Question

If the sum of $p,q$ & $r$ terms of $A.P.$ is a, b,& c respectively then prove $\frac{a}{p}\left(q-r\right)+\frac{b}{q}\left(r-p\right)+\frac{c}{r}\left(p-q\right)=0$

Answer 1

Given 9 p =a A = first term

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

$Given\left|ar\right|\gamma ,sq=b$

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

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

$Now\frac{a}{p}\left(q-r\right)=\left(\frac{1}{2}(p-1)D+A\right)\left(q-r\right)$....(1)

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

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

Adding (1) , ( 2) and (3) , we get

$LHS=\frac{a}{p}\left(q-r\right)+\frac{b}{q}\left(r-p\right)+\frac{c}{r}\left(p-q\right)$

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

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

$\frac{1}{2}D\left(0\right)+0=0=RHS$