If the product of three numbers in GP be 216 and their sum is 19, then the numbers are

Explanation

Let the three numbers be ` (a)/(r) ,a, ar`
Given , `(a)/(r) xx a xx ar = 216`
`rArr a^(3) = 216` " " `rArr a = 6 `
Also , `(a)/(r) + a + ar = 19`
`rArr (6)/(r) + 6+ 6r = 19` " "` rArr 6r^(2) - 9r - 4r + 6 = 0 `
`rArr 3r (2r -3) -2(2r -3) = 0 ` " "`rArr (3r -2) (2r -3) = 0`
`rArr r = (2)/(3), (3)/(2)`
Taking ` r = (2)/(3), (a)/(r) = 9, a = 6 and ar = 4 `
Taking ` r = (3)/(2) , (1)/(r) = 4 , a = 6 and ar = 9 `
Hence , the numbers are 4, 6, 9 .