Question

Plss answer my question I will mark u as brainliest in turn​

Answer 1

         

Proving by the concept given below,

$If \ T_1,\ T_2,\ and \ T_3\ are three consecutive terms of a GP, \\ \\ then, \\ \\ \boxed{\frac{T_1-T_2}{T_1+T_2}=\frac{T_2-T_3}{T_2+T_3}} \\ \\ \\ because, \\ \\ let the consecutive terms of the GP be \ \ a,\ ar\ \ and , \ ar^2. \\ \\ So that, \\ \\ \boxed{\frac{a-ar}{a+ar}=\frac{ar-ar^2}{ar+ar^2}=\frac{1-r}{1+r}}$

Okay, let's prove.

$As \ \ \frac{1}{x+y},\ \frac{1}{2y},\ \frac{1}{y+z}\ \ are consecutive terms of the AP, \\ \\ \\ d=\frac{1}{2y}-\frac{1}{x+y}=\frac{1}{y+z}-\frac{1}{2y} \\ \\ \\ \Rightarrow\ \frac{x+y-2y}{2y(x+y)}=\frac{2y-(y+z)}{2y(y+z)} \\ \\ \\ \Rightarrow\ \frac{x-y}{2y(x+y)}=\frac{2y-y-z}{2y(y+z)} \\ \\ \\ \Rightarrow\ \frac{x-y}{2y(x+y)}=\frac{y-z}{2y(y+z)}$

$\Rightarrow\ \frac{1}{2y} \times \frac{x-y}{x+y}=\frac{1}{2y} \times \frac{y-z}{y+z} \\ \\ \\ \Rightarrow\ \frac{x-y}{x+y}=\frac{y-z}{y+z} \\ \\ \\ \\ \\ Here, \ \ \frac{x-y}{x+y}=\frac{y-z}{y+z}\ \ is in the form \ \ \frac{T_1-T_2}{T_1+T_2}=\frac{T_2-T_3}{T_2+T_3} \\ \\ \\ \therefore\ x,\ y,\ z\ \ are three consecutive terms of a GP. \\ \\ \\ Hence proved!$

$Plz ask me if you've any doubts. \\ \\ \\ Thank you. :-))$