If pth qth and rth term of an ap are a b and c respectively then show that (a-b)r+(b-c)p+(c-a)q=0

 solution:-

let 1st term=k

pth term of A.P=k +(p-1)d =a

qth term of A.P =k +( q-1)d =b

rth term of A.P=k +(r-1)d =c

L.H.S=

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

{k+(p-1)d}(q-r)+{k+(q-1)d}(r-p)+{k+(r-1)d}(p-q)

kq-kr+pqd-qd-pdr+dr+kr-kp+qdr-qr-qdp+dp+kp-kq+rdp-dp-rdq+dq(over all are zero after solving)

=0 proved