Question

If $a_1, a_2, a_3, \dots, a_{12}$ are in A.P. and

$\Delta_1 = \begin{vmatrix} a_1 a_5 & a_1 & a_2 \\ a_2 a_6 & a_2 & a_3 \\ a_3 a_7 & a_3 & a_4 \end{vmatrix}, \Delta_2 = \begin{vmatrix} a_2 a_{10} & a_2 & a_3 \\ a_3 a_{11} & a_3 & a_4 \\ a_4 a_{12} & a_4 & a_5 \end{vmatrix}$

then $\Delta_1 : \Delta_2$ is equal to __

Answer 1

1:1

Answer 2

The elements $a_1, a_2, a_3, \dots, a_{12}$ are in A.P. Let the common difference be $d$. Thus, $a_{n+1} - a_n = d$ for any $n$.

Let's analyze the determinant $\Delta_1$: $\Delta_1 = \begin{vmatrix} a_1 a_5 & a_1 & a_2 \\ a_2 a_6 & a_2 & a_3 \\ a_3 a_7 & a_3 & a_4 \end{vmatrix}$

Perform row operations to simplify the determinant:

1. $R_2 \to R_2 - R_1$
2. $R_3 \to R_3 - R_2$

The new determinant is: $\Delta_1 = \begin{vmatrix} a_1 a_5 & a_1 & a_2 \\ a_2 a_6 - a_1 a_5 & a_2 - a_1 & a_3 - a_2 \\ a_3 a_7 - a_2 a_6 & a_3 - a_2 & a_4 - a_3 \end{vmatrix}$

Since $a_n$ are in A.P., we have $a_2 - a_1 = d$, $a_3 - a_2 = d$, $a_4 - a_3 = d$. $\Delta_1 = \begin{vmatrix} a_1 a_5 & a_1 & a_2 \\ a_2 a_6 - a_1 a_5 & d & d \\ a_3 a_7 - a_2 a_6 & d & d \end{vmatrix}$

Now, perform another row operation: 3. $R_3 \to R_3 - R_2$

$\Delta_1 = \begin{vmatrix} a_1 a_5 & a_1 & a_2 \\ a_2 a_6 - a_1 a_5 & d & d \\ (a_3 a_7 - a_2 a_6) - (a_2 a_6 - a_1 a_5) & 0 & 0 \end{vmatrix}$

Let $X = (a_3 a_7 - a_2 a_6) - (a_2 a_6 - a_1 a_5)$. The determinant simplifies to: $\Delta_1 = X \begin{vmatrix} a_1 & a_2 \\ d & d \end{vmatrix}$ $\Delta_1 = X (a_1 d - a_2 d) = X d (a_1 - a_2) = X d (-d) = -X d^2$

Now we need to calculate $X$. $X = a_3 a_7 - 2 a_2 a_6 + a_1 a_5$. Let $a_n = a + (n-1)d$, where $a$ is the first term. $a_1 = a$ $a_2 = a+d$ $a_3 = a+2d$ $a_5 = a+4d$ $a_6 = a+5d$ $a_7 = a+6d$

Substitute these into the expression for $X$: $X = (a+2d)(a+6d) - 2(a+d)(a+5d) + a(a+4d)$ $X = (a^2 + 8ad + 12d^2) - 2(a^2 + 6ad + 5d^2) + (a^2 + 4ad)$ $X = a^2 + 8ad + 12d^2 - 2a^2 - 12ad - 10d^2 + a^2 + 4ad$

Combine like terms: $X = (1 - 2 + 1)a^2 + (8 - 12 + 4)ad + (12 - 10)d^2$ $X = 0a^2 + 0ad + 2d^2$ $X = 2d^2$.

So, $\Delta_1 = -(2d^2)d^2 = -2d^4$.

Now, let's analyze $\Delta_2$: $\Delta_2 = \begin{vmatrix} a_2 a_{10} & a_2 & a_3 \\ a_3 a_{11} & a_3 & a_4 \\ a_4 a_{12} & a_4 & a_5 \end{vmatrix}$

This determinant has the same structure as $\Delta_1$. The indices are just shifted. Applying the same row operations as for $\Delta_1$:

1. $R_2 \to R_2 - R_1$
2. $R_3 \to R_3 - R_2$

$\Delta_2 = \begin{vmatrix} a_2 a_{10} & a_2 & a_3 \\ a_3 a_{11} - a_2 a_{10} & d & d \\ a_4 a_{12} - a_3 a_{11} & d & d \end{vmatrix}$

Now, perform $R_3 \to R_3 - R_2$: $\Delta_2 = \begin{vmatrix} a_2 a_{10} & a_2 & a_3 \\ a_3 a_{11} - a_2 a_{10} & d & d \\ (a_4 a_{12} - a_3 a_{11}) - (a_3 a_{11} - a_2 a_{10}) & 0 & 0 \end{vmatrix}$

Let $Y = (a_4 a_{12} - a_3 a_{11}) - (a_3 a_{11} - a_2 a_{10})$. The determinant simplifies to: $\Delta_2 = Y \begin{vmatrix} a_2 & a_3 \\ d & d \end{vmatrix}$ $\Delta_2 = Y (a_2 d - a_3 d) = Y d (a_2 - a_3) = Y d (-d) = -Y d^2$

Now we need to calculate $Y$. $Y = a_4 a_{12} - 2 a_3 a_{11} + a_2 a_{10}$. Substitute $a_n = a + (n-1)d$: $a_2 = a+d$ $a_3 = a+2d$ $a_4 = a+3d$ $a_{10} = a+9d$ $a_{11} = a+10d$ $a_{12} = a+11d$

Substitute these into the expression for $Y$: $Y = (a+3d)(a+11d) - 2(a+2d)(a+10d) + (a+d)(a+9d)$ $Y = (a^2 + 14ad + 33d^2) - 2(a^2 + 12ad + 20d^2) + (a^2 + 10ad + 9d^2)$ $Y = a^2 + 14ad + 33d^2 - 2a^2 - 24ad - 40d^2 + a^2 + 10ad + 9d^2$

Combine like terms: $Y = (1 - 2 + 1)a^2 + (14 - 24 + 10)ad + (33 - 40 + 9)d^2$ $Y = 0a^2 + 0ad + 2d^2$ $Y = 2d^2$.

So, $\Delta_2 = -(2d^2)d^2 = -2d^4$.

Finally, we need to find the ratio $\Delta_1 : \Delta_2$: $\Delta_1 : \Delta_2 = (-2d^4) : (-2d^4) = 1 : 1$

The final answer is $\boxed{1:1}$.

Explanation of the solution:

1. Simplify Determinants using Row Operations: Both determinants $\Delta_1$ and $\Delta_2$ have a similar structure. By applying row operations $R_2 \to R_2 - R_1$ and then $R_3 \to R_3 - R_2$, two elements in the third row become zero. This reduces the determinant to a product of one element from the first column of the modified matrix and a $2 \times 2$ minor.
2. Utilize A.P. Properties: The differences $a_{n+1} - a_n = d$ are used directly in the row operations, simplifying the second and third columns. The $2 \times 2$ minor that results is of the form $\begin{vmatrix} a_i & a_{i+1} \\ d & d \end{vmatrix}$, which evaluates to $d(a_i - a_{i+1}) = d(-d) = -d^2$.
3. Calculate the Remaining Term: The non-zero element in the third row of the modified determinant is a second-order difference of terms like $a_n a_{n+k}$. For $\Delta_1$, this term is $X = a_3 a_7 - 2 a_2 a_6 + a_1 a_5$. For $\Delta_2$, it is $Y = a_4 a_{12} - 2 a_3 a_{11} + a_2 a_{10}$.
4. Evaluate Second Differences: Since $a_n$ is an A.P., $a_n = a+(n-1)d$. A product of two terms from an A.P., like $a_n a_{n+k}$, is a quadratic function of $n$. The second difference of a quadratic $Pn^2+Qn+R$ is $2P$. In our case, the coefficient of $n^2$ in $a_n a_{n+k}$ is $d^2$. Thus, both $X$ and $Y$ evaluate to $2d^2$.
5. Calculate Determinants: Both $\Delta_1$ and $\Delta_2$ simplify to $-(2d^2)d^2 = -2d^4$.
6. Find the Ratio: The ratio $\Delta_1 : \Delta_2$ is therefore $(-2d^4) : (-2d^4) = 1:1$.