Question
The sum of the first 6 terms of a geometric sequence is 9 times the sum of the first 3 terms. find the common ratio.
Answers
oobleck
a(r^6 - 1)/(r-1) = 9a(r^3 - 1)/(r-1)
multiply by (r-1)/a to get
r^6 - 1 = 9r^3 - 9
r^6 - 9r^3 + 8 = 0
(r^3-1)(r^3-8) = 0
r = 1 or 2
But r=1 cannot be used, so r=2
multiply by (r-1)/a to get
r^6 - 1 = 9r^3 - 9
r^6 - 9r^3 + 8 = 0
(r^3-1)(r^3-8) = 0
r = 1 or 2
But r=1 cannot be used, so r=2