若等差数列{an}的前n项和为Sn,且an-3=10(n>7),S7=14,Sn=72,则n=_.

问题描述:

若等差数列{an}的前n项和为Sn,且an-3=10(n>7),S7=14,Sn=72,则n=______.

设公差为d,
由题意得:an-3=a1+(n-4)d=10①,
S7=

7(a1+a7
2
=
7(2a1+6d)
2
=7(a1+3d)=14②,
Sn=
n(a1+an
2
=
n[2a1+(n−1)d]
2
=72③,
联立①②③,解得:a1=-
14
5
,d=
8
5
,n=12,
则n=12.
故答案为:12