Question

If the $p^{th},q^{th}\&r^{th}$ terms of a H.P. are $a,b$&$c$ respectively, then the value of , $ab(p-q)+bc(q-r)+ca(r-p)$ is :

• A. $0$
• B. $1$
• C. $-2$
• D. none of these

Answer 1

The correct option is A $0$

$p^{\text{th}}$ term of corresponding $A\cdot P=\frac{1}{a}$

$q^{\text{th}}$ term of corresponding $A\cdot P=\frac{1}{b}$

$r^{\text{th}}$ term of corresponding $A\cdot P=\frac{1}{c}$

$\frac{1}{a}=A+(p-1)D$

$\frac{1}{b}=A+(q-1)D$

$\frac{1}{c}=A+(r-1)D$

$\frac{1}{b}-\frac{1}{c}=(q-1)D-(r-1)D$

$=qD-D-rD+D$

$=D(q-r)$

$\frac{c-b}{bc}=D(q-r)$

$(q-r)bc=-\frac{(b-c)}{D}$

In fimilar way

$\begin{array}{cc}(r-p)ca& =\frac{-(c-a)}{D}\\ (p-q)ab& =\frac{-(a-b)}{D}\end{array}$

$(q-r)bc+(r-p)ca+(p-q)ab$

$=\frac{-1}{D}(b-c+c-a+a-b)=\frac{1}{D}\left(0\right)=0$

Option (A) is correct.