Obtain the sum of the first 56 terms of an A.P. whose 19th and 38th terms are 52 and 148 respectively.

Answer

To solve the problem of finding the sum of the first 56 terms of an Arithmetic Progression (A.P.) where the 19th term \( A_{19} \) is 52 and the 38th term \( A_{38} \) is 148, we can follow these steps:

### Step 1: Write the formulas for the 19th and 38th terms
The general formula for the \( n \)-th term of an A.P. is given by:
\[ A_n = A + (n-1)D \]
where \( A \) is the first term and \( D \) is the common difference.

For the 19th term:
\[ A_{19} = A + 18D = 52 \quad \text{(Equation 1)} \]

For the 38th term:
\[ A_{38} = A + 37D = 148 \quad \text{(Equation 2)} \]

### Step 2: Set up the equations
From the above, we have:
1. \( A + 18D = 52 \) (Equation 1)
2. \( A + 37D = 148 \) (Equation 2)

### Step 3: Subtract Equation 1 from Equation 2
Subtract Equation 1 from Equation 2 to eliminate \( A \):
\[ (A + 37D) - (A + 18D) = 148 - 52 \]
This simplifies to:
\[ 19D = 96 \]

### Step 4: Solve for \( D \)
Now, solve for \( D \):
\[ D = \frac{96}{19} \]

### Step 5: Substitute \( D \) back into Equation 1 to find \( A \)
Now substitute \( D \) back into Equation 1 to find \( A \):
\[ A + 18\left(\frac{96}{19}\right) = 52 \]
This simplifies to:
\[ A + \frac{1728}{19} = 52 \]
To eliminate the fraction, multiply everything by 19:
\[ 19A + 1728 = 988 \]
Now, isolate \( A \):
\[ 19A = 988 - 1728 \]
\[ 19A = -740 \]
\[ A = \frac{-740}{19} \]

### Step 6: Calculate the sum of the first 56 terms
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left( 2A + (n-1)D \right) \]
For \( n = 56 \):
\[ S_{56} = \frac{56}{2} \left( 2A + 55D \right) \]
Substituting the values of \( A \) and \( D \):
\[ S_{56} = 28 \left( 2\left(\frac{-740}{19}\right) + 55\left(\frac{96}{19}\right) \right) \]
This simplifies to:
\[ S_{56} = 28 \left( \frac{-1480 + 5280}{19} \right) \]
\[ S_{56} = 28 \left( \frac{3800}{19} \right) \]
Calculating this gives:
\[ S_{56} = \frac{28 \times 3800}{19} = \frac{106400}{19} = 5600 \]

### Final Answer
Thus, the sum of the first 56 terms of the A.P. is:
\[ \boxed{5600} \]