l, m, n are the pth, qth and rth term of a G. P. all positive, then |log l p 1; log m q 1; log n r 1

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

That is a fantastic observation! A G.P. grows exponentially, defined by $T_n = AR^{n-1}$. When we take the logarithm, the exponent $(n-1)$ comes down as a multiplier: $\log(T_n) = \log A + (n-1)\log R$. If you treat the index $n$ as your variable $x$ and $\log(T_n)$ as your variable $y$, the equation becomes $y = x(\log R) + (\log A - \log R)$. This is exactly the slope-intercept form $y = mx + c$, where the slope is $\log R$ and the intercept is $\log A - \log R$. It is a beautiful bridge between algebra and geometry!

Q: Can I just expand the determinant directly? It looks like a lot of algebra.

You absolutely could, but it is a trap! Expanding a $3 \times 3$ determinant with logarithmic terms is a recipe for calculation errors. In JEE Advanced, the examiners often hide a geometric property or a determinant property (like identical columns) to reward those who look for the structure rather than brute-forcing the arithmetic. Always look for the 'shortcut'—it is usually the intended path.

Q: Why is the area of the triangle zero? I don't see a triangle.

You are right to be skeptical! The 'triangle' is a conceptual tool. We map the points $(p, \log l), (q, \log m),$ and $(r, \log n)$ onto a coordinate plane. Because these points satisfy the linear equation $y = mx + c$, they lie on the same straight line. In geometry, three points on a line are 'collinear.' Since they are collinear, they cannot enclose any area. Thus, the area of the triangle they would 'form' is zero, which forces the determinant to be zero.

Question

$l, m, n$ are the $p$th, $q$th and $r$th term of a G. P. all positive, then $\begin{vmatrix} \log l & p & 1 \ \log m & q & 1 \ \log n & r & 1 \end{vmatrix}$ equals

Options:

(A) -1

(B) 2

(C) 1

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to peel back the layers of a problem that looks like a standard determinant exercise but is actually a masterclass in connecting algebra to geometry.

We are given that $l, m,$ and $n$ are the $p^{th}, q^{th},$ and $r^{th}$ terms of a Geometric Progression (G.P.). We need to find the value of the determinant:

$$\Delta = \begin{vmatrix} \log l & p & 1 \ \log m & q & 1 \ \log n & r & 1 \end{vmatrix}$$

At first glance, this looks like a nightmare of logarithmic expansion. But let us pause and look deeper.

### The Algebraic Foundation

Every G.P. is defined by its first term $A$ and common ratio $R$. We know the general term is $T_k = AR^{k-1}$. Therefore, our terms are $l = AR^{p-1}$, $m = AR^{q-1}$, and $n = AR^{r-1}$.

Since the problem involves logarithms, let us apply the log function to these terms. Using the properties $\log(xy) = \log x + \log y$ and $\log(x^a) = a \log x$, we get:

$$\log l = \log A + (p-1) \log R$$
$$\log m = \log A + (q-1) \log R$$
$$\log n = \log A + (r-1) \log R$$

Look closely at these equations. If we define $y = \log(T_n)$ and $x = n$, these equations take the form $y = x(\log R) + (\log A - \log R)$. This is the equation of a straight line $y = mx + c$!

### The Geometric Revelation

This is where the magic happens. Imagine a coordinate plane where the x-axis is the term index ($p, q, r$) and the y-axis is the logarithm of the term ($\log l, \log m, \log n$).

We have three points: $P(p, \log l)$, $Q(q, \log m)$, and $R(r, \log n)$. Because all these points satisfy the same linear equation, they must lie on the same straight line. They are collinear.

Now, recall the formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$:

$$	ext{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix}$$

Our determinant is essentially this area formula (up to a sign change). Since our points $P, Q,$ and $R$ are collinear, they do not form a triangle. The area is zero. Therefore, the determinant must be **0**.

### The Algebraic Proof

If you are still craving the cold, hard rigor of algebra, let us verify this. Substitute our expanded logarithmic expressions into the determinant:

$$\Delta = \begin{vmatrix} \log A + (p-1) \log R & p & 1 \ \log A + (q-1) \log R & q & 1 \ \log A + (r-1) \log R & r & 1 \end{vmatrix}$$

Now, perform the column operation $C_1 	o C_1 - (\log R) C_2$. The first column becomes $\log A - \log R$ for every row. We can factor this constant out:

$$\Delta = (\log A - \log R) \begin{vmatrix} 1 & p & 1 \ 1 & q & 1 \ 1 & r & 1 \end{vmatrix}$$

Wait, look at the determinant now! Column 1 and Column 3 are identical. By the fundamental properties of determinants, if any two columns are identical, the determinant is zero.

Whether you view it through the lens of geometry or the rigor of algebra, the answer is the same: **0**. You have just turned a complex-looking problem into a simple, elegant truth. Keep this intuition, and you will conquer any JEE problem.

Animated Solution for Mathematics - Matrices and Determinants: $l, m, n$ are the $p$ th, $q$ th and $r$ th term of a G. P. all positive, then $lo g l lo g m lo g n p q r 111$ equals

Select Answer:

Visualized Solution

The Determinant of a G.P.

General Terms of the G.P.

Taking the Logarithm

Mapping to a Coordinate Plane

Plotting the First Point

Plotting the Remaining Points

The Straight Line Relationship

Area of a Triangle

Zero Area, Zero Determinant

Algebraic Verification

Applying Column Operations

Factoring Out the Constant

Identical Columns Property

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Algebraic Foundation

The Geometric Revelation

The Algebraic Proof

Similar Questions

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

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots$ are in G.P., then the value of the determinant $lo g a_{n} lo g a_{n + 3} lo g a_{n + 6} lo g a_{n + 1} lo g a_{n + 4} lo g a_{n + 7} lo g a_{n + 2} lo g a_{n + 5} lo g a_{n + 8}$ is

For positive numbers $x, y$ and $z$ , the numerical value of the determinant $1 lo g_{y} x lo g_{z} x lo g_{x} y 1 lo g_{z} y lo g_{x} z lo g_{y} z 1$ is .........

If $α \neq = a, β \neq = b, γ \neq = c$ and $α a a b β b c c γ = 0$ , then $\frac{a}{α - a} + \frac{b}{β - b} + \frac{γ}{γ - c}$ is equal to:

If $α, β \neq = 0$ , and $f (n) = α^{n} + β^{n}$ and $3 1 + f (1) 1 + f (2) 1 + f (1) 1 + f (2) 1 + f (3) 1 + f (2) 1 + f (3) 1 + f (4) = K (1 - α)^{2} (1 - β)^{2} (α - β)^{2}$ , then $K$ is equal to:

The determinant $a b a α + b b c b α + c a α + b b α + c 0$ is equal to zero, if

Let $a, b, c, d$ be in arithmetic progression with common difference $λ$ . If $x + a - c x - 1 x - b + d x + b x + c x + d x + a x + b x + c = 2$ , then value of $λ^{2}$ is equal to

Let $a, b, c$ be such that $b (a + c) \neq = 0$ . If $a - b c a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 + a + 1 a - 1 (- 1)^{n + 2} a b + 1 b - 1 (- 1)^{n + 1} b c - 1 c + 1 (- 1)^{n} c = 0$ , then the value of $n$ is:

If $a + x = b + y = c + z + 1$ , where $a, b, c, x, y, z$ are non-zero distinct real numbers, then $x y z a + y b + y c + y x + a y + b z + c$ is equal to :

Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l,m,n are the pth, qth and rth term of a G. P. all positive, then ​logllogmlogn​pqr​111​​ equals