Question

Sum of the first p,q and r terms of an AP are 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. $pq$
• B. $0$
• C. $rp$
• D. $3$

Answer 1

The correct option is B $0$

Let 'A' be the first term and 'D' be the common difference of AP

Given that a,b and c be the sum of the first p,q,r terms respectively.

$\therefore a=S_p$

$a=\frac{p}{2}[2A+(p-1\left)D\right]$

$\frac{a}{p}=\frac{1}{2}[2A+(p-1\left)D\right]...........\left(1\right)$

$\therefore b=S_q$

$b=\frac{q}{2}[2A+(q-1\left)D\right]$

$\frac{b}{q}=\frac{1}{2}[2A+(q-1\left)D\right]...........\left(2\right)$

$\therefore c=S_r$

$c=\frac{r}{2}[2A+(p-1\left)D\right]$

$\frac{c}{r}=\frac{1}{2}[2A+(p-1\left)D\right]...........\left(3\right)$

Now,

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

$=\frac{1}{2}[2A+(p-1\left)D\right](q-r)+\frac{1}{2}[2A+(q-1\left)D\right](r-p)+\frac{1}{2}[2A+(r-1\left)D\right](p-q)$

$=\frac{1}{2}\left(2A\right)[q-r+r-p+p-q]+\frac{1}{2}D\left[\right(p-1\left)\right(q-r)+(q-1\left)\right(r-p)+(r-1\left)\right(p-q\left)\right]$

$=\frac{1}{2}\left(2A\right)\left[0\right]+\frac{1}{2}D[pq-pr-q+r+qr-pq-r+p+rp-rp-p+q]$

$=0$