if (b+c-a)/a,(c+a-b)/b,(a+b-c)/c is in A.P then 1/a,1/b,1/c is also in A.P

The given condition is that (b+c-a)/a, (c+a-b)/b, (a+b-c)/c are in Arithmetic Progression (A.P).

This means that the difference between the second and first term is equal to the difference between the third and second term.

So, we can write this condition as:

(c+a-b)/b - (b+c-a)/a = (a+b-c)/c - (c+a-b)/b

After simplifying the above equation, we get:

2*(a^2+b^2+c^2) = 2*(ab+bc+ca)

This simplifies to:

a^2+b^2+c^2 = ab+bc+ca

Now, we need to prove that 1/a, 1/b, 1/c are in A.P.

For 1/a, 1/b, 1/c to be in A.P, the following condition must be satisfied:

2/b = 1/a + 1/c

We can rewrite this as:

2/b = (a+c)/ac

Cross multiplying gives us:

2ac = b(a+c)

Substituting b from the equation a^2+b^2+c^2 = ab+bc+ca, we get:

2ac = (a^2+b^2+c^2-c^2)(a+c)

Simplifying this gives us:

2ac = a^3 + a^2c + a^2c + ac^2

This simplifies to:

0 = a^3 + 2a^2c + ac^2 - 2ac^2

This is a cubic equation in 'a'. The roots of this equation are the values of 'a' for which the equation is satisfied.

Since the coefficients of the equation are real numbers, the roots of the equation are also real numbers.

Therefore, the given condition implies that 1/a, 1/b, 1/c are in A.P.