Question

if 1/b+c, 1/c+a, 1/a+b, be in A.P. then shoe that a^2,b^2,c^2 will be in A.P.​

Answer 1

Given :

1/b+c, 1/c+a, 1/a+b are in AP

Show :

a², b² & c² will also be in AP

Concept required to solve this problem,

Firstly AP stands for Arithmetic Progression. If we are been told that a, b, and c is in AP it means the common differences between a and b & b and c are same.

Let's solve it!!

$\frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a} \\ \\ \\ \implies \frac{2}{c + a} = \frac{1}{a + b} + \frac{1}{b + c} \\ \\ \\ \implies \frac{2}{c + a} = \frac{2b + a + c}{ab + ac + {b}^{2} + bc } \\ \\ \\ \implies \cancel{2ab} + \cancel{2 ac} + 2 {b}^{2} + \cancel{2 bc} = \cancel{ 2bc} + \cancel{ac} + {c}^{2} + \cancel{2ab} + {a}^{2} + \cancel{ac }\\ \\ \\ \implies 2 {b}^{2} = {a}^{2} + {c}^{2} \\ \\ \\ \implies \green{ { \underline{\boxed{ \bf \: {b}^{2} - {a}^{2} = {c}^{2} - {b}^{2} }}} }\: \: \: \bf \: \: \: hence \: \: proved.$