Question

If x, y, z are in continued proportion prove that `(x + y)^2/(y + z)^2 = x/z`

Answer 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.

Answer 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`