If p^th, q^th and r^th terms of an A.P. are a, b, c respectively, then show that (a – b)r + (b – c)p + (c – a)q = 0.

Let a be the first term and d be the common difference of the given A.P.

p^th term of AP = [ a + (p - 1)d] = a …………(i)

q^th term of AP = [ a + (q - 1)d] = b …………(ii)

r^th term of AP = [ a + (r - 1)d] = c …………(iii)

To show: (a – b)r + (b – c)p + (c – a)q = 0

Consider the L.H.S. = (a – b)r + (b – c)p + (c – a)q

= r[{a + (p - 1)d} - {a + (q - 1)d}]

+ p[{a + (q - 1)d} - {a + (r - 1)d}]

+ q[{a + (q - 1)d} - {a + (p - 1)d}]

= [r(p - 1- q + 1)d] + [p(q - 1- r + 1)d] + [q(r - 1- p + 1)d]

= d[p - q]r + d[q - r]p + d[r - p]q

= d[pr – qr + pq – pr + qr - pq]

= 0

Therefore, L.H.S. = RH.S.

Hence, proved.