Question

if pth,qth,rth terms of an Ap are a,b,c respectively
prove that a(q-r)+b(r-p)+c(p-q)=0

Answer 1

ap=a
aq=b
ar=c
a+(p-1)d=a. equation 1
a+(q-1)d=b. eq.2
a+(r-1)d=c. eq.3
subtract 3 from 2
a+(q-1)d-a-(r-1)d=b-c
qd-d-rd+d=b-c
d(q-r)=b-c
q-r=b-c/d. -eq.4
subtract 1 from 3
a+(r-1)d-a-(p-1)d=c-a
rd - d - pd + d = c - a
d(r-p)=c-a
r-p = c-a/d. eq5
subtract 2 from 1
a+(p-1)d-a-(q-1)d=a-b
pd-d-qd+d=a-b
d(p-q)=a-b. -eq.6
taking LHS
a(q-r)+b(r-p)+c(p-q)
substitute the values of q-r,r-p and p-q from 4 ,5 and 6
a(b- c/d)+b(c-a/d)+c(a- b/d)
ab-ac+bc-ab+ac-bc/d
0/d
0
LHS=RHS
hence proved