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Let the three numbers in GP be a/r,a,ar
Sum of these numbers are:
a/r +a +ar = 38
a(1+r+r^2) = 38r ---- eq1
Product of these numbers are:
a^3 = 1728
(a)^3 = (12)^3
a = 12
Putting the value of a in eq1 you will get:
12(1+r+r^2) = 38r
6r^2 -13r+6 = 0
(3r-2)(2r-3) = 0
3r-2 = 0 or 2r-3 = 0
3r = 2 or 2r = 3
r = 2/3 or r = 3/2
So the numbers could be 8,12,18 or 18,12,8. When a = 12
Sum of these numbers are:
a/r +a +ar = 38
a(1+r+r^2) = 38r ---- eq1
Product of these numbers are:
a^3 = 1728
(a)^3 = (12)^3
a = 12
Putting the value of a in eq1 you will get:
12(1+r+r^2) = 38r
6r^2 -13r+6 = 0
(3r-2)(2r-3) = 0
3r-2 = 0 or 2r-3 = 0
3r = 2 or 2r = 3
r = 2/3 or r = 3/2
So the numbers could be 8,12,18 or 18,12,8. When a = 12
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answered 8 months ago
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