prove that:

Let A and D be the first term and common difference of A.P.

 

From the given information, we have:

A + (p– 1) D = a … (1)

 A + (q – 1) D = b … (2)

A + (r – 1) D = c … (3)

 

a (q – r) + b (r – p) + c (p – q)

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

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

= A (q – r + r – p + p – q) + D (pq – pr – q + r + qr – pq – r + p + pr – rq – p + q)

= A (0) + D (0)

= 0