Question

If ${{a}_{1}},{{a}_{2}},...,{{a}_{n}},{{a}_{n+1}},...$ are in G.P. and ${{a}_{i}}>0\text{ }\forall i$ then $\left| \begin{matrix}
   \log {{a}_{n}} & \log {{a}_{n+2}} & \log {{a}_{n+4}} \\
   \log {{a}_{n+6}} & \log {{a}_{n+8}} & \log {{a}_{n+10}} \\
   \log {{a}_{n+12}} & \log {{a}_{n+14}} & \log {{a}_{n+16}} \\
\end{matrix} \right|$ is equal to
(a) $0$
(b) $n\log {{a}_{n}}$
(c) $n\left( n+1 \right)\log {{a}_{n}}$
(d) none of these

Answer 1

Hint: We should use the general form of nth term of a G.P. and substitute each term in that form. We can then simplify the determinant by using the logarithmic identity $\log {{a}^{m}}=m\log a$. We then need to use basic arithmetic operations on the determinant to find its value.

Complete step by step answer:
We have the following determinant,
Det = $\left| \begin{matrix}
   \log {{a}_{n}} & \log {{a}_{n+2}} & \log {{a}_{n+4}} \\
   \log {{a}_{n+6}} & \log {{a}_{n+8}} & \log {{a}_{n+10}} \\
   \log {{a}_{n+12}} & \log {{a}_{n+14}} & \log {{a}_{n+16}} \\
\end{matrix} \right|$
Let us try to evaluate the following determinant.
Let a be the first term, and r be the common difference of the geometric progression.
Then the nth term of this G.P. is $a{{r}^{n-1}}$ .
We can now replace the value of each term in the determinant.
Det = $\left| \begin{matrix}
   \log a{{r}^{n-1}} & \log a{{r}^{n+1}} & \log a{{r}^{n+3}} \\
   \log a{{r}^{n+5}} & \log a{{r}^{n+7}} & \log a{{r}^{n+9}} \\
   \log a{{r}^{n+11}} & \log a{{r}^{n+13}} & \log a{{r}^{n+15}} \\
\end{matrix} \right|$
We can simplify this determinant as
Det = $\left| \begin{matrix}
   \log {{r}^{n-1}} & \log {{r}^{n+1}} & \log {{r}^{n+3}} \\
   \log {{r}^{n+5}} & \log {{r}^{n+7}} & \log {{r}^{n+9}} \\
   \log {{r}^{n+11}} & \log {{r}^{n+13}} & \log {{r}^{n+15}} \\
\end{matrix} \right|$
We know that property of logarithms that $\log {{a}^{m}}=m\log a$, so by using this property we can express the determinant as
Det = \[\left| \begin{matrix}
   \left( n-1 \right)\log r & \left( n+1 \right)\log r & \left( n+3 \right)\log r \\
   \left( n+5 \right)\log r & \left( n+7 \right)\log r & \left( n+9 \right)\log r \\
   \left( n+11 \right)\log r & \left( n+13 \right)\log r & \left( n+15 \right)\log r \\
\end{matrix} \right|\]
 We can now take the term $\log r$ as common from each column. Hence, we get
Det = \[{{\left( \log r \right)}^{3}}\left| \begin{matrix}
   \left( n-1 \right) & \left( n+1 \right) & \left( n+3 \right) \\
   \left( n+5 \right) & \left( n+7 \right) & \left( n+9 \right) \\
   \left( n+11 \right) & \left( n+13 \right) & \left( n+15 \right) \\
\end{matrix} \right|\]
We know that we can perform arithmetic operations on rows or columns, without affecting the value of the determinant.
To simplify this determinant, let us perform the following operation ${{C}_{3}}={{C}_{3}}-{{C}_{2}}$.
We now get
Det = \[{{\left( \log r \right)}^{3}}\left| \begin{matrix}
   \left( n-1 \right) & \left( n+1 \right) & 2 \\
   \left( n+5 \right) & \left( n+7 \right) & 2 \\
   \left( n+11 \right) & \left( n+13 \right) & 2 \\
\end{matrix} \right|\]
Let us now perform the operation ${{C}_{2}}={{C}_{2}}-{{C}_{1}}$.
We will get the following determinant,
Det = \[{{\left( \log r \right)}^{3}}\left| \begin{matrix}
   \left( n-1 \right) & 2 & 2 \\
   \left( n+5 \right) & 2 & 2 \\
   \left( n+11 \right) & 2 & 2 \\
\end{matrix} \right|\]
We know by the property of determinants that if the values in two determinants are exactly the same, then the value of the determinant is zero.
Hence, we have
\[\left| \begin{matrix}
   \left( n-1 \right) & 2 & 2 \\
   \left( n+5 \right) & 2 & 2 \\
   \left( n+11 \right) & 2 & 2 \\
\end{matrix} \right|=0\]
Thus, we get
Det = ${{\left( \log r \right)}^{3}}\times 0$
And so, the value of this determinant is
Det = 0.

So, the correct answer is “Option a”.

Note: This determinant involves complex terms. So, we must be careful while handling these terms. We can also solve this problem by using the fact that if a, b, c are in G.P. then the terms log a, log b, log c will be in arithmetic progression (A.P.)