Question

If the $p^{th},q^{th},r^{th}$ terms of an A.P. are $a,b,c$ respectively, show that $(q-r)a+(r-p)b+(p-q)c=0$.

Answer 1

$a_p=A+(p-1)Da=A+(p-1)D........\left(i\right)a_q=A+(q-1)Db=A+(q-1)D........\left(ii\right)a_r=A+(r-1)D.........\left(iii\right)c=A+(r-1)D.........\left(iii\right)$

subtracting $\left(i\right)$ from $\left(ii\right)$

$\Rightarrow (a-b)=(p-q)D.....\left(iv\right)$

subtracting $\left(ii\right)$ from $\left(iii\right)$

$\Rightarrow (b-c)=(q-r)D.....\left(v\right)$

Dividing $\left(iv\right)$ by $\left(v\right)$

$\Rightarrow \frac{a-b}{b-c}=\frac{(p-q)}{(q-r)}\Rightarrow a(q-r)-b(q-r)=b(p-q)-c(p-q)\Rightarrow a(q-r)+b(r-q)-b(p-q)+c(p-q)=0\Rightarrow a(q-r)+b(r-p)+c(p-q)=0$

Hence proved.