Question

find three number in GP whose sum is 19 and product is 216

Answer 1

Let the numbers be a/r, a,ar 

a/r + a +r =19 
a/r*a*ar=219 
a^3= 216 
a=6 
a+ar+ar^2 = 19r 

6+6r+6r^2=19r 
6r^2-13r+6 =0 
6r^2-9r-4r+6=0 
3r(2r-3)-2(2r-3)=0 

(2r-3)(3r-2)=0 
r=3/2 OR r=2/3 
plug a & r 
when r=3/2 
a/r = 6/(3/2) =4 
a=6 
ar = 9 
4,6,9 
OR r=2/3 
a/r = 6/(2/3) =9 
a= 6 
ar = 4 
9,6,4 
plz mark as brilliant

Answer 2

Let three numbers in GP be a/r,a, and ar.
a/r +a+ar=19........(1)
a/r×a×ar=216
a^3=216
a=6.......(2)
from (1)6/r+6+6r=19
6+6r+6r^2=19r
6r^2-13r+6=0
6^2-9r-4r+6=0
3r(2r-3)-2(2r-3)=0
(3r-2)(2r-3)=0
r=2/3,3/2.
case1): a=6,r=2/3. then numbers :9,6,4
case2):a=6,r=3/2 then numbers 4,6and 9.