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Mathematics

If the mean proportion between x and z is y, find the mean proportion between x2 + y2 and y2 + z2.

Ratio Proportion

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Answer

Given,

Mean proportion between x and z is y.

∴xy=yz\therefore \dfrac{x}{y} = \dfrac{y}{z}

⇒ y2 = xz.

Let a be the mean proportion between x2 + y2 and y2 + z2.

∴x2+y2a=ay2+z2⇒a2=(x2+y2)(y2+z2)⇒a2=x2y2+x2z2+y4+y2z2⇒a2=(xy)2+(xz)2+(y2)2+(yz)2⇒a2=(xy)2+(xz)2+(xz)2+(yz)2⇒a2=(xy)2+2(xz)2+(yz)2⇒a2=(xy)2+2(xz)(xz)+(yz)2⇒a2=(xy)2+2y2(xz)+(yz)2⇒a2=(xy+yz)2⇒a=(xy+yz)2=(xy+yz).\therefore \dfrac{x^2 + y^2}{a} = \dfrac{a}{y^2 + z^2} \\[1em] \Rightarrow a^2 = (x^2 + y^2)(y^2 + z^2) \\[1em] \Rightarrow a^2 = x^2y^2 + x^2z^2 + y^4 + y^2z^2 \\[1em] \Rightarrow a^2 = (xy)^2 + (xz)^2 + (y^2)^2 + (yz)^2 \\[1em] \Rightarrow a^2 = (xy)^2 + (xz)^2 + (xz)^2 + (yz)^2 \\[1em] \Rightarrow a^2 = (xy)^2 + 2(xz)^2 + (yz)^2 \\[1em] \Rightarrow a^2 = (xy)^2 + 2(xz)(xz) + (yz)^2 \\[1em] \Rightarrow a^2 = (xy)^2 + 2y^2(xz) + (yz)^2 \\[1em] \Rightarrow a^2 = (xy + yz)^2 \\[1em] \Rightarrow a = \sqrt{(xy + yz)^2} = (xy + yz).

Hence, mean proportion of x2 + y2 and y2 + z2 = (xy + yz).

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