Question

The following is known about an AP:

$S_p=a,S_q=b,S_r=c$

What is the value of

$E=\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)$? [4 MARKS]

Answer 1

Concept: 2 Marks
Application: 2 Marks

$Given,S_p=a,S_q=b,S_r=c$
Let A and D be the first term and the common difference of the A.P.

$S_p=a=\frac{p}{2}(2A+(p-1\left)D\right)$
$\Rightarrow \frac{a}{p}=A+\frac{(p-1)D}{2}...\left(i\right)$

$S_q=b=\frac{q}{2}(2A+(q-1\left)D\right)$
$\Rightarrow \frac{b}{q}=A+\frac{(q-1)D}{2}...\left(ii\right)$

$S_r=c=\frac{r}{2}(2A+(r-1\left)D\right)$
$\Rightarrow \frac{c}{r}=A+\frac{(r-1)D}{2}...\left(iii\right)$

$E=\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)$
$=(A+\frac{(p-1)D}{2})(q-r)+(A+\frac{(q-1)D}{2})(r-p)+(A+\frac{(r-1)D}{2})(p-q)$ [From (i), (ii) and (iii)]
$=A(q-r+r-p+p-q)+\frac{D}{2}\left[\right(p-1\left)\right(q-r)+(q-1\left)\right(r-p)+(r-1\left)\right(p-q\left)\right]$
$=A\left(0\right)+\frac{D}{2}\left(0\right)$
$=0$