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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.

 Jun 25, 2020
edited by Firebolt  Jun 25, 2020
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avatar+122 
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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.

 

We can rewrite these sequences as \(k, ak, a^2k\) and \(k, bk, b^2k\) , since they are geometric sequences. 

We now have:

\(a^2k-b^2k = 3(ak-bk)\), which we can factor to become:

\(k(a^2-b^2)=3k(a-b)\).

We can rewrite:

\(k(a+b)(a-b)=3k(a-b)\)

Dividing by \(k(a-b)\) on both sides, we can get \(a+b=3\)

Since \(a\) is the common ratio of the first sequence, and \(b\) is the common ratio of the second, our answer is simply 3 .

 Jun 25, 2020

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