Question

Find the value of the expression (-101)+102+(-103)+104+...+(-199)+200. (with steps)

Answer 1

 hello : 
(-101)+102+(-103)+104+...+(-199)+200
=(  (-101)+(-103) +....+ (-199) ) +( (102) + ( 104) +....+(200))
let : A = (  (-101)+(-103) +....+ (-199) )
       B = ( (102) + ( 104) +....+(200))
note : the sum n term of arithemtic sequence 
S= n/2(u1 + un)
 un = u1 +(n-1) d       u1 : the first term      d : the common diference
in A :  u1= -101      d = -2       n = 49... 
in B : u1 =102       d=2        n= 49
A = 49/2(-101-199) =-7350
B=49/2(102+200)=4949
(-101)+102+(-103)+104+...+(-199)+200 = A+B =-2401