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Question
If x, y, z are in continued proportion prove that `(x + y)^2/(y + z)^2 = x/z`
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Solution 1
∵ x, y, z are in continued proportion
`∴ x/y = y/z`
`=> y^2 = zx` ...(1)
`=> (x + y)/y = (y + z)/z` ...(By componendo)
`=> (x + y)/(y + z) = y/z` ...(By alternendo)
`=> (x + y)^2/(y + z)^2 = y^2/z^2` ...(squaring both sides)
`=> (x + y)^2/(y + z)^2 = (zx)/z^2` ...[from (1)]
`=> (x + y)^2/(y + z)^2 = x/z`
Hence Proved.
Solution 2
Since, x, y, z are in continued proportion
`therefore x/y = y/z = k`
⇒ y = zk and x = yk = zk2
`L.H.S = (c+y)^2/(y+z)^2`
`(zk^2 + zk)^2/(zk+z)^2`
`(z^2k^4 + z^2k^2 + 2z^2k^3)/(z^2k^2+z^2+2z^2k)`
`(z^2k^2(k^2+1+2k))/(z^2(k^2+1+2k))`
k2
R.H.S. = `x/z`
`= zk^2/z = k^2`
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