Add the following:
$(-100)+(-92)+(-84)+.....+92$

For the given A.P., $a=(-100)$
$d=(-92)-(-100)=8$ and $T_n=92$
$T_n=a+(n-1)d$
$\Rightarrow$ $92=-100+(n-1)8$
$\Rightarrow$ $192=8(n-1)$
$\Rightarrow$ $24=n-1$
$\Rightarrow$ $n=25$
To find the sum of 25 terms
Now, $S_n=\frac{1}{2}n[2a+(n-1\left)d\right]$
$⟹S_{25}=\frac{1}{2}\left(25\right)\left[2\right(-100)+(25-1\left)\right(8\left)\right]=\frac{1}{2}\times 25\times (-8)=-100$

Hence, the sum is $-100$.