G.P

A G.P has first term and common ratio both equal to $ \displaystyle \alpha ,\alpha >1$.Given that the sum of the first 12 terms is 28 times the sum of the first 6 terms , find the exact value of  $ \displaystyle \alpha $.
Hence evaluate $ \displaystyle {{\log }_{3}}(\frac{3}{2}{{\alpha }^{2}}+{{\alpha }^{4}}+{{\alpha }^{6}}+...+{{\alpha }^{{58}}})$ giving your answer in the form of $ \displaystyle A-{{\log }_{3}}B$ , where A and B are positive integers to be determined.

Solution

$ \displaystyle a=r=\alpha \ \ \ ,\ \ \ \alpha >1$

$ \displaystyle {{S}_{{12}}}=28\ {{S}_{6}}$

$ \displaystyle \frac{{a({{r}^{{12}}}-1)}}{{r-1}}=28\times \frac{{a({{r}^{6}}-1)}}{{r-1}}$

$ \displaystyle \ \ \ \ \ \ {{r}^{6}}+1=28$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ {{r}^{6}}\ =27$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ r\ \ =\sqrt{3}\ \ \ \ (\because \alpha >1)$

$ \displaystyle \ \ \ \ \ \ \ \therefore \alpha =\sqrt{3}$

$ \displaystyle {{\log }_{3}}(\frac{3}{2}{{\alpha }^{2}}+{{\alpha }^{4}}+{{\alpha }^{6}}+...+{{\alpha }^{{58}}})$

$ \displaystyle ={{\log }_{3}}(\frac{9}{2}+{{3}^{2}}+{{3}^{3}}+{{3}^{4}}+...+{{3}^{{29}}})$

$ \displaystyle ={{\log }_{3}}(\frac{9}{2}+\frac{{9({{3}^{{28}}}-1)}}{2})$

$ \displaystyle ={{\log }_{3}}\frac{{{{3}^{{30}}}}}{2}$

$ \displaystyle ={{\log }_{3}}{{3}^{{30}}}-{{\log }_{3}}2$

$ \displaystyle =30-{{\log }_{3}}2$