Question

If y is the mean proportional between x and
z; show that xy + yz is the mean proportional
between x2 + y2 and y2 + z2.​

Answer 1

Since y is the mean proportional between x and z

Therefore, y²=xz

Now we have to prove that xy + yz is mean proportional between x² + y² and y² + z²

${x}^{2}+{y}^{2} :xy + yz = xy + yz: {y}^{2} + {z}^{2} \\ \frac{{x}^{2}+{y}^{2}}{xy + yz } = \frac{xy + yz }{{y}^{2} + {z}^{2}} \\ ( xy + yz) ^{2} = ( {x}^{2} + {y}^{2} )( {y}^{2} + {z}^{2} ) \\ LHS \\ ( xy + yz) ^{2} \\ [y(x + z)] ^{2} \\ {y}^{2} (x + z) ^{2} \\ xz(x + z) ^{2} \: \: \: \: \: \: \: \: = > \: \: {y }^{2} = xz \\ RHS \\( {x}^{2} + {y}^{2} )( {y}^{2} + z ^{2} ) \\ ( {x}^{2} + xz)(xz + {z}^{2} ) \\ x(x + z)z(x + z) \\ xz( {x + z})^{2} \\LHS = RHS$

Hence proved

Answer 2

Given : If y is the mean proportion between x and z .

Solution :

According to the question,:-

If y is the mean proportion between x and z i.e. x:y::y:z

$So, y^2=xzy \:$

$We \: have \: to \: show, \\ (xy+yz)^2=(x^2+y^2)(y^2+z^2)(xy+yz) \:$

Taking LHS,

$(xy+yz)^2=x^2y^2+2xy^2z+y^2z^2(xy+yz) \:$

$(xy+yz)^2=x^2xz+2x(xz)z+(xz)z^2(xy+yz) \:$

$xz+2x(xz)z+(xz)z \:$

$(xy+yz)^2=x^3z+2x^2z^2+xz^3(xy+yz) \:$

$(xy+yz)^2=xz(x^2+2xz+z^2)(xy+yz) \:$

Taking RHS,

$(x^2+y^2)(y^2+z^2)=x^2y^2+x^2z^2+y^4+z^2y^2)$

$(x^2+y^2)(y^2+z^2)=x^2(xz)+x^2z^2+(xz)^2+z^2(xz)$

$</p><p>(x^2+y^2)(y^2+z^2)=x^3z+x^2z^2+x^2z^2+xz^3(x </p><p>$

$(x^2+y^2)(y^2+z^2)=x^3z+2x^2z^2+xz^3$

$(x^2+y^2)(y^2+z^2)=xz(x^2+2xz+z^2)$

LHS=RHS