Answer:
(i)
If 1/a, 1/b, 1/c are in AP
1/b – 1/a = 1/c – 1/b
Consider LHS,
1/b – 1/a = (a-b)/ab
= c(a-b)/abc
Consider RHS,
1/c – 1/b = (b-c)/bc
= a(b-c)/bc [by multiplying with ‘a’ on both the numerator and denominator]
(b+c)/a, (c+a)/b, (a+b)/c are in AP
(ii)
If bc, ca, ab are in AP
ca – bc = ab – ca
c (a-b) = a (b-c)
1/a, 1/b, 1/c are in AP
1/b – 1/a = 1/c – 1/b
c (a-b) = a (b-c)
Hence, the given terms are in AP