Question

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

Answer 1

Let $t$ and $d$ be the first term and the common difference of the A.P. respectively,
The $n^{th}$ term of an A.P. is given by $a_n=t+(n-1)d$ Therefore,
$a_p=t+(p-1)d=a...\left(1\right)a_q=t+(q-1)d=b...\left(2\right)a_r=t+(r-1)d=c...\left(3\right)$
Subtracting equation (2) from (1), we obtain
$(p-1-q+1)d=a-b\Rightarrow (p-q)d=a-b\therefore d=\frac{a-b}{p-q}...\left(4\right)$
Subtracting equation (4) from (3), we obtain
$(q-1-r+1)d=b-c\Rightarrow (q-r)d=b-c\Rightarrow d=\frac{b-c}{q-r}...\left(5\right)$
Equating both the values of $d$ obtained in (4) and (5), we obtain
$\frac{a-b}{p-q}=\frac{b-c}{q-r}\Rightarrow (a-b)(q-r)=(b-c)(p-q)\Rightarrow aq-bq-ar+br=bp-bq-cp+cq\Rightarrow (-aq+ar)+(bp-br)+(-cp+cq)=0\Rightarrow -a(q-r)-b(r-p)-c(p-q)=0\Rightarrow a(q-r)+b(r-p)+c(p-q)=0$
Thus, the given result is proved.