Question Number 7911 by ashis786 last updated on 24/Sep/16
$$\mathrm{If}\:{a}>\mathrm{0},\:{b}>\mathrm{0},\:{c}>\mathrm{0}\:\mathrm{are}\:\mathrm{respectively}\:\mathrm{the} \\ $$$${p}^{\mathrm{th}} ,\:{q}^{\mathrm{th}} ,\:{r}^{\mathrm{th}} \:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{GP},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{determinant}\:\begin{vmatrix}{\mathrm{log}\:{a}}&{{p}}&{\mathrm{1}}\\{\mathrm{log}\:{b}}&{{q}}&{\mathrm{1}}\\{\mathrm{log}\:{c}}&{{r}}&{\mathrm{1}}\end{vmatrix}\:\mathrm{is} \\ $$
Answered by prakash jain last updated on 24/Sep/16
$$\mathrm{Let}\:\mathrm{first}\:\mathrm{term}\:\mathrm{of}\:\mathrm{GP}\:\mathrm{be}\:{xy}\:\mathrm{with}\:\mathrm{common}\:\mathrm{ratio}\:{y} \\ $$$${a}={xy}^{{p}} \Rightarrow\mathrm{log}\:{a}=\mathrm{log}\:{x}+{p}\mathrm{log}\:{y} \\ $$$${b}={xy}^{{q}} \Rightarrow\mathrm{log}\:{b}=\mathrm{log}\:{x}+{q}\mathrm{log}\:{y} \\ $$$${c}={xy}^{{r}} \Rightarrow\mathrm{log}\:{c}=\mathrm{log}\:{x}+{r}\mathrm{log}\:{y} \\ $$$$\begin{vmatrix}{\mathrm{log}\:{a}}&{{p}}&{\mathrm{1}}\\{\mathrm{log}\:{b}}&{{q}}&{\mathrm{1}}\\{\mathrm{log}\:{c}}&{{r}}&{\mathrm{1}}\end{vmatrix}=\begin{vmatrix}{\mathrm{log}\:{x}+{p}\mathrm{log}\:{y}}&{{p}}&{\mathrm{1}}\\{\mathrm{log}{x}+{q}\mathrm{log}\:{y}}&{{q}}&{\mathrm{1}}\\{\mathrm{log}\:{x}+{r}\mathrm{log}\:{y}}&{{r}}&{\mathrm{1}}\end{vmatrix} \\ $$$$=\begin{vmatrix}{\mathrm{log}\:{x}}&{{p}}&{\mathrm{1}}\\{\mathrm{log}{x}}&{{q}}&{\mathrm{1}}\\{\mathrm{log}\:{x}}&{{r}}&{\mathrm{1}}\end{vmatrix}+\begin{vmatrix}{{p}\mathrm{log}\:{y}}&{{p}}&{\mathrm{1}}\\{{q}\mathrm{log}\:{y}}&{{q}}&{\mathrm{1}}\\{{r}\mathrm{log}\:{y}}&{{r}}&{\mathrm{1}}\end{vmatrix} \\ $$$$=\mathrm{log}\:{x}\begin{vmatrix}{\mathrm{1}}&{{p}}&{\mathrm{1}}\\{\mathrm{1}}&{{q}}&{\mathrm{1}}\\{\mathrm{1}}&{{r}}&{\mathrm{1}}\end{vmatrix}+\mathrm{log}\:{y}\begin{vmatrix}{{p}}&{{p}}&{\mathrm{1}}\\{{q}}&{{q}}&{\mathrm{1}}\\{{r}}&{{r}}&{\mathrm{1}}\end{vmatrix} \\ $$$$\mathrm{C1}\:=\mathrm{C3}\:\left(\mathrm{in}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{determinant}\right)\:\:\:\mathrm{and}\:\mathrm{C1}=\mathrm{C2}\:\mathrm{in}\:\mathrm{2}^{\mathrm{nd}} \\ $$$$=\left(\mathrm{log}\:{x}\right)×\mathrm{0}+\mathrm{log}\:{y}×\mathrm{0}=\mathrm{0} \\ $$
Commented by Rasheed Soomro last updated on 24/Sep/16
$${Also}\:{like}\:{your}\:{way}\:{of}\:{correction}.\:\:\: \\ $$
Commented by ashis786 last updated on 24/Sep/16
$${thank}\:{you}\:{so}\:{much}… \\ $$
Commented by sandy_suhendra last updated on 24/Sep/16
$${if}\:{the}\:{first}\:{term}\:{of}\:{GP}\:={x}\:{and}\:{the}\:{ratio}={y} \\ $$$${so}\:\:\:{a}={the}\:{p}^{{th}} \:{term}\:{of}\:{GP}\:=\:{xy}^{\:{p}−\mathrm{1}} \: \\ $$$${isn}'{t}\:{it}? \\ $$
Commented by prakash jain last updated on 24/Sep/16
$$\mathrm{Thanks}.\:\mathrm{Updated}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{start}\:\mathrm{with} \\ $$$$\mathrm{first}\:\mathrm{term}\:{xy}. \\ $$