# Geometric series involving logarithms

## 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
$log_{3}(\frac{3}{2} a^{2}+ a^{4}+...+ a^{58})$

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?

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jedishrfu
Mentor
If you replace a^6 by b then you'll notice you have a quadratic b^2 - 28b + 27 = 0 and you should be able to solve for b and by extension for a.

Ok, now I have a=sqrt(3)... But how about the second part? :/

Ok, now I have a=sqrt(3)... But how about the second part? :/
You have the value of a, just plug it in the second part. It's a geometric progress inside the log. :)

But it doesn't look like one if I plugged in...
I got 9/2 + 9 + 27+...

ehild
Homework Helper
The sum is that of a geometric progression + something.
(1/2 a2+a2+a4+....+a58))

ehild

• 1 person
I get ##30-\log_3 {2}##. Is this correct?

ehild
Homework Helper
I get ##30-\log_3 {2}##. Is this correct?
Yes.

ehild