Let a₁, a₂, a₃, ....., a₁₀ be in G.P. with aᵢ 0 for i = 1, 2, ....., 10 and S be the set of pairs

Q: Wait, why does the log of a GP become an AP? I thought logs were for multiplication.

That is a brilliant observation! Logs do turn multiplication into addition, which is exactly why they are so powerful here. If $a_i = aR^{i-1}$, then $\log(a_i) = \log(a) + (i-1)\log(R)$. If you look at this, it is a linear equation in terms of $i$. The term $\log(a)$ is a constant, and $(i-1)\log(R)$ is a multiple of the index. This is the definition of an Arithmetic Progression with a first term of $\log(a)$ and a common difference of $\log(R)$. You are not just doing math; you are changing the 'language' of the sequence to make it easier to solve!

Q: Can I just expand the determinant directly? It looks like a standard $3 \times 3$ matrix.

Technically, yes, you could expand it, but that is the 'brute force' trap that JEE examiners set for you. If you expand it, you will be dealing with massive algebraic expressions involving $r$ and $k$ that will take you ten minutes and likely lead to a calculation error. The beauty of this problem lies in recognizing the symmetry. In JEE Advanced, if you see a determinant with a pattern, there is almost always a property-based shortcut. Always look for column or row operations before you start expanding!

Q: Does the value of $a$ or $R$ matter? The answer seems to be 'Infinitely many' regardless of the GP terms.

You have hit the nail on the head! The specific values of $a$ and $R$ do not matter because the determinant structure relies on the *differences* between terms. As long as the sequence is a GP, the differences of the logs will always be the constant $d = \log(R)$. Because the determinant evaluates to zero based on the *structure* of the columns (making them identical), the actual values of $a$ and $R$ cancel out or become irrelevant. This is the elegance of the problem—it is a universal truth for any GP!

Question

Let $a_1, a_2, a_3, ....., a_{10}$ be in G.P. with $a_i > 0$ for $i = 1, 2, ....., 10$ and $S$ be the set of pairs $(r, k), r, k \in N$ (the set of natural numbers) for which $\begin{vmatrix} \log_e a_1^r a_2^k & \log_e a_2^r a_3^k & \log_e a_3^r a_4^k \ \log_e a_4^r a_5^k & \log_e a_5^r a_6^k & \log_e a_6^r a_7^k \ \log_e a_7^r a_8^k & \log_e a_8^r a_9^k & \log_e a_9^r a_{10}^k \end{vmatrix} = 0$. Then the number of elements in S, is :

Options:

(A) Infinitely many

(B) 4

(C) 10

(D) 2

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking the Determinant

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of algebra. You see a $3 	imes 3$ determinant filled with logarithms, powers, and indices, and your instinct might be to run.

But I want you to take a deep breath. In the world of JEE Advanced, a complex-looking determinant is rarely a test of your ability to perform tedious arithmetic. It is a test of your ability to see the hidden symmetry. Let us peel back the layers of this problem together.

### Phase 1: The Logarithmic Transformation

We are given a Geometric Progression (G.P.) $a_1, a_2, \dots, a_{10}$ with $a_i > 0$. Let the first term be $a$ and the common ratio be $R$. Thus, the general term is $a_i = a \cdot R^{i-1}$.

Now, look at the elements inside our determinant: $\log_e(a_i^r a_{i+1}^k)$. Using the fundamental laws of logarithms, we know that $\log_e(x^m y^n) = m \log_e x + n \log_e y$. This allows us to rewrite every single element as $r \log_e a_i + k \log_e a_{i+1}$.

Here is the magic: let $L_i = \log_e a_i$. Since $a_i$ is a G.P., $L_i = \log_e(a \cdot R^{i-1}) = \log_e a + (i-1) \log_e R$.

Notice that this is a linear expression in $i$. This means that the sequence $L_1, L_2, \dots, L_{10}$ is an Arithmetic Progression (A.P.) with a common difference $d = \log_e R$. We have just transformed a multiplicative structure into an additive one.

### Phase 2: Constructing the Matrix

Now, let us substitute these $L_i$ terms into our determinant $D$. The matrix looks like this:

$$D = \begin{vmatrix} r L_1 + k L_2 & r L_2 + k L_3 & r L_3 + k L_4 \ r L_4 + k L_5 & r L_5 + k L_6 & r L_6 + k L_7 \ r L_7 + k L_8 & r L_8 + k L_9 & r L_9 + k L_{10} \end{vmatrix}$$

It still looks intimidating, doesn't it? But look at the columns. Column 1 contains terms with indices $(1, 2)$, $(4, 5)$, and $(7, 8)$.

Column 2 shifts these indices by one, and Column 3 shifts them again. This is not random; this is a pattern. In mathematics, patterns are invitations to simplify.

### Phase 3: The Power of Column Operations

We are going to perform two operations that will collapse this matrix. First, let us apply $C_2 	o C_2 - C_1$.

Look at the first row: $(r L_2 + k L_3) - (r L_1 + k L_2)$. Rearranging this, we get $r(L_2 - L_1) + k(L_3 - L_2)$.

Since $L_i$ is an A.P., we know that $L_2 - L_1 = d$ and $L_3 - L_2 = d$. Therefore, the element becomes $rd + kd = (r+k)d$. If you repeat this for the other rows, you will find that every single element in the new Column 2 is $(r+k)d$.

Now, let us apply the second operation: $C_3 	o C_3 - C_2$. Look at the first row again: $(r L_3 + k L_4) - (r L_2 + k L_3)$.

Again, this simplifies to $r(L_3 - L_2) + k(L_4 - L_3)$, which is $rd + kd = (r+k)d$. Just like before, every element in the new Column 3 becomes $(r+k)d$.

### Phase 4: The Grand Conclusion

Look at what we have created. Our determinant now has two identical columns:

$$D = \begin{vmatrix} r L_1 + k L_2 & (r+k)d & (r+k)d \ r L_4 + k L_5 & (r+k)d & (r+k)d \ r L_7 + k L_8 & (r+k)d & (r+k)d \end{vmatrix}$$

By the fundamental properties of determinants, if any two columns are identical, the value of the determinant is **zero**. This result is independent of $r$ and $k$.

It does not matter what natural numbers you choose for $r$ and $k$; the determinant will always be zero. Therefore, the set $S$ contains all possible pairs $(r, k) \in \mathbb{N} 	imes \mathbb{N}$. Since there are infinitely many natural numbers, there are infinitely many such pairs.

You see? We didn't need to calculate a single complex value. We simply needed to trust the structure. When you face the JEE Advanced paper, remember this: look for the symmetry, trust the properties, and let the math reveal its own beauty.

Select Answer:

Visualized Solution

Analyze the Geometric Progression

Apply Logarithmic Properties

Transform G.P. to A.P. using Logarithms

Structure the Determinant

Apply First Column Operation

Verify $C_{2} - C_{1}$ for all Rows

Apply Second Column Operation

Observe Identical Columns

Determine the Number of Elements in S

The Sigma Insight: Properties of Determinants

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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