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An infinite geometric series has first term -3/2 and sums to twice the common ratio. Find the sum of all possible values for the common ratio.

 Jul 23, 2019
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An infinite geometric series has first term -3/2 and sums to twice the common ratio.

Find the sum of all possible values for the common ratio.

 

\(\text{Let $a_1=a=-\dfrac{3}{2} $} \\ \text{Let the common ratio $=r$} \\ \text{Let the sum of the infinite geometric series $= \dfrac{a}{1-r} $ } \)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a}{1-r}} &=& \mathbf{2r} \\\\ a &=& 2r(1-r) \\ a &=& 2r-2r^2 \\ 2r^2-2r+a &=& 0 \quad &|\quad a = -\dfrac{3}{2} \\ 2r^2-2r-\dfrac{3}{2} &=& 0 \quad &|\quad \cdot 2 \\ 4r^2-4r-3 &=& 0 \\\\ r &=& \dfrac{4\pm \sqrt{4^2-4\cdot 4\cdot (-3)} }{2\cdot 4} \\ &=& \dfrac{4\pm \sqrt{4^2+4^2\cdot 3} }{2\cdot 4} \\ &=& \dfrac{4\pm \sqrt{4^2\cdot 4} }{2\cdot 4} \\ &=& \dfrac{4\pm 4\cdot 2 }{2\cdot 4} \\ &=& \dfrac{1\pm 2 }{2} \\\\ r_1 &=& \dfrac{1+ 2 }{2} \\ \mathbf{ r_1 } &=& \mathbf{ \dfrac{3}{2} } \\\\ r_2 &=& \dfrac{1- 2 }{2} \\ \mathbf{ r_2 } &=& \mathbf{ -\dfrac{1}{2} } \\ \hline \end{array} \)

 

The sum of all possible values for the common ratio \(\dfrac{3}{2}-\dfrac{1}{2} = \mathbf{1}\)

 

laugh

 Jul 23, 2019

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