This is in AP.
Maximum sum means we do not need to consider negative numbers and can stop at $0.$
First find number of terms using the formula $\large a{_n} = a + (n-1)d$
Here,
$a = 44,$
$d = 42-44 = -2,$
$a{_n} = 0.$
Therefore, $0 = 44 + (n-1)(-2)$
$\Rightarrow n = 23.$
Now, sum of n terms of AP is given by: $S{_n} =\dfrac{n}{2}[a+a{_n}] =\dfrac{23}{2}[44+0] = 506.$
Option C is correct!