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## Homework Statement

A geometric series has first term and common ratio both equal to ##a##, where ##a>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 a.

Hence, evaluate

[itex]log_{3}(\frac{3}{2} a^{2}+ a^{4}+...+ a^{58})[/itex]

Giving your answer in the form of ##A-log_{3} B##, where ##A## and ##B## are positive integers to be determined.

## Homework Equations

Geometric series, logarithms

## The Attempt at a Solution

For the first part, I have to write them as

##S_{12} =28 S_{6}##. Then I have to apply the formula for geometric sum...

Then I arrived at ##a^{13} -28a^{7} +27a=0##

Factoring give me ##a(a^{12}-28a^{6}+27)=0##

How should I solve for a in this case?