If `(1)/(a),(1)/(b),(1)/(c)` are also in A.P. then prove that : (i) `((b+c))/(a),((c+a))/(b),((a+b))/(c)` are also in A.P. (ii) `((b+c-a))/(a),((c+a-b))/(b),((a+b-c))/(c)` are also in A.P.

Explanation

To prove that if \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in Arithmetic Progression (A.P.), then \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\) and \(\frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}\) are also in A.P., we will follow these steps:

### Part (i): Proving \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\) are in A.P.

1. **Given Condition**: Since \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P., we can write:
\[ 2 \cdot \frac{1}{b} = \frac{1}{a} + \frac{1}{c} \]
...