Question

If the sum of first $p$ terms, first $q$ terms and first $r$ terms of an A.P. be $x$ ,$y$ and $z$ respectively, then $\frac{x}{p}(q-r)+\frac{y}{q}(r-p)+\frac{z}{r}(p-q)$ is

Answer 1

Given,

$S_p=x$

$S_q=y$

$S_r=z$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ ; $a$ is the first term of A.P. & $d$ is the common difference.

$\therefore S_p=\frac{p}{2}[2a+(p-1\left)d\right]=x$-----------(1)

$S_q=\frac{q}{2}[2a+(q-1\left)d\right]=y$--------------(2)

$S_r=\frac{r}{2}[2a+(r-1\left)d\right]=z$---------------(3)

From (1), we have

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

From (2), we have

$\frac{y}{q}=a+(q-1)\frac{d}{2}$

From (3), we have

$\frac{r}{z}=a+(r-1)d$

Now,

$\frac{x}{p}(q-r)+\frac{y}{q}(r-p)+\frac{z}{r}(p-q)$

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

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

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

Hence, this is the answer.