Question

if 1/2y is the arithmetic mean between 1/y-x and 1/y-z prove that y is the Geometric mean between x and z

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \sf{ \frac{1}{2y} \: is \: the \: arithmetic \: mean \: between \: \frac{1}{y - x} \: and \: \frac{1}{y - z} \: \: }$

TO PROVE

y is the Geometric mean between x and z

FORMULA TO BE IMPLEMENTED

1. If c is the arithmetic mean between a and b then

2c = a + b

2. If c is the Geometric mean between a and b then

$\sf{ {c}^{2} = ab \: }$

CALCULATION

Here it is given that

$\displaystyle \sf{ \frac{1}{2y} \: is \: the \: arithmetic \: mean \: between \: \frac{1}{y - x} \: and \: \frac{1}{y - z} \: \: }$

So

$\displaystyle \sf{ \frac{2}{2y} \: \: = \frac{1}{y - x} + \frac{1}{y - z} \: }$

$\implies \: \displaystyle \sf{ \frac{1}{y} \: \: = \frac{1}{y - x} + \frac{1}{y - z} \: }$

$\implies \: \displaystyle \sf{ \frac{1}{y} \: \: - \frac{1}{y - x} = \frac{1}{y - z} \: }$

$\implies \: \displaystyle \sf{ \frac{y - x - y}{y(y - x)} = \frac{1}{y - z} \: }$

$\implies \: \displaystyle \sf{ \frac{- x }{ {y}^{2} - xy} = \frac{1}{y - z} \: }$

$\implies \: \displaystyle \sf{ {y}^{2} - xy = - xy + xz\: }$

$\implies \: \displaystyle \sf{ {y}^{2} = xz\: }$

Therefore y is the Geometric mean between x and z

Hence proved

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