$1-6+2-7+3-8+\cdots\ to\ 100\ terms$
$=\underbrace{1-6}_{-5}+\underbrace{2-7}_{-5}+\underbrace{3-8}_{-5}+\cdots +\underbrace{50-55}_{-5}$
$= (1+2+\cdots + 50) - (6+7+\cdots +55)$
$= (1+2+3+4+5)+\underbrace{(6+7+\cdots +50) - (6+7+\cdots +50)}_0-(51+52+53+54+55)$
$= (1+2+3+4+5)-(51+52+53+54+55) $
$=\cfrac{5}{2}\Bigl[2(1)+(5-1)(1)\Bigr] - \cfrac{5}{2}\Bigl[2(51)+(5-1)(1)\Bigr]$
$\left( \because Sum\ of\ n\ terms\ of\ A.P.\ sequence\ is,\ \cfrac{n}{2}\Bigl[2a+(n-1)d\Bigr]\right)$
$=\cfrac{5}{2}(6) - \cfrac{5}{2}(106)$
$=5(3)-5(53)$
$=15-265$
$=-250$