The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is `(27)/(19)`. Then, the common ratio of this series is

Options

A. `(4)/(9)`
B. `(2)/(3)`
C. `(2)/(9)`
D. `(1)/(3)`

Answer

B

Explanation

Let the GP be `a, ar, ar^(2), ar^(3),...,oo`, where `a gt 0 and 0 lt r lt 1`
Then, according the problem, we have
`3 = (a)/(1 -r)`
and `(27)/(19) = a^(3) + (ar)^(3) + (ar^(2))^(3) + (ar^(3))^(3) + ..`
`rArr (27)/(19) = (a^(3))/(1 -r^(3)) " " [ :' S_(oo) = (a)/(1 -r)]`
`rArr (27)/(19) = ((3(1 -r))^(3))/(1 -r^(3)) [ :' 3 = (a)/(1 -r) rArr a = 3 (1 - r)]`
`rArr (27)/(19) (27 (1 -r) (1 + r^(2) - 2r))/((1 -r) (1 + r + r^(2))) " " [ :' (1 -r)^(3) = (1 - r) (1 -r)^(2)]`
`rArr r^(2) + r + 1 = 19 (r^(2) - 2r + 1)`
`rArr 18r^(2) - 39 r + 18 = 0`
`rArr 6r^(2) - 13r + 6 = 0`
`rArr (3r -2) (2r -3) = 0`
`:. r = (2)/(3) or r = (3)/(2) " (reject) " [ :' 0lt r lt 1]`