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Question
Find two numbers whose mean proportional is 16 and the third proportional is 128.
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Solution
Let x and y be two numbers
Their mean proportion = 16
and third proportion = 128
∴ `sqrt(xy)` = 16
⇒ xy = 256
⇒ x = `(256)/y` ....(i)
and `y^2/x` = 128
⇒ x = `y^2/(128)` ....(ii)
From (i) and (ii)
`(256)/y = y^2/(128)`
⇒ y3 = 256 x 128
= 32768
⇒ y3 = (32)3
⇒ y = 32
∴ x = `(256)/y`
= `(256)/(32)`
= 8
∴ Numbers are 8, 32.
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