Results 1 to 8 of 8

Math Help - common ratio of the seqence

  1. #1
    Junior Member
    Joined
    Jan 2011
    Posts
    47

    common ratio of the seqence

    "the length of the sides in a right triangle form 3 consecutive terms of a geometric sequence. Find the common ratio of the sequence"


    i don't know how to start this problem. please give me an idea how to answer this problem.

    Thanks In Advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68
    Quote Originally Posted by jam2011 View Post
    "the length of the sides in a right triangle form 3 consecutive terms of a geometric sequence. Find the common ratio of the sequence"
    a = short side of triangle (1st term)
    r = ratio

    a^2 + (ar)^2 = (ar^2)^2 (pythagorean theorem)
    a^2 + a^2 r^2 = a^2 r^4
    a^2 r^4 - a^2 r^2 - a^2 = 0
    a^2(r^4 - r^2 - 1) = 0

    Take over...
    Last edited by mr fantastic; March 3rd 2011 at 02:22 AM. Reason: Fixed end quote tag.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    rcs
    rcs is offline
    Senior Member rcs's Avatar
    Joined
    Jul 2010
    From
    iligan city. Philippines
    Posts
    455
    Thanks
    2
    does (r^4 - r^2 - 1) factorable? because (r^2 - 1 ) (r^2 + 1) does not fit to it... how is it done?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68
    Use the quadratic formula.
    The Quadratic Formula Explained
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2011
    Posts
    68
    I guess you substitute r^2 for t(or whatever you'd like to call it), and then solve it like any other quadratic.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    rcs
    rcs is offline
    Senior Member rcs's Avatar
    Joined
    Jul 2010
    From
    iligan city. Philippines
    Posts
    455
    Thanks
    2
    i got it thanks alot
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2011
    Posts
    47
    thanks a lot...Bless You all
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68
    Quote Originally Posted by scounged View Post
    I guess you substitute r^2 for t(or whatever you'd like to call it), and then solve it like any other quadratic.
    Yes; or this way:
    (r^2)^2 - r^2 - 1 = 0

    r^2 = [1 +- SQRT(1 + 4)] / 2

    r = SQRT{[1 +- SQRT(5)] / 2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find the common ratio?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 3rd 2013, 10:31 PM
  2. common ratio
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 10th 2011, 08:25 AM
  3. Common ratio problems
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 3rd 2010, 08:12 PM
  4. [SOLVED] common ratio
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 13th 2009, 09:41 AM
  5. Finding common ratio and more?
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: January 4th 2009, 04:23 PM

Search Tags


/mathhelpforum @mathhelpforum