What is the value of 10-8+6-4+2-0+(-2)-(-4)+(-6)-(-8)+(-10)-(-12)+(-1

Here you go:

The sequence is \(10-8+6-4+2-0+(-2)-(-4)+(-6)-(-8)+(-10)-(-12)+(-14)-(-16)+(-18)-(-20)\).

Notice that the odd numbered terms (1st, 3rd, 5th...) form arithmetic progression with common difference of -4 and the even numbered terms (2nd, 4th...) form arithmetic progression with common difference of 4:

The sum of the odd numbered terms is \(10+6+2+(-2)+(-6)+(-10)+(-14)+(-18)=10+6+2-2-6-10-14-18=-32\);

The sum of the even numbered terms is \(-8-4-0-(-4)-(-8)-(-12)-(-16)-(-20)=-8-4-0+4+8+12+16+20=48\);

Their sum is \(-32+48=16\).

Though I wouldn't recommend to solve this question this way. It's better if you notice that we have 8 pairs:
10-8=2;
6-4=2;
2-0=2;
(-2)-(-4)=2;
(-6)-(-8)=2;
(-10)-(-12)=2;
(-14)-(-16)=2;
(-18)-(-20)=2;

So, the sum of each pair is 2, which makes the whole sum equal to 8*2=16.

Hope it's clear.