Question

Sum of the first $p,q$ and $r$ terms of an $A.P$ are $a,b$ and $c$, respectively.
Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$

Answer 1

$\to$ By sum of terms in A.P, formula

$a=\frac{p}{2}(2a^1+(p-1\left)d^1\right)$

$b=\frac{q}{2}(2a^1+(q-1\left)d^1\right)$

$c=\frac{r}{2}(2a^1+(r-1\left)d^1\right)$

$\Rightarrow \frac{a}{p}=a^1+\frac{(p-1)}{2}{d}^1$

$\Rightarrow \frac{a}{p}(q-r)=a^1(q-r)+\frac{(p-1)(q-r)d^1}{2}$

$\frac{b}{q}(r-p)=a^1(r-p)+\frac{(q-1)(r-p)}{2}{d}^1$

$\frac{e}{r}(p-q)=a^1(p-q)+\frac{(r-1)(p-q)}{2}{d}^1$

Adding all three

$=a(p+q+r-p-q-r)+\frac{d^1}{2}(pq-q+r-pr+qr-rp-pq+pr-qr+q-p)$

$=0$