The sum of the first $p, q,$ and $r$ terms of an $A.P.$ are $a, b,$ and $c,$ respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$.

Let $A$ be the first term and $D$ be the common difference of the $A.P.$
According to the given information:
$S_{p} = \frac{p}{2}[2A + (p-1)D] = a \Rightarrow \frac{a}{p} = A + \frac{(p-1)D}{2} \dots (1)$
$S_{q} = \frac{q}{2}[2A + (q-1)D] = b \Rightarrow \frac{b}{q} = A + \frac{(q-1)D}{2} \dots (2)$
$S_{r} = \frac{r}{2}[2A + (r-1)D] = c \Rightarrow \frac{c}{r} = A + \frac{(r-1)D}{2} \dots (3)$
Now,consider the expression $\frac{a}{p}(q-r) + \frac{b}{q}(r-p) + \frac{c}{r}(p-q)$.
Substituting the values from $(1), (2),$ and $(3)$:
$= [A + \frac{(p-1)D}{2}](q-r) + [A + \frac{(q-1)D}{2}](r-p) + [A + \frac{(r-1)D}{2}](p-q)$
$= A(q-r+r-p+p-q) + \frac{D}{2}[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]$
$= A(0) + \frac{D}{2}[pq - pr - q + r + qr - qp - r + p + rp - rq - p + q]$
$= 0 + \frac{D}{2}[0] = 0$.
Thus,the result is proved.