Results 1 to 5 of 5

Math Help - common ratio

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    common ratio

    Given that m, -4, m+15 are the fourth, sixth and eighth terms of a geometric progression that has a first term which is positive, find the common ratio.

    \frac{-4}{m}=2r
    r=\frac{-2}{m}

    but answer is -\frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1
    Quote Originally Posted by Punch View Post
    Given that m, -4, m+15 are the fourth, sixth and eighth terms of a geometric progression that has a first term which is positive, find the common ratio.

    \frac{-4}{m}=2r
    shouldn't it be r^2 instead of 2r
    r=\frac{-2}{m}

    but answer is -\frac{1}{2}
    ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by abhishekkgp View Post
    ...
    sorry,

    \frac{-4}{m}=r^2

    r=-\frac{2}{\sqrt{m}}

    but ans still differs from given ans
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,706
    Thanks
    625
    Hello, Punch!

    \text{Given that: }\:m,\:-4,\:m+15\,\text{ are the fourth, sixth and eighth terms of a}
    \text{geometric progression that has a positive first term, find the common ratio.}

    We have: . \begin{Bmatrix}a_4 &=& ar^3 &=&  m & [1] \\ a_6 &=& ar^5 &=& -4 & [2] \\ a_8 &=& ar^7 &=& m+15 & [3] \end{Bmatrix}

    \begin{array}{ccccccccccc}\text{Divide [2] by [1]:} & \dfrac{ar^5}{ar^3} &=& -\dfrac{4}{m} & \Rightarrow & r^2 &=& -\dfrac{4}{m} & [5] \\ \\[-3mm] \text{Divide [3] by [2]:} & \dfrac{ar^7}{ar^5} &=& \dfrac{m+15}{-4} & \Rightarrow & r^2 &=& \dfrac{m+15}{-4} & [5] \end{array}

    Equate [4] and [5]: . -\frac{4}{m} \:=\:\frac{m+15}{-4} \quad\Rightarrow\quad m^2 + 15m - 16 \:=\:0

    . . (m - 1)(m + 16) \:=\:0 \quad\Rightarrow\quad m \:=\:1,\:\text{-}16


    m = 1:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{1} \quad\text{No real solution}

    m = \text{-}16:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{\text{-}16} \quad\Rightarrow\quad r \:=\:\pm\tfrac{1}{2}


    \text{If }m = \text{-}16,\:r = \tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& \text{-}64 \end{Bmatrix} \;\;\text{ but }a_1\text{ is positive.}

    \text{If }m = \text{-}16,\:r = \text{-}\tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& +64 \end{Bmatrix} \;\;\text{ There!}


    Therefore, the common ratio is: . r \,=\,\text{-}\tfrac{1}{2}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by Soroban View Post
    Hello, Punch!


    We have: . \begin{Bmatrix}a_4 &=& ar^3 &=&  m & [1] \\ a_6 &=& ar^5 &=& -4 & [2] \\ a_8 &=& ar^7 &=& m+15 & [3] \end{Bmatrix}

    \begin{array}{ccccccccccc}\text{Divide [2] by [1]:} & \dfrac{ar^5}{ar^3} &=& -\dfrac{4}{m} & \Rightarrow & r^2 &=& -\dfrac{4}{m} & [5] \\ \\[-3mm] \text{Divide [3] by [2]:} & \dfrac{ar^7}{ar^5} &=& \dfrac{m+15}{-4} & \Rightarrow & r^2 &=& \dfrac{m+15}{-4} & [5] \end{array}

    Equate [4] and [5]: . -\frac{4}{m} \:=\:\frac{m+15}{-4} \quad\Rightarrow\quad m^2 + 15m - 16 \:=\:0

    . . (m - 1)(m + 16) \:=\:0 \quad\Rightarrow\quad m \:=\:1,\:\text{-}16


    m = 1:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{1} \quad\text{No real solution}

    m = \text{-}16:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{\text{-}16} \quad\Rightarrow\quad r \:=\:\pm\tfrac{1}{2}


    \text{If }m = \text{-}16,\:r = \tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& \text{-}64 \end{Bmatrix} \;\;\text{ but }a_1\text{ is positive.}

    \text{If }m = \text{-}16,\:r = \text{-}\tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& +64 \end{Bmatrix} \;\;\text{ There!}


    Therefore, the common ratio is: . r \,=\,\text{-}\tfrac{1}{2}

    that was a lot of work, i understood it fully! thanks.

    but just to point out a minor mistake, you should have wrote: substitute in [5] because there isnt an equation [4]
    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 of the seqence
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: March 4th 2011, 08:00 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