Using properties of determinant , find the value of ` |{:( 2y +4, 5y+7, 8y+a),(3y+5,6y+8,9y+b),(4y+6, 7y+9,10y+c):}|` ,if a,b and c are in AP.

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 2y + 4 & 5y + 7 & 8y + a \\ 3y + 5 & 6y + 8 & 9y + b \\ 4y + 6 & 7y + 9 & 10y + c \end{vmatrix} \]

given that \( a, b, c \) are in arithmetic progression (AP), we proceed as follows:

### Step 1: Understand the condition of AP
Since \( a, b, c \) are in AP, we have the relation:
\[ b - a = c - b \implies 2b = a + c \implies 2b - a - c = 0 \]
This will be useful later.

### Step 2: Write the determinant
We can express the determinant \( D \) as:
\[ D = \begin{vmatrix} 2y + 4 & 5y + 7 & 8y + a \\ 3y + 5 & 6y + 8 & 9y + b \\ 4y + 6 & 7y + 9 & 10y + c \end{vmatrix} \]

### Step 3: Simplify the determinant
We can simplify the determinant by performing row operations. We will multiply the second row by 2 and divide the entire determinant by 2 to keep it equivalent:
\[ D = \frac{1}{2} \begin{vmatrix} 2y + 4 & 5y + 7 & 8y + a \\ 2(3y + 5) & 2(6y + 8) & 2(9y + b) \\ 4y + 6 & 7y + 9 & 10y + c \end{vmatrix} \]
This results in:
\[ D = \frac{1}{2} \begin{vmatrix} 2y + 4 & 5y + 7 & 8y + a \\ 6y + 10 & 12y + 16 & 18y + 2b \\ 4y + 6 & 7y + 9 & 10y + c \end{vmatrix} \]

### Step 4: Subtract rows
Now, we will perform the operation \( R_2 \leftarrow R_2 - R_1 - R_3 \):
\[ D = \frac{1}{2} \begin{vmatrix} 2y + 4 & 5y + 7 & 8y + a \\ (6y + 10 - (2y + 4) - (4y + 6)) & (12y + 16 - (5y + 7) - (7y + 9)) & (18y + 2b - (8y + a) - (10y + c)) \\ 4y + 6 & 7y + 9 & 10y + c \end{vmatrix} \]

### Step 5: Simplify the second row
Calculating the second row:
1. First element: \( 6y + 10 - 2y - 4 - 4y - 6 = 0 \)
2. Second element: \( 12y + 16 - 5y - 7 - 7y - 9 = 0 \)
3. Third element: \( 18y + 2b - 8y - a - 10y - c = 2b - a - c \)

Thus, we have:
\[ D = \frac{1}{2} \begin{vmatrix} 2y + 4 & 5y + 7 & 8y + a \\ 0 & 0 & 2b - a - c \\ 4y + 6 & 7y + 9 & 10y + c \end{vmatrix} \]

### Step 6: Evaluate the determinant
Since the second row is entirely zeros except for one element, we can conclude:
\[ D = 0 \quad \text{(because the determinant of a matrix with a row of zeros is zero)} \]

### Conclusion
Thus, the value of the determinant is:
\[ \boxed{0} \]