Question

If $a,b,c$ are $pth,qth$ and $rth$ terms respectively of an A.P prove that:
$p(b-c)+q(c-a)+r(a-b)=0$

Answer 1

$Given,a,b,carethe{p}^{th},q^{th},r^{th}termsofanA.Prespectively.\therefore (b-a)=(c-b)samecommondifference.\Rightarrow (a-b)=(b-c)⟶\left(1\right)Now,LHS=p(b-c)+q(c-a)+r(a-b)=a(b-c)+b(c-a)+c(a-b)(wecanreplacedp,q,rbya,b,casa,b,carethe{p}^{th},q^{th},r^{th}termsofanA.Prespectively)=ab-ac+bc-ba+ca-cb(sinceab=ba,bc=cb,ca=ac)=0=RHSHence,p(b-c)+q(c-a)+r(a-b)=0henceproved.$