Question

17已知在等差数列{an}中, a2=3, a5=-3 (1)求数列{an}的通项公式; (2)设？

Answer 1

解：因为a5=a2+（5-2）d
=a2+3d
所以d=（a5-a2）/3
=（-3-3）/3
=（-6）/3
=-2
所以a1=a2-d=3-（-2）=5
所以数列an的通项公式为an=a1+（n-1）d=5+（n-1）x（-2）=5-2n+2=7-2n
即an=7-2n

Answer 2

a2=a1+d=3
a5=a1+4d=-3
a5-a2＝3d＝-3-3=-6

d＝-2
a2＝3=a1+d
a1=3+2=5
an＝a1+(n-1)d＝5-2(n-1))＝5-2n+2=7-2n

Answer 3

a5-a2=3d=-6，d=-2，a1=a2-d=5，an=5+（n-1）*-2=-2n+7

Answer 4

an=a1+(n-1)d
a2=3
a1+d=3 (1)
a5=-3
a1+4d=-3 (2)
(2)-(1)
3d=-6
d=-2
from (1)
a1-2=3
a1=5
an = a1+(n-1)d = 5-2(n-1) = 7-2n