Question

If $\frac{1}{x+y},\frac{1}{2y},\frac{1}{y+z}$ are the three consecutive terms of an A.P., then $x,y,z$ are the three consecutive term of :

• A. $anA.P.$
• B. $aG.P.$
• C. $anH.P.$
• D. None of these

Answer 1

The correct option is A $anA.P.$

$\frac{1}{2y}-\frac{1}{x+y}=\frac{1}{y+2}-\frac{1}{2y}$

$\frac{x+y-2y}{x+y}=\frac{2y-y-z}{y+z}$

$\frac{x-y}{x+y}=\frac{y-z}{y+z}$

using componendo -> dividendo

$\frac{x-y+x+y}{x-y-x-y}=\frac{y-z+y+z}{y-z-y-z}$

$\frac{2x}{-2y}=\frac{2y}{-2z}$

$\frac{x}{y}=\frac{y}{z}$

$y^2=xz$

Hence $x,y,z$ are in $G\cdot P$.