1. Key Concepts and Formulas

This problem beautifully combines the properties of Geometric Progressions (G.P.) with the properties of logarithms and determinants. To solve it efficiently, we need to recall these fundamental concepts:

• Geometric Progression (G.P.): A sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
  • If $a_1, a_2, a_3, \dots, a_k, \dots$ are in G.P. with first term $A$ and common ratio $R$, then the $k$-th term is given by: $a_k = A \cdot R^{k-1}$
• Logarithm Properties:
  • Product Rule: $\log_b (MN) = \log_b M + \log_b N$
  • Power Rule: $\log_b (M^k) = k \log_b M$
  • (The base of the logarithm doesn't matter for this problem, as long as it's consistent and valid.)
• Determinant Properties:
  • If any two rows or any two columns of a determinant are identical, then the value of the determinant is zero.
  • The value of a determinant remains unchanged if we apply an operation of the form $R_i \to R_i + k R_j$ (or $C_i \to C_i + k C_j$).
  • If a row or column consists of elements in Arithmetic Progression (A.P.), specific row/column operations can often simplify the determinant significantly.

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2. Analyzing the Terms of the G.P. under Logarithm

Let the given G.P. be $a_1, a_2, \dots, a_n, \dots$ with first term $A$ and common ratio $R$. Then, any term $a_k$ can be written as $a_k = A \cdot R^{k-1}$.

Now, let's take the logarithm of a general term $a_k$: $\log a_k = \log (A \cdot R^{k-1})$ Using the logarithm product rule, $\log(MN) = \log M + \log N$: $\log a_k = \log A + \log (R^{k-1})$ Using the logarithm power rule, $\log(M^k) = k \log M$: $\log a_k = \log A + (k-1) \log R$

This is a crucial observation! If we let $\log A = x$ and $\log R = y$, then: $\log a_k = x + (k-1)y$ This means that the sequence $\log a_1, \log a_2, \log a_3, \dots$ forms an Arithmetic Progression (A.P.) with first term $x$ and common difference $y$.

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3. Expressing the Determinant Elements in Terms of the A.P.

The elements of our given determinant are $\log a_n, \log a_{n+1}, \dots, \log a_{n+8}$. Let's explicitly write these terms using our A.P. form $\log a_k = x + (k-1)y$:

• $\log a_n = x + (n-1)y$
• $\log a_{n+1} = x + (n+1-1)y = x + ny$
• $\log a_{n+2} = x + (n+2-1)y = x + (n+1)y$
• $\log a_{n+3} = x + (n+3-1)y = x + (n+2)y$
• $\log a_{n+4} = x + (n+4-1)y = x + (n+3)y$
• $\log a_{n+5} = x + (n+5-1)y = x + (n+4)y$
• $\log a_{n+6} = x + (n+6-1)y = x + (n+5)y$
• $\log a_{n+7} = x + (n+7-1)y = x + (n+6)y$
• $\log a_{n+8} = x + (n+8-1)y = x + (n+7)y$

Now, substitute these expressions into the determinant: \Delta = \left| {\begin{matrix} {x+(n-1)y} & {x+ny} & {x+(n+1)y} \\ {x+(n+2)y} & {x+(n+3)y} & {x+(n+4)y} \\ {x+(n+5)y} & {x+(n+6)y} & {x+(n+7)y} \cr } \right|

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4. Simplifying the Determinant Using Column Operations

Our goal is to simplify this determinant. Since the elements are in A.P. along each row (and also down each column), applying column or row operations that subtract adjacent terms will be very effective. This is because the difference between consecutive terms in an A.P. is constant (the common difference).

Let's apply the following column operations:

1. $C_2 \to C_2 - C_1$ (Replace Column 2 with Column 2 minus Column 1)
2. $C_3 \to C_3 - C_2$ (Replace Column 3 with Column 3 minus Column 2)

Why these operations? We want to simplify the terms. Subtracting consecutive terms of an A.P. will yield the common difference, making the elements much simpler.

Let's perform $C_2 \to C_2 - C_1$:

• Element $(1,2)$: $(x+ny) - (x+(n-1)y) = y$
• Element $(2,2)$: $(x+(n+3)y) - (x+(n+2)y) = y$
• Element $(3,2)$: $(x+(n+6)y) - (x+(n+5)y) = y$

So, the new second column becomes $\begin{pmatrix} y \\ y \\ y \end{pmatrix}$.

Now, let's perform $C_3 \to C_3 - C_2$:

• Element $(1,3)$: $(x+(n+1)y) - (x+ny) = y$
• Element $(2,3)$: $(x+(n+4)y) - (x+(n+3)y) = y$
• Element $(3,3)$: $(x+(n+7)y) - (x+(n+6)y) = y$

So, the new third column becomes $\begin{pmatrix} y \\ y \\ y \end{pmatrix}$.

The determinant now transforms to: \Delta = \left| {\begin{matrix} {x+(n-1)y} & {y} & {y} \\ {x+(n+2)y} & {y} & {y} \\ {x+(n+5)y} & {y} & {y} \cr } \right|

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5. Evaluating the Simplified Determinant

Observe the transformed determinant. We can clearly see that Column 2 ($C_2$) and Column 3 ($C_3$) are identical.

According to the properties of determinants, if any two columns (or rows) are identical, the value of the determinant is zero.

Therefore, $\Delta = 0$.

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6. Conclusion and Key Takeaways

The value of the determinant is 0.

• The correct option is (B).

Key Takeaways for JEE Aspirants:

1. Logarithms of a G.P. form an A.P.: This is a fundamental property that frequently appears in problems involving both sequences and series and logarithms. Always look for this connection.
   • If $a_k = A \cdot R^{k-1}$, then $\log a_k = \log A + (k-1)\log R$.
2. Determinants with A.P. elements: When determinant elements are in A.P., operations like $R_i \to R_i - R_{i-1}$ or $C_i \to C_i - C_{i-1}$ are extremely powerful. Applying these operations twice will often result in a row or column of zeros, or two identical rows/columns, directly leading to a determinant value of zero.
3. Strategic Use of Properties: The ability to identify and apply the correct properties (G.P., Logarithm, Determinant) in a strategic sequence is what makes these problems solvable efficiently. Don't just calculate blindly; look for patterns and simplifications.