+95 Let
a + ar + ar^2 + ar^3 + \dotsb
be an infinite geometric series. The sum of the series is 9. The sum of the cubes of all the terms is 36. Find the common ratio.
+130458 We have these two equations
a / ( 1 - r) = 9 → a = 9(1 - r) → a^3 = 729( 1 -r)^3 (1)
a^3 / (1 - r^3) = 36 → a^3 = 36 * (1 -r^3) (2)
Set (1) = (2)
729 ( 1 -r)^3 = 36 (1 - r^3)
729 [ ( 1 -r) (1- r)^2 ] = 36 / [ (1-r) (1 + r + r^2)]
729 (1 - r)^2 = 36 ( 1 + r + r^2)
729 ( r^2 - 2r + 1) = 36 ( 1 + r + r^2)
729r^2 - 1458r + 729 = 36 + 36r + 36r^2
693r^2 -1494r + 693 = 0
Solving for r produces
r = [83 - 8sqrt 15 ] / 77 ≈ .67553
and
r = [ 83 + 8sqrt 15 ] / 77 ≈ 1.4803
Since, in an infinite series
-1 < r < 1....then only the first solution is good
