Question

The pth, qth and rth 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 be the first term and d be the common difference of A.P.

Then $a_p=A+(p-1)d=a\dots \dots \left(1\right)$

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

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

Subtracting (2) from (1), we get

$(p-q)d=a-b\Rightarrow p-q=\frac{a-b}{d}$

Subtracting (3) from (2), we get

$(q-r)d=b-c\Rightarrow q-r=\frac{b-c}{d}$

Subtracting (1) from (3), we get

$(r-p)d=c-a\Rightarrow r-p=\frac{c-a}{d}$

Now,

$(q-r)a+(r-p)b+(p-q)c=(\frac{b-c}{d}a+\frac{(c-a)b}{d}+\frac{(a-b)c}{d})$

$=\frac{ab-ac-bc+bc-ab+ac-bc}{d}$

$=\frac{0}{d}=0$

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