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(#M40002958) TCS question Keep an EYE

find the smallest number in GP whose sum is 38 and product is 1728

Asked In TCS Nitish Kumar (13 years ago)

Solved saurav kumar sinha Read Solution (2)

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saurav kumar sinha  (13 years ago)

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

Is this solution Helpfull? Yes (17) | No   

ALKA  (13 years ago)

38r^2+1728r+1728=0

solve for r?

Is this solution Helpfull? Yes | No (7)  

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