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, $pth,\left(2q\right)th,and\left(3r\right)th$ terms of an H.P., is

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

Answer 1

The correct option is B $0$

Given ,$\frac{1}{x}=a+(p-1)d$ ....(1)
$\frac{1}{y}=a+(2q-1)d$ .....(2)
$\frac{1}{z}=a+(3r-1)d$ ....(3)
Subtracting (1) from (2), we get
$\frac{1}{y}-\frac{1}{x}=(2q-p)d$ ....(4)
Subtracting (2) from (3), we get
$\frac{1}{z}-\frac{1}{y}=(3r-2q)d$ .....(5)
Dividing (4) by (5),
$\frac{\frac{1}{y}-\frac{1}{x}}{\frac{1}{z}-\frac{1}{y}}=\frac{(2q-p)}{(3r-2q)}$
$\Rightarrow (3r-2q)(\frac{1}{y}-\frac{1}{x})=(2q-p)(\frac{1}{z}-\frac{1}{y})$
Now, consider $\left|\begin{array}{ccc}yz& zx& xy\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|$
$=xyz\left|\begin{array}{ccc}\frac{1}{x}& \frac{1}{y}& \frac{1}{z}\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|$
$C_2\to C_2-C_1,{C_3\to C}_3-C_2$
$=xyz\left|\begin{array}{ccc}\frac{1}{x}& \frac{1}{y}-\frac{1}{x}& \frac{1}{z}-\frac{1}{y}\\ p& 2q-p& 3r-2q\\ 1& 0& 0\end{array}\right|$
$=xyz\left[\right(3r-2q\left)\right(\frac{1}{y}-\frac{1}{x})-(2q-p\left)\right(\frac{1}{z}-\frac{1}{y}\left)\right]$
$=0$