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

(N/A) Let $A$ be the first term and $D$ be the common difference of the $A.P.$
The $n^{\text{th}}$ term of an $A.P.$ is given by $a_n = A + (n-1)D$.
Therefore,
$a_p = A + (p-1)D = a$ $(1)$
$a_q = A + (q-1)D = b$ $(2)$
$a_r = A + (r-1)D = c$ $(3)$
Subtracting $(2)$ from $(1)$:
$(p-q)D = a-b \Rightarrow D = \frac{a-b}{p-q}$ $(4)$
Subtracting $(3)$ from $(2)$:
$(q-r)D = b-c \Rightarrow D = \frac{b-c}{q-r}$ $(5)$
Equating $(4)$ and $(5)$:
$\frac{a-b}{p-q} = \frac{b-c}{q-r}$
Cross-multiplying:
$(a-b)(q-r) = (b-c)(p-q)$
$aq - ar - bq + br = bp - bq - cp + cq$
Rearranging the terms to one side:
$aq - ar - bq + br - bp + bq + cp - cq = 0$
$a(q-r) + b(r-p) + c(p-q) = 0$
Thus,the result is proved.