Question

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

Answer 1

Let A be the first term and D the common difference of A.P.
$T_p=a=A+(p-1)D=(A-D)+pD$ $\left(1\right)$
$T_q=b=A+(q-1)D=(A-D)+qD$ ..$\left(2\right)$
$T_r=c=A+(r-1)D=(A-D)+rD$ ..$\left(3\right)$
Here we have got two unknowns A and D which are to be eliminated.
We multiply $\left(1\right),\left(2\right)$ and $\left(3\right)$ by $q-r,r-p$ and $p-q$ respectively and add:
$a(q-r)+b(r-p)+c(p-q)$
$=(A-D)[q-r+r-p+p-q]+D\left[p\right(q-r)+q(r-p)+r(p-q\left)\right]=0$.