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Let \(k, a_2, a_3 \) and \( k, b_2, b_3\) be nonconstant geometric sequences with different common ratios. If \(a_3-b_3=2(a_2-b_2),\) then what is the sum of the common ratios of the two sequences?

 Dec 27, 2018

Best Answer 

 #2
avatar+99117 
+3

I am gong to change the sequences as follows

 

\(k,\;\;kr_1,\;\;kr_1^2 \qquad and \qquad k,\;\;kr_2,\;\;kr_2^2\\~\\ a_3-b_3\\=k(r_1)^2-k(r_2)^2\\ =k[(r_1)^2-k(r_2)^2]\\=k[(r_1)-(r_2)][(r_1)+(r_2)]\)

 

 

\(a_2-b_2\\ =kr_1-kr_2\\ =k(r_1-r_2)\)

 


\(a_3-b_3=2(a_2-b_2)\\ k[(r_1)-(r_2)][(r_1)+(r_2)]=2[k(r_1-r_2)]\\\\ (r_1)+(r_2)=2\\ \)

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 Dec 28, 2018
 #1
avatar+33 
0

can anyone help?

 Dec 27, 2018
 #2
avatar+99117 
+3
Best Answer

I am gong to change the sequences as follows

 

\(k,\;\;kr_1,\;\;kr_1^2 \qquad and \qquad k,\;\;kr_2,\;\;kr_2^2\\~\\ a_3-b_3\\=k(r_1)^2-k(r_2)^2\\ =k[(r_1)^2-k(r_2)^2]\\=k[(r_1)-(r_2)][(r_1)+(r_2)]\)

 

 

\(a_2-b_2\\ =kr_1-kr_2\\ =k(r_1-r_2)\)

 


\(a_3-b_3=2(a_2-b_2)\\ k[(r_1)-(r_2)][(r_1)+(r_2)]=2[k(r_1-r_2)]\\\\ (r_1)+(r_2)=2\\ \)

Melody Dec 28, 2018
 #3
avatar+98005 
+1

Very nice, Melody!!!

 

cool cool cool

CPhill  Dec 28, 2018
 #4
avatar+99117 
+1

Thanks Chris :)

Melody  Dec 28, 2018

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