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find the smallest number in GP whose sum is 38 and product is 1728
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- Let x,y,z be the numbers in geometric progression?
y^2=xz
x+y+z=38
xyz=1728
xyz = xzy = y^2y = y^3 = 1728
y = 12
y^2=xz=144
z=144/x
x+y+z = x+12+144/x = 38
x^2+12x+144=38x
x^2-26x+144=0
(x-18)(x-8)=0
x=8,18
If x =8, z = 38-8-12=18
The numbers are 8,12, 18
Their sum is 38
Their product is 1,728
The smallest number is 8 - 13 years agoHelpfull: Yes(17) No(0)
- 38r^2+1728r+1728=0
solve for r? - 13 years agoHelpfull: Yes(0) No(7)
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