1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Using two sums to find geometric sequence

  1. Nov 5, 2012 #1
    Hi. I'm currently tutoring this student with High school math, and I'm completely stumped on this question that he was asked on his test. I'm hoping the community can help me help my student!

    1. The problem statement, all variables and given/known data

    The student was presented with two sums of a geometric sequence (eg, Sum of the first 5 terms and sum of the first 8 terms, both of which indicate that the sequence is not convergent), and he's supposed to find the first term of the sequence as well as the common factor.


    2. Relevant equations

    Sn = a (r^n - 1) / (r-1)
    tn = (t1)(r^(n-1))

    3. The attempt at a solution

    I've Googled and looked through my notes but I can't find anything with this type of example! I thought that maybe I could do a substitution but that didn't really work out...
     
  2. jcsd
  3. Nov 5, 2012 #2
    What are the sums of the sequences?
     
  4. Nov 5, 2012 #3
    I don't know the exact sums, 56 and 84 or something similar? All I need are some pointers as to what to do and I should be able to figure out the rest :) thanks!
     
  5. Nov 5, 2012 #4
    Well,

    Sum(1st 5 terms)=a1(r^4)/(r-1)
    Sum(1st 8 terms)=a1(r^7)/(r-1)

    Sum(1st 5 terms)/Sum(1st 8 terms)=r^-3 so you can find r this way.

    Then just put back into either one of the Sum equations and solve for a1.
     
  6. Nov 5, 2012 #5
    That makes sense. Thanks so much for your help; I really appreciate it!
     
  7. Nov 5, 2012 #6
    Uhh, nvm. It seemed to make sense but when I tested it, it doesn't seem to work! I made my own sequence: 1, 2.5, 6.25, 15.625, 39.0625, so r=2.5. If I divided Sum(5) = 64.4375 by Sum(3) = 9.75, then square root that answer, I would end up with 2.571. Will the r that comes out not be exact?

    EDIT: Unless I followed your instructions incorrectly, I think I understand why r isn't exact (perhaps you can confirm for me?). The larger the number of terms between the two given sums, the closer the calculated r value becomes to the true r value, right?
     
    Last edited: Nov 6, 2012
  8. Nov 6, 2012 #7
    Ah, I wrote my equations wrong,

    Sum(1st 5)=a*((r^5)-1)/(r-1)
    Sum(1st 8)=a*((r^8)-1)/(r-1)

    Sum(1st 5)/Sum(1st 8)=((r^5)-1)/((r^8)-1)

    If you can solve that for r then you have, though it doesn't look like it'll be so easy to do.
     
  9. Nov 6, 2012 #8
    Thanks for your help! I had a feeling this question was deceptively easy... My student is in Grade 11 math, so this question seems inappropriate for their level of math if this is really the only way to go about it.
     
  10. Nov 6, 2012 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In klawlor419's defense, there can be two different formulas for the sum of a geometric series, through the "nth" term depending upon whether you start indexing with "0" or "1" since in the first, the "sum of n terms" will be the sum through the n-1 term.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook