Let `A_1, A_2, A_3` be the three A. P. with the common difference d and having their first terms as A, A+1, A+2, respectively. Let a, b, c be the` 7^(th), 9^(th), 17^(th)` terms of `A_1, A_2, A_3`, respectively such that `|(a,7,1),(2b,17,1),(c,17,1)|+70=0` If a = 29, the sum of first 20 terms of an AP whose first term is `c–a–b` and common difference is `d/12`,is equal to____

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the necessary equations.

### Step 1: Define the terms of the A.P.s
Let the three arithmetic progressions (A.P.s) be defined as follows:
- For \( A_1 \): First term = \( A \), Common difference = \( d \)
- For \( A_2 \): First term = \( A + 1 \), Common difference = \( d \)
- For \( A_3 \): First term = \( A + 2 \), Common difference = \( d \)

The \( n^{th} \) term of an A.P. can be expressed as:
\[ T_n = \text{First term} + (n-1) \times \text{Common difference} \]

### Step 2: Calculate the specific terms
- The \( 7^{th} \) term of \( A_1 \):
\[ a = A + (7-1)d = A + 6d \]
- The \( 9^{th} \) term of \( A_2 \):
\[ b = (A + 1) + (9-1)d = A + 1 + 8d \]
- The \( 17^{th} \) term of \( A_3 \):
\[ c = (A + 2) + (17-1)d = A + 2 + 16d \]

### Step 3: Set up the determinant equation
We are given the equation:
\[ |(a, 7, 1), (2b, 17, 1), (c, 17, 1)| + 70 = 0 \]
This implies:
\[ |(a, 7, 1), (2b, 17, 1), (c, 17, 1)| = -70 \]

### Step 4: Calculate the determinant
The determinant can be calculated as follows:
\[ \text{Det} = a \cdot (17 \cdot 1 - 1 \cdot 17) - 7 \cdot (2b \cdot 1 - 1 \cdot c) + 1 \cdot (2b \cdot 17 - 17 \cdot c) \]
This simplifies to:
\[ 0 - 7(2b - c) + (34b - 17c) \]
Thus:
\[ -7(2b - c) + (34b - 17c) = -70 \]

### Step 5: Substitute \( a = 29 \)
Given \( a = 29 \), we can substitute this into our equations:
\[ 29 = A + 6d \quad \text{(1)} \]
From equation (1), we can express \( A \):
\[ A = 29 - 6d \]

### Step 6: Substitute \( b \) and \( c \)
Using the expression for \( A \):
\[ b = (29 - 6d) + 1 + 8d = 30 + 2d \quad \text{(2)} \]
\[ c = (29 - 6d) + 2 + 16d = 31 + 10d \quad \text{(3)} \]

### Step 7: Substitute into the determinant equation
Substituting equations (2) and (3) into the determinant equation:
\[ -7(2(30 + 2d) - (31 + 10d)) + (34(30 + 2d) - 17(31 + 10d)) = -70 \]
This simplifies to:
\[ -7(60 + 4d - 31 - 10d) + (1020 + 68d - 527 - 170d) = -70 \]
\[ -7(29 - 6d) + (493 - 102d) = -70 \]
\[ -203 + 42d + 493 - 102d = -70 \]
Combining like terms gives:
\[ -60d + 290 = -70 \]
\[ -60d = -360 \quad \Rightarrow \quad d = 6 \]

### Step 8: Find \( A \), \( b \), and \( c \)
Substituting \( d \) back into the equations:
\[ A = 29 - 6 \times 6 = 29 - 36 = -7 \]
\[ b = 30 + 2 \times 6 = 30 + 12 = 42 \]
\[ c = 31 + 10 \times 6 = 31 + 60 = 91 \]

### Step 9: Calculate \( c - a - b \)
\[ c - a - b = 91 - 29 - 42 = 20 \]

### Step 10: Find the sum of the first 20 terms of the new A.P.
The first term of the new A.P. is \( 20 \) and the common difference is \( \frac{d}{12} = \frac{6}{12} = \frac{1}{2} \).

The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 20 \):
\[ S_{20} = \frac{20}{2} \times (2 \times 20 + (20-1) \times \frac{1}{2}) = 10 \times (40 + 9.5) = 10 \times 49.5 = 495 \]

### Final Answer
The sum of the first 20 terms of the A.P. is \( \boxed{495} \).