n - 1}}} & {\log a{r^n}} & {\log a{r^{n + 1}}} \cr
{\log a{r^{n + 2}}} & {\log a{r^{n + 3}}} & {\log a{r^{n + 4}}} \cr
{\log a{r^{n + 5}}} & {\log a{r^{n + 6}}} & {\log a{r^{n + 7}}} \cr
} } \right|$$
Operating $${C_3} - {C_2}$$ and $${C_2} - {C_1}$$ and using
$$\log m - \log n = \log {m \over n}\,\,\,\,$$ we get
$$ = \left| {\matrix{
{\log a{r^{n - 1}}} & {\log r} & {\log r} \cr
{\log a{r^{n + 2}}} & {\log r} & {\log r} \cr
{\log a{r^{n + 5}}} & {\log r} & {\log r} \cr
} } \right| $$
$$=0$$ (two columns being identical)