Question

已知等差数列{an}中,a1+a7=20,a3+a9=18,求数列{an}的通项公式及前n项和

Answer 1

等差数列an=a1+(n-1)d

Sn=na1+n(n-1)d/2

a1+a7=20 a1+a1+6d=20

a3+a9=18 a1+2d+a1+8d=18

联合上面两式可解得a1=23/2 d=-1/2

an=23/2 - (n-1)/2

Sn=23n/2 - n(n-1)/4