(i) a²(b + c), b²(c + a), c²(a + b) are also in A.P.

If b²(c + a) – a²(b + c) = c²(a + b) – b²(c + a)

b²c + b²a – a²b – a²c = c²a + c²b – b²a – b²c

Given, b – a = c – b

And since a, b, c are in AP,

c(b² – a² ) + ab(b – a) = a(c² – b² ) + bc(c – b)

(b – a) (ab + bc + ca) = (c – b) (ab + bc + ca)

Upon cancelling, ab + bc + ca from both sides

b – a = c – b

2b = c + a [which is true]

Hence, given terms are in AP

(ii) b + c – a, c + a – b, a + b – c are in A.P.

If (c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)

Then, b + c – a, c + a – b, a + b – c are in A.P.

Let us consider LHS and RHS

(c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)

2a – 2b = 2b – 2c

b – a = c – b

And since a, b, c are in AP,

b – a = c – b

Hence, given terms are in AP.

(iii) bc – a², ca – b², ab – c² are in A.P.

If (ca – b²) – (bc – a²) = (ab – c²) – (ca – b²)

Then, bc – a², ca – b², ab – c² are in A.P.

Let us consider LHS and RHS

(ca – b²) – (bc – a²) = (ab – c²) – (ca – b²)

(a – b² – bc + a²) = (ab – c² – ca + b²)

(a – b) (a + b + c) = (b – c) (a + b + c)

a – b = b – c

And since a, b, c are in AP,

b – c = a – b

Hence, given terms are in AP