If $$\{a_i\}_{i=1}^{n}$$, where $$n$$ is an even integer, is an arithmetic progression with common difference $$1$$, and $$\sum_{i=1}^{n} a_i = 192$$, $$\sum_{i=1}^{n/2} a_{2i} = 120$$, then $$n$$ is equal to

Let $$\{a_i\}_{i=1}^{n}$$ be an AP with common difference $$d = 1$$ and first term $$a_1 = a$$.

So $$a_i = a + (i-1)$$.

Using the first condition: $$\sum_{i=1}^{n} a_i = 192$$

$$ \sum_{i=1}^{n} [a + (i-1)] = na + \frac{n(n-1)}{2} = 192 \quad \cdots (1) $$

Using the second condition: $$\sum_{i=1}^{n/2} a_{2i} = 120$$

The even-indexed terms are $$a_2, a_4, a_6, \ldots, a_n$$ (since $$n$$ is even).

$$a_{2i} = a + (2i - 1)$$

$$ \sum_{i=1}^{n/2} [a + (2i-1)] = \frac{n}{2} \cdot a + \sum_{i=1}^{n/2}(2i-1) $$

$$ = \frac{na}{2} + \sum_{i=1}^{n/2}(2i-1) = \frac{na}{2} + \left(\frac{n}{2}\right)^2 = \frac{na}{2} + \frac{n^2}{4} = 120 \quad \cdots (2) $$

Note: $$\sum_{i=1}^{m}(2i-1) = m^2$$, so with $$m = n/2$$, the sum is $$n^2/4$$.

From equation (1): $$na + \frac{n(n-1)}{2} = 192$$

From equation (2): $$\frac{na}{2} + \frac{n^2}{4} = 120$$

Multiply equation (2) by 2:

$$ na + \frac{n^2}{2} = 240 \quad \cdots (3) $$

Subtract equation (1) from equation (3):

$$ \frac{n^2}{2} - \frac{n(n-1)}{2} = 240 - 192 $$

$$ \frac{n^2 - n^2 + n}{2} = 48 $$

$$ \frac{n}{2} = 48 $$

$$ n = 96 $$

The answer is Option C: 96.