If the $p$th, $q$th and $r$th terms of an AP are $a,b$ and $c$ respectively, the value of $a(q -r) + b(r -p) + c(p -q)$ is

Given : $p$th, $q$th and $r$th terms of an AP are $a,b$ and $c$

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

Let $m$ and $d$ be the first term and common difference of given AP. then,

$p^{th}$ term $=m+\left(p-1\right)d=a$…….(1)

$q^{th}$ term $=m+\left(q-1\right)d=b$…….(2)

and $r^{th}$ term $=m+\left(r-1\right)d=c$………(3)

On multiplying $\left(q-r\right)$,$\left(r-p\right)$ and $\left(p-q\right)$ in Equation (1) (2) & (3) respectively, we get 

$=a\left(q-r\right)+b\left(r-p\right)+c\left(p-q\right)$

$=\left(q-r\right)\left(m+\left(p-1\right)d\right)+\left(r-p\right)\left(m+\left(q-1\right)d\right)+\left(p-q\right)\left(m+\left(r-1\right)d\right)$

$=m\left(q-r+r-p+p-q\right)+d\left[\left(p-1\right)\left(q-r\right)+\left(q-1\right)\left(r-p\right)+\left(r-1\right)\left(p-q\right)\right]$

$=m\times 0+d\left[pq-pr-q+r+qr-pq-r+p+pr-qr-p+q\right]$

$=d\left(0\right)$

$=0$

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

Hence, option c) is correct answer.