$\begin{align}  & \log a+\log ar+\log a{{r}^{2}}+\cdot \cdot \cdot +\log a{{r}^{n-1}} \\ & 1st\text{ term }=\log a \\ & 2nd\text{ term }=\log ar \\ & {{T}_{n}}=\log a{{r}^{n-1}}----(i) \\ & \text{Common difference }=\log ar-\log a=\log r \\ & For\text{ }A.P\text{ sequence} \\ & {{T}_{n}}=a+(n-1)d \\ & {{T}_{n}}=\log a+(n-1)\log r \\ & {{T}_{n}}=\log a+n\log r-\log r \\ & {{T}_{n}}=\log a+\log {{r}^{n}}-\log r \\ & {{T}_{n}}=\log a+\log \frac{{{r}^{n}}}{r}=\log a+\log {{r}^{n-1}} \\ & {{T}_{n}}=\log a{{r}^{n-1}}---(ii) \\ & \text{Since (i) and (ii) are equal, the sequence is an A}\text{.P} \\ & \text{Sum of the sequence} \\ & {{S}_{n}}=\tfrac{n}{2}\left[ {{T}_{1}}+{{T}_{n}} \right] \\ & {{S}_{n}}=\tfrac{n}{2}\left[ \log a+\log a{{r}^{n-1}} \right]=\tfrac{n}{2}\left[ \log (a\times a{{r}^{n-1}}) \right] \\ & {{S}_{n}}=\tfrac{n}{2}[\log {{a}^{2}}{{r}^{n-1}}]=\log {{({{a}^{2}}{{r}^{n-1}})}^{\tfrac{n}{2}}}=\log {{a}^{n}}{{r}^{\tfrac{n(n-1)}{2}}} \\\end{align}$