Question

If $p^{th},q^{th},r^{th}$ terms of an A.P. are $a,b,c$ respectively, then the value of $a(q-r)+b(r-p)+c(p-q)$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $0$

We know that,
$T_p=a,T_q=b,T_r=c$
Assuming the first term of the A.P. to be $k$ and the common difference to be $d$, then
$k+(p-1)d=a\Rightarrow p=\frac{a-k}{d}+1\cdots \left(1\right)k+(q-1)d=b\Rightarrow q=\frac{b-k}{d}+1\cdots \left(2\right)k+(r-1)d=c\Rightarrow r=\frac{c-k}{d}+1\cdots \left(3\right)$

Therefore, using equation $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get
$p-q=\frac{a-b}{d}q-r=\frac{b-c}{d}r-p=\frac{c-a}{d}$

Now, $a(q-r)+b(r-p)+c(p-q)$
$=a\left(\frac{b-c}{d}\right)+b\left(\frac{c-a}{d}\right)+c\left(\frac{a-b}{d}\right)=\frac{(ab-ac+bc-ba+ca-cb)}{d}=0$