Question

The value of the determinant $\left|\begin{array}{ccc}bc& ca& ab\\ p& q& r\\ 1& 1& 1\end{array}\right|$, where a,b and c are respectively the pth, qth and rth terms of a H.P., is

• A. $0$
• B. $abc$
• C. $pqr$
• D. none of these

Answer 1

The correct option is A $0$

Since, $a,b,c$ are in H.P.

$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

$T_p=A+(p-1)d$
$\Rightarrow \frac{1}{a}=A+(p-1)d$ ....(i)
$T_q=A+(q-1)d$
$\Rightarrow \frac{1}{b}=A+(q-1)d$ ....(ii)
$T_r=A+(r-1)d$
$\Rightarrow \frac{1}{c}=A+(r-1)d$ ...(iii)
Subtracting (ii) from (i), we get
$\frac{1}{a}-\frac{1}{b}=(p-q)d$
$\Rightarrow d=\frac{\frac{1}{a}-\frac{1}{b}}{p-q}$
Subtracting (iii) from (ii), we get
$\frac{1}{b}-\frac{1}{c}=(q-r)d$
$\Rightarrow d=\frac{\frac{1}{b}-\frac{1}{c}}{q-r}$

$\Rightarrow \frac{\frac{1}{a}-\frac{1}{b}}{p-q}=\frac{\frac{1}{b}-\frac{1}{c}}{q-r}$
$\Rightarrow c\frac{(b-a)}{p-q}=a\frac{(c-b)}{q-r}$ .....(iv)
Now, $\left|\begin{array}{ccc}bc& ca& ab\\ p& q& r\\ 1& 1& 1\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$=\left|\begin{array}{ccc}c(b-a)& a(c-b)& ab\\ p-q& q-r& r\\ 0& 0& 1\end{array}\right|$
$=c(b-a)(q-r)-a(p-q)(c-b)$
$=0$ ...... (by (iv))