Simple but tricky

Let the first term and common difference be $a=a_1$ and $d$ respectively. Then, we have:

$\begin{aligned} \sum_{i=1}^{25} a_i = \frac{25(2a+24d)}{2} & = 725 \\ \Rightarrow a + 12d & = 29 \end{aligned}$

Now we have:

$\begin{aligned} \sum_{r=0}^{12} a_{2r+1} & = a_1+a_3+a_5+...+a_{25} \\ & = a + (a+2d)+(a+4d)+...+(a+24d) \\ & = 13a + 2\times \frac{12(12+1)}{2}d \\ & = 13(a+12d) = 13 \times 29 = \boxed{377} \end{aligned}$

$a_1+a_2+...a_{25}=725$ $\frac{25}{2}(a_1+a_{25}=725$ $a_1+a_{25}=58$

Now we have to calculate $a_1+a_3+...+a_{25}$ For this we can do $(a_1+a_{25})+(a_3+a_{23})+...+(a_{11}+a_{15})+a_{13}$ Which can also be written as $(a_1+a_{25})+(a_1+a_{25})+...+(a_1+a_{25})+\frac{(a_1+a_{25})}{2}$

Which is $\frac{13}{2}(a_1+a_{25})=\frac{13}{2}58=377$ So the correct answer is $\boxed{377}$