$34+32+30+....+10$
This sequence is an AP.
Here, $a=34,d=32-34=-2.$
Let the number of terms of the AP be $n.$
We know that
$a_n=a+(n-1)d.$

$\Rightarrow 10=34+(n-1)(-2)$

$\Rightarrow (n-1)(-2)=-24$

$\Rightarrow n-1=\frac{24}{2}=12$
$\Rightarrow n=13$ 
Also, we know that the sum to $n$ terms of an AP with first term $a$ and last term $l$ is given by
     $S_n=\frac{n}{2}(a+l).$

$\therefore S_{13}=\frac{13}{2}(34+10)$

    $S_{13}=286$

Hence, the required sum is 286.