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 have an AP with $$a_1 = 8$$, sum of first four terms = 50, and sum of last four terms = 170. We need the product of the middle two terms.

First, we use the sum of the first four terms to find the common difference $$d$$.

The first four terms are: $$a_1, a_2, a_3, a_4 = 8, 8+d, 8+2d, 8+3d$$.

$$ S_4 = 4(8) + d(0+1+2+3) = 32 + 6d = 50 $$

$$ 6d = 18 \implies d = 3 $$

Next, we use the sum of the last four terms to find $$n$$ (number of terms).

The last four terms are $$a_{n-3}, a_{n-2}, a_{n-1}, a_n$$.

$$ a_k = 8 + (k-1) \times 3 = 5 + 3k $$

Sum of last four terms:

$$ \sum_{k=n-3}^{n} (5 + 3k) = 4 \times 5 + 3(n-3+n-2+n-1+n) = 20 + 3(4n - 6) = 20 + 12n - 18 = 2 + 12n $$

$$ 2 + 12n = 170 \implies 12n = 168 \implies n = 14 $$

From this, we identify the middle two terms.

With $$n = 14$$ terms (even), the middle two terms are $$a_7$$ and $$a_8$$:

$$ a_7 = 8 + 6 \times 3 = 8 + 18 = 26 $$

$$ a_8 = 8 + 7 \times 3 = 8 + 21 = 29 $$

Now, we compute the product.

$$ a_7 \times a_8 = 26 \times 29 = 754 $$

The product of the middle two terms is 754.