If the sum of the first four terms of an AP is 40 and the sum of the first fourteen terms of an AP is 280. Find the sum of first n terms of the A.P.

Answer

To solve the problem, we need to find the sum of the first \( n \) terms of an arithmetic progression (AP) given the sums of the first 4 and 14 terms. Let's break down the solution step by step.

### Step 1: Understand the formula for the sum of the first \( n \) terms of an AP
The formula for the sum of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the number of terms.

### Step 2: Write the equations based on the given information
We are given:
1. The sum of the first 4 terms (\( S_4 \)) is 40:
\[ S_4 = \frac{4}{2} \times (2a + (4 - 1)d) = 40 \]
Simplifying this, we get:
\[ 2 \times (2a + 3d) = 40 \implies 2a + 3d = 20 \quad \text{(Equation 1)} \]

2. The sum of the first 14 terms (\( S_{14} \)) is 280:
\[ S_{14} = \frac{14}{2} \times (2a + (14 - 1)d) = 280 \]
Simplifying this, we get:
\[ 7 \times (2a + 13d) = 280 \implies 2a + 13d = 40 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( 2a + 3d = 20 \) (Equation 1)
2. \( 2a + 13d = 40 \) (Equation 2)

We can subtract Equation 1 from Equation 2:
\[ (2a + 13d) - (2a + 3d) = 40 - 20 \]
This simplifies to:
\[ 10d = 20 \implies d = 2 \]

### Step 4: Substitute \( d \) back to find \( a \)
Now that we have \( d = 2 \), we can substitute this value back into Equation 1 to find \( a \):
\[ 2a + 3(2) = 20 \implies 2a + 6 = 20 \implies 2a = 14 \implies a = 7 \]

### Step 5: Find the sum of the first \( n \) terms
Now that we have \( a = 7 \) and \( d = 2 \), we can find the sum of the first \( n \) terms:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Substituting the values of \( a \) and \( d \):
\[ S_n = \frac{n}{2} \times (2 \times 7 + (n - 1) \times 2) \]
\[ S_n = \frac{n}{2} \times (14 + 2n - 2) = \frac{n}{2} \times (2n + 12) \]
\[ S_n = n \times (n + 6) \]

### Final Result
Thus, the sum of the first \( n \) terms of the AP is:
\[ S_n = n^2 + 6n \]