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