The pth, qth and rth terms of an A.P. are a, b and c respectively.Show that a(q – r) + b(r-p) + c(p – q) = 0

Question

Philoid · Accepted Answer

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

Tp = a = A + (p − 1) D = (A − D) + pD …..(1) 1/2

Tq = b = A + (q – 1) D = (A − D) + qD …..(2) 1/2

Tr = c = A + (r − 1) D = (A − D) + rD …..(3) 1/2

Here we have got two unknowns A and D which are to be eliminated.

We multiply (1), (2) and (3) by q − r, r − p and p − q respectively and add:

a (q – r) = (A – D )(q – r) + D p(q – r) 1/2

b(r – p) = (A – D) (r – p) + Dq (r – p) 1/2

c(p – q) = (A – D) (p – q) + Dr (p – q) 1/2

a(q − r) + b(r − p) + c(p − q) 1

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

= (A – D)(0) + D[pq – pr + qr – pq + rp – rq ) 1

= 0

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