等差数列{an},a2=8,前9项和为153.(1)求{an}的通项公式an(2)若bn=2*an,求{bn}的前n项和

Question

等差数列{an},a2=8,前9项和为153.
(1)求{an}的通项公式an
(2)若bn=2*an,求{bn}的前n项和

Answer 1

(1)S9=9(a2+a8)/2=153
又因为a2=8,解得a8=26
公差d=(a8-a2)/6=3
a1=a2-d=5
所以an=3n+2
(2)bn=2an=6n+4
可知数列{bn}是首项为b1=10,公差为6的等差数列
{bn}的前n项和Tn=(6n+4+10)n/2=(3n+7)n=3n^2+7n