Discussion
The numbers \(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\) and \(a_{6}\) are in G.P. If \(a_{1} = 2\) and the common ratio \(r = \frac{1}{2}\) , then the value of
\[ \begin{vmatrix} a_{1} & a_{2} & 1 \\ a_{3} & a_{4} & 1 \\ a_{5} & a_{6} & 1 \end{vmatrix} \]
is equal to
Answer: E
The determinant is 0.
In a G.P., \(a_{2} = ra_{1}\), \(a_{4} = ra_{3}\) and \(a_{6} = ra_{5}\). So the second column is \(r\) times the first column, and a determinant with two proportional columns is zero:
\[ \begin{vmatrix} a_{1} & ra_{1} & 1 \\ a_{3} & ra_{3} & 1 \\ a_{5} & ra_{5} & 1 \end{vmatrix} = r\begin{vmatrix} a_{1} & a_{1} & 1 \\ a_{3} & a_{3} & 1 \\ a_{5} & a_{5} & 1 \end{vmatrix} = 0 \]
The values \(a_{1} = 2\) and \(r = \frac{1}{2}\) do not matter; any G.P. gives 0, so the non-zero options are wrong.
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