If $$a_1, a_2, a_3 \ldots$$ and $$b_1, b_2, b_3 \ldots$$ are A.P. and $$a_1 = 2, a_{10} = 3, a_1b_1 = 1 = a_{10}b_{10}$$ then $$a_4b_4$$ is equal to

We are given that $$a_1, a_2, a_3, \ldots$$ and $$b_1, b_2, b_3, \ldots$$ are in A.P. with $$a_1 = 2$$, $$a_{10} = 3$$, $$a_1 b_1 = 1$$, and $$a_{10} b_{10} = 1$$.

Since $$a_{10} = a_1 + 9d_a$$, substituting $$a_{10}=3$$ and $$a_1=2$$ gives $$3 = 2 + 9d_a$$, which yields $$d_a = \frac{1}{9}$$.

Using the conditions $$a_1 b_1 = 1$$ and $$a_{10} b_{10} = 1$$, we find $$b_1 = \frac{1}{a_1} = \frac{1}{2}$$ and $$b_{10} = \frac{1}{a_{10}} = \frac{1}{3}$$.

Now, since $$b_{10} = b_1 + 9d_b$$, we have $$\frac{1}{3} = \frac{1}{2} + 9d_b$$, leading to $$9d_b = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$$ and hence $$d_b = -\frac{1}{54}$$.

To determine $$a_4$$ and $$b_4$$, note that $$a_4 = a_1 + 3d_a = 2 + 3 \times \frac{1}{9} = 2 + \frac{1}{3} = \frac{7}{3}$$ and $$b_4 = b_1 + 3d_b = \frac{1}{2} + 3 \times \left(-\frac{1}{54}\right) = \frac{1}{2} - \frac{1}{18} = \frac{9 - 1}{18} = \frac{8}{18} = \frac{4}{9}$$.

Therefore, $$a_4 b_4 = \frac{7}{3} \times \frac{4}{9} = \frac{28}{27}$$.

The correct answer is Option A: $$\frac{28}{27}$$.