等差数列{an}中,a1+a2=20,a3+a4=80,则S10=_.

问题描述:

等差数列{an}中,a1+a2=20,a3+a4=80,则S10=______.

∵a1+a2=a1+(a1+d)=2a1+d=20,
a3+a4=(a1+2d)+(a1+3d)=2a1+5d=80,
∴d=15,a1=

5
2

∴S10=a1×10+
10×11
2
d
=700
故答案为:700