作业帮 > 数学 > 作业

等差数列{an}的前n项和为Sn,已知a10=30,a20=50,

来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/11/12 11:20:28
等差数列{an}的前n项和为Sn,已知a10=30,a20=50,
(1)求通项an
(2)若Sn=80,求n
(3)设数列{bn}满足log2bn=an-12,求数列{bn}的前n项和Tn
等差数列{an}的前n项和为Sn,已知a10=30,a20=50,
(1)∵等差数列{an}的前n项和为Sn,a10=30,a20=50,


a1+9d=30
a1+19d=50,解得a1=12,d=2,
∴an=12+(n-1)×2=2n+10.
(2)∵a1=12,d=2,
∴Sn=12n+
n(n-1)
2×2=n2+11n,
∵Sn=80,∴n2+11n=80,
解得n=5.或n=-16(舍),
故n=5.
(3)∵log2bn=an-12=2n-2,
∴bn=22n-2=4n-1
∴{bn}是以1为首项,4为公比的等比数列,
∴Tn=
1-4n
1-4=
1
3(4n-1).