(i) Given a, b, c are in A.P
⇒ \(\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}\) are also in A.P.
[On dividing each term by a, b, c]
⇒ \(\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}\) are also in A.P.
(ii) b + c, c + a, a + b will be in A.P.
If (c + a) − (b + c) = (a + b) − (c + a)
i.e if a − b = b − c
i.e if 2b = a + c
i.e if a, b, c are in A.P
Thus, a, b, c are in A.P ⇒ b + c, c + a, a + b are in A.P.