Question

1-2+3-4+5-6+7.....+179-180=?

Answer 1

this can be splited into two Arithmetic progression:
1+3+5...+179  and -2-4-6...-180
Now, we will find sum of them separately using summation formula of AP
S_{n}Sn​ = n/2{2a+(n-1)d}
where,S_{n}Sn​ = sum of n terms
d = common difference
a = first term of AP
First, we will no of terms i.e. n of both APs
a) 1+3+5...+179 179 = a+(n-1)d
179 = 1+(n-1)2
179-1 = (n-1)2
178/2 = n-1
89+1 = n
n= 90
S_{n}Sn​ (a)= n/2{2a+(n-1)d}                          
                          = 90/2{2*1 +(90-1)2}                          
                          = 45{2+89*2}                          
                          =45{2+178} 
                         = 45*180 = 8100
b) -2-4-6...-180 = -(2+4+6...+180)
180 = a+(n-1)d
180 = 2+(n-1)2
180-2 = (n-1)2
178/2 = n-1
89+1 = n
n= 90
S_{n}Sn​ (b)= n/2{2a+(n-1)d}
                          = 90/2{2*2 +(90-1)2} 
                          = 45{4+89*2}
                          =45{4+178}
                          = 45*182 
                          = 8190
Now, 1+3+5...+179-2-4-6...-180        
S_{n}(a)Sn​(a) +S_{n}(b)Sn​(b) = 8100 - 8190  = -90

Ans is -90

Answer 2

S =   1 - 2 + 3 - 4 + 5 - 6 + ___ + 179 - 180

+   S =       1 - 2 + 3 -  4 + 5 - 6 + ___ + 179 - 180

----------------------------------------------------------------

2S =           1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - ___ - 1 - 1 - 1 - 180

2S = -180

S =  $\frac{-180}{2}$

S = -90