(1)a=2,k=10(2)
(1)设该等差数列为{a n},则a 1=a,a 2=4,a 3=3a,由已知有a+3a=8,得a 1=a=2,公差d=4-2=2,所以S k=ka 1+
·d=2k+
×2=k 2+k.由S k=110,得k 2+k-110=0,解得k=10或k=-11(舍去),故a=2,k=10.
(2)由(1)S n=
=n(n+1),则b n=
=n+1,故b n +1-b n=(n+2)-(n+1)=1,即数列{b n}是首项为2,公差为1的等差数列,所以T n=
=
(1)a=2,k=10(2)
(1)设该等差数列为{a n},则a 1=a,a 2=4,a 3=3a,由已知有a+3a=8,得a 1=a=2,公差d=4-2=2,所以S k=ka 1+
·d=2k+
×2=k 2+k.由S k=110,得k 2+k-110=0,解得k=10或k=-11(舍去),故a=2,k=10.
(2)由(1)S n=
=n(n+1),则b n=
=n+1,故b n +1-b n=(n+2)-(n+1)=1,即数列{b n}是首项为2,公差为1的等差数列,所以T n=
=