Question 809009
1 - 2 + 3 - 4 + ... + 65 - 66 + 67

let us break it up into two sequences

both are arithmetic progressions

1+3+5..........67

a=1,d=2 Tn=67

67=1+(n-1)2
67=1+2n-2
2n=68
n=34

Sn = n/2(2a+(n-1)d)
=(34/2(2*1+(37-1)2)
=17*(2+36*2)
=17*74
=1258


and -2,-4,-6......-66

a=-2, d=-2 Tn= -66

-66=-2+(n-1)*-2

-66=-2-2n+2

n=33

S33= 33/2( 2*-2+(33-1)*-2)

S33=33/2(-4-64)

S33=33/2 * -68

=33*-34
=-1122

1258-1122

=136