Question

If a₁a₂a₃ …….. a₉ are in A.P. then the value of \(\begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \\ \end{vmatrix}\) is

• A. \(\frac{9}{2}(a_1+a_9)\)
• B. a₁ + a₉
• C. log_e(log_ee)
• D. 1

Answer 1

The Correct Option is C

If \(a_1, a_2, a_3, ..., a_9\) are in A.P., then we need to find the value of the determinant:

\(\begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{vmatrix}\)

Since the terms are in A.P., we can write \(a_n = a_1 + (n-1)d\), where d is the common difference.

So, \(a_2 = a_1 + d\), \(a_3 = a_1 + 2d\), \(a_4 = a_1 + 3d\), \(a_5 = a_1 + 4d\), \(a_6 = a_1 + 5d\), \(a_7 = a_1 + 6d\), \(a_8 = a_1 + 7d\), \(a_9 = a_1 + 8d\).

Substituting these into the determinant:

\(\begin{vmatrix} a_1 & a_1+d & a_1+2d \\ a_1+3d & a_1+4d & a_1+5d \\ a_1+6d & a_1+7d & a_1+8d \end{vmatrix}\)

We can perform row operations without changing the value of the determinant. Apply \(R_2 \rightarrow R_2 - R_1\) and \(R_3 \rightarrow R_3 - R_1\):

\(\begin{vmatrix} a_1 & a_1+d & a_1+2d \\ 3d & 3d & 3d \\ 6d & 6d & 6d \end{vmatrix}\)

Since the second and third rows are proportional (i.e., \(R_3 = 2R_2\)), the determinant is 0.

Thus, (C) \(\log_e(\log_e e)\) is zero.

Answer 2

Let the common difference of the A.P. be $ d $. Then $ a_n = a_1 + (n-1)d $. The determinant is:

$$ \begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{vmatrix} = \begin{vmatrix} a_1 & a_1+d & a_1+2d \\ a_1+3d & a_1+4d & a_1+5d \\ a_1+6d & a_1+7d & a_1+8d \end{vmatrix} $$

Apply $ R_2 \to R_2 - R_1 $ and $ R_3 \to R_3 - R_1 $:

$$ \begin{vmatrix} a_1 & a_1+d & a_1+2d \\ 3d & 3d & 3d \\ 6d & 6d & 6d \end{vmatrix} $$

Since $ R_3 = 2R_2 $, the rows are linearly dependent, and the determinant is 0.

However, 0 is not one of the choices. Let's simplify by replacing $ R_2 $ by $ R_2 - R_1 $ and $ R_3 $ by $ R_3 - R_2 $. Then the determinant is:

$$ \begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 - a_1 & a_5 - a_2 & a_6 - a_3 \\ a_7 - a_4 & a_8 - a_5 & a_9 - a_6 \end{vmatrix} $$

Since $ a_4 - a_1 = 3d $, $ a_5 - a_2 = 3d $, $ a_6 - a_3 = 3d $, and $ a_7 - a_4 = 3d $, $ a_8 - a_5 = 3d $, $ a_9 - a_6 = 3d $,

$$ \begin{vmatrix} a_1 & a_2 & a_3 \\ 3d & 3d & 3d \\ 3d & 3d & 3d \end{vmatrix} = 0 $$

It seems there might be a typo in the question.

If the question intended the determinant to be:

$$ \begin{vmatrix} a_1 & a_2 & a_3 \\ a_2 & a_3 & a_4 \\ a_3 & a_4 & a_5 \end{vmatrix} $$

Then the answer is 0.