Find $y$ such that $x$ and $z$ have to follow progressions.

Relevant wikis:

• Geometric Progression
• Harmonic Progressions

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From the condition for geometric progression, we have

$xz = y^2$

And from the condition of harmonic progression,

$\begin{aligned} & \dfrac{1}{x+3} + \dfrac{1}{z+3} = \dfrac{2}{y+3} \\ \implies & \dfrac{x+z+3}{xz+3x+3z+9} = \dfrac{2}{y+3} \\ \implies & (x+z+6)(y+3) = 2(y^2 + 3x+3z+9) \qquad \qquad \qquad \small{\color{#3D99F6} \text{As } xz = y^2} \\ \implies & xy+3x+yz+3z+6y+18 = 2y^2 + 6x+6z+18 \\ \implies & xy+yz-3x-3z = 2y^2 - 6y \\ \implies & (y-3)(x+z) = 2y(y-3) \\ \implies & (y-3)(x+z-2y) = 0 \\ \implies & \boxed{y=3} \\ & \text{ or } \\ & x+z-2y = 0 \qquad (\small{\color{#3D99F6} \text{Not true } \forall \ x,y,z \in \mathbb{R} \text{ as they do not follow an arithmetic progression}}) \end{aligned}$