Question

If $a,b,c$ are $pth,qth$ and $rth$ terms of a GP, then $\left[\begin{array}{ccc}\log a& p& 1\\ \log b& q& 1\\ \log c& r& 1\end{array}\right]$ is equal to -

• A. $0$
• B. $1$
• C. $\log abc$
• D. $pqr$

Answer 1

The correct option is A $0$

As $a,b,c$ are $pth,qth,rth$ term of $G.P,,$ then

Let common ratio of $G.P.$ is $R$ and first term is $A$

$\therefore a=A{R}^{p-1},b=A{R}^{q-1},c=A{R}^{r-1}$

Let $\Delta =\left|\begin{array}{ccc}\log a& p& 1\\ \log b& q& 1\\ \log c& r& 1\end{array}\right|\Delta =\left|\begin{array}{ccc}\log a& p& 1\\ \log b& q& 1\\ \log c& r& 1\end{array}\right|$

Applying $R_2\to R_2-R_1;R_3\to R_3-R_1,$ we get

$\Delta =\left|\begin{array}{ccc}\log a& p& 1\\ \log b-\log a& q-p& 0\\ \log c-\log a& r-p& 0\end{array}\right|$

Expanding along $C_3,$ we get

$\Delta =(r-p)(\log b-\log a)-(q-p)(\log c-\log a)$

$=(r-p)\left(\log \frac{b}{a}\right)-(q-p)\left(\log \frac{c}{a}\right)$

$=(r-p)\left(\log \frac{A{R}^{q-1}}{A{R}^{p-1}}\right)-(q-p)\left(\log \frac{A{R}^{r-1}}{A{R}^{p-1}}\right)$

$=(r-p)\left(\log R^{q-p}\right)-(q-p)\left(\log R^{(r-p)}\right)=(r-p)(q-p)\log R-(q-p)(r-p)\log R=0$