题目内容
已知等差数列的前三项依次为a,4,3a,前n项和为Sn,且Sk=110.
(1)求a及k的值;
(2)设数列{bn}的通项bn=,证明数列{bn}是等差数列,并求其前n项和Tn.
(1)求a及k的值;
(2)设数列{bn}的通项bn=,证明数列{bn}是等差数列,并求其前n项和Tn.
(1)a=2,k=10(2)
(1)设该等差数列为{an},则a1=a,a2=4,a3=3a,由已知有a+3a=8,得a1=a=2,公差d=4-2=2,所以Sk=ka1+·d=2k+×2=k2+k.由Sk=110,得k2+k-110=0,解得k=10或k=-11(舍去),故a=2,k=10.
(2)由(1)Sn==n(n+1),则bn==n+1,故bn+1-bn=(n+2)-(n+1)=1,即数列{bn}是首项为2,公差为1的等差数列,所以Tn==
(2)由(1)Sn==n(n+1),则bn==n+1,故bn+1-bn=(n+2)-(n+1)=1,即数列{bn}是首项为2,公差为1的等差数列,所以Tn==
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