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find the sum of the first 6 terms of the geometric sequence for which a^2=0.7 and a^3=0.49
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\(\dfrac{a^{3}}{a^{2}} = a = \dfrac{0.49}{0.70} = 0.70\) \(a^{1} = 1\) Are you SURE you mean \(a^{3}\), or do you mean \(a_{3}\)?
I meant to put both of the numbers as subscripts. Sorry?
We should know, because it is a geometric sequence, that \(a_{2} * r = a_{3}\). Use this to solve for the common ratio, r.
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