Question

If the sum of first p terms, first q terms and first r terms of an A.P . be a, b and c respectively, then $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)$ is equal to

• A. $0$
• B. $2$
• C. $pqr$
• D. $\frac{8abc}{pqr}$

Answer 1

The correct option is A $0$

We have $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

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

$\therefore \frac{2a}{p}=2A+(p-1)d.....\left(i\right)$ similarely

$\therefore \frac{2b}{q}=2A+(q-1)d.....\left(ii\right)$

$\therefore \frac{2c}{r}=2A+(r-1)d.....\left(iii\right)$

Multiply (i), (ii) and (iii) by $q-r$, $r-p$ and $p-q$ respectively and add

$\therefore \sum \frac{a}{p}(q-r)=0$