The sum of an infinite G.P. is 57 and the sum of their cubes is `9457 ,` find the G.P.

Options

A. `1//3`
B. `2//3`
C. `1//6`
D. none of these

Answer

B

Explanation

Let a be the first term and r the common ratio of the G.P. Then, the sum is given by
`a/(1-r)=57` (1)
Sum of the cubes is 9747. Hence,
`a^(3)+a^(3)r^(3)+a^(3)r^(6)+…=9747`
`rArra^(3)/(1-r^(3))=9747` (2)
dividing the cube of (1) by (2), we get
`(a^(3))/((1-r^(3)))((1-r^(3)))/(a^(3))=((57)^(3))/(9747)`
or `(1-r^(3))/((1-r)^(3))=19`
or `(1+r+r^(2))/((1-r)^(2))=19`
or `18r^(2)-39r+18=0`
or (3r-2)(6r-9)=0
or r=2//3[`becauserne3//2`, because 0`ltabsrlt1` for an infinite G.P]