Question

$\frac{1}{x+y},\frac{1}{2y},\frac{1}{y+z}areinA.P.$. Prove that x, y, z are the consecutive terms of G.P, then we have to prove that $y^2=xz$

Answer 1

Given $\frac{1}{x+y},\frac{1}{2y},\frac{1}{y+z}areinA.P.$
To prove that x, y, z are the consecutive terms of G.P, then we have to prove that $y^2=xz$
Since, $\frac{1}{x+y},\frac{1}{2y},\frac{1}{y+z}areinA.P.$
then, $\frac{1}{2y}=\frac{\frac{1}{x+y}+\frac{1}{y+z}}{2}$
Also we can calculate the common difference
$\frac{1}{2y}-\frac{1}{x+y}=d\Rightarrow \frac{x-y}{x+y}=2yd---\left(1\right)$
$\frac{1}{y+z}-\frac{1}{2y}=d\Rightarrow \frac{y-z}{y+z}=2yd---\left(2\right)$
from (1) and (2) we get $\frac{x-y}{x+y}=\frac{y-z}{y+z}$
$\Rightarrow (x-y)(y+z)=(x+y)(y-z)\Rightarrow xy+xz-y^2-yz=xy-xz+y^2-yz\Rightarrow 2y^2-2xz=0\Rightarrow 2(y^2-xz)=0\Rightarrow y^2-xz=0\Rightarrow y^2=xz$
Hence the terms x,y,z are the consecutive terms of a G.P