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Question

If pth, qth and rth term of an AP are a, b, c respectively, then show that

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


Solution

Let A = first term of the AP.
and 
Let d = common difference of the AP

Now
a = A+(p-1).d.......(1)
b = A+(q-1).d.......(2)
c = A+(r-1).d........(3)

Subtracting 2nd from 1st , 3rd from 2nd and 1st from 3rd we get

a-b = (p-q).d......(4)
b-c = (q-r).d........(5)
c-a = (r-p).d.......(6)

multiply 4,5,6 by c,a,b respectively we have

c.(a-b) = c.(p-q).d......(4)
a.(b-c) = a.(q-r).d........(5)
b.(c-a) = b.(r-p).d.......(6)

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

Now since d is common difference it should be non zero 

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

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