+1577 Let
a + ar + ar^2 + ar^3 + \dotsb
be an infinite geometric series. The sum of the series is 3. The sum of the squares of all the terms is 10. Find the common ratio.
+130081 a / ( 1-r) = 3
a^2 / (1 - r^2) = 10
[ a / (1-r) ] * [ a/ (1+ r) ] =10
[ 3] * [ a / ( 1 + r) ] =10
a/ [ (1+ r)] =10/3
a = (10/3)(1 + r)
So
(10/3)(1 + r) / (1-r) = 3
(10/3) (1+r) = 4 (1 - r)
10/3 + (10/3)r = 4 -4r
(10/3 + 4)r = 4 -10/3
r = [ 4 - 10/3 ]/ [ 4 + 10/3] = [ 2] / [ 22] = 1 / 11
