Question

If the sums of p, q and r terms of an A.P. be a, b and c respectively, then prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$.

Answer 1

$S_p=\frac{p}{2}[2A+(p-1\left)d\right]=a$
$\therefore \frac{2a}{p}=2A+(p-1)d$ $\left(1\right)$
$\frac{2b}{q}=2A+(q-1)d$ .$\left(2\right)$
$\frac{2c}{r}=2A+(r-1)d$ .$\left(3\right)$
Multiply $\left(1\right),\left(2\right)$ and $\left(3\right)$ by $q-r,r-p$ and $p-q$ respectively and add
$\therefore \sum \frac{a}{p}(q-r)=0$.