Question

已知等差数列{an}满足：a3=7,a5+a7=26,{an}前n项和为sn,求an及sn

Answer 1

a3=7
a5+a7=2a6=26
a6=13
a6-a3=6=5d-2d=3d, d=2
a1+2d=7=a1+4 a1=3
an=3+2(n-1)=2n+1
∴Sn=n(3+2n+1)/2=n²+2n
O(∩_∩)O，希望对你有帮助

Answer 2

a5+a7=2(a6),a6=13
an=a1+nd
a6=a1+6d 13=a1+6d
a3=a1+3d 7=a1+3d
解得a1和d，通过等差数列的求和公式和通项公式就可以求an和sn了

Answer 3

a5+a7=26得a6=13.公差d=（13-7）/(6-3)=2.所以an=a3+(n-3)d=2n+1,sn=n(3+2n+1)/2=n(n+2)

Answer 4

a5+a7=a3+2d+a3+4d=2a3+6d=14+6d=26
所以d=2 a1=a3-2d=3
所以an=3+2(n-1)=2n+1
sn=n(3+2n+1)/2=n^2+2n

Answer 5

an=1+2n
sn=(3+2n+1)n/2=2n+n²

Answer 6

a3=a1+2d=7,a5+a7=a1+4d+a1+6d=2a1+10d=26以上两个方程联立可得a1=3，d=2。所以an=2n+1,sn=n^2+2n