Find three numbers in gp whose sum is 38 and whose product is 1728
Answers
Answer:
Step-by-step explanation:
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
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Answer:
Let a/r, a, and ar be the three numbers in GP.
Sum, a/r + a + ar = 38 …(i)
Product, (a/r)a(ar) = 1728
a³= 1728
Taking cube root
a = 12
Substitute a in (i)
(12/r) + 12 + 12r = 38
(12/r) + 12r = 26
((1/r) + r) = 26/12
(r² + 1)/ r = 13/6
6r²-13r+6 = 0
Solving using the quadratic formula, we get
r = 2/3or 3/2
The numbers will be 18, 12, 8 or 18, 12, 8.
The greatest number is 18.