hello :
<span>(-101)+102+(-103)+104+...+(-199)+200
=( </span>(-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
Answer 25 r 32
Step by step: hopefully you can see the photo
Answer:
y>0.6
Step-by-step explanation:
Simplify
5y+3>6
Subtract 3 from both sides.
5y>3
Divide
y>0.6
