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=______.
答
∵{an}是等差数列,a1=13,d=-2,
∴sn=na1+
d=13n+n(n−1) 2
×(-2)=-n2+14n,n(n−1) 2
∵Sn=40,
∴-n2+14n=40,
解得n=4或n=10,
故答案为4或10.