If $x, y, z$ are all positive and are the $p$ th, $q$ th and $r$ th terms of a geometric progression…
If $x, y, z$ are all positive and are the $p$ th, $q$ th and $r$ th terms of a geometric progression respectively, then the value of the determinant
$\left|\begin{array}{lll}
\log x & p & 1 \\
\log y & q & 1 \\
\log z & r & 1
\end{array}\right| \text { equals }$
$\log x y z$
$(p-1)(q-1)(r-1)$
$pqr$
0
Solution
Let $a$ and $R$ be the first term and common ratio of a GP.
$\begin{array}{ll}
\therefore & T_p=a R^{p-1}=x \\
& T_q=a R^{q-1}=y \\
\text { and } & T_r=a R^{r-1}=z
\end{array}$
and $T_r=a R^{r-1}=z$
$\begin{array}{rlrl}
\Rightarrow & \log x & =\log a+(p-1) \log R \\
& & \log y & =\log a+(q-1) \log R \\
\text { and } & & \log z & =\log a+(r-1) \log R
\end{array}$
$\begin{aligned}
& \therefore\left|\begin{array}{lll}
\log x & p & 1 \\
\log y & q & 1 \\
\log z & r & 1
\end{array}\right|=\left|\begin{array}{lll}
\log a+(p-1) \log R & p & 1 \\
\log a+(q-1) \log R & q & 1 \\
\log a+(r-1) \log R & r & 1
\end{array}\right| \\
& =\left|\begin{array}{lll}
\log a & p & 1 \\
\log a & q & 1 \\
\log a & r & 1
\end{array}\right|+\left|\begin{array}{ccc}
(p-1) \log R & p & 1 \\
(q-1) \log R & q & 1 \\
(r-1) \log R & r & 1
\end{array}\right| \\
& =\log a\left|\begin{array}{lll}
1 & p & 1 \\
1 & q & 1 \\
1 & r & 1
\end{array}\right|+\log R\left|\begin{array}{ccc}
p-1 & p-1 & 1 \\
q-1 & q-1 & 1 \\
r-1 & r-1 & 1
\end{array}\right| \\
& \left(C_2 \rightarrow C_2-C_3\right) \\
&
\end{aligned}$
$=0+0=0$
( $\because$ two columns are identical)