Isn't Everything Related?

Let the $n$ th term of the arithmetic progression be $t_{n} = A + (n - 1)d$ for some reals $A,d$ .

If $d = 0$ then $a = b = c$ , in which case the given expression equals $a(q - r + r - p + p - q) = 0$ .

So now suppose $d \ne 0$ .

Then $t_{q} - t_{r} = (A + (q - 1)d) - (A + (r - 1)d) \Longrightarrow b - c = (q - r)d \Longrightarrow q - r = \dfrac{b - c}{d}$ .

Similarly $c - a = (r - p)d \Longrightarrow r - p = \dfrac{c - a}{d}$ and $a - b = (p - q)d \Longrightarrow p - q = \dfrac{a - b}{d}$ .

Then for $d \ne 0$ we have that $a(q - r) + b(r - p) + c(p - q) =$

$a \times \dfrac{(b - c)}{d} + b \times \dfrac{(c - a)}{d} + c \times \dfrac{(a - b)}{d} =$

$\dfrac{1}{d} \times (a(b - c) + b(c - a) + c(a - b)) = \dfrac{1}{d}(ab - ac + bc - ab + ca - cb) = 0$ .

So for any reals $A,d$ the given expression is equal to $\boxed{0}$ .

let the arithmetic progression be $2,4,6,8,10,12$ . We let $p^{th}=a=2, q^{th}=b=4$ and $r^{th}=c=6$ .

Now,

$a(q-r)+b(r-p)+c(p-q)=2(4-6)+4(6-2)+6(2-4)=-4+16-12=0$