1．由等差数列的前n项和的公式特点,
可设Sn=An²+Bn,
则S10=10²A+10B=100,S100=100²A+100B=10,
两式相减得,(100²-10²)A+(100-10)B= -(100-10),
∴110A+B= -1,
∴S110=110²A+110B=110(110A+B)= -110..
2．设等比数列的公比为q,(q≠0)
则由题意,S4=2S2,
即S4-S2=S2,
∴q²S2=S2,
(q²-1)(a1+a1q)=0
a1(q+1)(q-1)(1+q)=0
∵a1≠0,
∴q=±1,
当q=1时,an=a3=2,
当q= -1时,an=a3q^(n-3)=2×(-1)^(n-3)=2×(-1)^(n-1).