• Support PF! Buy your school textbooks, materials and every day products Here!

Geometric series involving logarithms

  • Thread starter sooyong94
  • Start date
  • #1
173
2

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?
 

Answers and Replies

  • #2
11,816
5,442
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.
 
  • #3
173
2
Ok, now I have a=sqrt(3)... But how about the second part? :/
 
  • #4
3,812
92
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. :)
 
  • #5
173
2
But it doesn't look like one if I plugged in...
I got 9/2 + 9 + 27+...
 
  • #6
ehild
Homework Helper
15,494
1,876
The sum is that of a geometric progression + something.
(1/2 a2+a2+a4+....+a58))

ehild
 
  • Like
Likes 1 person
  • #7
173
2
I get ##30-\log_3 {2}##. Is this correct?
 
  • #8
ehild
Homework Helper
15,494
1,876

Related Threads on Geometric series involving logarithms

Replies
4
Views
690
Replies
2
Views
5K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
6
Views
6K
Replies
5
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
13
Views
4K
  • Last Post
Replies
7
Views
1K
Top