Question

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

• A. $p+q+r$
• B. $0$
• C. $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$
• D. none of these

Answer 1

Answer: (B)

$0$

$T_p=A+(p-1)d$

$\Rightarrow \frac{1}{a}=A+(p-1)d$ ....(1)
$T_q=A+(q-1)d$

$\Rightarrow \frac{1}{b}=A+(q-1)d$ ....(2)
$T_r=A+(r-1)d$

$\Rightarrow \frac{1}{c}=A+(r-1)d$ ...(3)
Subtracting (1) from (2), we get

$\frac{1}{a}-\frac{1}{b}=(p-q)d$

$\Rightarrow d=\frac{\frac{1}{a}-\frac{1}{b}}{p-q}$
Subtracting (2) from (3), we get

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

$\Rightarrow d=\frac{\frac{1}{b}-\frac{1}{c}}{q-r}$
$\Rightarrow \frac{\frac{1}{a}-\frac{1}{b}}{p-q}=\frac{\frac{1}{b}-\frac{1}{c}}{q-r}$
$\Rightarrow c\frac{(b-a)}{p-q}=a\frac{(c-b)}{q-r}$ .....(4)
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 (4))