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=______.
答
∵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+
d=70010×11 2
故答案为:700