1. ## Questions

if a^2,b^2,c^2 are in A.P. then 1/b+c,1/c+a,1/a+b ARE In??
Will it be AP,GP,HP or none..??

2. Hello, cnpranav!

If $\displaystyle a^2,b^2,c^2$ are in A.P. then: .$\displaystyle \frac{1}{b+c},\;\frac{1}{c+a},\;\frac{1}{a+b}$ .are in ...

Will it be AP, GP, HP or none?
Since $\displaystyle a^2,b^2,c^2$ are in Arithmetic Progression,

we have: .$\displaystyle \begin{array}{ccccccccc} b^2-a^2 \:=\: d &\Rightarrow & (b-a)(b+a) \:=\:d & \Rightarrow & a+b \:=\:\frac{d}{b-a} \\ c^2-b^2 \:=\:d & \Rightarrow & (c-b)(c+b) \:=\:d & \Rightarrow & b+c \:=\:\frac{d}{c-b} \end{array}$
. . . . . . . $\displaystyle \begin{array}{ccccccccc}c^2-a^2 \:=\:2d & \Rightarrow & (c-a)(c+a) \:=\:2d & \Rightarrow & c+a \:=\:\frac{2d}{c-a} \end{array}$

$\displaystyle \text{Then: }\;\left\{\frac{1}{b+c},\;\frac{1}{c+a},\;\frac{1} {a+b}\right\}\;\text{ becomes: }\;\underbrace{\frac{c-b}{d}}_{u_1},\;\underbrace{\frac{c-a}{2d}}_{u_2},\;\underbrace{\frac{b-a}{d}}_{u_3}$

We find that: .$\displaystyle \begin{array}{ccccc}u_2-u_1 &=&\dfrac{c-a}{2d} - \dfrac{c-b}{d} &=& \dfrac{2b-a-c}{2d} \\ \\[-4mm] u_3-u_2 &=& \dfrac{b-a}{d} - \dfrac{c-a}{2d} &=& \dfrac{2b-a-c}{2d} \end{array}$

$\displaystyle u_1,u_2,u_3\,\text{ have a }common\;di\!f\!\!ference\;\hdots\; \text{ They are in }{\color{blue}A.P.}$

3. Originally Posted by Soroban
Hello, cnpranav!

Since $\displaystyle a^2,b^2,c^2$ are in Arithmetic Progression,

we have: .$\displaystyle \begin{array}{ccccccccc} b^2-a^2 \:=\: d &\Rightarrow & (b-a)(b+a) \:=\:d & \Rightarrow & a+b \:=\:\frac{d}{b-a} \\ c^2-b^2 \:=\:d & \Rightarrow & (c-b)(c+b) \:=\:d & \Rightarrow & b+c \:=\:\frac{d}{c-b} \end{array}$
. . . . . . . $\displaystyle \begin{array}{ccccccccc}c^2-a^2 \:=\:2d & \Rightarrow & (c-a)(c+a) \:=\:2d & \Rightarrow & c+a \:=\:\frac{2d}{c-a} \end{array}$

$\displaystyle \text{Then: }\;\left\{\frac{1}{b+c},\;\frac{1}{c+a},\;\frac{1} {a+b}\right\}\;\text{ becomes: }\;\underbrace{\frac{c-b}{d}}_{u_1},\;\underbrace{\frac{c-a}{2d}}_{u_2},\;\underbrace{\frac{b-a}{d}}_{u_3}$

We find that: .$\displaystyle \begin{array}{ccccc}u_2-u_1 &=&\dfrac{c-a}{2d} - \dfrac{c-b}{d} &=& \dfrac{2b-a-c}{2d} \\ \\[-4mm] u_3-u_2 &=& \dfrac{b-a}{d} - \dfrac{c-a}{2d} &=& \dfrac{2b-a-c}{2d} \end{array}$

$\displaystyle u_1,u_2,u_3\,\text{ have a }common\;di\!f\!\!ference\;\hdots\; \text{ They are in }{\color{blue}A.P.}$

thanks a ton