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If, for positive real numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$, the numbers $\mathrm{x}+\mathrm{y}, 2 \mathrm{y}$ and $\mathrm{y}+\mathrm{z}$ are in harmonic progression, then which one of the following is correct?
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Verified Answer
The correct answer is:
$\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in geometric progression
As given:
$\mathrm{x}+\mathrm{y}, 2 \mathrm{y}$ and $\mathrm{y}+\mathrm{z}$ are in harmonic progression. $2 y=\frac{(x+y)(y+z)}{x+y+y+z}$
$\Rightarrow \quad \frac{1}{2 y}=\frac{1}{x+y}+\frac{1}{y+z}$
$\Rightarrow \quad y(x+2 y+z)=\left(x y+x z+y^{2}+y z\right)$
$\Rightarrow x y+2 y^{2}+y z=x y+x z+y^{2}+y z$
$\Rightarrow \quad y^{2}=x z$
$\Rightarrow \mathrm{x}, \mathrm{y}, \mathrm{z}$ are in geometric progession.
$\mathrm{x}+\mathrm{y}, 2 \mathrm{y}$ and $\mathrm{y}+\mathrm{z}$ are in harmonic progression. $2 y=\frac{(x+y)(y+z)}{x+y+y+z}$
$\Rightarrow \quad \frac{1}{2 y}=\frac{1}{x+y}+\frac{1}{y+z}$
$\Rightarrow \quad y(x+2 y+z)=\left(x y+x z+y^{2}+y z\right)$
$\Rightarrow x y+2 y^{2}+y z=x y+x z+y^{2}+y z$
$\Rightarrow \quad y^{2}=x z$
$\Rightarrow \mathrm{x}, \mathrm{y}, \mathrm{z}$ are in geometric progession.
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