Question

If $a$, b,c are the $p^{th},q^{th},r^{th}$ terms in $H.P$. then $\left|\begin{array}{ccc}bc& p& 1\\ ca& q& 1\\ ab& r& 1\end{array}\right|$ =

• A. $1$
• B. $0$
• C. $abc$
• D. $a^2+b^2+c^2$

Answer 1

The correct option is B $0$

Expanding along $C_1,$ we get

$\left|\begin{array}{ccc}bc& p& 1\\ ca& q& 1\\ ab& r& 1\end{array}\right|=bc(q-r)-ca(p-r)+ab(p-q)$

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

Now given that $a,b,c$ are the $pth,qth$ and $rth$ terms respectively of an $H.P.$

Hence $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ will be the

$pth,qth$ and $rth$ terms respectively of an $A.P.$

Let $x$ and $y$ be the first term and common difference of the corresponding $A.P.$

$\therefore \frac{1}{a}=x+(p-1)y$ ...(1)

$\frac{1}{b}=x+(q-1)y$ ...(2)

$\frac{1}{c}=x+(r-1)y$ ...(3)

Subtracting (2) from (1), we get

$\frac{1}{a}-\frac{1}{b}=(p-q)y\Rightarrow \frac{(b-a)}{y}=(p-q)ab$

Similarly (2)-(3) gives $\frac{(c-a)}{y}=(q-r)bc$

and (3)-(1) gives $\frac{(a-c)}{y}=(r-p)ac.$

Adding them we get

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

$=\frac{1}{y}(b-a+c-b+a-c)=0$