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\(k, a_2, a_3\) and \(k, b_2, b_3\) are both nonconstant geometric sequences with different common ratios. We have \(a_3-b_3=3(a_2-b_2).\). Find the sum of the common ratios of the two sequences.

 

 

Im not quite sure how to do this problem.

 Dec 25, 2019
edited by Guest  Dec 25, 2019
edited by Guest  Dec 25, 2019

Best Answer 

 #1
avatar+106887 
+2

\(k, rk, r^2k   \qquad    \qquad      k, Rk, R^2k      \)

 

\(r^2k -  R^2k = 3(rk-Rk)\\ r^2 -  R^2 = 3(r-R)\\ (r-R)(r+R) = 3(r-R)\\ r+R=3 \)

 

The sum of the ratios is 3

 Dec 25, 2019
 #1
avatar+106887 
+2
Best Answer

\(k, rk, r^2k   \qquad    \qquad      k, Rk, R^2k      \)

 

\(r^2k -  R^2k = 3(rk-Rk)\\ r^2 -  R^2 = 3(r-R)\\ (r-R)(r+R) = 3(r-R)\\ r+R=3 \)

 

The sum of the ratios is 3

Melody Dec 25, 2019
 #2
avatar+106516 
+1

Nice, Melody.....!!!

 

cool cool cool

CPhill  Dec 25, 2019

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