Let $a_1, a_2, a_3, \ldots$ be an A.P. and $g_1 = a_1, g_2, g_3, \ldots$ be an increasing G.P. If $a_1 = a_2 + g_2 = 1$ and $a_3 + g_3 = 4$, then $a_{10} + g_5$ is equal to:

Let the common difference of the A.P. be $d$ and the common ratio of the G.P. be $r$.

Given $a_1 = 1$ and $g_1 = a_1 = 1$.

From $a_2 + g_2 = 1$, we have:
$(a_1 + d) + g_1 r = 1$
$1 + d + r = 1 \Rightarrow d = -r$

From $a_3 + g_3 = 4$, we have:
$(a_1 + 2d) + g_1 r^2 = 4$
$1 + 2d + r^2 = 4$
$2d + r^2 = 3$

Substituting $d = -r$ into the equation:
$-2r + r^2 = 3$
$r^2 - 2r - 3 = 0$
$(r - 3)(r + 1) = 0$

Since the G.P. is increasing and $g_1 = 1$, the common ratio $r$ must be greater than $1$. Thus, $r = 3$.

Then, $d = -3$.

We need to find $a_{10} + g_5$:
$a_{10} = a_1 + 9d = 1 + 9(-3) = -26$
$g_5 = g_1 r^4 = 1(3)^4 = 81$

$a_{10} + g_5 = -26 + 81 = 55$

Answer: $55$