If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is:

Options

A. `(1)/(6)`
B. `(1)/(5)`
C. `(1)/(4)`
D. `(1)/(7)`

Answer

To solve the problem, we need to find the common difference \( d \) of an arithmetic progression (A.P.) where the first term \( a = 3 \) and the sum of the first 25 terms is equal to the sum of the next 15 terms.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- First term \( a = 3 \)
- Let the common difference be \( d \).
- We need to find \( d \) such that the sum of the first 25 terms equals the sum of the next 15 terms.

2. **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. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

3. **Calculate the Sum of the First 25 Terms \( S_{25} \):**
\[ S_{25} = \frac{25}{2} \left(2 \cdot 3 + (25 - 1)d\right) = \frac{25}{2} \left(6 + 24d\right) = 25 \left(3 + 12d\right) = 75 + 300d \]

4. **Calculate the Sum of the First 40 Terms \( S_{40} \):**
\[ S_{40} = \frac{40}{2} \left(2 \cdot 3 + (40 - 1)d\right) = 20 \left(6 + 39d\right) = 120 + 780d \]

5. **Calculate the Sum of the Next 15 Terms \( S_{15} \):**
The sum of the next 15 terms (from the 26th to the 40th term) can be expressed as:
\[ S_{15} = S_{40} - S_{25} \]
Thus,
\[ S_{15} = (120 + 780d) - (75 + 300d) = 45 + 480d \]

6. **Set Up the Equation:**
According to the problem, \( S_{25} = S_{15} \):
\[ 75 + 300d = 45 + 480d \]

7. **Solve for \( d \):**
Rearranging the equation gives:
\[ 75 - 45 = 480d - 300d \]
\[ 30 = 180d \]
\[ d = \frac{30}{180} = \frac{1}{6} \]

### Final Answer:
The common difference \( d \) of the A.P. is \( \frac{1}{6} \).

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