l, m, n are the p^th , q^th and r^th term of a G.P all positive, then\,| \beginmatrix \log \,l & p & 1 \log \,m & q & 1 \log \,n & r & 1 \endmatrix |\,equals

Sequences and Series PYQ — JEE Main 2002 · Shift 0 · Q111 with Solution

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Correct answer: D

Let the G.P. have first term $a>0$ and common ratio $t>0$ . Then its $k^{\text{th}}$ term is $T_k=a t^{k-1}.$

So, $l=T_p=a t^{p-1},\qquad m=T_q=a t^{q-1},\qquad n=T_r=a t^{r-1}.$

Since all terms are positive, their logarithms are defined.

$\log l=\log a+(p-1)\log t,$ $\log m=\log a+(q-1)\log t,$ $\log n=\log a+(r-1)\log t.$

Rewrite each as $\log l=(\log a-\log t)+p\log t,$ $\log m=(\log a-\log t)+q\log t,$ $\log n=(\log a-\log t)+r\log t.$

Let $A=\log t,\qquad B=\log a-\log t.$ Then $\log l=Ap+B,\qquad \log m=Aq+B,\qquad \log n=Ar+B.$

The determinant is

\log l & p & 1\\ \log m & q & 1\\ \log n & r & 1 \end{vmatrix} =\begin{vmatrix} Ap+B & p & 1\\ Aq+B & q & 1\\ Ar+B & r & 1 \end{vmatrix}.$$ Now observe that the first column is $$C_1=A C_2 + B C_3,$$ where $C_2=\begin{bmatrix}p\\q\\r\end{bmatrix}$ and $C_3=\begin{bmatrix}1\\1\\1\end{bmatrix}$. Thus the three columns are linearly dependent. Hence the determinant must be zero. So, $$\Delta=0.$$ --- 4. **Check options** The value is $0$. Therefore the correct option is: $$\boxed{\text{D}}$$