Question

If the sum of first p, q, r term of an A.P are a, b, c respectively. Prove that
$\frac{a}{P}\left(q-r\right)+\frac{b}{q}\left(r-p\right)+\frac{c}{r}\left(p-q\right)=0$

Answer 1

Given sum of first p, q, r term of an A.P

are a, b, c respectively.

Sum of first p terms, when A is the first term and D is the common difference

Given $s_p=a$

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

Similarly $s_q=b$

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

and $s_r=c$

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

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

Multply by $q-r$, we get $\frac{a}{p}(q-r)=(A+\frac{(p-1)}{2}D)(q-r)$ ……..$\left(1\right)$

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

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

Adding $\left(1\right),\left(2\right)$ and $\left(3\right)$

$\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=A(q-r+r-p+p-q)+\frac{D}{2}[pq-pr-q+r+rq-pq-r+p+rp-rq-p-q]$

$=0$.