common ratio

**Punch** · June 10th 2011, 01:01 AM

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 $-\frac{1}{2}$

**abhishekkgp** · June 10th 2011, 02:34 AM

Originally Posted by Punch

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 $-\frac{1}{2}$

...

**Punch** · June 10th 2011, 06:06 AM

Originally Posted by abhishekkgp

...

sorry,

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

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

but ans still differs from given ans

**Soroban** · June 10th 2011, 08:20 AM

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: . $r \,=\,\text{-}\tfrac{1}{2}$

**Punch** · June 10th 2011, 08:25 AM

Originally Posted by Soroban

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]

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