Question

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

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

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:

$a(q-r)+b(r-p)+c(p-q)=(A-D)[q-r+r-p+p-q]+D\left[p\right(q-r)+q(r-p)+r(p-q\left)\right]=0$.

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