If p , q and r terms of an AS are a,b and c respectively,prove that a(q-r)+b(r-p)+c(p-q) =0

Question

This queston is from arithemetic sequence and series

Answer 1

If the series has x as the first term and the common difference is y, the series is x, x+y, ..., x+(n-1)y where n is the nth term.

If n=p, then the pth term is x+(p-1)y; the qth term x+(q-1)y; and the rth term x+(r-1)y.

So a=x+(p-1)y, b=x+(q-1)y, c=x+(r-1)y, or a=x+py-y, b=x+qy-y, c=x+ry-y.

And b-c=(q-r)y; a-b=(p-q)y; c-a=(r-p)y. Also: a(b-c)=a(q-r)y; c(a-b)=c(p-q)y; b(c-a)=b(r-p)y.

But a(b-c)+c(a-b)+b(c-a)=ab-ac+ac-bc+bc-ab=0. Therefore a(q-r)y+b(r-p)y+c(p-q)y=0.

Because y is a common factor it can be removed leaving a(q-r)+b(r-p)+c(p-q)=0 QED