Question

f a,b,c are the pth , qth,and rth term of an A.P, then prove that a(q-r)+b(r-p)+c(p-q)=0

Answer 1

let x is the first term and d is the common difference
according to question ,
pth term =a
x+(p-1) d=a------------(1)
in the same way ,
x+(q-1) d=b -------------(2)
x+(r-1)d =c ---------------(3)
subtract (1)-(2)
(p-q) d=(a-b)
multiply with c in both side
c (p-q) d=c (a-b)

subtract (2)-(3) and multiply with a
a (q-r) d=a (b-c)
again ,
subtract (3)-(1) and multiply with b
b (r-p) d=b (c-a)

now add all finding equation ,
a (q-r) d+b (r-p) d+c (p-q) d=a (b-c)+b (c-a)+c (a-b)

d {a (q-r)+b (r-p)+c (p-q)}=ab-ac+bc-ba+ac-bc =0

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