S=0−1+2−3+4−5+6−7....+16−17+18−19+20S=−(1+3+5+7+...+19)+(2+4+6+....+20)To find S
Series:1+3+5+7+....+19a(firstterm)=1d(commondifference)=2∴an=a+(n−1)d19=1+(n−1)2⇒n=10∴10termsinseries.∴S1=n2[2a+(n−1)d]=102[2+(10−1)2]=100
To find S2
Series:2+4+6+...+20a=2,d=2an=a+(n−1)d⇒20=2+(n−1)2⇒n=10terms∴S2=n2[2a+(n−1)d]=102[2×2+(9)2]=110