Question

Find 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 $p^{th},(2q{)}^{th}$ and $(3r{)}^{th}$ terms of an H.P.

Answer 1

Given $x,y$ and $z$ are respectively the $p^{th},(2q{)}^{th}$ and $(3r{)}^{th}$ terms of a harmonic progression.
$\Rightarrow x=\frac{1}{A+(p-1)D},y=\frac{1}{A+(2q-1)D}$ and $z=\frac{1}{A+(3r-1)D}$
where $A,D$ are first term and common difference of corresponding Arithmetic progression.
$\left|\begin{array}{ccc}yz& zx& xy\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|=xyz\left|\begin{array}{ccc}1/x& 1/y& 1/z\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|$
$=xyz\left|\begin{array}{ccc}A+(p-1)D& A+(2q-1)D& A+(3r-1)D\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|$
applying $R_1\to R_1-A{R}_3$ gives
$=xyzD\left|\begin{array}{ccc}p-1& 2q-1& 3r-1\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|$
applying $R_1+R_3$ gives
$=xyzD\left|\begin{array}{ccc}p& 2q& 3r\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|=0$
$\therefore \left|\begin{array}{ccc}yz& zx& xy\\ p& 2q& 3r\\ 1& 1& 1\end{array}\right|=0$