Question

If a, b, c (all +ve) are the pth, qth and rth terms respectively of a geometric progression, then prove that $\left|\begin{array}{ccc}loga& p& 1\\ logb& q& 1\\ logc& r& 1\end{array}\right|$ = 0

Answer 1

Let A be the 1st term and R the common ratio of G.P., then
$a=T_p=A{R}^{p-1}$
$\therefore$ $\log a=\log A+(p-1)\log R.$
Similarly $b=T_q,c=T_r$ etc.

$\therefore △=\left|\begin{array}{ccc}logA+(p-1)\log R& p& 1\\ logA+(q-1)\log R& q& 1\\ logA+(r-1)\log R& r& 1\end{array}\right|$

Split into two determinants and in the first take log A common and in the second take log R common

$△=\log A\left|\begin{array}{ccc}1& p& 1\\ 1& q& 1\\ 1& r& 1\end{array}\right|+\log R\left|\begin{array}{ccc}p-1& p& 1\\ q-1& q& 1\\ r-1& r& 1\end{array}\right|$

Apply $C_1-C_2+C_3$ in the second
$△=0+\log R\left|\begin{array}{ccc}0& p& 1\\ 0& q& 1\\ 0& r& 1\end{array}\right|=0$