Question

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 \(\left| \begin{array}{lll}a_{1} & a_{2} & 1 a_{3} & a_{4} & 1 a_{5} & a_{6} & 1 \end{array} \right|\) is equal to

• A. 1
• B. 2
• C. \(\frac{1}{2}\)
• D. 4
• E. 0

Hint:

If two columns (or rows) of a determinant are proportional, the determinant is zero.

Answer 1

Answer: (E)

Step 1: Concept:
• Given sequence: \[ a_1=2,\; a_2=1,\; a_3=\frac{1}{2},\; a_4=\frac{1}{4},\; a_5=\frac{1}{8},\; a_6=\frac{1}{16} \]
• Observe that the terms are in a geometric progression (GP).

Step 2: Key Observation:
• Columns of the determinant are: \[ C_1 = (a_1, a_3, a_5), \quad C_2 = (a_2, a_4, a_6) \]
• Using GP property: \[ a_3 = r^2 a_1,\quad a_5 = r^4 a_1 \] \[ \Rightarrow C_1 = a_1 (1, r^2, r^4) \]
• Similarly: \[ a_4 = r^2 a_2,\quad a_6 = r^4 a_2 \] \[ \Rightarrow C_2 = a_2 (1, r^2, r^4) \]
• Hence, \[ C_2 = \frac{a_2}{a_1} \cdot C_1 \]
• So, columns \(C_1\) and \(C_2\) are proportional.

Step 3: Final Answer:
• If any two columns of a determinant are proportional, the determinant is zero.
• Therefore, value of determinant = \(0\)
• Correct Option: (E)