\(4 = a + ar + ar^2 + ar^3 + \dotsb\)
\(10 = a^3 + a^3r^3 + a^3r^6 + \dotsb\) Find \(r\)
Infinite geometric series formula: a1 [first term] divided by (1 - r) [1 - common ratio] = \(a_1\over{(1 - r)}\)
Using this piece of information, we can rewrite the equations from earlier:
\(4 = {a\over{(1 - r)}}\); \(4 - 4r = a\)
\(10 = {a^3\over{1 - r^3}}\); \(10 - 10r^3=a^3=(4-4r)^3=-64r^3+192r^2-192r+64\)
\(10 - 10r^3 = -64r^3+192r^2-192r+64\); \(54r^3-192r^2+192r-54=0\); \(54r^3-54 - 192r^2+192r=0\)
Factor by grouping: This part is hard...
\(54(r^3-1)-192r(r-1)=0\)
\(r^3-1=(r-1)(r^2+r+1)\); remember this!!! Factoring difference/sum of cubes is important!!
\(54(r-1)(r^2+r+1)-192(r-1)=0\); \((r - 1)(54r^2+54r-138)=0\)
We know the common ratio cannot be 1, because then the series would not converge.
\(54r^2+54r-138=0\); divide by 54: \(r^2+r-{23\over9}=0\)
By quadratic formula:
\(r = {-1 \pm \sqrt{1+{92\over9}} \over 2}={-1\pm{\sqrt{101}\over3}\over2}=-{1\over2}\pm{\sqrt{101}\over6}\)
please reply if you catch any errors... because this number is ugly :/
