Divide all the 3 terms by abc . as we that when an ap is divided by any common factor, the resulting sequence is also in ap, so

a/bc, b/ca, c/ab

are also in ap.

Proof:

Let us assume that a^2, b^2, and c^2 are in arithmetic progression (AP) with a common difference of d.

Then we can write the following equations:

b^2 - a^2 = d ...(1)
c^2 - b^2 = d ...(2)

Simplifying equation (1), we get:

(b - a)(b + a) = d

Similarly, simplifying equation (2), we get:

(c - b)(c + b) = d

We can rearrange equation (1) to express b in terms of a:

b = √(a^2 + d) ...(3)

Similarly, we can rearrange equation (2) to express c in terms of b:

c = √(b^2 + d) ...(4)

Now, let's prove that a/bc, b/ca, and c/ab are in arithmetic progression.

1. Proving a/bc in AP:

To prove a/bc is in AP, we need to show that:

(b/ca) - (a/bc) = constant

Substituting the values of b and c from equations (3) and (4) respectively, we get:

(b/ca) - (a/bc) = (√(a^2 + d))/(√(b^2 + d) * √(c^2 + d)) - (√(b^2 + d))/(√(c^2 + d) * √(a^2 + d))

Taking a common denominator and simplifying, we get:

(b/ca) - (a/bc) = (a^2 - b^2)/(√(a^2 + d) * √(b^2 + d) * √(c^2 + d))

Using equation (1), we can substitute the value of (a^2 - b^2) as d:

(b/ca) - (a/bc) = d/(√(a^2 + d) * √(b^2 + d) * √(c^2 + d))

Since a^2, b^2, and c^2 are in AP, d is a constant. Thus, we can rewrite the equation as:

(b/ca) - (a/bc) = constant

Hence, we have proved that a/bc is in arithmetic progression.

2. Proving b/ca in AP:

To prove b/ca is in AP, we need to show that:

(c/ab) - (b/ca) = constant

Substituting the values of a and b from equations (3) and (4) respectively, we get:

(c/ab) - (b/ca) = (√(b^2 + d))/(√(a^2 + d) * √(b^2 + d) * √(c^2 + d)) - (√(c^2 + d))/(√(a^2 + d) * √(b^2 + d) * √(c^2 + d))

Taking a common denominator and simplifying, we get:

(c/ab) - (b/ca) = (b^2 - c^2)/(√(a^2 + d) *