s8 = 3·r7 and s4 = 3·r3 so s4 / s8 = 1 / r4 = 1 / 17 and r = 4√17
3 , 3 · 4√17 , 3 · 2√17 , 3 · 4√173
(Btw the 5th term is the next after the first to be rational: s5 = 51.)
Josh F.
tutor
This is a fascinating question because my initial thought was no, but it made me think harder about the question and it led me in this direction:
To begin with, my solution above is incomplete because although the question asks for “THE first 3 terms,” the given information allows us to find 2 distinct geometric series that satisfy the conditions.
The 2nd solution uses a - value for r, namely r = - 4th rt. Of 17. This will generate an ALTERNATING geometric sequence, still having the 1st term = 3, and still with the correct ratio of the 4th and 8th terms.
From there, if we want to allow complex ratios, we can find still 2 MORE sequences that satisfy the givens, whose r-values are both pure imaginary numbers, namely + or - 4th rt. of 17 i.
Geometric sequences with complex ratios are of considerable interest and use in mathematics, and graphing them in the complex plane leads to spirals and other concepts.
If we imagine a path of infinitely many steps in the complex plane, we can generate an infinite sequence with a finite sum, meaning the path will terminate at a specific point in the complex plane provided the complex number we are multiplying by has a norm < 1.
Finally, returning to your question about quadratic formula, yes, we can use QF to find solutions, though it is not a very natural approach:
r^4 = 17
r^4 - 17 = 0
(r^2 + sqrt17)(r^2 - sqrt17) = 0
Set each quadratic factor = 0 and solve using quadratic formula.
Thanks for the cool question!
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02/11/21
Trill D.
Would there be a way to solve this problem using Quadratic Formula?02/11/21