题目: 等差数列{an}中,a1+a2=20,a3+a4=80,则S10=______. 答案: ∵a1+a2=a1+(a1+d)=2a1+d=20,a3+a4=(a1+2d)+(a1+3d)=2a1+5d=80,∴d=15,a1=52∴S10=a1×10+10×112d=700故答案为:700