Question

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

Answer 1

Let $A=$first term of the A.P

and

$d=$common difference of the A.P

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 $\left(2\right)$ from $\left(1\right),\left(3\right)$ from $\left(2\right)$ and $\left(1\right)$ from $\left(3\right)$ 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 $\left(4\right),\left(5\right),\left(6\right)$ by $c,a,b$ respectively we have

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

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

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

Now,

$a(q-r)d+b(r-p)d+c(p-q)d=\left[a\right(q-r)+b(r-p)+c(p-q\left)\right]d=0$

Now since d is common difference it should be non zero

Hence,

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