The sum of first 7 terms of an A.P. is 49 and that of first 17 terms of it is 289. Find the sum of first n terms.

Answer

To solve the problem step-by-step, we will use the formula for the sum of the first n terms of an arithmetic progression (A.P.), which is given by:

\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]

where:
- \( S_n \) = sum of the first n terms,
- \( A \) = first term of the A.P.,
- \( D \) = common difference,
- \( n \) = number of terms.

### Step 1: Set up the equations using the given information

We know:
- The sum of the first 7 terms \( S_7 = 49 \)
- The sum of the first 17 terms \( S_{17} = 289 \)

Using the formula for \( S_n \):

1. For \( n = 7 \):
\[ S_7 = \frac{7}{2} \times (2A + (7 - 1)D) = 49 \]
Simplifying this, we get:
\[ \frac{7}{2} \times (2A + 6D) = 49 \]
Multiplying both sides by 2:
\[ 7(2A + 6D) = 98 \]
Dividing by 7:
\[ 2A + 6D = 14 \quad \text{(Equation 1)} \]

2. For \( n = 17 \):
\[ S_{17} = \frac{17}{2} \times (2A + (17 - 1)D) = 289 \]
Simplifying this, we get:
\[ \frac{17}{2} \times (2A + 16D) = 289 \]
Multiplying both sides by 2:
\[ 17(2A + 16D) = 578 \]
Dividing by 17:
\[ 2A + 16D = 34 \quad \text{(Equation 2)} \]

### Step 2: Solve the system of equations

Now we have two equations:
1. \( 2A + 6D = 14 \) (Equation 1)
2. \( 2A + 16D = 34 \) (Equation 2)

We can eliminate \( A \) by subtracting Equation 1 from Equation 2:
\[ (2A + 16D) - (2A + 6D) = 34 - 14 \]
This simplifies to:
\[ 10D = 20 \]
Dividing both sides by 10:
\[ D = 2 \]

### Step 3: Substitute \( D \) back to find \( A \)

Now substitute \( D = 2 \) back into Equation 1:
\[ 2A + 6(2) = 14 \]
This simplifies to:
\[ 2A + 12 = 14 \]
Subtracting 12 from both sides:
\[ 2A = 2 \]
Dividing by 2:
\[ A = 1 \]

### Step 4: Find the sum of the first n terms

Now that we have \( A = 1 \) and \( D = 2 \), we can find the sum of the first n terms \( S_n \):
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
Substituting \( A \) and \( D \):
\[ S_n = \frac{n}{2} \times (2(1) + (n - 1)(2)) \]
This simplifies to:
\[ S_n = \frac{n}{2} \times (2 + 2(n - 1)) \]
\[ = \frac{n}{2} \times (2 + 2n - 2) \]
\[ = \frac{n}{2} \times 2n \]
\[ = n^2 \]

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