If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer

To solve the problem, we need to find the sum of the first \( n \) terms of an arithmetic progression (A.P.) given the sums of the first 7 and 17 terms.

### Step-by-Step Solution:

1. **Use the formula for the sum of the first \( n \) terms of an A.P.**:
The sum of the first \( n \) terms \( S_n \) of an A.P. can be expressed as:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Set up the equations for the given sums**:
From the problem, we know:
- The sum of the first 7 terms \( S_7 = 49 \):
\[ S_7 = \frac{7}{2} \left(2a + (7-1)d\right) = 49 \]
Simplifying this, we get:
\[ 7(2a + 6d) = 98 \quad \Rightarrow \quad 2a + 6d = 14 \quad \text{(Equation 1)} \]

- The sum of the first 17 terms \( S_{17} = 289 \):
\[ S_{17} = \frac{17}{2} \left(2a + (17-1)d\right) = 289 \]
Simplifying this, we get:
\[ 17(2a + 16d) = 578 \quad \Rightarrow \quad 2a + 16d = 34 \quad \text{(Equation 2)} \]

3. **Solve the system of equations**:
Now we have two equations:
- Equation 1: \( 2a + 6d = 14 \)
- Equation 2: \( 2a + 16d = 34 \)

We can subtract Equation 1 from Equation 2:
\[ (2a + 16d) - (2a + 6d) = 34 - 14 \]
This simplifies to:
\[ 10d = 20 \quad \Rightarrow \quad d = 2 \]

4. **Substitute \( d \) back to find \( a \)**:
Now substitute \( d = 2 \) back into Equation 1:
\[ 2a + 6(2) = 14 \quad \Rightarrow \quad 2a + 12 = 14 \quad \Rightarrow \quad 2a = 2 \quad \Rightarrow \quad a = 1 \]

5. **Find the sum of the first \( n \) terms**:
Now that we have \( a = 1 \) and \( d = 2 \), we can express the sum of the first \( n \) terms:
\[ S_n = \frac{n}{2} \left(2(1) + (n-1)(2)\right) \]
Simplifying this:
\[ S_n = \frac{n}{2} \left(2 + 2(n-1)\right) = \frac{n}{2} \left(2 + 2n - 2\right) = \frac{n}{2} \cdot 2n = n^2 \]

Thus, the sum of the first \( n \) terms of the A.P. is:
\[ \boxed{n^2} \]