Question

If pth, qth and rth termof an AP are a, b and c respectively, then show that (a-b)r + (b-c)p +(c-a)q =0

Answer 1

LHS = (a-b)r + (b-c)p + (c-a)q

=(a-b)r + (a-b)p + (c-a)q

=(a-b)(r+p) + (c-a)q

=(a-b)2q + (c-a)q

=q(2a-2b+c-a)

=q(a+c-2b)

=q(2b-2b)

=q(0)

=0

=RHS

Answer 2

Pth term = a

Qth term = b

Rth term = c

So,

A + (p-1)d = a

A + (q-1)d = b

A + (r-1)d = c

Now,

a - b will be equal to :-

$a + (p - 1)d - a - (q - 1)d \\ \\ \\ = (p - 1)d - (q - 1)d \\ \\ \\ = d(p - 1 - q + 1) \\ \\ \\ = d(p - q)$

Now,

b - c will be equal to :-

$a + (q - 1)d - a - (r - 1)d \\ \\ \\ = (q - 1)d - (r - 1)d \\ \\ \\ = d(q - r)$

Now,

c - a will be equal to :-

$a + (r - 1)d - a - (p - 1)d \\ \\ \\ =(r - 1)d -(p - 1)d \\ \\ \\ = d(r - p)$

Now,

(a- b) r + (b-c) p + (c-a) q

We will put the values of (a-b) , (b-c) and (c-a) in the above equation.

So,

It will become :-

$d(p - q)r + d(q - r)p + d(r - p)q \\ \\ \\ = d(pr - qr + pq - pr + qr - pq) \\ \\ \\ = d(0) \\ \\ = 0 \\ \\ = rhs$

Hence,

Proved!!