Question

已知等差数列 的前几项和为sn=t*n^2+(t-9)n+t-3/2 求数列 的通项公式

Answer 1

sn=na1+n(n-1)d/2=d(n^2)/2+(a1-d/2)n=t*n^2+(t-9)n+t-3/2,所以:d/2=t,a1-d/2=t-9,t-3/2=0.解得：t=3/2,d=3,a1=-6,所以an=3n-9.