Question

The value of $\left|\begin{array}{ccc}yz& zx& xy\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|,$ where $x,y,z$ are respectively the $pth,2qth$ and $3rth$ terms of any $H.P.,$ is

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

Answer 1

Answer: (B)

$0$

$x=\frac{1}{a+(p-1)d}$

$y=\frac{1}{a+(2q-1)d}$

$z=\frac{1}{a+(3r-1)d}$

Let $a=1=d$

$\Rightarrow x=\frac{1}{p}y=\frac{1}{2q}z=\frac{1}{3r}$

$\Rightarrow △=xyz\left|\begin{array}{ccc}\frac{1}{x}& \frac{1}{y}& \frac{1}{z}\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|$

$\Rightarrow △=xyz\left|\begin{array}{ccc}p& 2q& 3r\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|$

$\Rightarrow △=0$