等差数列{an}中,Sn=40,a1=13,d=-2时,n=_.

问题描述:

等差数列{an}中,Sn=40,a1=13,d=-2时,n=______.

∵{an}是等差数列,a1=13,d=-2,
∴sn=na1+

n(n−1)
2
d=13n+
n(n−1)
2
×(-2)=-n2+14n,
∵Sn=40,
∴-n2+14n=40,
解得n=4或n=10,
故答案为4或10.