等差数列{an}中,已知a10=30,a20=50,sn=242,求n
问题描述：
等差数列{an}中,已知a10=30,a20=50,sn=242,求n
答
a10=a1+9d=30
a20=a1+19d=50
相减得
10d=20
d=2
a1=12
Sn=na1+n(n-1)*d/2
=12n+n(n-1)*2/2=242
12n+n^2-n=242
n^2+11n-242=0
(n+22)(n-11)=0
n=11