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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