If a₁, a₂, a₃, ..., aₙ, ... are in G.P., then the determinant Δ = |log aₙ log aₙ₊₁ log aₙ₊₂; log aₙ₊

Q: Wait, why does taking the log turn a G.P. into an A.P.?

Great question! A G.P. is defined by multiplication: $a_k = ar^{k-1}$. When you apply the logarithm, the product rule $\log(xy) = \log(x) + \log(y)$ and the power rule $\log(x^n) = n\log(x)$ transform the multiplication into addition. Specifically, $\log(a_k) = \log(a) + (k-1)\log(r)$. Since $\log(a)$ and $\log(r)$ are constants, this is exactly the form of an A.P. term, $A + (k-1)D$.

Q: Can we just expand the determinant directly to solve it?

Technically, yes, you could expand it, but it would be an algebraic nightmare! In JEE Advanced, the goal is to recognize the pattern. Whenever you see a determinant with logs of G.P. terms, it is almost always a test of your ability to use row/column operations to simplify the matrix, not your ability to do brute-force expansion.

Q: What if the common ratio $r$ is 1?

If $r=1$, then all terms in the G.P. are equal to $a$. Consequently, all terms in the determinant are $\log(a)$. A determinant where all elements are identical (or where all rows/columns are identical) is always zero. So, the result holds true even in that trivial case!

Question

If $a_1, a_2, a_3, \dots, a_n, \dots$ are in G.P., then the determinant $\Delta = \begin{vmatrix} \log a_n & \log a_{n+1} & \log a_{n+2} \ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{vmatrix}$ is equal to

Options:

(A) $1$

(B) $0$

(C) $4$

(D) $2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a sequence $a_1, a_2, a_3, \dots$ which is a Geometric Progression. This means any term $a_k$ can be written as $a_k = ar^{k-1}$, where $a$ is the first term and $r$ is the common ratio.

Now, consider the determinant $\Delta$ where every entry is a logarithm of a G.P. term. Applying the logarithm to our general term, we get:

$$\log(a_k) = \log(ar^{k-1})$$

Using the fundamental properties of logarithms, this splits into $\log(a) + \log(r^{k-1})$, which further simplifies to:

$$\log(a_k) = \log(a) + (k-1)\log(r)$$

This is the key insight. We have transformed a multiplicative sequence into an additive one, where every entry in our determinant is of the form $\log(a) + (	ext{index}-1)\log(r)$.

### The Power of Column Operations

Given the linear structure derived above, the determinant $\Delta$ can be expressed as:

$$\Delta = \begin{vmatrix} \log a + (n-1)\log r & \log a + n\log r & \log a + (n+1)\log r \ \log a + (n+2)\log r & \log a + (n+3)\log r & \log a + (n+4)\log r \ \log a + (n+5)\log r & \log a + (n+6)\log r & \log a + (n+7)\log r \end{vmatrix}$$

When you see a determinant where the entries are in an Arithmetic Progression, your first instinct should be to simplify using column operations. Let us perform the operation $C_2 	o C_2 - C_1$.

For any row, the $\log(a)$ terms will cancel out, and the difference in the coefficients of $\log(r)$ will be exactly one. For example, in the first row, we get:

$$(\log a + n\log r) - (\log a + (n-1)\log r) = \log r$$

This result holds for every row. Consequently, our determinant now has a second column consisting entirely of $\log(r)$.

### The Final Cancellation

We proceed by performing the operation $C_3 	o C_3 - C_2$. Again, the $\log(a)$ terms vanish, and the difference in the coefficients of $\log(r)$ is once more exactly one.

The third column also becomes a column of $\log(r)$. Now, observe the resulting matrix:

$$\Delta = \begin{vmatrix} \log a + (n-1)\log r & \log r & \log r \ \log a + (n+2)\log r & \log r & \log r \ \log a + (n+5)\log r & \log r & \log r \end{vmatrix}$$

We have two identical columns. A fundamental property of determinants states that if any two rows or columns are identical, the value of the determinant is zero.

And just like that, the intimidating wall of logarithms collapses. The final value of the determinant is **0**.

Animated Solution for Mathematics - Matrices and Determinants: If $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ 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

Select Answer:

Visualized Solution

The Given Determinant

General Term of a G.P.

Applying Logarithm to the General Term

Expanding the Logarithmic Expression

Analyzing the Determinant Elements

The Structure of the Determinant

First Column Operation: $C_{2} \to C_{2} - C_{1}$

Updating the Determinant after $C_{2} \to C_{2} - C_{1}$

Second Column Operation: $C_{3} \to C_{3} - C_{2}$

Updating the Determinant after $C_{3} \to C_{3} - C_{2}$

Final Evaluation of the Determinant

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Power of Column Operations

The Final Cancellation

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