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  1. #1
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    common ratio

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

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

    but answer is $\displaystyle -\frac{1}{2}$
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  2. #2
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    Quote Originally Posted by Punch View Post
    Given that $\displaystyle m, -4, m+15$ are the fourth, sixth and eighth terms of a geometric progression that has a first term which is positive, find the common ratio.

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

    but answer is $\displaystyle -\frac{1}{2}$
    ...
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  3. #3
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    Quote Originally Posted by abhishekkgp View Post
    ...
    sorry,

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

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

    but ans still differs from given ans
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  4. #4
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    Hello, Punch!

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

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

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

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

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


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

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


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

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


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

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Punch!


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

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

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

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


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

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


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

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


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

    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]
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