Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant.
• First Term ($a$): The initial term of the A.P.
• Common Difference ($d$): The constant difference between consecutive terms.
• Sum of the first $n$ terms of an A.P. ($S_n$): Given by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$.

Step-by-Step Solution

Step 1: Understand the problem statement and identify given information. We are given that the first term of an A.P. is $a = 3$. We are also told that the sum of its first 25 terms ($S_{25}$) is equal to the sum of its next 15 terms. We need to find the common difference ($d$).

Step 2: Express the sum of the "next 15 terms" mathematically. The "next 15 terms" refers to the terms from the $26^{th}$ term to the $40^{th}$ term (since $25 + 15 = 40$). The sum of these terms can be calculated as the sum of the first 40 terms ($S_{40}$) minus the sum of the first 25 terms ($S_{25}$). So, the sum of the next 15 terms is $S_{40} - S_{25}$.

Step 3: Formulate the equation based on the given condition. The problem states that the sum of the first 25 terms is equal to the sum of its next 15 terms. Therefore, we can write the equation: $S_{25} = S_{40} - S_{25}$ Rearranging this equation, we get: $2S_{25} = S_{40}$

Step 4: Calculate $S_{25}$ using the sum formula. Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n=25$ and $a=3$: $S_{25} = \frac{25}{2}[2(3) + (25-1)d]$ $S_{25} = \frac{25}{2}[6 + 24d]$ We can simplify this by factoring out 2 from the bracket: $S_{25} = \frac{25}{2} \cdot 2[3 + 12d]$ $S_{25} = 25(3 + 12d)$

Step 5: Calculate $S_{40}$ using the sum formula. Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n=40$ and $a=3$: $S_{40} = \frac{40}{2}[2(3) + (40-1)d]$ $S_{40} = 20[6 + 39d]$

Step 6: Substitute the expressions for $S_{25}$ and $S_{40}$ into the equation from Step 3 and solve for $d$. We have the equation $2S_{25} = S_{40}$. Substituting the expressions from Step 4 and Step 5: $2 \cdot [25(3 + 12d)] = 20[6 + 39d]$ $50(3 + 12d) = 20(6 + 39d)$ Divide both sides by 10 to simplify: $5(3 + 12d) = 2(6 + 39d)$ Distribute the constants: $15 + 60d = 12 + 78d$ Now, rearrange the terms to group $d$ terms on one side and constant terms on the other: $15 - 12 = 78d - 60d$ $3 = 18d$ Solve for $d$: $d = \frac{3}{18}$ $d = \frac{1}{6}$

Common Mistakes & Tips

• Misinterpreting "next 15 terms": Ensure you correctly identify that the sum of the next 15 terms is $S_{40} - S_{25}$, not $S_{15}$ or the sum starting from the $25^{th}$ term.
• Algebraic errors: Be meticulous with algebraic manipulations, especially when expanding brackets and rearranging terms to avoid sign errors or calculation mistakes.
• Simplification: Look for opportunities to simplify the equation by dividing common factors on both sides, which reduces the chance of arithmetic errors.

Summary

The problem requires us to find the common difference of an Arithmetic Progression given its first term and a relationship between the sums of certain terms. By expressing the sum of the "next 15 terms" as the difference between the sum of the first 40 terms and the sum of the first 25 terms, we formed an equation. Substituting the formula for the sum of an A.P. and solving the resulting linear equation for the common difference $d$ yielded the answer.

The final answer is $\boxed{\frac{1}{6}}$, which corresponds to option (D).