The $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an $A.P.$ are $a, b, c,$ respectively. Show that $(q-r) a+(r-p) b+(p-q) c=0$

Let $t$ and $d$ be the first term and the common difference of the $A.P.$ respectively. 

The $n^{th}$ term of an $A.P.$ is given by, $a_{n}=t+(n-1) d$

Therefore,

$a_{p}=t+(p-1) d=a$        .........$(1)$

$a_{q}=t+(q-1) d=b$        .........$(2)$

$a_{r}=t+(r-1) d=c$        .........$(3)$

Subtracting equation $(2)$ from $(1),$ we obtain

$(p-1-q+1) d=a-b$

$\Rightarrow(p-q) d=a-b$

$\therefore d=\frac{a-b}{p-q}$           .........$(4)$

Subtracting equation $(3)$ from $(2),$ we obtain

$(q-1-r+1) d=b-c$

$\Rightarrow(q-r) d=b-c$

$\Rightarrow d=\frac{b-c}{q-r}$          .........$(5)$

Equating both the values of $d$ obtained in $(4)$ and $(5),$ we obtain

$\frac{a-b}{p-q}=\frac{b-c}{q-r}$

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

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

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

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

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

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

Thus, the given result is proved.