Let $$a_1 = 8, a_2, a_3, \ldots, a_n$$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is _____.

We need to find the product of the middle two terms of an A.P. $$a_1 = 8, a_2, a_3, \ldots, a_n$$.

Since the sum of the first four terms is 50, we have $$a_1 + a_2 + a_3 + a_4 = 50$$ which simplifies to $$4a_1 + 6d = 50$$. Substituting $$a_1 = 8$$ gives $$32 + 6d = 50$$ and hence $$d = 3$$.

Because the sum of the last four terms is 170, it follows that $$a_{n-3} + a_{n-2} + a_{n-1} + a_n = 170$$. Writing these in terms of the last term leads to $$4a_n - 6d = 170$$, so $$4a_n - 18 = 170$$ and therefore $$a_n = 47$$.

Now, using the nth term formula $$a_n = a_1 + (n-1)d$$ and substituting $$a_n = 47$$, $$a_1 = 8$$, and $$d = 3$$ yields $$47 = 8 + 3(n-1)$$. From this we get $$39 = 3(n-1)$$, giving $$n = 14$$.

Since $$n = 14$$ is even, the middle two terms are $$a_7$$ and $$a_8$$. Using $$a_k = a_1 + (k-1)d$$, we find $$a_7 = 8 + 6(3) = 26$$ and $$a_8 = 8 + 7(3) = 29$$.

Finally, the product of these middle terms is $$a_7 \times a_8 = 26 \times 29 = 754$$, so the required product is 754.