If $a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots$ are in G.P., then the determinant $$\Delta=\left|\begin{array}{lll} \log a_{n} & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{array}\right| \text{ is equal to}$$

JEE Main · 2005 · 08 May 2005 · Offline · Determinants

a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots

are in G.P., then the determinant

Δ = lo g a_{n} lo g a_{n + 3} lo g a_{n + 6} lo g a_{n + 1} lo g a_{n + 4} lo g a_{n + 7} lo g a_{n + 2} lo g a_{n + 5} lo g a_{n + 8} is equal to

(A)1
(B)0✓ Correct
(C)4
(D)2

Answer: (B)

Solution

Since the terms are in G.P.,

lo g a_{k} = lo g a_{1} + (k - 1) lo g r

, so each entry of the determinant is a linear expression in its index. Moving down the rows, corresponding entries increase by the constant

3 lo g r

, which makes the three rows themselves an arithmetic progression: the second row is the average of the first and third. A determinant in which one row is the mean of the other two has linearly dependent rows, so its value is

0

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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