If a, b, c are in A.P., then show that:

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


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


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

(i)


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


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


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


b - a = c - b


And since a, b, c are in AP,


b - a = c - b


Hence given terms are in AP


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


Therefore, (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.


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


(a - b)(a + b + c) = (b - c)(a + b + c)


a - b = b - c


b - a = c - b


And since a, b, c are in AP,


b - a = c - b


Hence given terms are in AP


3