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 $a=$ First term of the AP

and

$d=$ Common difference of the AP

Now,

$a=A+(p-1)d....\left(1\right)$

$b=A+(q-1)d....\left(2\right)$

$c=A+(r-1)d....\left(3\right)$

Subtracting (2) from (1), (3) from (2) and (1) from (3), we get

$a-b=(p-q)d....\left(4\right)$

$b-c=(q-r)d....\left(5\right)$

$c-a=(r-p)d....\left(6\right)$

Multiply (4),(5),(6) by c,a,b respectively, we have

$c(a-b)=c(p-q)d....\left(7\right)$

$a(b-c)=a(q-r)d....\left(8\right)$

$b(c-a)=b(r-p)d....\left(9\right)$

Adding (7),(8) and (9) we get

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

$=ca-cb+ab-ac+bc-ba$

$=0$

Now since d is common difference it should be non zero

Hence,

$a(q-r)+b(r-p)+c(p-q)=0$