Question

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

Answer 1

Let the first term be $x$ and common difference be $d$

$\therefore a=p\frac{2x+(p-1)d}{2}$

$\therefore b=q\frac{2x+(q-1)d}{2}$

$\therefore c=r\frac{2x+(r-1)d}{2}$

$\therefore \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)$

$=(x+\frac{(p-1)d}{2})(q-r)+(x+\frac{(q-1)d}{2})(r-p)+(x+\frac{(r-1)d}{2})(p-q)$

$=\frac{d}{2}\left[\right(pq-pr-q+r)+(qr-qp-r+p)+(rp-rq-p+q\left)\right]$

$\frac{d}{2}\times 0=0$