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?

hashtagmath Dec 27, 2018

#2**+3 **

I am gong to change the sequences as follows

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

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

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

Melody Dec 28, 2018

#2**+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