Answers:
(i)
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
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)
Cancelling ab + bc + ca from both sides,
b – a = c – b
2b = c + a
Hence, the given terms are in AP
(ii)
If (c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)
b + c – a, c + a – b, a + b – c are in A.P.
Consider LHS and RHS,
(c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)
2a – 2b = 2b – 2c
b – a = c – b
a, b, c are in AP,
b – a = c – b
Hence, the given terms are in AP.