已知等差数列{an}满足a1=3.a4+a8=26,{an}的前n项和为Sn.求an及sn
问题描述：
已知等差数列{an}满足a1=3.a4+a8=26,{an}的前n项和为Sn.求an及sn
答
∵a1=3,a4+a8=26,而此数列是等差数列，
∴a1=3
a1+3d+a1+7d=26
a1=3，d=2
∴an=3+(n-1)*2
Sn=na1+n(n-1)d/2
=3n+n(n-1)*d/2
答
a4+a8=a1 +3d +a1 +7d =26
得 d=2
an = 2n+1
sn =(a1+an)*n/2
= n（n+2）
答
设公差为d
∵a1=3,a4+a8=26,
∴a4+a8=a1+3d+a1+7d=2a1+10d=26
可得d=2,
即得an=a1+(n-1)d=2n+1
则Sn=a1n+n(n-1)d/2=n2+2n