The p^th, q^th, and r^th terms of an A.P. are a, b, and c respectively. Show that (q − r)a + (r − p)b + (p − q)c = 0.

Let the A.P. be represented as:

a_p = A + (p − 1)D = a

a_q = A + (q − 1)D = b

a_r = A + (r − 1)D = c

LHS: (q − r)a + (r − p)b + (p − q)c

= (q − r)[A + (p − 1)D] + (r − p)[A + (q − 1)D] + (p − q)[A + (r − 1)D]

= (q − r)A + (q − r)(p − 1)D + (r − p)A + (r − p)(q − 1)D + (p − q)A + (p − q)(r − 1)D

= A[q − r + r − p + p − q] + D[qp − q − rp + r + rq − r − pq + p + pr − p − qr + q]

= 0 + 0 = RHS

Hence proved: (q − r)a + (r − p)b + (p − q)c = 0.