Question

已知｛an｝是等差数列，其中a1=25，前四项和S4=82． （1）求数列｛an｝的通项公式an；（答案）

求解

Answer 1

sn = (a1 + an) * n /2==> s4 = (25 + a4) * 4/2==> 82 = (25 + a4) * 2==> 41= 25 +a4==> a4 =16 a4 = a1 +3d = 25 + 3d==> 16 = 25 +3d==> 3d = -9==> d = -3 an = a1 + (n-1)d==> an = 25 + (n-1) * (-3)==> an = 25 - 3n + 3==> an = 28 - 3n

Answer 2

解：s4=4x25+(4x3÷2)d=82,d=-3an=25+(n-1)x(-3)=-3n+28