The sum of the first fifteen terms of an A.P. is 105 and the sum of the next fifteen terms is 780. Find the common difference of A.P.:

Options

A. 4
B. 3
C. 6
D. 5

Answer

To solve the problem, we need to find the common difference of an arithmetic progression (A.P.) given the sum of the first 15 terms and the sum of the next 15 terms.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The sum of the first 15 terms, \( S_{15} = 105 \)
- The sum of the next 15 terms (terms 16 to 30), \( S_{30} - S_{15} = 780 \)
- Therefore, \( S_{30} = S_{15} + 780 = 105 + 780 = 885 \)

2. **Use the Formula for the Sum of the First n Terms of an A.P.:**
The formula for the sum of the first n terms of an A.P. is:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

3. **Set Up the Equations:**
- For the first 15 terms:
\[ S_{15} = \frac{15}{2} \times (2a + 14d) = 105 \]
- For the first 30 terms:
\[ S_{30} = \frac{30}{2} \times (2a + 29d) = 885 \]

4. **Simplify the Equations:**
- From the first equation:
\[ 15(2a + 14d) = 210 \implies 2a + 14d = 14 \quad \text{(Equation 1)} \]
- From the second equation:
\[ 15(2a + 29d) = 885 \implies 2a + 29d = 59 \quad \text{(Equation 2)} \]

5. **Subtract Equation 1 from Equation 2:**
\[ (2a + 29d) - (2a + 14d) = 59 - 14 \]
This simplifies to:
\[ 15d = 45 \]

6. **Solve for d:**
\[ d = \frac{45}{15} = 3 \]

### Conclusion:
The common difference \( d \) of the A.P. is **3**.