Question

If ${{\rm{a}}_1}{\rm{,}}{{\rm{a}}_2}{\rm{, }}....{{\rm{a}}_{\rm{n}}}{\rm{,}}{{\rm{a}}_{{\rm{n + 1}}}},....$ are in GP and ${{\rm{a}}_{\rm{i}}} > 0\forall {\rm{i}}$, then\left| {\begin{array}{*{20}{c}} {\log {{\rm{a}}_{\rm{n}}}}&{\log {{\rm{a}}_{{\rm{n + 2}}}}}&{\log {{\rm{a}}_{{\rm{n + 4}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 6}}}}}&{\log {{\rm{a}}_{{\rm{n + 8}}}}}&{\log {{\rm{a}}_{{\rm{n + 10}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 12}}}}}&{\log {{\rm{a}}_{{\rm{n + 14}}}}}&{\log {{\rm{a}}_{{\rm{n + 16}}}}} \end{array}} \right| is equal to
A.$0$
B.${\rm{n}}\log {{\rm{a}}_{\rm{n}}}$
C.${\rm{n(n + }}1)\log {{\rm{a}}_{\rm{n}}}$
D.None of these

Answer 1

Here, we have to use the concept of the Geometric progression (A.P.) as the series given is in GP form. Geometric progression (A.P.) is the sequence of numbers such that the common ratio between the consecutive numbers remains constant. So, in this question, we have to apply the basic matrix operation to simplify the given matrix, and then by solving the matrix we will be able to get the value of the matrix.

Complete step-by-step answer:
It is given that ${{\rm{a}}_1}{\rm{,}}{{\rm{a}}_2}{\rm{, }}....{{\rm{a}}_{\rm{n}}}{\rm{,}}{{\rm{a}}_{{\rm{n + 1}}}},....$is in GP
let a be the first term of the G.P. and r is the common ratio.
We know that the G.P. series is in the form of${\rm{a}},{\rm{ar}},{\rm{a}}{{\rm{r}}^2},{\rm{a}}{{\rm{r}}^3},{\rm{a}}{{\rm{r}}^4},.........$
Therefore, we can write${{\rm{a}}_{\rm{n}}}{\rm{ = a}}{{\rm{r}}^{\rm{n}}}$or${{\rm{a}}_{{\rm{n + 1}}}} = {\rm{a}}{{\rm{r}}^{{\rm{n + 1}}}}$.
So the matrix becomes

{\log {{\rm{a}}_{\rm{n}}}}&{\log {{\rm{a}}_{{\rm{n + 2}}}}}&{\log {{\rm{a}}_{{\rm{n + 4}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 6}}}}}&{\log {{\rm{a}}_{{\rm{n + 8}}}}}&{\log {{\rm{a}}_{{\rm{n + 10}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 12}}}}}&{\log {{\rm{a}}_{{\rm{n + 14}}}}}&{\log {{\rm{a}}_{{\rm{n + 16}}}}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\log {\rm{a}}{{\rm{r}}^{\rm{n}}}}&{\log {\rm{a}}{{\rm{r}}^{{\rm{n + 2}}}}}&{\log {\rm{a}}{{\rm{r}}^{{\rm{n + 4}}}}}\\\ {\log {\rm{a}}{{\rm{r}}^{{\rm{n + 6}}}}}&{\log {\rm{a}}{{\rm{r}}^{{\rm{n + 8}}}}}&{\log {\rm{a}}{{\rm{r}}^{{\rm{n + 10}}}}}\\\ {\log {\rm{a}}{{\rm{r}}^{{\rm{n + 12}}}}}&{\log {\rm{a}}{{\rm{r}}^{{\rm{n + 14}}}}}&{\log {\rm{a}}{{\rm{r}}^{{\rm{n + 16}}}}} \end{array}} \right|$$ Now we have to simplify the above matrix, we get $$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\log {{\rm{a}}_{\rm{n}}}}&{\log {{\rm{a}}_{{\rm{n + 2}}}}}&{\log {{\rm{a}}_{{\rm{n + 4}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 6}}}}}&{\log {{\rm{a}}_{{\rm{n + 8}}}}}&{\log {{\rm{a}}_{{\rm{n + 10}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 12}}}}}&{\log {{\rm{a}}_{{\rm{n + 14}}}}}&{\log {{\rm{a}}_{{\rm{n + 16}}}}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\log {{\rm{r}}^{\rm{n}}}}&{\log {{\rm{r}}^{{\rm{n + 2}}}}}&{\log {{\rm{r}}^{{\rm{n + 4}}}}}\\\ {\log {{\rm{r}}^{{\rm{n + 6}}}}}&{\log {{\rm{r}}^{{\rm{n + 8}}}}}&{\log {{\rm{r}}^{{\rm{n + 10}}}}}\\\ {\log {{\rm{r}}^{{\rm{n + 12}}}}}&{\log {{\rm{r}}^{{\rm{n + 14}}}}}&{\log {{\rm{r}}^{{\rm{n + 16}}}}} \end{array}} \right|$$ Now we have to apply the basic property of log function i.e.$$\log {{\rm{a}}^{\rm{b}}} = {\rm{b}}\log {\rm{a}}$$. Therefore, we get $$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\log {{\rm{a}}_{\rm{n}}}}&{\log {{\rm{a}}_{{\rm{n + 2}}}}}&{\log {{\rm{a}}_{{\rm{n + 4}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 6}}}}}&{\log {{\rm{a}}_{{\rm{n + 8}}}}}&{\log {{\rm{a}}_{{\rm{n + 10}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 12}}}}}&{\log {{\rm{a}}_{{\rm{n + 14}}}}}&{\log {{\rm{a}}_{{\rm{n + 16}}}}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{\rm{n}}\log {\rm{r}}}&{{\rm{(n + 2)}}\log {\rm{r}}}&{{\rm{(n + 4)}}\log {\rm{r}}}\\\ {{\rm{(n + 6)}}\log {\rm{r}}}&{{\rm{(n + 8)}}\log {\rm{r}}}&{{\rm{(n + 10)}}\log {\rm{r}}}\\\ {{\rm{(n + 12)}}\log {\rm{r}}}&{{\rm{(n + 14)}}\log {\rm{r}}}&{{\rm{(n + 16)}}\log {\rm{r}}} \end{array}} \right|$$ Now taking$$\log {\rm{r}}$$common from all the terms of the matrix, we get $$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\log {{\rm{a}}_{\rm{n}}}}&{\log {{\rm{a}}_{{\rm{n + 2}}}}}&{\log {{\rm{a}}_{{\rm{n + 4}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 6}}}}}&{\log {{\rm{a}}_{{\rm{n + 8}}}}}&{\log {{\rm{a}}_{{\rm{n + 10}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 12}}}}}&{\log {{\rm{a}}_{{\rm{n + 14}}}}}&{\log {{\rm{a}}_{{\rm{n + 16}}}}} \end{array}} \right| = {(\log {\rm{r)}}^3}\left| {\begin{array}{*{20}{c}} {\rm{n}}&{{\rm{(n + 2)}}}&{{\rm{(n + 4)}}}\\\ {{\rm{(n + 6)}}}&{{\rm{(n + 8)}}}&{{\rm{(n + 10)}}}\\\ {{\rm{(n + 12)}}}&{{\rm{(n + 14)}}}&{{\rm{(n + 16)}}} \end{array}} \right|$$ Now we have to perform some determinant function i.e. $${{\rm{C}}_3}{\rm{ = }}{{\rm{C}}_3} - {{\rm{C}}_2}$$and$${{\rm{C}}_2}{\rm{ = }}{{\rm{C}}_2} - {{\rm{C}}_1}$$where C is the denoted for columns. Therefore, we get $$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\log {{\rm{a}}_{\rm{n}}}}&{\log {{\rm{a}}_{{\rm{n + 2}}}}}&{\log {{\rm{a}}_{{\rm{n + 4}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 6}}}}}&{\log {{\rm{a}}_{{\rm{n + 8}}}}}&{\log {{\rm{a}}_{{\rm{n + 10}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 12}}}}}&{\log {{\rm{a}}_{{\rm{n + 14}}}}}&{\log {{\rm{a}}_{{\rm{n + 16}}}}} \end{array}} \right| = {(\log {\rm{r)}}^3}\left| {\begin{array}{*{20}{c}} {\rm{n}}&2&2\\\ {{\rm{n}} - {\rm{6}}}&2&2\\\ {{\rm{n}} - {\rm{12}}}&2&2 \end{array}} \right|$$ And according to the properties of the determinants if any two rows or columns of the determinants are equal then the value of the determinant is zero. Therefore, we get $$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\log {{\rm{a}}_{\rm{n}}}}&{\log {{\rm{a}}_{{\rm{n + 2}}}}}&{\log {{\rm{a}}_{{\rm{n + 4}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 6}}}}}&{\log {{\rm{a}}_{{\rm{n + 8}}}}}&{\log {{\rm{a}}_{{\rm{n + 10}}}}}\\\ {\log {{\rm{a}}_{{\rm{n + 12}}}}}&{\log {{\rm{a}}_{{\rm{n + 14}}}}}&{\log {{\rm{a}}_{{\rm{n + 16}}}}} \end{array}} \right| = {(\log {\rm{r)}}^3} \times 0 = 0$$ Hence, 0 is the value of the determinant. **So, option A is the correct option.** **Note:** We always have to keep in mind the basic A.P. and G.P series A.P. series is $${\rm{a}},{\rm{a}} + {\rm{d}},{\rm{a}} + 2{\rm{d}},{\rm{a}} + 3{\rm{d,}}............$$ $${{\rm{n}}^{th}}{\rm{term }} = {\rm{ a + }}\left( {{\rm{n - 1}}} \right){\rm{d}}$$ Where a is the first term and d is the common difference. G.P. series is $${\rm{a}},{\rm{ar}},{\rm{a}}{{\rm{r}}^2},{\rm{a}}{{\rm{r}}^3},{\rm{a}}{{\rm{r}}^4},.........$$ $${{\rm{n}}^{{\rm{th}}}}{\rm{term = a}}{{\rm{r}}^{{\rm{n - 1}}}}$$ Where a is the first term of the G.P. and r is the common ratio.