Question

If $p^{th},q^{th}$ and $r^{th}$ terms of an A.P. are $a,b,c$ respectively, then show that
(i) $a(q-r)+b(r-p)+c(p-q)=0$
(ii) $(a-b)r+(b-c)p+(c-1)q=0$

Answer 1

(i) Let $A$ be the the first term and $d$ be the common difference of the given A.P.

Then,
$a=p^{th}$ term $⟹a=A+(p-1)D$ ...(i)
$b=q^{th}$ term $⟹b=A+(q-1)D$ ...(ii)
$c=r^{th}$ term $⟹c=A+(r-1)D$ ...(iii)

We have,
$a(q-r)+b(r-p)+c(p-q)$

$=\{A+(p-1\left)D\right\}(q-r)+\{A+(q-1\left)D\right\}(r-p)+\{A+(r-1\left)D\right\}(p-q)$ [Using (i), (ii) and (iii)]

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

= $A\times 0+D\left\{p\right(q-r)-q(r-p)+r(q-r)-(r-p)-(p-q\left)\right\}$

= $A\times 0+D\times 0=0$
(ii) On subtracting equation (ii) from equation (i), equation (iii) from equation (ii) and equation (i) from equation (iii), we get
$a-b=(p-q)D$, $(b-c)=(q-r)D$ and $c-a=(r-p)D$

$(a-b)r+(b-c)p+(c-a)q$

$(p-q)Dr+(q-r)Dp+(c-a)q$

$(p-q)Dr+(q-r)Dp+(r-p)Dq$

$=D\left\{\right(p-q)r+(q-r)p+(r-p\left)q\right\}$

$=D\times 0=0$