Want to ask us a question? Click here
Browse Questions
 Ad
0 votes

# If $a_1,a_2,a_3.....a_n$ are in G.P then the determinant $\Delta=\begin{vmatrix}\log a_n&\log a_{n+1}& \log a_{n+2}\\\log a_{n+3}&\log a_{n+4}&\log a_{n+5}\\\log a_{n+6}&\log a_{n+7}&\log a_{n+8}\end{vmatrix}$ is equal to

$(a)\;1\qquad(b)\;0\qquad(c)\;4\qquad(d)\;2$

Can you answer this question?

## 1 Answer

0 votes
$a_1,a_2,a_3$ are in G.P
$\therefore$ Using $a_n=ar^{n-1}$ we get the determinant as
$\begin{vmatrix}\log ar^{n-1}&\log ar^n&\log ar^{n+1}\\\log ar^{n+2}&\log ar^{n+3}&\log ar^{n+4}\\\log ar^{n+5}&\log ar^{n+6}&\log ar^{n+7}\end{vmatrix}$
Operating $C_3-C_2$ and $C_2-C_1$ and using $\log m-\log n=\log \large\frac{m}{n}$ we get,
$\Rightarrow \begin{vmatrix}\log ar^{n-1}&\log r&\log r\\\log ar^{n+2}&\log r&\log r\\\log a r^{n+5}&\log r&\log r\end{vmatrix}$
$\Rightarrow 0$
When two columns are identical.
Hence (b) is the correct answer.
answered Nov 22, 2013

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer