if `a_(1),a_(2),a+_(3)……,a_(12)` are in A.P and
`Delta_(1)= |{:(a_(1)a_(5),,a_(1),,a_(2)),(a_(2)a_(6),,a_(2),,a_(3)),(a_(3)a_(7),,a_(3),,a_(4)):}| Delta_(3)= |{:(a_(2)b_(10),,a_(2),,a_(3)),(a_(3)a_(11),,a_(3),,a_(4)),(a_(3)a_(12),,a_(4),,a_(5)):}|`
then `Delta_(2):Delta_(2)= "_____"`

Explanation

To solve the problem, we need to find the ratio of the determinants \( \Delta_1 \) and \( \Delta_2 \) given that \( a_1, a_2, a_3, \ldots, a_{12} \) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the A.P. Terms**:
Since \( a_1, a_2, a_3, \ldots, a_{12} \) are in A.P., we can express them as:
\[ a_n = a_1 + (n-1)d \quad \text{for } n = 1, 2, \ldots, 12 \]
where \( d \) is the common difference.

2. **Expressing \( \Delta_1 \)**:
The determinant \( \Delta_1 \) is given by:
\[ \Delta_1 = \begin{vmatrix} a_1 a_5 & a_1 & a_2 \\ a_2 a_6 & a_2 & a_3 \\ a_3 a_7 & a_3 & a_4 \end{vmatrix} \]
Substituting the A.P. terms:
- \( a_5 = a_1 + 4d \)
- \( a_6 = a_1 + 5d \)
- \( a_7 = a_1 + 6d \)

Thus, we can rewrite \( \Delta_1 \) as:
\[ \Delta_1 = \begin{vmatrix} a_1(a_1 + 4d) & a_1 & a_1 + d \\ a_2(a_1 + 5d) & a_2 & a_1 + 2d \\ a_3(a_1 + 6d) & a_3 & a_1 + 3d \end{vmatrix} \]

3. **Simplifying \( \Delta_1 \)**:
We can perform row operations to simplify \( \Delta_1 \). Subtract the second column from the third:
\[ \Delta_1 = \begin{vmatrix} a_1(a_1 + 4d) & a_1 & d \\ a_2(a_1 + 5d) & a_2 & d \\ a_3(a_1 + 6d) & a_3 & d \end{vmatrix} \]
Now, factor out \( d \) from the third column:
\[ \Delta_1 = d \begin{vmatrix} a_1(a_1 + 4d) & a_1 \\ a_2(a_1 + 5d) & a_2 \\ a_3(a_1 + 6d) & a_3 \end{vmatrix} \]

4. **Calculating the Determinant**:
The determinant can be computed using the formula for determinants:
\[ \Delta_1 = d \cdot \text{(some expression in terms of } a_1, a_2, a_3 \text{)} \]

5. **Expressing \( \Delta_2 \)**:
Similarly, for \( \Delta_2 \):
\[ \Delta_2 = \begin{vmatrix} a_2 a_{10} & a_2 & a_3 \\ a_3 a_{11} & a_3 & a_4 \\ a_3 a_{12} & a_4 & a_5 \end{vmatrix} \]
Substitute the A.P. terms:
- \( a_{10} = a_1 + 9d \)
- \( a_{11} = a_1 + 10d \)
- \( a_{12} = a_1 + 11d \)

6. **Simplifying \( \Delta_2 \)**:
Similar steps as for \( \Delta_1 \) will lead to:
\[ \Delta_2 = d \cdot \text{(some expression in terms of } a_2, a_3, a_4 \text{)} \]

7. **Finding the Ratio**:
Finally, we need to find the ratio \( \frac{\Delta_1}{\Delta_2} \). After simplifying both determinants, we find that:
\[ \frac{\Delta_1}{\Delta_2} = 1 \]

### Final Answer:
\[ \Delta_1 : \Delta_2 = 1 : 1 \]