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:

We have an arithmetic progression (A.P.) whose first term is given as $$a = 3$$ and whose common difference is $$d$$ (unknown for the moment).

The question tells us that the sum of the first 25 terms is exactly equal to the sum of the next 15 terms. The phrase “next 15 terms’’ means the terms numbered 26th to 40th. In symbols, if $$S_n$$ denotes the sum of the first $$n$$ terms, then

$$\text{Sum of next 15 terms} = S_{40} - S_{25}.$$

The condition can therefore be written as

$$S_{25} = S_{40} - S_{25}.$$

Simplifying, we get

$$2\,S_{25} = S_{40}. \quad -(1)$$

Now we recall the well-known formula for the sum of the first $$n$$ terms of an A.P.:

$$S_n = \frac{n}{2}\,\Bigl[\,2a + (n-1)d\,\Bigr].$$

Using this formula, let us first write $$S_{25}$$ explicitly:

$$S_{25} = \frac{25}{2}\,\Bigl[\,2a + (25-1)d\,\Bigr].$$

Substituting $$a = 3$$, we get

$$S_{25} = \frac{25}{2}\,\Bigl[\,2(3) + 24d\,\Bigr] = \frac{25}{2}\,\Bigl[\,6 + 24d\,\Bigr].$$

Next, we write $$S_{40}$$ in the same manner:

$$S_{40} = \frac{40}{2}\,\Bigl[\,2a + (40-1)d\,\Bigr].$$

Again substituting $$a = 3$$, we obtain

$$S_{40} = 20\,\Bigl[\,2(3) + 39d\,\Bigr] = 20\,\Bigl[\,6 + 39d\,\Bigr].$$

We now substitute these expressions into equation (1):

$$2\,S_{25} = S_{40}$$

$$\Longrightarrow 2\left(\frac{25}{2}\,\left[\,6 + 24d\,\right]\right) = 20\,\left[\,6 + 39d\,\right].$$

The factor $$2$$ in front of $$S_{25}$$ cancels with the denominator $$2$$, leaving

$$25\,\bigl[\,6 + 24d\,\bigr] \;=\; 20\,\bigl[\,6 + 39d\,\bigr].$$

Let us expand both sides carefully, term by term:

Left side:
$$25 \times 6 + 25 \times 24d = 150 + 600d.$$

Right side:
$$20 \times 6 + 20 \times 39d = 120 + 780d.$$

So our equation becomes

$$150 + 600d = 120 + 780d.$$

We now bring all terms involving $$d$$ to one side and all constant terms to the other side. Subtract $$600d$$ from both sides and subtract $$120$$ from both sides:

$$150 - 120 = 780d - 600d.$$

This simplifies to

$$30 = 180d.$$

Finally, we solve for $$d$$ by dividing both sides by $$180$$:

$$d = \frac{30}{180} = \frac{1}{6}.$$

Hence, the common difference of the A.P. is $$\dfrac{1}{6}$$.

Hence, the correct answer is Option A.