Let $a_1,a_2,a_3,..........and{b}_1{b}_2,b_3...........$ be arithmetic progression such that $a_1=25,b_1=75and{a}_{100}+b_{100}=100$, then

Given  $a_1,a_2,a_3,..........$ and $b_1{b}_2,b_3...........$ be two AP's.
Let $d_1,d_2$ be the common difference of the two AP's.
Then, the AP's can be written as 
$a_1,a_1+d_1,a_1+2d_1..........$
$b_1,b_1+d_2,b_1+2d_2..........$
Now, given $a_{100}+b_{100}=100$
$\Rightarrow a_1+99d_1+b_1+99d_2=100$
$\Rightarrow 100+99(d_1+d_2)=100$
$\Rightarrow d_1+d_2=0$
$\Rightarrow d_1=-d_2$
Hence, option A is correct.

Now, $a_n+b_n$
$=a_n+(n-1)d_1+b_1+(n-1)d_2$
$=a_1+b_1=100$
Hence, option B is correct.

Since, $a_n+b_n=100\forall n$
So, $a_1+b_1,a_2+b_2+....$ are in AP.
Hence, option C is also true.

Now $\sum _{r=1}^{100}(a_r+b_r))$

$=a_1+b_1,a_2+b_2+....+a_{99}+b_{99}+a_{100}+b_{100}$

$=100+100+....upto100times$
$=10^4$
Hence, option D is also true.