Q.Given that $a$, $b$ and $c$ are in A.P., evaluate $\Delta = \begin{vmatrix} 2y+4 & 5y+7 & 8y+a \\ 3y+5 & 6y+8 & 9y+b \\ 4y+6 & 7y+9 & 10y+c \end{vmatrix}$.

What is an Arithmetic Progression Determinant?

You already know what an arithmetic progression (AP) is — a sequence where each term differs from the previous one by a fixed number called the common difference $d$. For example, $2, 5, 8, 11, \dots$ is an AP with $d = 3$.

Now imagine you take three terms from an AP — any three, not necessarily consecutive — and arrange them in a $3 \times 3$ determinant like this:

If the numbers in each row (or each column) form an arithmetic progression, something remarkable happens: the determinant is always zero. That is the core idea.

The Intuition

Why should that be true? Think about what a determinant measures. Geometrically, a $3 \times 3$ determinant gives the volume of a parallelepiped formed by three vectors. If those vectors are "linearly dependent" — meaning one can be written as a combination of the others — the volume collapses to zero.

When numbers in a row (or column) are in AP, that row is of the form:

This is a linear function of the column index. The same linear pattern repeats across rows (or columns). That repetition creates a dependency: the second row is a linear combination of the first and third, or something similar. The determinant detects this dependency and returns zero.

The Precise Statement

Let's write it clearly. Suppose we have a determinant:

Case 1 — Rows in AP:

If $a_1, a_2, a_3$ are in AP, and $b_1, b_2, b_3$ are in AP, and $c_1, c_2, c_3$ are in AP, then $\Delta = 0$.

Case 2 — Columns in AP:

If $a_1, b_1, c_1$ are in AP, and $a_2, b_2, c_2$ are in AP, and $a_3, b_3, c_3$ are in AP, then $\Delta = 0$.

A Quick Proof (for the curious)

Take the row-AP case. Let the first row be $a,\ a+d_1,\ a+2d_1$, the second row $b,\ b+d_2,\ b+2d_2$, and the third row $c,\ c+d_3,\ c+2d_3$. Write the determinant:

Now perform column operations: $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$. This gives:

Factor $2$ from the third column:

Now the second and third columns are identical. A determinant with two equal columns is zero. Hence $\Delta = 0$. …