If a,b,c are pth,qth and rth terms of an A.P. ,prove that

p(b-c)+q(c-a)+r(a-b)=0

Solution:

We assume the first term of an A.P.is m, the tolerance = d

a = m + (p-1)*d

b = m + (q-1)*d

c = m + (r-1)*d

p(b-c) = p*(q-r)*d

q(c-a) = q*(r-p)*d

r(a-b) = r*(p-q)*d

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

= p*(q-r)*d + q*(r-p)*d +r*(p-q)*d

= (pq-pr+qr-pq+rp-qr)*d

= 0*d = 0

So we prove p(b-c)+q(c-a)+r(a-b)=0